MATH 55003–55103
Theory of Functions of a Real Variable I–II
Fall 2024–Spring 2025
Next class day: Friday, April 4, 2025
(Problem 10) [De Morgan] If \(X\) is a set and \(\mathcal {F}\) is a family of sets, show that \begin {equation*} X\setminus \Bigl (\bigcup _{F\in \mathcal {F}} F\Bigr ) = \bigcap _{F\in \mathcal {F}} (X\setminus F). \end {equation*}
(Problem 20) [De Morgan] If \(X\) is a set and \(\mathcal {F}\) is a family of sets, show that \begin {equation*} X\setminus \Bigl (\bigcap _{F\in \mathcal {F}} F\Bigr ) = \bigcup _{F\in \mathcal {F}} (X\setminus F). \end {equation*}
(Problem 30) If \(f:A\to B\) is a function, and \(E\), \(F\subseteq B\), show that \begin {equation*} f^{-1}(E\cup F)=f^{-1}(E)\cup f^{-1}(F). \end {equation*}
(Problem 40) If \(f:A\to B\) is a function, and \(E\), \(F\subseteq B\), show that \begin {equation*} f^{-1}(E\cap F)=f^{-1}(E)\cap f^{-1}(F). \end {equation*}
(Problem 50) If \(f:A\to B\) is a function, and \(E\), \(F\subseteq B\), show that \begin {equation*} f^{-1}(E\setminus F)=f^{-1}(E)\setminus f^{-1}(F). \end {equation*}
(Problem 60) If \(f:A\to B\) is a function, and we define \(f(E)=\{f(e):e\in E\}\) for all \(E\subseteq A\), are the analogues to the above formulas true?
[Definition: Relation] Let \(X\) be a set. A subset \(R\) of the Cartesian product \(X\times X\) is called a relation on \(X\); we write \(xRy\) if \((x,y)\in R\).
[Definition: Reflexive relation] A relation \(R\) is reflexive if \(xRx\) for all \(x\in X\).
[Definition: Transitive relation] A relation \(R\) is transitive if \(xRy\) and \(yRz\) implies \(xRz\).
[Definition: Symmetric relation] A relation \(R\) is symmetric if \(xRy\) if and only if \(yRx\).
[Definition: Equivalence relation] A relation \(R\) is an equivalence if it is reflexive, symmetric, and transitive.
[Definition: Partial ordering] A relation \(R\) is a partial ordering if it is reflexive, transitive, and as far from symmetric as possible: if \(xRy\) and \(yRx\) then \(x=y\).
[Definition: Totally ordered] Let \(R\) be a partial ordering on a set \(X\) and let \(E\subseteq X\). We say that \(E\) is totally ordered if, for every \(x\), \(y\in E\), either \(xRy\) or \(yRx\).
[Definition: Equivalence class] The equivalence class of an element \(x\) of a set \(X\) with respect to an equivalence relation \(R\) on \(X\) is \(R_x=\{y\in X:x R y\}\), and \(X/R=\{R_x:x\in X\}\).
[Definition: Partition] Let \(X\) be a set and let \(\mathcal {P}\) be a collection of subsets of \(X\). If, for every \(x\in X\), there is exactly one \(P\in \mathcal {P}\) that satisfies \(x\in P\), we say that \(\mathcal {P}\) is a partition of \(X\).
(Problem 70) Let \(R\) be an equivalence relation. Show that \(X/R\) is a partition of \(X\).
(Problem 80) Let \(\mathcal {P}\) be a partition of \(X\). Define \(R=\bigcup _{P\in \mathcal {P}} P\times P\). Show that \(R\) is an equivalence relation and that \(xRy\) if and only if there is a \(P\in \mathcal {P}\) with \(x\), \(y\in P\).
[Definition: Choice function] Let \(\mathcal {F}\) be a nonempty family of nonempty sets and let \(X=\bigcup _{F\in \mathcal {F}} F\). (We do not require that \(\mathcal {F}\) be a partition of \(X\).) A choice function on \(\mathcal {F}\) is a function \(f:\mathcal {F}\to X\) such that \(f(F)\in F\) for each \(F\in \mathcal {F}\).
Zermelo’s axiom of choice. If \(\mathcal {F}\) is a nonempty collection of nonempty sets, then there exists a choice function on \(\mathcal {F}\).
(Problem 90) Suppose that \(f:X\to Y\) is a function from one metric space to another. Suppose that \(x\in X\) and that \(f\) is continuous at \(x\) in the sequential sense: for all sequences \(\{x_n\}_{n=1}^\infty \subset X\) that satisfy \(x_n\to x\), we have that \(f(x_n)\to f(x)\). Use the axiom of choice to show that \(f\) is continuous in the \(\varepsilon \)-\(\delta \) sense as well. Be sure to explain where you must use a choice axiom.
[Definition: Upper bound] Let \(R\) be a partial ordering on a set \(X\) and let \(E\subseteq X\). If \(x\in X\) and \(eRx\) for all \(e\in E\), then \(x\) is an upper bound on \(E\).
[Definition: Maximal element] Let \(R\) be a partial ordering on a set \(X\) and let \(x\in X\). If \(\{y\in X:xRy\}=\{x\}\), then \(x\) is said to be maximal.
[Homework 1.1] Let \(\mathcal {F}\) be a family of sets.
Zorn’s lemma. Let \(X\) be a nonempty partially ordered set. Assume that, if \(E\subseteq X\) is totally ordered, then \(E\) has an upper bound in \(X\) (not necessarily in \(E\)). Then \(X\) contains a maximal element.
[Definition: Cartesian product of a parameterized collection of sets] If \(\{E_\lambda \}_{\lambda \in \Lambda }\) is a parameterized collection of sets, then the Cartesian product \(\prod _{\lambda \in \Lambda } E_\lambda \) is defined to be the set of functions \(f\) from \(\Lambda \) to \(\bigcup _{\lambda \in \Lambda } E_\lambda \) such that \(f(\lambda )\in E_\lambda \) for all \(\lambda \in \Lambda \).
(Problem 100) When is the Cartesian product of a parameterized collection of sets equal to the set of choice functions on the family of sets? Can you modify the axiom of choice so that it is equivalent to the statement that the Cartesian product of of a parameterized collection of nonempty sets is nonempty?
[Definition: Field] A field is a set \(F\) together with two functions from \(F\times F\) to itself (denoted \(a+b\) and \(ab\) rather than \(s((a,b))\) and \(p((a,b))\)) that satisfy the axioms
\(a+b=b+a\) for all \(a\), \(b\in F\),
\(a+(b+c)=(a+b)+c\) for all \(a\), \(b\), \(c\in F\),
There is a \(0\in F\) such that \(a+0=a\) for all \(a\in F\),
If \(a\in F\) then there is a \(-a\in F\) such that \(a+(-a)=0\),
\(ab=ba\) for all \(a\), \(b\in F\),
\(a(bc)=(ab)c\) for all \(a\), \(b\), \(c\in F\),
There is a \(1\in F\) such that \(1a=a\) for all \(a\in F\),
If \(a\in F\) and \(a\neq 0\) then there is a \(1/a\in F\) such that \(a(1/a)=1\),
\(1\neq 0\),
\(a(b+c)=ab+ac\) for all \(a\), \(b\), \(c\in F\),
(Problem 110) Show from the axioms that \(0a=0\) for all \(a\in F\).
Because \(0\) is the additive identity, \(0+0=0\) and so \(0a=(0+0)a\). By the distributivity of multiplication over addition (and commutativity of multiplication), \((0+0)a=0a+0a\). Thus \(0a=0a+0a\). We add \(-(0a)\) to both sides to see that \(0=0a\), as desired.
[Definition: Ordered field] A field \(F\) is an ordered field if
\(F\) is a field,
\(F\) is totally ordered with respect to some relation on \(F\) (which we will call \(\leq \)),
If \(a\geq 0\) then \((-a)\leq 0\),
\(a-b\geq 0\) if and only if \(a\geq b\),
If \(a\geq 0\) and \(b\geq 0\) then both \(a+b\geq 0\) and \(ab\geq 0\).
(Problem 120) Show that \(\Q \) and \(\R \) (as defined in undergraduate analysis/advanced calculus) are both ordered fields. If you have taken complex analysis, show that \(\C \) is not an ordered field.
[Definition: Complete ordered field] An ordered field \(F\) is complete if, whenever \(E\subset F\) is a nonempty subset with an upper bound, then \(E\) has a least upper bound.
[Definition: Absolute value] If \(x\) is an element of an ordered field, we define \(|x|=x\) if \(x\geq 0\) and \(|x|=-x\) if \(x\leq 0\).
The Triangle inequality. If \(x\) and \(y\) are real numbers (or rational numbers), then \(|x+y|\leq |x|+|y|\).
[Definition: Extended real numbers] We introduce the symbols \(\infty =+\infty \) and \(-\infty \) denoting two objects that are not in \(\R \), and let the extended real numbers be \(\R \cup \{-\infty ,\infty \}\). (The extended real numbers are not a field!)
We extend the relation \(\leq \) by stating \(-\infty \leq r\leq \infty \) for all \(r\in \R \).
We extend the operation \(+\) by writing \begin {gather*} \infty +r=\infty ,\quad -\infty +r=-\infty ,\gatherbreak \infty +\infty =\infty ,\quad -\infty +(-\infty )=-\infty \end {gather*} for all \(r\in \R \). (\(\infty +(-\infty )\) and \((-\infty )+\infty \) are not defined.)
If \(a\), \(b\in \R \cup \{-\infty ,\infty \}\), then \begin {gather*} (a,b)=\{r\in \R \cup \{-\infty ,\infty \}:a<r<b\}, \gatherbreak {} [a,b)=\{r\in \R \cup \{-\infty ,\infty \}:a\leq r< b\}, \\ (a,b]=\{r\in \R \cup \{-\infty ,\infty \}:a< r\leq b\}, \gatherbreak {} [a,b]=\{r\in \R \cup \{-\infty ,\infty \}:a\leq r\leq b\} . \end {gather*} We may write \(\R \cup \{-\infty ,\infty \}=[-\infty ,\infty ]\).
(Problem 130) Show that every subset of the extended real numbers has a supremum in the extended real numbers. (A similar argument shows that every subset of the extended real numbers has an infimum in the extended real numbers.)
Let \(X\subset [-\infty ,\infty ]\). There are then four cases:
\(X=\emptyset \) or \(X=\{-\infty \}\). Then \(-\infty \) is an upper bound for \(X\), because \(x\leq -\infty \) for all \(x\in X\). Furthermore, if \(y\in [-\infty ,\infty ]\) then \(-\infty \leq y\), and so \(-\infty \leq y\) for all upper bounds \(y\) of \(X\). Thus \(-\infty \) is the least upper bound on \(X\).
For every \(r\in \R \) there is an \(x\in X\) with \(x>r\). In particular there is an \(x\in X\) with \(x>0>-\infty \), and so \(-\infty \) is not an upper bound for \(X\). Furthermore, no element of \(\R \) is an upper bound for \(X\). However, \(\infty \) is an upper bound for \([-\infty ,\infty ]\), so is an upper bound for \(X\); as it is the only upper bound
None of the preceding cases is valid. Thus, there is some \(r\in R\) such that \(x\leq r\) for all \(x\in X\). Furthermore, \(X\neq \emptyset \) and \(X\neq \{-\infty \}\). In this case \(\infty \notin X\) (because \(\infty \not <r\)). Thus, \(X\not \subseteq \{-\infty ,\infty \}\), and so \(X\cap \R \neq \emptyset \). \(r\) is an upper bound for \(X\cap \R \), so by the completeness axiom \(X\cap \R \) has a (real) least upper bound \(s\). If \(X=X\cap \R \) we are done; otherwise, \(X\cap \R =X\setminus \{-\infty \}\). \(s\) is an upper bound for \(X\) because \(s>-\infty \), and is the least upper bound for \(X\) because it is the least upper bound for a subset of \(X\). This completes the proof.
[Definition: Inductive set] Suppose that \(F\) is a field and that \(E\subseteq F\). Suppose furthermore that \(1\in E\) and that, if \(x\in E\), then \(x+1\in E\). Then we say that \(E\) is inductive.
[Definition: Natural numbers] If \(F\) is a field, we define the natural numbers \(\N _F\) in \(F\) as the intersection of all inductive subsets of \(F\).
(Problem 140) If \(F\) is an ordered field, show that \(1\in \N _F\) and that \(f\notin \N _F\) for all \(f<1\).
Let \(S=\{x\in F:x\geq 1\}\). Then \(1\in S\). Furthermore, if \(x\in F\) then \(x\geq 1\). Because \(1>0\), we have that \(x+1>x\geq 1\) and so \(x+1\in S\). Thus \(S\) is inductive, and so \(\N _F\subseteq S\). Because \(\geq \) is a partial ordering, if \(f<1\) then \(f\not \geq 1\) and so \(f\notin S\), so \(f\notin \N _F\), as desired.
(Problem 141) If \(n\), \(m\in \N _F\), show that \(n+m\in \N _F\). Phrase your proof in terms of inductive sets rather than in terms of induction as done in undergraduate mathematics.
[Chapter 1, Problem 8] If \(F\) is an ordered field and \(n\in \N _F\), then \(\N _F\cap (n,n+1)=\emptyset \).
[Chapter 1, Problem 9] If \(F\) is an ordered field and \(n\), \(m\in \N _F\) with \(n>m\), then \(n-m\in \N _F\).
Theorem 1.1. Let \(F\) be an ordered field and let \(E\subseteq \N _F\). Suppose that \(E\neq \emptyset \). Then \(E\) has a smallest element.
(Problem 150) Prove Theorem 1.1. For bonus points, prove this in a general ordered field; the proof in your book is only valid in complete ordered fields.
Corollary. Let \(\Z _F=\{n-m:n,m\in \N _F\}\). If \(S\subseteq \Z _F\) and \(S\) is bounded above (respectively, below) then \(S\) contains a maximal (respectively, minimal) element.
The Archimedean property. An ordered field \(F\) has the Archimedean property if, for every \(a\in F\), there is a \(n\in \N _F\) with \(n>a\).
(Problem 160) Let \(F\) be a complete ordered field. Prove that \(F\) has the Archimedean property.
Suppose that \(F\) does not have the Archimedian property. Then there is some \(a\in F\) fuch that \(n\leq a\) for all \(n\in \N _F\).Thus \(a\) is an upper bound for \(\N _F\). Because \(F\) is complete, there is a least upper bound \(c\) of \(\N _F\). Consider \(c-1\). Then \(c-1<c\), and so \(c-1\) is not an upper bound for \(\N _F\), and so there is some \(m\in \N _F\) such that \(c-1<m\). But then \(c<m+1\), and so \(c\) is not an upper bound for \(\N _F\). This is a contradiction.
[Potential homework problem] Let \(F\) and \(R\) be two complete ordered fields. Show that there is a unique bijection \(\varphi :F\to R\) that satisfies
\(\varphi (a+b)=\varphi (a)+\varphi (b)\),
\(\varphi (ab)=\varphi (a)\varphi (b)\),
If \(a\leq b\) then \(\varphi (a)\leq \varphi (b)\).
Thus (up to isomorphism) there is only one complete ordered field.
[Definition: Dense] Let \(F\) be an ordered field. A subset \(S\) of \(F\) is dense if, whenever \(a\), \(b\in F\) with \(a<b\), there is a \(s\in S\) with \(a<s<b\).
Theorem 1.2. The rational numbers \(\Q \) are dense in the real numbers \(\R \).
(Problem 170) Prove Theorem 1.2.
Since \(b-a>0\), we have that \(\frac {2}{b-a}>0\). Thus there is a \(q\in \N \) such that \(q>\frac {1}{b-a}\) (and thus \(1/q<b-a\)) by the Archimedean property.Let \(S=\{n\in \Z :n<qb\}\). \(S\) is clearly bounded above, so by (the corollary to) Theorem 1.1 we have that \(S\) contains a maximal element \(p\).
Thus \(p<qb\). Because \(q\in \N \), we have that \(q>0\) and so \(\frac {p}{q}<b\). Furthermore, \(p+1\) is not in \(S\) and so \(p+1\geq qb\), that is, \(\frac {p+1}{q}\geq b\). But then \(\frac {p}{q}=\frac {p+1}{q}-\frac {1}{q}\geq b-\frac {1}{q}>b-(b-a)=a\), and so \(a<\frac {p}{q}<b\), as desired.
The axiom of countable choice. Let \(X\) be a set. For each \(n\in \N \), let \(E_n\subseteq X\). Then there is a sequence \(\{x_n\}_{n=1}^\infty \) such that, for each \(n\in \N \), we have that \(x_n\in E_n\).
The axiom of dependent choice. Let \(X\) be a set and let \(R\) be a relation on \(X\). Suppose that for each \(x\in X\), there is at least one \(y\in X\) such that \(xRy\).
If \(x\in X\), then there is a sequence (a function from \(\N \) to \(X\)) such that \(x_1=x\) and such that \(x_nRx_{n+1}\) for all \(n\in \N \).
The Bolzano-Weierstrauß theorem. If \(\{x_n\}_{n=1}^\infty \) is a bounded sequence of points in \(\R \), then there is a subsequence \(\{x_{n_k}\}_{k=1}^\infty \) that converges.
(Bonus Problem 171) Use the axiom of dependent choice to prove the Bolzano-Weierstrauß theorem. Can you do this using only the axiom of countable choice?
(Bonus Problem 172) Show that the axiom of dependent choice implies the axiom of countable choice.
(Bonus Problem 180) Use Zorn’s lemma to prove the axiom of dependent choice.
(Problem 181) Choose a standard inductive proof from undergraduate analysis and rephrase it in terms of inductive sets.
[Definition: Finite] A subset of \(\N \) is finite if it is bounded above. An arbitrary set \(S\) is finite if there is a bijection \(f\) from \(S\) to a finite subset of \(\N \).
(Memory 190) If \(S\) is finite, then there is a \(m\in \N \) and a bijection \(f:S\to \{k\in \N :k\leq m\}\).
(Memory 200) If \(S\) is a set, \(T\) is a finite set, and there exists either
An injection \(g:S\to T\),
A surjection \(h:T\to S\),
then \(S\) is also finite.
[Definition: Countable] A set \(S\) is countable if there exists an injection \(g:S\to \N \).
(Memory 210) \(S\) is countable if and only if there exists a surjection \(h:\N \to S\).
(Memory 220) All finite sets are countable.
(Memory 221) If \(S\) is a set, \(T\) is a countable set, and there exists either
An injection \(g:S\to T\),
A surjection \(h:T\to S\),
then \(S\) is also countable.
[Definition: Countably infinite] A set \(S\) is countably infinite if it is countable but not finite.
(Memory 230) \(S\) is countably infinite if and only if there exists a bijection \(f:S\to \N \).
(Memory 231) The Cartesian product of finitely many countable sets is countable.
(Memory 240) \(\Q \) is countable.
(Memory 250) The countable union of countable sets is countable.
[Definition: Uncountable] A set is uncountable if it is not countable.
(Memory 260) The real numbers are uncountable. In fact, if \(I\subseteq \R \) is an interval, then either \(I=\emptyset \), \(I=\{a\}\) is a single point, or \(I\) is uncountable.
(Problem 270) Let \(S\) be a set. Let \(2^S\) (or \(P(S)\)) denote the set of all subsets of \(S\). Show that there does not exist a surjection \(f:S\to 2^S\).
[Definition: Open interval] A subset \(I\) of \(\R \) is an open interval if there exist \(a\), \(b\in [-\infty ,\infty ]\) with \(a\leq b\) such that \(I=(a,b)=\{r\in \R :a<r<b\}\).
(Problem 280) Is the empty set an open interval?
[Definition: Open set] A subset \(\mathcal {O}\) of \(\R \) is open if it is the union of open intervals.
Proposition 1.9. In \(\R \) (but not in other metric spaces), every open set may be written as a union of countably many open intervals that are pairwise-disjoint.
[Definition: Closed set] A set in \(\R \) is closed if its complement is open.
[Definition: Closure] If \(E\subset \R \), then \(\overline E=\bigcap _{F:F\text { closed}, E\subseteq F} F\).
(Memory 290) If \(E\subseteq \R \) then \(\overline E\) is closed. If \(E\subseteq \R \) is closed then \(E=\overline E\).
(Memory 300) If \(E\subseteq \R \) then \(\overline E=\{r\in \R :\)if \(\varepsilon >0\) then there is an \(e\in E\) with \(|r-e|<\varepsilon \}\).
The nested set theorem. If \(\{F_n\}_{n=1}^\infty \) is a sequence of sets such that \(F_n\supseteq F_{n+1}\), \(F_n\) is compact, and \(F_n\neq \emptyset \), then \(\bigcap _{n=1}^\infty F_n\neq \emptyset \).
[Definition: \(\sigma \)-algebra] Let \(X\) be a set and let \(\mathcal {A}\subseteq 2^X\). We say that \(\mathcal {A}\) is a \(\sigma \)-algebra of subsets of \(X\), or a \(\sigma \)-algebra over \(X\), if
(Problem 310) Show that \(2^X\) and \(\{X,\emptyset \}\) are both \(\sigma \)-algebras over \(X\).
(Problem 320) Suppose that \(\mathcal {A}\) is a \(\sigma \)-algebra and that \(\mathcal {B}\subseteq \mathcal {A}\) is countable. Show that \(\bigcap _{S\in \mathcal {B}}S\) is in \(\mathcal {A}\).
(Problem 321) Let \(\mathcal {G}\) be a collection of \(\sigma \)-algebras over \(X\). Let \(\mathcal {A}=\bigcap _{\mathcal {B}\in \mathcal {G}} \mathcal {B}\). Show that \(\mathcal {A}\) is also a \(\sigma \)-algebra over \(X\).
(Problem 322) Is the same true for unions of \(\sigma \)-algebras?
Proposition 1.13. Let \(\mathcal {F}\subseteq 2^X\). Let \begin {equation*} \mathcal {S}=\{\mathcal {A}\subseteq 2^X:\mathcal {F}\subseteq \mathcal {A},\>\mathcal {A}\text { is a $\sigma $-algebra}\}. \end {equation*} Let \(\mathcal {A}_\mathcal {F}=\bigcap _{\mathcal {A}\in \mathcal {S}} \mathcal {A}\). Then \(\mathcal {A}_{\mathcal {F}}\) is a \(\sigma \)-algebra.
Furthermore, \(\mathcal {F}\subseteq \mathcal {A}_{\mathcal {F}}\) and if \(\mathcal {A}\) is a \(\sigma \)-algebra with \(\mathcal {F}\subseteq \mathcal {A}\) then \(\mathcal {A}_{\mathcal {F}}\subseteq \mathcal {A}\).
We call \(\mathcal {A}_{\mathcal {F}}\) the smallest \(\sigma \)-algebra that contains \(\mathcal {F}\).
(Problem 330) Prove Proposition 1.13.
(Problem 340) Let \(\mathcal {A}\) be a \(\sigma \)-algebra on \(X\) and let \(\{E_n\}_{n=1}^\infty \subseteq \mathcal {A}\). Let \begin {equation*} \limsup _{n\to \infty } E_n=\{x\in X:x\in E_n\text { for infinitely many values of }n\}. \end {equation*} Show that \(\limsup _{n\to \infty } E_n\in \mathcal {A}\).
The statement that \(x\in E_n\) for infinitely many values of \(n\) is equivalent to the statement that, for all \(m\in \N \), there is a \(n\geq m\) such that \(x\in E_n\).The statement that there is a \(n\geq m\) such that \(x\in E_n\) is equivalent to the statment that \(x\in \bigcup _{n=m}^\infty E_n\).
Thus, the statement that \(x\in E_n\) for infinitely many values of \(n\) is equivalent to the statement that, for all \(m\in \N \), \(x\in \bigcup _{n=m}^\infty E_n\).
But for any sequence of sets \(\{S_m\}_{m=1}^\infty \), \(x\in S_m\) for all \(m\in \N \) if and only if \(x\in \bigcap _{m=1}^\infty S_m\).
Thus, \begin {equation*} \limsup _{n\to \infty } E_n = \bigcap _{k=1}^\infty \Bigl [\bigcup _{n=k}^\infty E_n\Bigr ] \end {equation*} and the result that \(\limsup _{n\to \infty } E_n\in \mathcal {A}\) follows immediately from the definition of a \(\sigma \)-algebra and from Problem 320.
(Problem 350) Let \(\mathcal {A}\) be a \(\sigma \)-algebra on \(X\) and let \(\{E_n\}_{n=1}^\infty \subseteq \mathcal {A}\). Let \begin {equation*} \liminf _{n\to \infty } E_n=\{x\in X:x\notin E_n\text { for at most finitely many values of }n\}. \end {equation*} Show that \(\liminf _{n\to \infty } E_n\in \mathcal {A}\).
A similar argument to that of the previous problem yields that \begin {equation*} \liminf _{n\to \infty } E_n = \bigcup _{k=1}^\infty \Bigl [\bigcap _{n=k}^\infty E_n\Bigr ] \end {equation*} and so again the result follows from the definition of a \(\sigma \)-algebra and from Problem 320.
[Definition: Borel sets] The collection \(\mathcal {B}\) of Borel subsets of \(\R \) is the smallest \(\sigma \)-algebra containing all open subsets of \(\R \).
[We will assume that all students saw all material in this section in advanced calculus or real analysis.]
[We will assume that all students saw all material in this section in advanced calculus or real analysis.]
[Definition: Characteristic function] Let \(E\subset \R \) be a set. The characteristic function of \(E\) is defined to be \begin {equation*} \chi _E(x)=\begin {cases}1,&x\in E,\\0,&x\in \R \setminus E.\end {cases} \end {equation*}
[Definition: Jordan content] Let \(E\subset \R \) be a bounded set. If \(\chi _E\) is Riemann integrable, then we say that \(E\) is Jordan measurable and that its Jordan content is \(\mathcal {J}(E)=\int _{-\infty }^\infty \chi _E\).
(Problem 360) Suppose that \(\{E_n\}_{n=1}^m\) is a finite collection of pairwise disjoint Jordan measurable sets. Show that \(\cup _{n=1}^m E_n\) is also Jordan measurable and that \begin {equation*} \mathcal {J}\Bigl (\bigcup _{n=1}^m E_n\Bigr )=\sum _{n=1}^m \mathcal {J}(E_n). \end {equation*}
(Problem 370) Find a bounded set \(E\) that is not Jordan measurable, but such that \(E=\bigcup _{n=1}^\infty E_n\), where \(\{E_n\}_{n=1}^\infty \) is a sequence of pairwise disjoint Jordan measurable sets.
Let \(E=\Q \cap [0,1]\). Then \(E\) is countable and so is the union of countably many pairwise disjoint singleton sets. It is an elementary real analysis argument to show that singleton sets are Jordan measurable but that \(E\) is not.
[Definition: Measure] If \(X\) is a set and \(\M \) is a \(\sigma \)-algebra of subsets of \(X\), then we call \((X,\M )\) a measurable space. A measure on a measurable space \((X,\M )\) is a function \(\mu \) such that:
\(\mu :\mathcal {M}\to [0,\infty ]\).
\(\mu (\emptyset )=0\),
If \(\{E_k\}_{k=1}^\infty \subseteq \mathcal {M}\) and \(E_k\cap E_j=\emptyset \) for all \(j\neq k\), then \begin {equation*} \mu \Bigl (\bigcup _{k=1}^\infty E_k\Bigr )=\sum _{k=1}^\infty \mu (E_k). \end {equation*}
(Recall that if \(\mathcal {M}\) is a \(\sigma \)-algebra then \(\emptyset \in \mathcal {M}\).)
[Chapter 2, Problem 1] If \(\mu \) is a measure on a \(\sigma \)-algebra \(\mathcal {M}\), and if \(A\), \(B\in \M \) with \(A\subseteq B\), then \(\mu (A)\leq \mu (B)\).
[Chapter 2, Problem 2] We may replace the second condition by the condition “\(\mu (E)<\infty \) for at least one \(E\in \mathcal {M}\)”.
[Chapter 2, Problem 3] If \(\mu \) is a measure on a \(\sigma \)-algebra \(\mathcal {M}\), and if \(\{E_k\}_{k=1}^\infty \subseteq \mathcal {M}\), then \begin {equation*} \mu \Bigl (\bigcup _{k=1}^\infty E_k\Bigr )\leq \sum _{k=1}^\infty \mu (E_k). \end {equation*}
[Chapter 2, Problem 4] Let \((X,\M )\) be a measurable space. If \(E\in \M \), let \(c(E)=\infty \) if \(E\) is infinite and \(c(E)=\#E\) if \(E\) is finite. Then \(c\) is a measure on \((X,\M )\).
(Problem 380) Let \((X,\M )\) be a measurable space. There are functions from \(\M \) to \(\{0,\infty \}\) that are measures. Find two of them. (These are the trivial measures on \(\M \).)
[Definition: Length] Let \(I\subseteq \R \) be an interval, so \(I=(a,b)\), \([a,b]\), \([a,b)\), or \((a,b]\) for some \(a\), \(b\in [-\infty ,\infty ]\) with \(a\leq b\). We define \(\ell (I)=b-a\).
[Definition: Outer measure] Let \(A\subseteq \R \). The Lebesgue outer measure of \(A\) is \begin {align*} m^*(A)=\inf \Bigl \{\sum _{k=1}^\infty \ell (I_k):{}&A\subseteq \bigcup _{k=1}^\infty I_k,\alignbreak \text { each $I_k$ is a bounded open interval}\Bigr \}. \end {align*}
(Problem 390) Let \(A\subseteq B\subseteq \R \). Show that \(m^*(A)\leq m^*(B)\).
(Problem 391) Let \(S\subset \R \) be a finite set. Show that \(m^*(S)=0\).
(Problem 400) Let \(S\subset \R \) be a countably infinite set. Show that \(m^*(S)=0\). In particular, \(m^*(\Q )=0\).
Proposition 2.1. The outer measure of an interval in \(\R \) is its length.
(Problem 410) Let \(I=[a,b]\) be a closed bounded interval (that is, \(a\) and \(b\) are both real numbers). Show that \(m^*(I)=\ell (I)=b-a\).
(Problem 420) Prove Proposition 2.1.
Proposition 2.2. Outer measure is translation invariant: if \(E\subseteq \R \) and \(r\in \R \), then \begin {equation*} m^*(E)=m^*\bigl (\{e+r:e\in E\}\bigr ). \end {equation*}
(Problem 430) Prove Proposition 2.2.
Let \(\{I_k\}_{k=1}^\infty \) be a countable sequence of bounded open intervals that satisfies \(E\subseteq \bigcup _{k=1}^\infty I_k\). There are real numbers \(a_k\), \(b_k\) with \(a_k\leq b_k\) and \(I_k=(a_k,b_k)\).Let \(J_k=(a_k+r,b_k+r)\).
If \(f\in E_r=\{e+r:e\in E\}\), then \(f=e+r\) for some \(e\in E\). Because \(E\subseteq \bigcup _{k=1}^\infty I_k\), we have that \(e\in I_k\) (that is, \(a_k<e<b_k\)) for at least one value of \(k\). Then \(a_k+r<e+r<b_k+r\), so \(f\in J_k\). Thus \(E_r\subset \bigcup _{k=1}^\infty J_k\). But \(\{J_k:k\in \N \}\) is a countable collection of bounded open intervals, and so by definition of Lebesgue outer measure \begin {equation*} m^*(E_r)\leq \sum _{k=1}^\infty \ell (J_k)=\sum _{k=1}^\infty (b_k-a_k)=\sum _{k=1}^\infty \ell (I_k). \end {equation*} Taking the infimum of both sides over the set of all such \(\{I_k\}_{k=1}^\infty \), we see that \begin {equation*} m^*(E_r)\leq m^*(E). \end {equation*} But \(E=(E_r)_{-r}\), and so by the previous argument \(m^*(E)=m^*((E_r)_{-r})\leq m^*(E_r)\). Thus \(m^*(E)=m^*(E_r)\), as desired.
(Problem 431) If \(E\subseteq \R \) and \(r\in \R \), then \begin {equation*} |r|m^*(E)=m^*\bigl (\{re:e\in E\}\bigr ). \end {equation*}
Proposition 2.3. If for each \(k\in \N \) we have a set \(E_k\subseteq \R \), then \begin {equation*} m^*\Bigl (\bigcup _{k=1}^\infty E_k\Bigr )\leq \sum _{k=1}^\infty m^*(E_k). \end {equation*}
(Problem 440) Prove Proposition 2.3.
If \(\sum _{k=1}^\infty m^*(E_k)=\infty \) we are done because \(m^*(S)\in [0,\infty ]\) for all \(S\subseteq \R \). Therefore we assume that \(\sum _{k=1}^\infty m^*(E_k)<\infty \). In particular, \(m^*(E_k)<\infty \) for all \(k\in \N \).Let \(\varepsilon >0\). By definition of infimum and by definition of \(m^*\), for each \(k\), there is a sequence \(\{I_{k,\ell }\}_{\ell =1}^\infty \) of bounded open intervals that satisfies \begin {equation*} E_k\subseteq \bigcup _{\ell =1}^\infty I_{k,\ell }, \quad \sum _{\ell =1}^\infty \ell (I_{k,\ell })\leq m^*(E_k)+2^{-k}\varepsilon . \end {equation*} Now, \(\mathcal {B}=\{I_{k,\ell }:k,\ell \in \N \}\) is the union of countably many collections of countably many bounded open intervals, and so by Problem 250 \(\mathcal {B}\) is a countable set of bounded open intervals.
Furthermore, if \(x\in \bigcup _{k=1}^\infty E_k\) then \(x\in E_k\) for some \(k\), and so \(x\in I_{k,\ell }\) for some \(\ell \). Thus \(E_k\subset \bigcup _{I\in \mathcal {B}}I\).
Thus by definition of \(m^*\), \begin {equation*} m^*\Bigl (\bigcup _{k=1}^\infty E_k\Bigr ) \leq \sum _{I\in \mathcal {B}} \ell (I). \end {equation*} By definition of \(\mathcal {B}\) \begin {equation*} m^*\Bigl (\bigcup _{k=1}^\infty E_k\Bigr ) \leq \sum _{k=1}^\infty \sum _{\ell =1}^\infty \ell (I_{k,\ell }) . \end {equation*} By definition of \(I_{k,\ell }\), \begin {equation*} m^*\Bigl (\bigcup _{k=1}^\infty E_k\Bigr )\leq \sum _{k=1}^\infty \bigl (m^*(E_k)+2^{-k}\varepsilon \bigr ) =\varepsilon +\sum _{k=1}^\infty m^*(E_k). \end {equation*} Because this is true for all \(\varepsilon >0\), we must have that \begin {equation*} m^*\Bigl (\bigcup _{k=1}^\infty E_k\Bigr )\leq \sum _{k=1}^\infty m^*(E_k) \end {equation*} as desired.
[Chapter 2, Problem 9] If \(A\), \(B\subseteq \R \) and \(m^*(A)=0\), then \(m^*(A\cup B)=m^*(A)+m^*(B)\).
[Chapter 2, Problem 10] Let \(A\), \(B\subset \R \) be bounded. Suppose that \(\inf \{|a-b|:a\in A,b\in B\}>0\). Show that \(m^*(A\cup B)=m^*(A)+m^*(B)\).
[Definition: Rationally equivalent] If \(r\), \(s\in \R \), we say that \(r\) and \(s\) are rationally equivalent if \(r-s\in \Q \).
(Problem 450) Show that rational equivalence is an equivalence relation.
Let \(R\) denote rational equivalence.
Suppose \(rRs\). Then \(r-s\in \Q \). Because \(\Q \) is a field, \(-(r-s)=s-r\in \Q \) and so \(sRr\). Thus \(R\) is symmetric.
Suppose \(r\in \R \). Then \(r-r=0\in \Q \) and so \(rRr\). Thus \(R\) is reflexive.
Suppose \(rRs\) and \(sRt\). Then \(r-s\in \Q \) and \(s-t\in \Q \). Because \(\Q \) is a field, \(r-t=(r-s)+(s-t)\in \Q \) and so \(rRt\). Thus \(R\) is transitive.
Because \(R\) is symmetric, reflexive, and transitive, it is an equivalence relation.
(Problem 451) Let \(R\) denote rational equivalence. Let \([-1,1]/R\) be the set of equivalence classes in \([-1,1]\) under \(R\). This is a collection of (pairwise disjoint) sets.
If \(S\in [-1,1]/R\), is \(S\) finite, countably infinite, or uncountable?
Let \(x\in S\). Then \(S=\{y\in [-1,1]:xRy\}=\{y\in [-1,1]:y=x+q\) for some \(q\in \Q \}\). \(S\) is clearly infinite. Conversely, \(S\subset \{x+q:q\in \Q \}\). This set is countable because it has a natural bijection to the countable set \(\Q \). Thus \(S\) is countably infinite.
(Problem 452) Is the collection of equivalence classes \([-1,1]/R\) finite, countably infinite, or uncountable?
We have that \begin {equation*} [-1,1]=\bigcup _{S\in [-1,1]/R} S. \end {equation*} The right hand side is uncountable but each \(S\) on the left hand side is countable. The countable union of countable sets is countable, and so we must have that \([-1,1]/R\) is uncountable.
(Problem 453) Let \(\varphi \) be a choice function on \([-1,1]/R\) and let \(V=\varphi ([-1,1]/R)\). Is \(V\) countable or uncountable?
The elements of \([-1,1]/R\) are pairwise disjoint because \(R\) is an equivalence relation. Thus, if \(\varphi (S)=\varphi (T)\), then \(\varphi (T)\in S\cap T\) because \(\varphi \) is a choice function and so \(\varphi (S)\in S\), \(\varphi (T)\in T\). Thus \(S\cap T\neq \emptyset \) and so \(S=T\). Thus \(\varphi \) is a bijection from \([-1,1]/R\) to \(V\). Because \([-1,1]/R\) is uncountable, so is \(V\).
(Problem 460) If \(q\) is rational, define \(V_q=\{v+q:v\in V\}\). Show that if \(q\), \(p\in \mathbb {Q}\cap [-2,2]\) then either \(q=p\) or \(V_q\cap V_p=\emptyset \).
Let \(p\), \(q\in \mathbb {Q}\cap [-2,2]\) and suppose that \(V_q\cap V_p\neq \emptyset \). Let \(x\in V_q\cap V_p\). Then \(x=v+q=w+p\) for some \(v\), \(w\in V\) by definition of \(V_q\). We must have that \(v=\varphi (S)\), \(w=\varphi (T)\) for some \(S\), \(T\in [-1,1]/R\) by definition of \(V\).Thus \(v-w=p-q\in \Q \) and so \(vRw\). Because \(\varphi \) is a choice function, \(v=\varphi (S)\in S\). Because \(vRw\), we must also have \(w\in S\). But \(w=\varphi (T)\in T\). As in the previous problem this implies \(S=T\), and so \(v=\varphi (S)=\varphi (T)=w\). Thus \(0=v-w=p-q\) and so \(p=q\), as desired.
(Problem 470) Show that \begin {equation*} [-1,1]\subseteq \bigcup _{q\in [-2,2]\cap {\Q }} V_q\subseteq [-3,3]. \end {equation*}
(Problem 480) Show that \(m^*(V)>0\).
By Proposition 2.1 and Problem 390, we have that \begin {equation*} 2=m^*([-1,1])\leq m^*\Bigl (\bigcup _{q\in [-2,2]\cap {\Q }} V_q\Bigr ). \end {equation*} By Proposition 2.3, and because the rational numbers are countable, we have that \begin {equation*} 2\leq \sum _{q\in [-2,2]\cap {\Q }} m^*(V_q). \end {equation*}By Proposition 2.2, we have that \(m^*(V_p)=m^*(V_q)\) for all \(p\), \(q\in \Q \). Thus \begin {equation*} 2\leq \sum _{q\in [-2,2]\cap {\Q }} m^*(V). \end {equation*} The right hand side is either zero (if \(m^*(V)=0\)) or \(\infty \) (if \(m^*(V)>0\)). The first possibility is precluded by the positive lower bound, and thus we must have that \(m^*(V)>0\).
(Problem 490) Let \(\{q_k\}_{k=1}^\infty \) be a sequence that contains each rational number in \([-2,2]\) exactly once and contains no other numbers. Show that \(\sum _{k=1}^\infty m^*(V_{q_k}) \neq m^*\Bigl (\bigcup _{k=1}^\infty V_{q_k}\Bigr )\).
(Problem 500) Show that there exist two disjoint sets \(A\) and \(B\) such that \(m^*(A\cup B) \neq m^*(A)+m^*(B)\).
[Definition: Measurable set] Let \(E\subseteq \R \). We say that \(E\) is Lebesgue measurable (or just measurable) if, for all \(A\subseteq \R \), we have that \begin {equation*} m^*(A)=m^*(A\cap E)+m^*(A\setminus E). \end {equation*}
(Problem 501) Show that \(E\subseteq \R \) is measurable if and only if, for all \(A\subseteq \R \), we have that \begin {equation*} m^*(A)\geq m^*(A\cap E)+m^*(A\setminus E). \end {equation*}
This is an easy consequence of Proposition 2.3.
(Problem 510) Show that the complement of a measurable set is measurable.
(Problem 520) Show that the empty set is measurable.
Proposition 2.10. Let \(E\subseteq \R \) be a measurable set. Let \(y\in \R \). Define \(E+y=\{e+y:e\in E\}\). Show that \(E+y\) is also measurable.
(Problem 530) Prove Proposition 2.10.
(Problem 531) Let \(E\subseteq \R \) be a measurable set. Let \(r\in \R \). Define \(rE=\{re:e\in E\}\). Show that \(rE\) is also measurable.
Proposition 2.4. Let \(E\subset \R \). Suppose that \(m^*(E)=0\). Then \(E\) is measurable.
(Problem 540) Prove Proposition 2.4.
Proposition 2.5. The union of finitely many measurable sets is measurable.
(Problem 550) Prove Proposition 2.5.
Let \(S=\{n\in \N :\)the union of \(n\) measurable sets is measurable\(\}\). It is trivially true that \(1\in S\). We will now show that \(S\) is inductive and thus \(S=\N \); this will complete the proof.Suppose that \(n\in S\). Let \(\{E_k\}_{k=1}^{n+1}\) be a set of \(n+1\) measurable sets. Let \(F=\bigcup _{k=1}^n E_k\); because \(n\in S\) we have that \(F\) is measurable. Let \(G=E_{n+1}\); by assumption \(G\) is measurable. We need only show that \(F\cup G=\bigcup _{k=1}^{n+1}\) is measurable; because \(\{E_k\}_{k=1}^{n+1}\) was arbitrary this will imply that \(n+1\in S\), which will show that \(S\) is inductive and thus complete the proof.
Let \(A\subseteq \R \). Applying the measurability of \(F\) yields that \begin {align*} m^*(A) &= m^*(A\cap F)+m^*(A\cap F^C) \end {align*}
where the superscript \(C\) denotes the complement. Applying the measurability of \(G\) yields that \begin {align*} m^*(A) &= m^*(A\cap F)+m^*(G\cap (A\cap F^C)) \\&\qquad +m^*(( A\cap F^C)\cap G^C) . \end {align*}
Set intersection is associative and so \(( A\cap F^C)\cap G^C=A\cap (F^C\cap G^C)\). Recall that \(F^C=\R \setminus F\). By De Morgan’s laws (Problem 10), \begin {gather*} F^C\cap G^C=(\R \setminus F)\cap (\R \setminus G) \gatherbreak =\R \setminus (F\cup G)=(F\cup G)^C. \end {gather*} Thus \begin {align*} m^*(A) &= m^*(A\cap F)+m^*(G\cap (A\cap F^C)) \\&\qquad +m^*(A\cap (F\cup G)^C) . \end {align*}
By Proposition 2.2, \begin {equation*} m^*(A\cap F)+m^*(G\cap (A\cap F^C)) \geq m^*((A\cap F)\cup (G\cap (A\cap F^C))) . \end {equation*} Again using associativity of intersection, \begin {equation*} G\cap (A\cap F^C)=A\cap (G\cap F^C). \end {equation*} Because unions distribute over intersections, \begin {equation*} (A\cap F)\cup (A\cap (G\cap F^C)) = A\cap (F\cup (G\cap F^C)). \end {equation*} But \(F\cup (G\cap F^C)=F\cup G\) and so \begin {align*} m^*(A) &\geq m^*(A\cap (F\cup G))+m^*(A\cap (F\cup G)^C) . \end {align*}
By Proposition 2.2, \(m^*(A)\leq m^*(A\cap (F\cup G))+m^*(A\cap (F\cup G)^C)\), and so we must have that \begin {align*} m^*(A) &= m^*(A\cap (F\cup G))+m^*(A\cap (F\cup G)^C) \end {align*}
for any set \(A\). Thus \(F\cup G\) is measurable, as desired.
Proposition 2.6. If \(\{E_k\}_{k=1}^n\) is a collection of finitely many measurable pairwise disjoint sets, then \begin {equation*} m^*\Bigl (\bigcup _{k=1}^n E_k\Bigr )=\sum _{k=1}^n m^*(E_k). \end {equation*} More generally, if \(A\subseteq \R \) then \begin {equation*} m^*\Bigl (\bigcup _{k=1}^n (A\cap E_k)\Bigr )=\sum _{k=1}^n m^*(A\cap E_k). \end {equation*}
(Problem 560) Prove Proposition 2.6.
Let \(A\subseteq \R \). We prove by induction. The base case \(n=1\) is trivially true.Suppose that the proposition is true for some \(n\), that is, that \begin {equation*} m^*\Bigl (\bigcup _{k=1}^n (A\cap E_k)\Bigr )=\sum _{k=1}^n m^*(A\cap E_k) \end {equation*} whenever \(\{E_k\}_{k=1}^n\) is a collection of \(n\) measurable sets. Let \(\{E_k\}_{k=1}^{n+1}\) be a collection of \(n+1\) measurable sets.
Because \(E_{n+1}\) is measurable, we have that \begin {align*} m^*(A\cap \bigcup _{k=1}^{n+1} E_k ) &= m^*\Bigl (\Bigl (A\cap \bigcup _{k=1}^{n+1} E_k \Bigr )\cap E_{n+1} \Bigr ) \alignbreak +m^*\Bigl (\Bigl (A\cap \bigcup _{k=1}^{n+1} E_k \Bigr )\setminus E_{n+1}\Bigr ). \end {align*}
Now, \begin {equation*} \Bigl (A\cap \bigcup _{k=1}^{n+1} E_k \Bigr )\cap E_{n+1} = A\cap \Bigl (\bigcup _{k=1}^{n+1} E_k\cap E_{n+1} \Bigr ) =A\cap E_{n+1} \end {equation*} because the \(E_k\)s are pairwise disjoint.
Similarly, \begin {align*} \Bigl (A\cap \bigcup _{k=1}^{n+1} E_k \Bigr )\setminus E_{n+1} &=\Bigl (A\cap \bigcup _{k=1}^{n+1} E_k \Bigr )\cap E_{n+1}^C \alignbreak = A\cap \Bigl (\bigcup _{k=1}^{n+1} E_k\cap E_{n+1}^C \Bigr ) = A\cap \Bigl (\bigcup _{k=1}^{n} E_k\Bigr ) . \end {align*}
Thus \begin {align*} m^*(A\cap \bigcup _{k=1}^{n+1} E_k ) &= m^*(A\cap E_{n+1}) +m^*\Bigl (A\cap \Bigl (\bigcup _{k=1}^{n} E_k\Bigr )\Bigr ). \end {align*}
By our induction hypothesis, the second term is equal to \begin {equation*} \sum _{k=1}^n m^*(A\cap E_k) \end {equation*} and so \begin {align*} m^*(A\cap \bigcup _{k=1}^{n+1} E_k ) &= \sum _{k=1}^{n+1} m^*(A\cap E_k). \end {align*}
A standard inductive argument completes the proof.
(Problem 561) Show that the intersection of finitely many measurable sets is measurable.
Let \(\{E_k\}_{k=1}^n\) be a collection of finitely many measurable sets.By Problem 510, each \(E_k^C=\R \setminus E_k\) is measurable.
By Proposition 2.4, \(\bigcup _{k=1}^n E_k^C\) is measurable.
Again by Problem 510, \(\R \setminus \left (\bigcup _{k=1}^n E_k^C\right )\) is measurable.
Finally, by Problem 10, \begin {equation*} \R \setminus \left (\bigcup _{k=1}^n E_k^C\right )=\bigcap _{k=1}^n (\R \setminus E_k^C)=\bigcap _{k=1}^n E_k \end {equation*} and so \(\bigcap _{k=1}^n E_k\) is measurable.
Proposition 2.13. Let \(\{E_k\}_{k=1}^\infty \) be a sequence of pairwise disjoint measurable sets. Then \begin {equation*} m^*\Bigl (\bigcup _{k=1}^\infty E_k\Bigr )=\sum _{k=1}^\infty m^*(E_k). \end {equation*}
[Chapter 2, Problem 26]If \(A\subseteq \R \) and \(\{E_k\}_{k=1}^\infty \) is as in Proposition 2.13, then \begin {equation*} m^*\Bigl (\bigcup _{k=1}^\infty (A\cap E_k)\Bigr )=\sum _{k=1}^\infty m^*(A\cap E_k). \end {equation*}
The inequality \begin {equation*} m^*\Bigl (\bigcup _{k=1}^\infty (A\cap E_k)\Bigr )\leq \sum _{k=1}^\infty m^*(A\cap E_k) \end {equation*} is Proposition 2.3.By Problem 390, if \(n\in \N \) then \begin {equation*} m^*\Bigl (\bigcup _{k=1}^\infty (A\cap E_k)\Bigr )\geq m^*\Bigl (\bigcup _{k=1}^n (A\cap E_k)\Bigr ). \end {equation*} By Proposition 2.6 we have that \begin {equation*} m^*\Bigl (\bigcup _{k=1}^\infty (A\cap E_k)\Bigr )\geq \sum _{k=1}^n m^*(A\cap E_k). \end {equation*} Taking the supremum over \(n\) yields that \begin {equation*} m^*\Bigl (\bigcup _{k=1}^\infty (A\cap E_k)\Bigr )\geq \sum _{k=1}^\infty m^*(A\cap E_k) \end {equation*} as desired.
(Problem 570) Let \(\{E_k\}_{k=1}^\infty \) be a sequence of measurable sets. For each \(n\), let \begin {equation*} F_n=E_n\setminus \bigcup _{k=1}^{n-1} E_k, \qquad G_n=\bigcup _{k=1}^n E_k. \end {equation*} Show that
Each \(F_n\) is measurable.
If \(m\in \N \) then \(G_m=\bigcup _{n=1}^m F_n\).
By Proposition 2.5, \(\bigcup _{k=1}^{n-1} E_k\) is measurable. By Problem 510, the complement \(\R \setminus \bigcup _{k=1}^{n-1} E_k\) is measurable. By Problem 561, \(E_n\cap \Bigl (\R \setminus \bigcup _{k=1}^{n-1} E_k\Bigr )\) is measurable. It is straightforward to establish that this final expression is equal to \(F_n\).Because \(F_n\subseteq E_n\subseteq G_m\) whenever \(n\leq m\), we have that \(G_m\supseteq \bigcup _{n=1}^m F_n\). Conversely, let \(e\in G_m\). Then \(e\in E_k\) for some \(k\leq m\). Let \(n\) be the smallest natural number with \(e\in E_n\). Then \(e\notin \bigcup _{k=1}^{n-1} E_k\), and so \(e\in F_n\). Thus \(G_m\subseteq \bigcup _{n=1}^m F_n\). This completes the proof.
Each \(G_n\) is measurable.
If the \(E_n\)s are pairwise disjoint then \(E_n=F_n\).
If \(n< m\) then \(F_n\cap F_m=\emptyset \).
If \(n< m\) then \(G_n\subseteq G_m\).
\(\bigcup _{n=1}^\infty E_n =\bigcup _{n=1}^\infty F_n =\bigcup _{n=1}^\infty G_n\).
Proposition 2.7. The union of countably many measurable sets is measurable.
(Problem 590) Prove Proposition 2.7.
Let \(\{E_k\}_{k=1}^\infty \) be a sequence of measurable sets and let \(F_n\), \(G_n\) be as in the previous problem. It suffices to show that \(E=\bigcup _{n=1}^\infty F_n\) is measurable.Let \(A\subseteq \R \). By Problem 501, it suffices to show that \begin {equation*} m^*(A)\geq m^*(A\cap E)+m^*(A\setminus E) \end {equation*} for all such \(A\).
If \(m^*(A)=\infty \) there is nothing to prove, so assume that \(m^*(A)<\infty \). By Problem 2.26, \begin {equation*} m^*(A\cap E)=m^*\Bigl (A\cap \bigcup _{k=1}^\infty F_k\Bigr ) =\sum _{k=1}^\infty m^*(A\cap F_k). \end {equation*} The left hand side is at most \(m^*(A)<\infty \), and the right hand side is a sum of nonnegative real numbers, and so the sum must converge. In particular, we must have that \(\sum _{k=n+1}^\infty m^*(A\cap F_k)\to 0\) as \(n\to \infty \). Thus \begin {equation*} m^*(A\cap E)=\lim _{n\to \infty }\sum _{k=1}^n m^*(A\cap F_k). \end {equation*} By Proposition 2.6 and by the previous two problems, \begin {equation*} m^*(A\cap E)=\lim _{n\to \infty }m^*(A\cap G_n). \end {equation*} Because \(G_n\) is measurable and \(m^*(A)<\infty \), \begin {equation*} m^*(A\cap E)=\lim _{n\to \infty }m^*(A)-m^*(A\setminus G_n). \end {equation*} But \(G_n\subseteq E\) and so \(A\setminus G_n\supseteq A\setminus E\), and so \begin {equation*} m^*(A\cap E)\leq m^*(A)-m^*(A\setminus E). \end {equation*} Rearranging completes the proof.
[Homework 2.1] The collection of Borel sets \(\mathcal {B}\) is the smallest \(\sigma \)-algebra that contains \(\{(-\infty ,a):a\in \mathbb {R}\}\).
[Homework 3.1] If \(A\), \(B\subseteq \R \) and \(\sup A\leq \inf B\), then \(m^*(A\cup B)=m^*(A)+m^*(B)\).
(Problem 591) If \(a\in \R \) then \((-\infty ,a)\) is measurable. (Note that we use Homework 3.1 to prove this, and so you may not use this fact to do Homework 3.1.)
Theorem 2.9. The collection \(\mathcal {M}\) of Lebesgue measurable subsets of \(\R \) is a \(\sigma \)-algebra on \(\R \) and contains the Borel sets.
(Problem 600) Show that the collection of Lebesgue measurable subsets of \(\R \) is a \(\sigma \)-algebra on \(\R \).
(Problem 610) Prove Theorem 2.9.
(Problem 611) Let \(\mathcal {M}\) denote the collection of Lebesgue measurable subsets of \(\R \). Then \((\R ,\mathcal {M})\) is a measurable space, and \(m^*\big \vert _{\mathcal {M}}\) is a measure on \((\R ,\mathcal {M})\).
[Definition: Lebesgue measure] We let \(m=m^*\big \vert _{\mathcal {M}}\) and refer to \(m\) as the Lebesgue measure.
(Problem 620) Let \(E\subseteq \R \) be measurable. Show that \(E=\bigcup _{n=1}^\infty F_n\), where each \(F_n\) is bounded and measurable and where \(F_n\cap F_m=\emptyset \) if \(n\neq m\).
Let \(E_1=(-1,1)\) and let \(E_n=(-n,n)\setminus (1-n,n-1)\) for all \(n\geq 2\). Because the measurable sets form a \(\sigma \)-algebra and contain the Borel sets, we have that each \(E_n\) is measurable. Clearly each \(E_n\) is bounded. Furthermore, the \(E_n\)s are pairwise disjoint. Let \(F_n=E_n\cap E\). Then \(F_n\subseteq E_n\), and so \(F_n\) is bounded and the \(F_n\)s are pairwise disjoint. The fact that \(E=\bigcup _{n=1}^\infty F_n\) follows from the fact that \(R=\bigcup _{n=1}^\infty E_n\), and the fact that the \(F_n\)s are measurable follows from the measurability of \(E\) and from Problem 561.
[Definition: \(G_\delta \)-set] A set \(G\subseteq \R \) is a \(G_\delta \)-set if there is a sequence \(\{\mathcal {O}_n\}_{n=1}^\infty \) of countably many open sets such that \(G=\bigcap _{n=1}^\infty \mathcal {O}_n\).
[Definition: \(F_\sigma \)-set] A set \(F\subseteq \R \) is a \(F_\sigma \)-set if there is a sequence \(\{\mathcal {C}_n\}_{n=1}^\infty \) of countably many closed sets such that \(F=\bigcup _{n=1}^\infty \mathcal {C}_n\).
(Problem 630) Show that all \(F_\sigma \) sets and all \(G_\delta \) sets are measurable.
Theorem 2.11. Let \(E\subseteq \R \) and let \(E^C=\R \setminus E\). The following statements are equivalent:
Recall [Problem 510]: (v) and (vi) are equivalent.
(Problem 640) Show that (ii) and (iv) are equivalent.
(Problem 650) Show that (i) and (iii) are equivalent.
(Problem 651) Show that (v) implies (i) in the special case where \(m^*(E)<\infty \).
(Problem 660) Show that (v) implies (i) in general.
(Problem 670) Show that (iii) implies (iv).
(Problem 680) Show that (iv) implies (vi).
Theorem 2.12. Let \(E\subset \R \) be a measurable set with finite measure. Let \(\varepsilon >0\). Show that there is a collection of finitely many pairwise disjoint open intervals \(\{I_k\}_{k=1}^n\) such that \begin {equation*} m\Bigl (E\setminus \bigcup _{k=1}^n I_k\Bigr ) +m\Bigl (\bigcup _{k=1}^n I_k\setminus E\Bigr )<\varepsilon . \end {equation*}
(Problem 690) Prove Theorem 2.12.
Let \(\mathcal {O}\) be as in Theorem 2.11 with \(\varepsilon \) replaced by \(\varepsilon /3\). By Proposition 1.9, there is a sequence of pairwise disjoint open intervals with \(\mathcal {O}=\bigcup _{k=1}^\infty I_k\).Observe that by Problem 390, Proposition 2.13, and Proposition 2.1, \begin {equation*} m(E)\leq m(\mathcal {O})=\sum _{k=1}^\infty m(I_k)=\sum _{k=1}^\infty \ell (I_k)<m(E)+\varepsilon /2. \end {equation*} In particular each \(I_k\) is bounded. There is also an \(n\in \N \) such that \(\sum _{k=1}^n \ell (I_k)>m(E)-\varepsilon /3\).
Then \begin {equation*} E\setminus \bigcup _{k=1}^n I_k \subseteq \mathcal {O}\setminus \bigcup _{k=1}^n I_k =\bigcup _{k=n+1}^\infty I_k \end {equation*} and so \begin {equation*} m\Bigl (E\setminus \bigcup _{k=1}^n I_k\Bigr )\leq \sum _{k=n+1}^\infty \ell (I_k) =\sum _{k=1}^\infty \ell (I_k) - \sum _{k=1}^n \ell (I_k) < m(E)+\varepsilon /3-(m(E)-\varepsilon /3)=2\varepsilon /3. \end {equation*} Conversely, \begin {equation*} \bigcup _{k=1}^n I_k\setminus E \subseteq \mathcal {O}\setminus E \end {equation*} and so \begin {equation*} m\Bigl (\bigcup _{k=1}^n I_k\setminus E\Bigr ) \leq m(\mathcal {O}\setminus E)<\varepsilon /3. \end {equation*} This completes the proof.
(Problem 700) Why do we need the assumption that \(E\) has finite outer measure?
(Problem 710) Give an example of a measurable set \(E\subset \R \) with finite measure and a \(\varepsilon >0\) such that there is no collection \(\{I_k\}_{k=1}^n\) of finitely many pairwise disjoint open intervals with \(\bigcup _{k=1}^n I_k\subseteq E\) and with \(m\Bigl (E\setminus \bigcup _{k=1}^n I_k\Bigr )<\varepsilon \). Note: The empty set is an open interval, and the empty collection is a finite collection of open intervals.
Let \(E=[0,1]\setminus \Q \). There are no nonempty open intervals that are subsets of \(E\), and so \(m\Bigl (E\setminus \bigcup _{k=1}^n I_k\Bigr )=m(E)=1\) for all finite collections \(\{I_k\}_{k=1}^n\) of open intervals contained in \(E\).
(Problem 720) Give an example of a measurable set \(E\subset \R \) with finite measure and a \(\varepsilon >0\) such that there is no collection \(\{I_k\}_{k=1}^n\) of finitely many pairwise disjoint open intervals with \(\bigcup _{k=1}^n I_k\supseteq E\) and with \(m^*\Bigl (\bigcup _{k=1}^n I_k\setminus E\Bigr )<\varepsilon \).
Let \(E=[0,1]\cap \Q \). Suppose \(E\subseteq \bigcup _{k=1}^n I_k\) where each \(I_k\) is an open interval. Then \(\overline {E}\subseteq \overline {\bigcup _{k=1}^n I_k} = \bigcup _{k=1}^n \overline {I_k}\), where \(\overline {E}\) denotes the closure of \(E\). But \(\overline {E}=[0,1]\) and \(\overline {I_k}\setminus I_k\) contains exactly two points, so \(\bigcup _{k=1}^n I_k\) contains all but finitely many points of \([0,1]\). Thus \(\sum _{k=1}^n\ell (I_k)\geq m^*([0,1])=1\). Because \(E\) has measure zero and is measurable, \(m^*\Bigl (\bigcup _{k=1}^n I_k\setminus E\Bigr )=m^*\Bigl (\bigcup _{k=1}^n I_k\Bigr )\geq 1\), and so no such collection can have measure less than \(\varepsilon \) for any \(\varepsilon <1\).
[Chapter 2, Problem 19] Suppose that \(E\) is not measurable but does have finite outer measure. Show that there is an open set \(\mathcal {O}\) containing \(E\) such that \(E\subset \mathcal {O}\) but such that \(m^*(\mathcal {O}\setminus E)\neq m^*(\mathcal {O})-m^*(E)\).
[Chapter 2, Problem 20] In the definition of measurable set, it suffices to check for sets \(A\) that are bounded open intervals; that is, \(E\subseteq \R \) is measurable if and only if, for every \(a<b\), we have that \begin {equation*} b-a=m^*((a,b)\cap E)+m^*((a,b)\setminus E). \end {equation*}
(Memory 721) Let \(X\), \(Y\) be two metric spaces. Let \(f:X\to Y\) and let \(f_n:X\to Y\) for each \(n\in \N \). Suppose that each \(f_n\) is continuous and that \(f_n\to f\) uniformly on \(X\). Then \(f\) is continuous.
(Memory 722) A sequence of functions \(f_n:X\to Y\) is uniformly Cauchy if, for every \(\varepsilon >0\), there is a \(K\in \N \) such that if \(m\), \(n\in \N \) with \(m\geq n\geq K\), then \(d(f_n(x),f_m(x))<\varepsilon \) for all \(x\in X\). Suppose that \(\{f_n\}_{n=1}^\infty \) is uniformly Cauchy and that \(Y\) is complete. Then \(f_n\to f\) uniformly for some function \(f:X\to Y\).
(Memory 723) (The intermediate value theorem.) Suppose that \(a<b\) and that \(f:[a,b]\to \R \) is continuous. If \(f(a)\leq y\leq f(b)\), then there is an \(x\in [a,b]\) such that \(f(x)=y\).
(Memory 724) (Definition of interval) A set \(I\subseteq \R \) is an interval if, whenever \(a<b<c\) and \(a\), \(c\in I\), we also have that \(b\in I\). Then \(\{I\subseteq \R :I\) is an interval\(\}\) is the union of the following ten collections of sets:
\(\{\emptyset \}\)
\(\{\R \}\)
\(\{(a,b):a<b\) and \(a\), \(b\in \R \}\)
\(\{[a,b:a<b\) and \(a\), \(b\in \R \}\)
\(\{[a,b):a<b\) and \(a\), \(b\in \R \}\)
\(\{(a,b]:a<b\) and \(a\), \(b\in \R \}\)
\(\{(a,\infty ):a\in \R \}\)
\(\{[a,\infty ):a\in \R \}\)
\(\{(-\infty ,b):b\in \R \}\)
\(\{(-\infty ,b]:b\in \R \}\)
[Definition: Relatively open] Let \((X,d)\) be a metric space and let \(Y\subset X\). Then \((Y,d\big \vert _{Y\times Y})\) is also a metric space. If \(G\subseteq Y\) is open in \((Y,d\big \vert _{Y\times Y})\), then we say that \(G\) is relatively open in \(Y\).
(Memory 725) If \((X,d)\) is a metric space and \(G\subseteq Y\subseteq X\), then \(G\) is relatively open in \(Y\) if and only if \(G=Y\cap U\) for some \(U\subseteq X\) that is open in \((X,d)\).
[Definition: Connected] Let \((X,d)\) be a metric space. We say that \((X,d)\) is disconnected if there exist two sets \(A\), \(B\subseteq X\) such that
\(A\) and \(B\) are both open in \((X,d)\),
\(X=A\cup B\),
\(\emptyset =A\cap B\),
\(A\neq \emptyset \neq B\).
If no such \(A\) and \(B\) exist then we say that \((X,d)\) is connected.
(Memory 726) Let \((X,d)\) be a metric space and let \(Y\subseteq X\). Then the metric space \((Y,d\big \vert _{Y\times Y})\) is disconnected if and only if there exist two sets \(A\), \(B\subseteq X\) such that
\(A\) and \(B\) are both open in \((X,d)\),
\(Y\subseteq A\cup B\),
\(\emptyset =A\cap B\),
\(A\cap Y\neq \emptyset \neq B\cap Y\).
(Memory 727) A subset of \(\R \) is connected if and only if it is an interval.
(Memory 728) If \((X,d)\) is a connected metric space and \(f:X\to Y\) is continuous, then \(f(X)\) is also connected.
(Memory 729) By Problem 400 and Proposition 2.4, if \(E\subset \R \) is countable, then \(E\) is measurable and has measure zero.
Proposition 2.19. There is a set of measure zero that is uncountable.
(Problem 730) In this problem we begin the construction of an uncountable set of measure zero. We define the points \(a_{k,n}\) and \(b_{k,n}\), for \(n\in \N _0\) and for \(1\leq k\leq 2^n\), as follows. \begin {align*} a_{1,0}&=0,&b_{1,0}&=1,\\ a_{2\ell -1,n}&=a_{\ell ,n-1}, & b_{2\ell -1,n}&=\frac {2}{3}a_{\ell ,n-1}+\frac {1}{3}b_{\ell ,n-1}, \\ a_{2\ell ,n}&=\frac {1}{3}a_{\ell ,n-1}+\frac {2}{3}b_{\ell ,n-1}, &b_{2\ell ,n}&=b_{\ell ,n-1}. \end {align*}
Show that:
We prove by induction. Our base case is \(n=0\). In this case \(2^n=1\), so in (a) \(k\) must also be 1 and so (a) is true by inspection. (b) and (c) are vacuously true as there are no \(k\) that satisfy \(1\leq k<k+1\leq 2^0=1\).Now suppose that the statements are all true for some \(n-1\geq 0\). Let \(1\leq k\leq 2^n\).
If \(k\) is even, then \(k=2\ell \) for some \(1\leq \ell \leq 2^{n-1}\), and so \begin {align*} b_{k,n}-a_{k,n}&=b_{2\ell ,n}-a_{2\ell ,n} =b_{\ell ,n-1} - \biggl (\frac {1}{3}a_{\ell ,n-1}+\frac {2}{3}b_{\ell ,n-1}\biggr ) \alignbreak =\frac {1}{3}(b_{\ell ,n-1}-a_{\ell ,n-1}) \end {align*}
and thus (a) holds by the induction hypothesis. If in addition \(k+1\leq 2^n\), then \(\ell <2^{n-1}\) and so \(\ell +1\leq 2^{n-1}\), and so \begin {equation*} a_{k+1,n}-b_{k,n} = a_{2(\ell +1)-1,n}-b_{2\ell ,n} = a_{\ell +1,n-1}-b_{\ell ,n-1}. \end {equation*} By assumption (b) and (c) hold for \(n-1\), and so \begin {equation*} a_{k+1,n}-b_{k,n} = a_{\ell +1,n-1}-b_{\ell ,n-1}\geq 3^{1-n}>3^{-n}. \end {equation*}
If \(k\) is odd, then \(k=2\ell -1\) for some \(1\leq \ell \leq 2^{n-1}\). Then \begin {align*} b_{k,n}-a_{k,n} &=b_{2\ell -1,n}-a_{2\ell -1,n} \alignbreak =\biggl (\frac {2}{3}a_{\ell ,n-1}+\frac {1}{3}b_{\ell ,n-1}\biggr )-a_{\ell ,n-1} \alignbreak =\frac {1}{3}(b_{\ell ,n-1}-a_{\ell ,n-1}) \end {align*}
and thus (a) holds by the induction hypothesis. If in addition \(k+1\leq 2^n\), then \(\ell \leq 2^{n-1}\) and so \begin {align*} a_{k+1,n}-b_{k,n} &= a_{2\ell ,n}-b_{2\ell -1,n} \alignbreak = \biggl (\frac {1}{3}a_{\ell ,n-1}+\frac {2}{3}b_{\ell ,n-1}\biggr ) - \biggl (\frac {2}{3}a_{\ell ,n-1}+\frac {1}{3}b_{\ell ,n-1}\biggr ) \alignbreak =\frac {1}{3}(b_{\ell ,n-1}-a_{\ell ,-1}). \end {align*}
By assumption (a) holds for \(n-1\), and so this must equal \(3^{-n}\). This completes the proof.
(Problem 731) If \(n\in \N _0\) and \(1\leq k\leq 2^n\), let \(F_{k,n}=[a_{k,n},b_{k,n}]\) be the closed interval of length \(3^{-n}\) with endpoints at \(a_{k,n}\) and \(b_{k,n}\). Show that \(F_{j,n}\cap F_{k,n}=\emptyset \) if \(j\neq k\).
(Problem 740) Define \(F_n=\bigcup _{k=1}^{2^n} [a_{k,n},b_{k,n}]=\bigcup _{k=1}^{2^n} F_{k,n}\). Show that \(F_{n}\subseteq F_{n-1}\) for all \(n\in \N \), and that \(F_n\setminus F_{n-1}\) may be written as the union of \(2^{n-1}\) open intervals each of length \(3^{-n}\).
Let \(n\geq 1\). Observe that \begin {align*} F_{2\ell -1,n}\cup F_{2\ell ,n} &= [a_{2\ell -1,n},b_{2\ell -1,n}]\cup [a_{2\ell ,n},b_{2\ell ,n}] \\&= \phantom {{}_2}[a_{\ell ,n-1},b_{2\ell -1,n}]\cup [a_{2\ell ,n},b_{\ell ,n-1}] \alignbreak \subset [a_{\ell ,n-1},b_{\ell ,n-1}]=F_{\ell ,n-1}. \end {align*}Thus, if \(1\leq \ell \leq 2^{n-1}\), then there are at least two distinct values of \(k\), namely \(k=2\ell -1\) and \(k=2\ell \), that satisfy \(1\leq k\leq 2^n\) and \(F_{k,n}\subset F_{\ell ,n-1}\).
Because there are only \(2^n=2\cdot 2^{n-1}\) such values of \(k\), and there are \(2^{n-1}\) such values of \(\ell \), we must have that each \(F_{k,n}\) is contained in a \(F_{\ell ,n-1}\). (This may also be seen by choosing \(\ell =k/2\) if \(k\) is even and \(\ell =(k+1)/2\) if \(k\) is odd, and observing that \(F_{k,n}\subset F_{\ell ,n-1}\) by the above analysis.) Thus \(F_n=\bigcup _{k=1}^{2^n} F_{k,n} \subset \bigcup _{\ell =1}^{2^{n-1}} F_{\ell ,n-1}=F_{n-1}\), as desired.
Again because there are only \(2^n=2\cdot 2^{n-1}\) such values of \(k\), and there are \(2^{n-1}\) such values of \(\ell \), we must have that each \(F_{\ell ,n-1}\) contains \(F_{k,n}\) for exactly two values of \(k\), namely \(k=2\ell \) and \(k=2\ell -1\). (This may also be seen by observing that if \(1\leq j\leq 2^n\) and \(j\notin \{2\ell -1,2\ell \}\), then either \(j>2\ell \) or \(j<2\ell -1\). In the first case \(a_{j,n}> b_{2\ell ,n}=b_{\ell ,n-1}\), while in the second case \(b_{j,n}<a_{2\ell -1,n}=a_{\ell ,n-1}\); in either case \([a_{j,n},b_{j,n}]\) is clearly disjoint from \([a_{\ell ,n-1},b_{\ell ,n-1}]\).)
Thus \(F_{\ell ,n-1}\setminus F_n=F_{\ell ,n-1}\setminus (F_{2\ell -1,n}\cup F_{2\ell ,n})\), which by the above analysis is equal to the interval \((b_{2\ell -1,n},a_{2\ell ,n})\). There are \(2^{n-1}\) such intervals (one for each \(\ell \)) and by the above analysis, because \(2\ell -1\) is odd we have that the interval is of length \(3^{-n}\).
(Problem 750) Let \(C=\bigcap _{n=0}^\infty F_n\). The set \(C\) is called the Cantor set. Show that \(m(F_n)=(2/3)^n\) for all \(n\in \N _0\) and that \(m(C)=0\).
Each \(F_{k,n}\) is a closed interval, and so is closed. The union of finitely many closed sets is closed, and so each \(F_n\) is closed. The intersection of an arbitrary collection of closed sets is closed, and so \(C\) must be closed.For each \(n\), \(C\subset F_n\), and so \(m^*(C)\leq m^*(F_n)\). But each \(F_{k,n}\) has length (thus measure) \(3^{-n}\), the \(F_{k,n}\)s are disjoint for distinct \(k\), and there are \(2^n\) intervals \(F_{k,n}\) in \(F_n\); thus \(m(F_n)=\sum _{k=1}^n m(F_{k,n})=(2/3)^n\). Recalling from undergraduate analysis that \((2/3)^n\to 0\) as \(n\to \infty \), we have that \(m^*(C)=0\). Sets of measure zero are measurable by Proposition 2.4, and so \(m(C)\) exists and equals zero.
[Homework 3.1b] If \(E\subseteq \R \) and \(m^*(E)<\infty \), and if we define \(f\) by \(f(x)=m(E\cap (-\infty ,x))\), then \(f:\R \to \R \) is continuous.
(Problem 751) Let \(\Lambda _k(x)=\frac {m(F_k\cap (-\infty ,x))}{m(F_k)}\). Then \(\Lambda _k\) is continuous and nondecreasing. Sketch the graphs of \(\Lambda _0\), \(\Lambda _1\), and \(\Lambda _2\).
(Problem 760) Suppose that \(x\notin F_n\). Show that \(\Lambda (x)=2^{-n}|\{k\in \{1,2,\dots ,2^n\}:b_{k,n}<x\}|\).
(Problem 770) Show that \(\Lambda _n(\R \setminus F_n)=\{i2^{-n}:0\leq i\leq 2^n,i\in \Z \}\) and that, if \(m\geq n\), then \(\Lambda _n(x)=\Lambda _m(x)\) for all \(x\notin F_n\).
(Problem 780) Show that \(\{\Lambda _k\}_{k=1}^\infty \) is uniformly Cauchy.
[Definition: The Cantor function] Let \(\Lambda (x)=\lim _{k\to \infty } \Lambda _k(x)\).
(Problem 790) Show that \(\Lambda \) exists and is continuous, nondecreasing, and surjective \(\Lambda :[0,1]\to [0,1]\).
We have that \(\Lambda _k:\R \to \R \) is uniformly Cauchy and \(\R \) is complete. Thus by Problem 722, the sequence \(\{\Lambda _k\}_{k=1}^\infty \) converges uniformly to some function \(\Lambda :\R \to \R \). Thus \(\Lambda \) exists.By Homework 3.1b, each \(\Lambda _k\) is continuous, so by Problem 721, \(\Lambda \) must also be continuous.
Suppose \(x<y\). Clearly \(\Lambda _k(x)\leq \Lambda _k(y)\) for all \(k\in \N \), and so we must have that \(\Lambda _k(x)\leq \Lambda _k(y)\) as well.
Finally, observe that \(\Lambda _k(0)=0\) and \(\Lambda _k(1)=1\) for all \(k\in \N \). Thus \(\Lambda (0)=0\) and \(\Lambda (1)=1\). By the intermediate value theorem (Problem 723), if \(0<y<1\) then \(y=\Lambda (x)\) for some \(x\in (0,1)\), and so \(\Lambda \) is surjective \([0,1]\to [0,1]\).
[Definition: Almost everywhere] Suppose that \(E\subseteq \R \) is a set. If a property \(P\) is true for every \(x\in E\setminus E_0\), where \(m^*(E_0)=0\), we say that \(P\) is true almost everywhere on \(E\).
(Problem 800) Show that \(\Lambda '(x)=0\) for every \(x\in \R \setminus C\), and thus for almost every \(x\in \R \).
(Problem 810) Show that \(\Lambda ([0,1]\setminus C)\) is countable.
Note that \(0\), \(1\in C\), and so \([0,1]\setminus C=(0,1)\setminus C\).Because \(C\) is closed, we must have that \((0,1)\setminus C\) is open. Thus by Proposition 1.9, \((0,1)\setminus C=\bigcup _{k=1}^\infty I_k\), where the \(I_k\)s are (possibly empty) open intervals. Because \(I_k\subset (0,1)\setminus C\subset \R \setminus C\), we have that \(\Lambda '=0\) on \(I_k\) and so \(\Lambda \) is constant on each \(I_k\). Because each \(\Lambda (I_k)\) is empty or a single point, and there are countably many \(I_k\)s, we must have that \(\Lambda ((0,1)\setminus C)=\bigcup _{k=1}^\infty \Lambda (I_k)\) is countable.
Alternatively, observe that if \(x\notin C\) then \(x\notin F_n\) for some \(n\). Then \(\Lambda _m(x)=\Lambda _n(x)\) for all \(m\geq n\), and so \(\Lambda (x)=\lim _{m\to \infty } \Lambda _m(x)=\Lambda _n(x)\in \Lambda _n(\R \setminus F_n)=\{i2^{-n}:0\leq i\leq 2^n, i\in \Z \}\). In particular \(\Lambda (x)\) is rational. Thus \(\Lambda ((0,1)\setminus \C )\subset \Q \) and so must be countable.
(Problem 820) Show that \(C\) is uncountable.
We know that \(\Lambda ([0,1])=[0,1]\) is uncountable. But \(\Lambda ([0,1]\setminus C)\) is countable, and \([0,1]=\Lambda ([0,1])=\Lambda (C)\cup \Lambda ([0,1]\setminus C)\). Thus \(\Lambda (C)\) must be uncountable, for if it were countable then \(\Lambda (C)\cup \Lambda ([0,1]\setminus C)\) would be countable, which is a contradiction. But if \(C\) were countable then \(\Lambda (C)\) would also be countable, so \(C\) must be uncountable.
Theorem 2.17. If \(E\subseteq \R \) has positive outer measure, then there is an \(A\subseteq E\) that is not measurable.
(Problem 830) In this problem we begin the proof of Theorem 2.17. Let \(F\subset [-1,1]\) and let \(V_{q_k}\) be as in Problem 490. Show that either \(F\cap V_{q_k}\) is not measurable or \(m^*(F\cap V_{q_k})=0\).
(Problem 840) Suppose that \(m^*(F)>0\). Show that we must have that \(m^*(F\cap V_{q_k})>0\) for at least one value of \(k\).
Recall that \([-1,1]\subseteq \bigcup _{k=1}^\infty V_{q_k}\). Thus \(F=F\cap [-1,1]=F\cap \bigcup _{k=1}^\infty V_{q_k}=\bigcup _{k=1}^\infty F\cap V_{q_k}\).Thus by Proposition 2.3, we must have that \(m^*(F)\leq \sum _{k=1}^\infty m^*(F\cap V_{q_k})\). Because the left hand side is positive, at least one of the summands on the right hand side must be positive.
(Problem 841) Prove Theorem 2.17.
Proposition 2.22. There is a measurable set that is not a Borel set.
(Problem 850) In this problem we begin the proof of Proposition 2.22. Let \(\Lambda \) be the Cantor-Lebesgue function and let \(f(x)=x+\Lambda (x)\). Show that \(f\) is continuous, strictly increasing, and surjective \(\R \to \R \).
[Chapter 2, Problem 47] If \(f:\R \to \R \) is continuous and strictly increasing, and if \(B\) is a Borel set, then \(f(B)\) is also a Borel set.
(Problem 860) Show that \(f(C)\) has positive measure.
(Problem 870) Prove Proposition 2.22.
Because \(m(f(C))>0\), by Theorem 2.17 there is an \(A\subseteq f(C)\) that is not measurable.Let \(B=f^{-1}(A)\cap C\). Then \(B\subseteq C\), and \(m^*(C)=0\), so \(m^*(B)=0\); thus \(B\) is measurable by Proposition 2.4.
We claim that \(f(B)=A\). Because \(B\subseteq f^{-1}(A)\), by definition \(f(B)\subseteq A\). Conversely, suppose that \(a\in A\). Then \(a\in f(C)\) because \(A\subseteq f(C)\), so \(a=f(b)\) for some \(b\in C\). Then \(b\in f^{-1}(A)\) by definition, so \(b\in f^{-1}(A)\cap C=B\). Thus \(b\in B\) and so \(a=f(b)\in f(B)\); because \(a\) was arbitrary we have that \(A\subseteq f(B)\). Thus \(A=f(B)\).
If \(B\) were Borel, then by Problem 2.47 we would have that \(f(B)\) was Borel and therefore measurable. But \(f(B)=A\) is not measurable, and so \(B\) is not Borel. We have seen that \(B\) is measurable, and so \(B\) must be a set that is measurable but not Borel.
(Problem 880) Prove part (ii) of Theorem 2.15 without using part (i).
(Problem 890) Find a sequence \(\{B_k\}_{k=1}^\infty \) such that \(B_k\) is measurable and \(B_k\supseteq B_{k+1}\) for all \(k\in \N \), but such that \begin {equation*} m\Bigl (\bigcap _{k=1}^\infty B_k\Bigr )\neq \lim _{n\to \infty } m(B_n). \end {equation*}
Take \(B_k=(k,\infty )\). Then \(m(B_n)=\infty \) for all \(k\) and so \(\lim _{n\to \infty } m(B_n)=\infty \), but \(\bigcap _{k=1}^\infty B_k=\emptyset \) and so \(m\bigl (\bigcap _{k=1}^\infty B_k)=0\).
The Borel-Cantelli Lemma. Let \(\{E_k\}_{k=1}^\infty \) be a sequence of measurable sets. Suppose that \(\sum _{k=1}^\infty m(E_k)<\infty \). Then \(|\{k\in \N :x\in E_k\}|<\infty \) for almost every \(x\in \R \).
(Problem 900) Prove the Borel-Cantelli lemma. Start by writing a formula for the set of \(x\in \R \) such that \(|\{k\in \N :x\in E_k\}|=\infty \) using unions and intersections. Explain carefully why your formula is true.
We claim that \(|\{k\in \N :x\in E_k\}|=\infty \) if and only if, for all \(n\in \N \), there is a \(k\in \N \) with \(k\geq n\) such that \(x\in E_k\).To prove the claim, we will prove that the two negations are equivalent. Suppose first that \(|\{k\in \N :x\in E_k\}|\neq \infty \). Then the set \(\{k\in \N :x\in E_k\}\) is finite, and so it contains a largest element \(K\). Then \(n=K+1\in \N \), but there is no \(k\in \N \) with \(k\geq n\) such that \(x\in E_k\).
Suppose to the contrary that for some \(n\), there does not exist a \(k\geq n\) with \(x\in E_k\). Then \(x\notin E_k\) for all \(k\geq n\). Thus \(x\in E_k\) for at most \(n-1\) possible values of \(k\), and so \(x\in E_k\) for only finitely many \(k\).
Let \begin {align*} A&=\{x\in \R :|\{k\in \N :x\in E_k\}|=\infty \} \\&=\{x\in \R :\text {for all }n\in \N \text { there exists a } k\geq n\text { such that } x\in E_k\} . \end {align*}
For any fixed \(n\in \N \), the set \begin {equation*} \{x\in \R :\text {there exists a } k\geq n\text { such that } x\in E_k\} = \bigcup _{k=n}^\infty E_k. \end {equation*} Thus \begin {equation*} A=\Bigl \{x\in \R :\text {for all }n\in \N , \>x\in \bigcup _{k=n}^\infty E_k\Bigr \} =\bigcap _{n=1}^\infty \Bigl (\bigcup _{k=n}^\infty E_k\Bigr ) . \end {equation*}
Let \(B_n=\bigcup _{k=n}^\infty E_k\). Then \(B_n=E_n\cup B_{n+1}\supseteq B_{n+1}\) for all \(n\). Furthermore, \(m(B_1)\leq \sum _{k=1}^\infty m(E_k)<\infty \), and so by Theorem 2.15ii, \begin {equation*} m(A)=m \Bigl (\bigcap _{n=1}^\infty B_n\Bigr )=\lim _{n\to \infty } m(B_n). \end {equation*} But \begin {equation*} m(B_n)\leq \sum _{k=n}^\infty m(E_k). \end {equation*} Because \(\sum _{k=1}^\infty m(E_k)<\infty \), and each \(m(E_k)\geq 0\), we have that the series converges absolutely and so \(\lim _{n\to \infty } \sum _{k=n}^\infty m(E_k)=0\). Thus \(m(A)=0\), as desired.
[Definition: Measurable function] Let \(E\subseteq \R \) be measurable and let \(f:E\to [-\infty ,\infty ]\). Suppose that for every \(c\in \R \) the set \begin {equation*} \{x\in E:f(x)>c\}=f^{-1}((c,\infty ]) \end {equation*} is measurable. Then we say that \(f\) is a measurable function (or that \(f\) is measurable on \(E\)).
Proposition 3.3. Let \(E\subseteq \R \) be measurable and let \(f:E\to \R \) be continuous. Then \(f\) is measurable.
(Problem 910) Prove Proposition 3.3.
(Problem 920) Let \(f:\R \to [-\infty ,\infty ]\). Suppose that \(\lim _{y\to x} f(y)=f(x)\) for all \(x\in \R \). Show that \(f\) is measurable.
[Chapter 3, Problem 24] A monotonic function defined on a measurable set is measurable.
Proposition 3.1. Let \(E\subseteq \R \) be measurable and let \(f:E\to [-\infty ,\infty ]\). The following statements are equivalent.
Furthermore, if any of these conditions is true, then \(f^{-1}(\{c\})\) is measurable for all \(c\in [-\infty ,\infty ]\).
(Problem 930) Prove Proposition 3.1.
[Chapter 3, Problem 4] If \(f^{-1}(\{c\})\) is measurable for all \(c\in [-\infty ,\infty ]\), is it necessarily the case that \(f\) is measurable?
(Problem 931) Let \(E\subseteq \R \) be measurable and let \(f:E\to [-\infty ,\infty ]\). Suppose that, for all \(c\in \R \), the set \(\{x\in E:c<f(x)<\infty \}=f^{-1}((c,\infty ))\) is measurable. Is \(f\) necessarily measurable? If not, what additional assumptions must be imposed to show that \(f\) is measurable?
\(f\) need not be measurable. Let \(A\) be a non-measurable set; such sets exist by Theorem 2.17. Let \begin {equation*} f(x)=\begin {cases}\infty ,&x\in A,\\ -\infty ,&x\notin A.\end {cases} \end {equation*} Then \(f^{-1}((c,\infty ))=\emptyset \) is measurable for all \(c\in \R \), but \(f^{-1}((0,\infty ])=f^{-1}(\{\infty \})=A\) is not measurable.However, if for all \(c\in \R \), the set \(\{x\in E:f(x)>c\}=f^{-1}((c,\infty ))\) is measurable, and if in addition the set \(f^{-1}(\{\infty \})\) is measurable, then \(f\) is measurable.
(Problem 932) Let \(\mathcal {A}\) be a \(\sigma \)-algebra over a set \(X\), let \(Y\in \mathcal {A}\), and define \(\mathcal {S}=\{S\cap Y:S\in \mathcal {A}\}\). Show that \(\mathcal {S}\) is a \(\sigma \)-algebra over \(Y\).
(Problem 940) Let \((X,\mathcal {A})\) be a measurable space (that is, \(\mathcal {A}\) is a \(\sigma \)-algebra over \(X\)). Let \(f:X\to Y\) be a function and let \(\mathcal {F}=\{S\subseteq Y:f^{-1}(S)\in \mathcal {A}\}\). Show that \(\mathcal {F}\) is a \(\sigma \)-algebra over \(Y\).
[Homework 2.1] The collection \(\mathcal {B}\) of Borel sets is the smallest \(\sigma \)-algebra containing \(\{(-\infty ,a):a\in \R \).
Proposition 3.2. If \(f\) is measurable, then \(f^{-1}(\mathcal {O})\) is measurable for all open sets \(\mathcal {O}\).
(Problem 950) Prove that in fact, if \(f\) is measurable, then \(f^{-1}(B)\) is measurable for all Borel sets \(B\).
(Problem 960) If \(g\) is measurable, is it true that \(g^{-1}(E)\) is measurable for all measurable sets \(E\)?
(Problem 970) If \(h\) is measurable, is it true that \(h^{-1}(B)\) is Borel for all Borel sets \(B\)?
No. Let \(A\) be a set that is measurable but not Borel; such an \(A\) must exist by Proposition 2.22. Let \begin {equation*} h(x)=\begin {cases}1,&x\in A,\\0,&x\notin A.\end {cases} \end {equation*} If \(E\subseteq [-\infty ,\infty ]\), then \(h^{-1}(E)\) is either \(\emptyset \) (if \(0\), \(1\notin E\)), \(A\) (if \(1\in E\), \(0\notin E\)), \(\R \setminus A\) (if \(0\in E\), \(1\notin E\)), or \(\R \) (if \(0\), \(1\in E\)). In any case \(h^{-1}(E)\) is measurable, and so \(h\) is a measurable function. However, \(h^{-1}(\{1\})=A\) is not Borel.
Proposition 3.5. Let \(E\subseteq \R \) be measurable and let \(f:E\to [-\infty ,\infty ]\).
(Problem 971) Prove Proposition 3.5, part (ii).
(Problem 980) Prove Proposition 3.5, part (i).
Let \(S=\{x\in E:f(x)\neq g(x)\}\). By assumption \(m^*(S)=0\), and so by Proposition 2.4 \(S\) and all of its subsets are measurable.Let \(c\in \R \). Then \begin {align*} \{x\in E:f(x)>c\}&= \bigl (\{x\in E:g(x)>c\}\cup \{x\in E:f(x)>c\geq g(x)\}\bigr )\setminus \{x\in E:g(x)>c\geq f(x)\}. \end {align*}
The first set \(S_1=\{x\in E:g(x)>c\}\) is measurable by assumption. The second set \(S_2=\{x\in E:f(x)>c\geq g(x)\}\subseteq \{x\in E:f(x)\neq g(x)\}=S\) has outer measure zero and thus is measurable. Thus \(S_1\cup S_2\) is measurable by Proposition 2.5. The third set \(S_3=\{x\in E:g(x)>c\geq f(x)\}\subseteq \{x\in E:g(x)\neq f(x)\}=S\) also has outer measure zero and thus is measurable, and so \((S_1\cup S_2)\setminus S_3\) is measurable by Problem 510 and Problem 561. Thus \(\{x\in E:f(x)>c\}\) is measurable for all \(c\in \R \), and so \(f\) is a measurable function.
(Problem 981) Let \(E\subseteq \R \). Show that \(E\) is measurable if and only if the characteristic function \(\chi _E\) is measurable.
[Chapter 3, Problem 6] Let \(E\subseteq \R \) be measurable. Let \(f:E\to [-\infty ,\infty ]\). Show that \(f\) is measurable on \(E\) if and only if the function \begin {equation*} g(x)=\begin {cases} f(x), &x\in E,\\ 0, &x\notin E\end {cases} \end {equation*} is measurable.
(Problem 982) Did we need the condition that \(E\) was measurable?
Theorem 3.6. Let \(E\subseteq \R \) be measureable, and let \(f\), \(g:E\to [-\infty ,\infty ]\) be measurable functions that are finite almost everywhere in \(E\)
If \(\alpha \), \(\beta \in \R \), then \(fg\) and \(\alpha f+\beta g\) are defined almost everywhere on \(E\) and are measurable on \(E\) in the sense of Proposition 3.5, that is, in the sense that any of the extensions of \(fg\) and \(\alpha f+\beta g\) to \(E\) are measurable.
(Problem 983) If \(f\) is measurable and \(\alpha \in \R \), then \(\alpha f\) is measurable.
(Problem 990) Suppose that \(f\) and \(g\) are measurable and finite almost everywhere. Show that \(f+g\) is measurable.
First observe that \(D=\{x\in E:f(x)+g(x)\) does not exist\(\}=\{x\in E:f(x)=\infty ,g(x)=-\infty \}\cup \{x\in E:f(x)=-\infty ,g(x)=\infty \}\) and so has measure zero.Let \(c\in \R \). Observe that if \(f(x)+g(x)\) exists and is greater than \(c\), then neither \(f(x)\) nor \(g(x)\) can equal \(-\infty \). Thus \(E=\bigcup _{q\in \Q } \{x\in E:f(x)>q\}\).
Now, if \(f(x)>q\) and \(g(x)>c-q\), then \(f(x)+g(x)>c\). Conversely, if \(f(x)+g(x)>c\), then \(f(x)>c-g(x)\), the left hand side is not \(-\infty \), and the right hand side is not \(+\infty \). By density of the rationals (if both sides are finite) or by the Archimedean property (if \(f(x)=\infty \) or \(g(x)=\infty \) or both), there is a \(q\in \Q \) with \(f(x)>q>c-g(x)\) and so \(g(x)>c-q\). We thus have that \begin {align*} \{x\in E:f(x)+g(x)>c\}&=\bigcup _{q\in \Q } \{x\in E:f(x)>q\}\alignbreak \cap \{x\in E:g(x)>c-q\}. \end {align*}
Because \(f\) and \(g\) are measurable, the two sets \(\{x\in E:f(x)>q\}\) and \(\{x\in E:g(x)>c-q\}\) are measurable. Thus their intersection is measurable. The rationals are countable, and the union of countably many measurable sets is measurable. Thus \(\{x\in E:f(x)+g(x)>c\}\) is measurable for all \(c\in \R \), as desired.
(Problem 1000) Suppose that \(f\) is measurable. Show that \(f^2\) is measurable. Then prove Proposition 3.5.
If \(f\) is measurable and \(c\in \R \), then \begin {align*} \{x\in E: f(x)^2\leq c\}&=\{x\in E:-\sqrt {c}\leq f(x)\leq \sqrt {c}\} \alignbreak = \{x\in E:f(x)\leq \sqrt {c}\}\cap \{x\in E:f(x)\geq -\sqrt {c}\} \end {align*}is the intersection of two measurable sets, and so is measurable. Thus \(f^2\) is measurable by Proposition 3.1.
Now, observe that \begin {equation*} fg=\frac {(f+g)^2-f^2-g^2}{2}. \end {equation*} Each of the three elements of the numerator is measurable by the previous argument, while their sum is measurable by Problem Problem 990 and so \(fg\) is measurable by Problem Problem 983.
(Problem 1010) Give an example a measurable function \(h\) and a continuous function \(g\) such that \(h\circ g\) is not measurable.
Proposition 3.7. If \(D\), \(E\subseteq \R \) are measurable, if \(h:E\to D\) is measurable, and if \(g:D\to \R \) is continuous, then \(g\circ h\) is measurable.
[Chapter 3, Problem 8iv] More generally, if \(E\subseteq \R \) is measurable, if \(D\subseteq \R \) is a Borel set, if \(h:E\to D\) is measurable, and if \(g:D\to \R \) is such that \(\{x\in D:g(x)>c\}\) is Borel for all \(c\in \R \), then \(g\circ h\) is measurable.
(Problem 1020) Prove Proposition 3.7.
(Problem 1021) If \(f\) is measurable, show that \(|f|\) is measurable.
(Problem 1022) Define \begin {equation*} f^+(x)=\begin {cases}f(x),&f(x)\geq 0,\\0,&f(x)\leq 0,\end {cases} \qquad f^-(x)=\begin {cases}0,&f(x)\geq 0,\\f(x),&f(x)\leq 0.\end {cases} \end {equation*} If \(f\) is measurable, show that \(f^+\) and \(f^-\) are measurable.
Proposition 3.8. Let \(E\subseteq \R \) be measurable and let \(f_1\), \(f_2,\dots ,f_k:E\to [-\infty ,\infty ]\) be finitely many measurable functions. Then \(f(x)=\max \{f_1(x),\dots ,f_k(x)\}\) is also measurable.
(Problem 1030) Prove Proposition 3.8.
[Definition: Pointwise convergence] Let \(E\) be a set and let \(f_n\), \(f:E\to [-\infty ,\infty ]\). If \(f_n(x)\to f(x)\) for all \(x\in E\), then we say that \(f_n\to f\) pointwise on \(E\).
[Definition: Almost everywhere convergence] Let \(E\subseteq \R \) be a set and let \(f_n\), \(f:E\to [-\infty ,\infty ]\). If \(f_n(x)\to f(x)\) for all \(x\in E\setminus D\), where \(m(D)=0\), then we say that \(f_n\to f\) almost everywhere on \(E\).
[Definition: Uniform convergence] Let \(E\subseteq \R \) be a set and let \(f_n\), \(f:E\to \R \). Suppose that for all \(\varepsilon >0\) there is a \(m\in \N \) such that, if \(n\geq m\), then \(|f_n(x)-f(x)|<\varepsilon \). Then we say that \(f_n\to f\) uniformly on \(E\).
(Problem 1031) Show that uniform convergence implies pointwise convergence and that pointwise convergence implies almost everywhere convergence.
(Problem 1032) Show that none of the reverse implications hold.
Proposition 3.9. Let \(\{f_n\}_{n=1}^\infty \) be a sequence of measurable functions on a measurable set \(E\). Suppose that \(f_n\to f\) almost everywhere on \(E\) for some \(f:E\to [-\infty ,\infty ]\). Then \(f\) is measurable.
(Problem 1040) Prove Proposition 3.9 by showing that \(\{x\in E:c\leq f(x)\}\) is measurable for all \(c\in \R \). Be sure to explain why your proof works even if \(f_n\), \(f\) are allowed to be infinite.
[Definition: Characteristic function] If \(A\subseteq \R \), then the characteristic function of \(A\), denoted \(\chi _A\), is defined by \begin {equation*} \chi _A(x)=\begin {cases}1,&x\in A,\\0,&x\notin A.\end {cases} \end {equation*}
[Definition: Simple function] A function \(\varphi \) is simple if its domain \(E\subseteq \R \) is measurable, if \(\varphi \) is measurable on \(E\), and if \(\{\varphi (x):x\in E\}\) is a set of of finitely many real numbers.
[Chapter 3, Problem 6] If \(E\) is measurable and \(f:E\to \R \), then \(f\) is measurable on \(E\) if and only if \begin {equation*} g(x)=\begin {cases}f(x),&x\in E,\\0,&x\notin E\end {cases} \end {equation*} is measurable on \(\R \).
(Problem 1041) A function \(\varphi \) with domain \(E\subseteq \R \) is simple if and only if \(\varphi =\psi \big \vert _E\) for some simple function \(\psi :\R \to \R \). Furthermore, we may require that \(\psi (x)=0\) for all \(x\notin E\).
(Problem 1042) The set of simple functions contains all of the characteristic functions and is closed under taking finite linear combinations.
(Problem 1050) Suppose that \(\varphi :\R \to \R \) is simple. Show that there is a unique list of numbers \(c_1<c_2<\dots <c_n\) and a unique list of nonempty measurable sets \(E_1\), \(E_2,\dots ,E_n\) such that \(\varphi =\sum _{j=1}^n c_j \chi _{E_j}\).
Furthermore, show that \(\R =\bigcup _{j=1}^n E_j\) and \(E_j\cap E_k=\emptyset \) for all \(j\neq k\).
We may write \(\varphi (\R )=\{c_1,\dots ,c_n\}\) because \(\varphi \) takes on finitely many values. As \(\varphi (\R )\) is a set, we may require the \(c_k\)s to be distinct. Any finite set can be ordered, so we may require \(c_1<c_2<\dots <c_n\). Let \(E_k=\varphi ^{-1}(\{c_k\})\); the given properties are straightforward to check.
(Problem 1051) If \(\varphi \) and \(\psi \) are simple functions with the same domain, show that \(\max (\varphi ,\psi )\) is also simple.
The simple approximation lemma. Let \(f:E\to \R \) be measurable and bounded. Let \(\varepsilon >0\). Then there are two simple functions \(\varphi _\varepsilon \) and \(\psi _\varepsilon \) with \begin {equation*} \psi _\varepsilon (x)-\varepsilon \leq \varphi _\varepsilon (x)\leq f(x)\leq \psi _\varepsilon (x)\leq \varphi _\varepsilon (x)+\varepsilon \end {equation*} for all \(x\in E\).
(Problem 1060) Prove the simple approximation lemma.
Let \(M\) be such that \(-M\leq f(x)\leq M\) for all \(x\in E\); such an \(M\) exists by definition of bounded function. Let \(N\in \N \) be such that \(N\varepsilon >M\); such an \(N\) exists by the Archimedean property.For each \(k\in \Z \), let \(D_k=f^{-1}([k\varepsilon ,(k+1)\varepsilon ))\) and let \(E_k=f^{-1}(((k-1)\varepsilon ,k\varepsilon ])\). If \(|k|\geq N+1\), then \(D_k\) and \(E_k\) are empty. Furthermore, \(E=\bigcup _{k=-N}^N D_k=\bigcup _{k=-N}^N E_k\) and \(D_k\cap D_j=\emptyset =E_k\cap E_j\) if \(j\neq k\).
Let \(\varphi _\varepsilon =\sum _{k=-N}^N k\varepsilon \chi _{D_k}\) and let \(\psi _\varepsilon =\sum _{k=-N}^N k\varepsilon \chi _{E_k}\). These functions are simple by construction.
If \(f(x)=k\varepsilon \) for some \(k\in \Z \), then \(x\in D_k\cap E_k\) and so \(\varphi _\varepsilon (x)=k\varepsilon =f(x)=\psi _\varepsilon (x)\), and thus the desired inequalities hold.
Otherwise, \(k\varepsilon <f(x)<(k+1)\varepsilon \) for some \(k\in \Z \), and so \(x\in D_k\cap E_{k+1}\). Thus \(\varphi (x)=k\varepsilon <f(x)<(k+1)\varepsilon =\psi _\varepsilon (x)\), and \(\psi _\varepsilon (x)=\varphi _\varepsilon (x)+\varepsilon \), and so the desired inequalities are again satisfied.
The simple approximation theorem. Let \(f:E\to [-\infty ,\infty ]\). Then \(f\) is measurable if and only if there is a sequence \(\{\varphi _n\}_{n=1}^\infty \) such that
(Problem 1061) Prove the easy direction; that is, suppose that such a sequence \(\{\varphi _n\}_{n=1}^\infty \) exists and show that \(f\) is measurable.
(Problem 1070) Prove the simple approximation theorem.
For each \(n\in \N \), \(k\in \N \), let \(E_{n,k}=\{x\in E:|f(x)|\geq k/2^n\}\). Define \begin {equation*} \psi _n(x)=\sum _{k=1}^{n2^n} \frac {1}{2^n} \chi _{E_{n,k}}. \end {equation*} Then \(\psi \) is simple and nonnegative.
If \(x\in E_{n,k}\) and \(k\leq n2^n\), then \(2k\leq n2^{n+1}\leq (n+1)2^{n+1}\), \(x\in E_{n+1,2k}\), and \(x\in E_{n+1,2k-1}\). Thus \(\sum _{k=1}^{(n+1)2^{n+1}} \frac {1}{2^{n+1}} \chi _{E_{n+1,k}}(x)\) has at least twice as many nonzero terms as \(\sum _{k=1}^{n2^n} \frac {1}{2^n} \chi _{E_{n,k}}(x)\), so \(\{|\psi _n(x)|\}_{n=1}^\infty \) is nondecreasing.
Finally, for any \(x\) and \(n\), either \(\psi _n(x)=n\) if \(|f(x)|\geq n\), or \(|f(x)|-1/2^n\leq \psi _n(x)\leq |f(x)|\) if \(|f(x)|\leq n\). It is clear that \(\psi _n(x)\to |f(x)|\) pointwise.
Now, define \(\varphi _n(x)=\psi (x)\sgn (f(x))\), that is, \begin {equation*} \varphi _n(x)=\begin {cases} \psi (x), & f(x)\geq 0,\\-\psi (x), & f(x)<0.\end {cases} \end {equation*} Then \(\varphi _n\) is also simple, \(|\varphi _n|=|\psi _n|\) so \(\{|\varphi _n(x)|\}_{n=1}^\infty \) is nondecreasing for all \(x\in E\), and \(\varphi _n\to f\) pointwise.
(Problem 1080) Let \(\varphi :\R \to \R \) be a simple function and let \(c_j\), \(E_j\) be as in Problem 1050. What do you expect \(\int _\R \varphi \,dm\) to equal?
(Memory 1081) (Tietze’s Extension Theorem in \(\R \)). Let \(F\subseteq \R \) be closed and let \(f:F\to \R \) be continuous. Then there is a function \(g:\R \to \R \) that is continuous on all of \(\R \) and satisfies \(g=f\) on \(F\).
(Memory 1082) Let \(F\) and \(D\) be two disjoint closed sets and let \(f:F\cup D\to \R \) be a function. Suppose that \(f\big \vert _F\) and \(f\big \vert _D\) are continuous on \(F\) and \(D\), respectively. Then \(f\) is continuous on \(F\cup D\).
Egoroff’s theorem. Let \(E\subseteq \R \) be measurable with \(m(E)<\infty \). Let \(\{f_n\}_{n=1}^\infty \) be a sequence of measurable functions on \(E\) that converges pointwise almost everywhere to some function \(f\) that is finite almost everywhere.
Then for every \(\varepsilon >0\) there is a closed set \(F\subseteq E\) with \(m(E\setminus F)<\varepsilon \) and such that \(f_n\to f\) uniformly on \(F\).
Lemma 3.10. Under the conditions of Egoroff’s theorem, if \(\mu >0\) and \(\delta >0\), then there is a measurable set \(A\subseteq E\) and a \(k\in \N \) such that \(m(E\setminus A)<\delta \) and such that \(|f_n(x)-f(x)|<\mu \) for all \(x\in A\) and all \(n\geq k\).
(Problem 1090) Prove Lemma 3.10. (Note that we will use Lemma 3.10 to prove Egoroff’s theorem, and so you may not use Egoroff’s theorem to prove Lemma 3.10.)
(Problem 1100) Use Lemma 3.10 to prove Egoroff’s theorem.
Let \(E_0=\{x\in E:|f(x)|<\infty ,f_n(x)\to f(x)\}\). By assumption \(\{x\in E:|f(x)|=\infty \}\) and \(\{x\in E:f_n(x)\not \to f(x)\}\) have measure zero, so \(E_0\) is measurable and \(m(E_0)=m(E)\).If \(\ell \in \N \), let \(\mu =1/\ell \) and let \(\delta =\varepsilon /2^{\ell +1}\), and let \(A=A_\ell \) and \(k=k_\ell \) be as in Lemma 3.10. Then \(A_\ell \subseteq E\), \(m(E\setminus A_\ell )<\varepsilon /2^{\ell +1}\), and \(|f_n(x)-f(x)|<1/\ell \) for all \(x\in A_\ell \) and all \(n\geq k_\ell \).
Let \(B=\bigcap _{\ell =1}^\infty A_\ell \). Then \(B\) is measurable because the sets \(A_\ell \) are measurable. Then \(B\subseteq E\) and \begin {align*} m(E\setminus B) &=m\Bigl (E\setminus \bigcap _{\ell =1}^\infty A_\ell \Bigr ) =m\Bigl (\bigcup _{\ell =1}^\infty E\setminus A_\ell \Bigr ) \alignbreak \leq \sum _{\ell =1}^\infty m(E\setminus A_\ell ) <\frac {\varepsilon }{2}. \end {align*}
We claim that \(f_n\to f\) uniformly on \(B\). Let \(\eta >0\). There is a \(\ell \in \N \) with \(1/\ell <\eta \). If \(n>k_\ell \), then \(|f_n(x)-f(x)|<1/\ell <\eta \) for all \(x\in A_\ell \), and thus all \(x\in B\) because \(B\subseteq A_\ell \). Thus \(f_n\to f\) uniformly on \(B\).
Finally, by Theorem 2.11, there is a closed set \(F\) with \(F\subseteq B\) and \(m(B\setminus F)<\varepsilon /2\). Then \(F\subseteq B\subseteq E\) so \(F\subseteq E\), \(m(E\setminus F)=m(E\setminus B)+m(B\setminus F)<\varepsilon \), and \(f_n\to f\) uniformly on \(F\) because \(F\subseteq B\) and \(f_n\to f\) uniformly on \(B\). This completes the proof.
(Problem 1110) Give an example of a sequence of measurable functions on an unbounded measurable set \(E\) that converges pointwise almost everywhere to some function \(f\) that is finite almost everywhere, but such that the conclusion of Egoroff’s theorem fails.
[Chapter 3, Problem 16] Let \(I\subseteq \mathbb {R}\) be a closed, bounded interval and let \(E\subseteq I\) be measurable. Show that, for each \(\varepsilon >0\), there exists a step function \(h:I\to \R \) and a measurable set \(F\subseteq I\) such that \(h=\chi _E\) on \(F\) and such that \(m(I\setminus F)<\varepsilon \).
Proposition 3.11. Let \(\varphi :\R \to \R \) be a simple function and let \(\varepsilon >0\). Then there is a continuous function \(g:\R \to \R \) such that \(m^*(\{x\in \R :g(x)\neq \varphi (x)\})<\varepsilon \).
(Problem 1120) Prove Proposition 3.11.
Lusin’s theorem. Let \(E\subseteq \R \) be a measurable set and let \(f:E\to [-\infty ,\infty ]\) be measurable and finite almost everywhere. If \(\varepsilon >0\), then there is a continuous function \(g:\R \to \R \) and a closed set \(F\subseteq E\) such that \(f=g\) on \(F\) and such that \(m(E\setminus F)<\varepsilon \).
(Problem 1130) Prove Lusin’s theorem in the case \(m(E)<\infty \).
Let \(\{\varphi _n\}_{n=1}^\infty \) be as in the simple approximation theorem, so \(\varphi _n\to f\) pointwise. For each \(n\), apply Proposition 3.11 to obtain a continuous function \(g_n:\R \to \R \) such that \(m(\{x\in \R :g_n\neq \varphi _n\}) <\frac {\varepsilon }{2^{n+1}}\).Let \(A_n=\{x\in \R :g_n\neq \varphi _n\}\). Then \(m(\bigcup _{n=1}^\infty A_n)<\varepsilon /2\), and \(g_n=\varphi _n\) on \(E\setminus \bigcup _{n=1}^\infty A_n\). Thus \(g_n\to f\) on \(E\setminus \bigcup _{n=1}^\infty A_n\).
By Egoroff’s theorem, there is a closed set \(F\subseteq E\setminus \bigcup _{n=1}^\infty A_n\) such that \(g_n\to f\) uniformly on \(F\) and such that \(m((E\setminus \bigcup _{n=1}^\infty A_n)\setminus F)<\varepsilon /2\). Thus, \(F\subseteq E\) is closed and \(m(E\setminus F)<\varepsilon \), and because \(f\) is the uniform limit of a sequence of continuous functions, \(f\) is continuous on \(F\).
By Tietze’s extension theorem, there is a continuous function \(g:\R \to \R \) such that \(f=g\) on \(F\). This completes the proof.
(Problem 1131) Prove Lusin’s theorem.
By the previous result, for each \(n\in \Z \), there is a closed set \(F_n\subset (n,n+1)\cap E\) such that \(f\) is continuous on \(F_n\) and such that \(m(E\cap (n,n+1)\setminus F_n)<\varepsilon /2^{|n|+2}\).It is elementary to show that \(f\) is continuous on \(\bigcup _{n\in \Z } F_n\), and \(m(E\setminus \bigcup _{n\in \Z } F_n)<\varepsilon \). The conclusion follows from Tietze’s theorem.
[Definition: Step function] If \([a,b]\subset \R \) is a closed and bounded interval, we say that \(\varphi :[a,b]\to \R \) is a step function if there are finitely many points \(a=x_0<x_1<\dots <x_n=b\) such that \(\varphi \) is constant on each of the intervals \((x_{k-1},x_k)\) for all \(1\leq k\leq n\).
(Problem 1132) \(\varphi :[a,b]\to \R \) is a step function if and only if there is a finite integer \(n\), finitely many (possibly degenerate!) closed intervals \(I_1\), \(I_2,\dots ,I_n\) with each \(I_k\subseteq [a,b]\), and finitely many real numbers \(a_k\) such that \(\varphi =\sum _{k=1}^n a_k\chi _{I_k}\).
[Definition: Integral of a step function] If \(\varphi :[a,b]\to \R \) is a step function and \(x_0\), \(x_1,\dots ,x_n\) are the numbers in the definition of step function, we define \begin {equation*} \int _a^b \varphi = \sum _{k=1}^n (x_k-x_{k-1}) \, \varphi \biggl (\frac {x_{k-1}+x_k}{2}\biggr ). \end {equation*}
[Definition: Riemann integrable] Let \([a,b]\subset \R \) be a closed and bounded interval and let \(f:[a,b]\to \R \) be bounded. We say that \(f\) is Riemann integrable on \([a,b]\) if \begin {multline*} \sup \biggl \{\int _a^b \varphi \biggm |\varphi :[a,b]\to \R \text { is a step function and } \varphi (x)\leq f(x)\text { for all }x\in [a,b]\biggr \} \\= \inf \biggl \{\int _a^b \psi \biggm |\psi :[a,b]\to \R \text { is a step function and } \psi (x)\geq f(x)\text { for all }x\in [a,b]\biggr \}. \end {multline*} If \(f\) is Riemann integrable we define \begin {equation*} \int _a^b f=\sup \biggl \{\int _a^b \varphi \biggm |\varphi :[a,b]\to \R \text { is a step function and } \varphi \leq f\biggr \}. \end {equation*}
[Definition: Integral of a simple function] Let \(E\subseteq \R \) be measurable with \(m(E)<\infty \) and let \(\varphi :E\to \R \) be simple. Let \(\varphi (E)=\{c_1,c_2,\dots ,c_n\}\); as in Problem 1050, \(\varphi =\sum _{k=1}^n c_k \,\chi _{E_k}\), where \(E_k=\varphi ^{-1}(\{c_k\})\). We define \begin {equation*} \int _E \varphi = \sum _{k=1}^n c_k \,m(E_k). \end {equation*}
(Problem 1133) Let \(\varphi :\R \to \R \) be simple. If \(E\subseteq \R \) is measurable with \(m(E)<\infty \), we define \(\int _E \varphi :\int _E (\varphi \big \vert _E)\) where \(\varphi \big \vert _E\) denotes the restriction of \(\varphi \) to \(E\). If \(\varphi =\sum _{k=1}^n c_k \,\chi _{E_k}\), show that \begin {equation*} \int _E \varphi = \sum _{k=1}^n c_k \,m(E_k\cap E). \end {equation*}
(Problem 1134) Let \(E\) be a measurable set. Let \(\{D_1,\dots ,D_\ell \}\) be a partition of \(E\): \(E=\bigcup _{j=1}^\ell D_j\) and \(D_j\cap D_k=\emptyset \) if \(j\neq k\). Suppose furthermore that each \(D_j\) is measurable. Let \(\varphi :E\to \R \) and suppose that \(\varphi \) is constant on each \(D_j\). Then \(\varphi \) takes on at most \(\ell \) values, so is simple. Let \(b_j\) be such that \(\varphi (x)=b_j\) for all \(x\in D_j\). Show that \begin {equation*} \int _E \varphi =\sum _{j=1}^\ell b_j\,m(D_j) \end {equation*} even if the \(D_j\)s are not as in Problem 1050.
Let \(n\), \(c_k\), and \(E_k\) be as in Problem 1050. If \(1\leq j\leq \ell \), then either \(D_j=\emptyset \) or \(D_j\) contains at least one point \(x\in E\). But then \(x\in E_k\) for some \(k\), and so \(b_j=\varphi (x)=c_k\) because \(x\in D_j\) and \(x\in E_k\). Thus \(b_j=c_k\) and so \(D_j\subseteq E_k\). Thus if \(D_j\) is not empty then \(D_j\subseteq E_k\) for some \(k\). Conversely, the \(E_k\)s are pairwise disjoint and so if \(D_j\subseteq E_k\) then \(D_j\cap E_r\subseteq E_k\cap E_r=\emptyset \) if \(k\neq r\).Thus we may write \begin {equation*} \sum _{j=1}^\ell b_j\,m(D_j) =\sum _{\substack {1\leq j\leq \ell \\ D_j=\emptyset }} b_j\,m(D_j) +\sum _{k=1}^n\sum _{\substack {1\leq j\leq \ell \\ D_j\neq \emptyset \\D_j\subseteq E_k}} b_j\,m(D_j) . \end {equation*} If \(D_j=\emptyset \) then \(m(D_j)=0\). Thus \begin {equation*} \sum _{j=1}^\ell b_j\,m(D_j) =\sum _{k=1}^n\sum _{\substack {1\leq j\leq \ell \\D_j\subseteq E_k}} b_j\,m(D_j) . \end {equation*} But if \(D_j\subseteq E_k\) then \(b_j=c_k\). So \begin {equation*} \sum _{j=1}^\ell b_j\,m(D_j) =\sum _{k=1}^n c_k\sum _{\substack {1\leq j\leq \ell \\D_j\subseteq E_k}} m(D_j) . \end {equation*} Because the \(D_j\)s are pairwise disjoint and measurable, \begin {equation*} \sum _{\substack {1\leq j\leq \ell \\D_j\subseteq E_k}} m(D_j) = m\Bigl (\bigcup _{\substack {1\leq j\leq \ell \\D_j\subseteq E_k}} D_j\Bigr ). \end {equation*} By assumption \(\bigcup _{\substack {1\leq j\leq \ell \\D_j\subseteq E_k}} D_j \subseteq E_k\), while \(E_k\subseteq E =\bigcup _{j=1}^\ell D_j\) and so \(E_k=\bigcup _{j=1}^\ell (D_j\cap E_k)\). But either \(D_j\cap E_k=\emptyset \) or \(D_j\subseteq E_k\), and so \(\bigcup _{\substack {1\leq j\leq \ell \\D_j\subseteq E_k}} D_j = E_k\). Thus \begin {equation*} \sum _{j=1}^\ell b_j\,m(D_j) =\sum _{k=1}^n c_k m(E_k) \end {equation*} as desired.
(Problem 1140) Suppose that \(E=[a,b]\) and that \(\varphi \) is a step function. Show that \(\varphi \) is also a simple function and that \(\int _E \varphi =\int _a^b \varphi \).
Lemma 4.1. If \(E_1\), \(E_2,\dots ,E_n\) are measurable, \(c_1\), \(c_2,\dots ,c_n\in \R \), \(\varphi =\sum _{k=1}^n c_k \,\chi _{E_k}\), and \(\bigcup _{k=1}^n E_k\subseteq E\) for some measurable set \(E\), then \(\int _E \varphi =\sum _{k=1}^n c_k\,m(E_k)\) even if the \(E_k\)s and \(c_k\)s are not as in Problem 1050.
(Problem 1150) Prove Lemma 4.1.
Let \(S=\{1,2,\dots ,n\}\). Recall that \(2^S\) is the set of all subsets of \(S\). Observe that \(2^S\) is also a finite set. If \(x\in E\), let \(\sigma (x)=\{k\in S: x\in E_k\}\); then \(\sigma \) is a well defined function.For each \(A\subseteq S\), let \begin {equation*} D_A=\Bigl (\bigcap _{k\in A} E_k\Bigr ) \cap \Bigl (\bigcap _{k\in S\setminus A} E\setminus E_k\Bigr ). \end {equation*} Then each \(D_A\) is measurable and \(x\in D_{\sigma (x)}\) for each \(x\in E\).
Furthermore, we claim that \(A\neq B\) then \(D_A\cap D_B=\emptyset \). To see this, observe that if \(k\in A\) and \(k\notin B\), then \(D_A\subseteq E_k\) and \(D_B\subseteq E\setminus E_k\), and so \(D_A\cap D_B=\emptyset \). Similarly if \(k\in B\) and \(k\notin A\) then \(D_A\cap D_B=\emptyset \). If \(A\neq B\) there is at least one such \(k\).
Thus \(\{D_A:A\in 2^S\}\) is a partition of \(E\). Furthermore, observe that \begin {equation*} \varphi (x)=\sum _{k=1}^n c_k \,\chi _{E_k}(x)=\sum _{k\in \sigma (x)} c_k \end {equation*} and so \(\varphi = \sum _{k\in A} c_k\) on \(D_A\), and in particular is constant on \(D_A\).
Thus by Problem 1134, \begin {equation*} \int _E \varphi = \sum _{A\in 2^S} \sum _{k\in A} c_k m(D_A). \end {equation*} Changing the order of summation, we see that \begin {equation*} \int _E \varphi = \sum _{k=1}^n c_k \sum _{\substack {A\in 2^S\\ A\owns k}} m(D_A). \end {equation*} The \(D_A\)s are a partition of \(E\), and each \(D_A\) is either a subset of \(E_k\) or a subset of \(E\setminus E_k\) (that is, disjoint from \(E_k\)); thus \(E_k=\bigcup _{\substack {A\in 2^S\\D_A\subseteq E_k}} D_A\) and \(m(E_k)=\sum _{\substack {A\in 2^S\\ A\owns k}} m(D_A)\). Thus \begin {equation*} \int _E \varphi = \sum _{k=1}^n c_k m(E_k) \end {equation*} as desired.
Proposition 4.2. Let \(\varphi \), \(\psi \) be simple functions defined on a set of finite measure \(E\).
(Problem 1160) Prove Proposition 4.2, part (i).
Because \(\varphi \) and \(\psi \) are simple, we may write \(\varphi =\sum _{k=1}^n a_k \,\chi _{D_k}\) and \(\psi =\sum _{k=1}^\ell b_k\,\chi _{S_k}\) for some real numbers \(a_k\), \(b_k\) and some measurable sets \(D_k\), \(S_k\subseteq E\).Define \begin {gather*} \widetilde a_k=\begin {cases}a_k, & 1\leq k\leq n,\\0,&n+1\leq k\leq n+\ell ,\end {cases} \gatherbreak \widetilde b_k=\begin {cases}0, & 1\leq k\leq n,\\b_{k-n},&n+1\leq k\leq n+\ell ,\end {cases} \gatherbreak E_k=\begin {cases}D_k, & 1\leq k\leq n,\\S_{k-n},&n+1\leq k\leq n+\ell .\end {cases} \end {gather*} Then \(\varphi =\sum _{k=1}^{n+\ell } \widetilde a_k\,\chi _{E_k}\), \(\psi =\sum _{k=1}^{n+\ell } \widetilde b_k\,\chi _{E_k}\), and \(\alpha \varphi +\beta \psi =\sum _{k =1}^{n+\ell } (\alpha \widetilde a_k+\beta \widetilde b_k)\,\chi _{E_k}\), and so by Lemma 4.1 \begin {align*} \alpha \int _E \varphi +\beta \int _E \psi &= \alpha \sum _{k=1}^{n+\ell } \widetilde a_k\,m(E_k) +\beta \sum _{k=1}^{n+\ell } \widetilde b_k\,m(E_k) \alignbreak = \sum _{k=1}^{n+\ell } (\alpha \widetilde a_k+\beta \widetilde b_k)\,m(E_k) =\int _E (\alpha \varphi +\beta \psi ) \end {align*}
as desired.
(Problem 1170) Prove Proposition 4.2, part (ii).
Let \(\eta =\psi -\varphi \). Then \(\eta \geq 0\) on \(E\). Furthermore, \(\eta \) is simple. Let \(\{c_1,\dots ,c_n\}=\eta (E)\) and let \(E_k=\eta ^{-1}(\{c_k\})\). Then by part (i) and by definition \begin {equation*} \int _E\psi -\int _E\varphi =\int _E (\psi -\varphi )=\int _E\eta = \sum _{k=1}^n c_k\,m(E_k). \end {equation*} But each \(m(E_k)\geq 0\) by definition of measure, and each \(c_k\geq 0\) because \(\eta \geq 0\) and so \(\eta (E)\subset [0,\infty )\). Thus \(\sum _{k=1}^n c_k\,m(E_k)\geq 0\), so \(\int _E \psi \geq \int _E \varphi \), as desired.
[Definition: Integral of a bounded function over a bounded set] Let \(E\subset \R \) be measurable with \(m(E)<\infty \) and let \(f:E\to [-M,M]\) be a bounded function. We say that \(f\) is Lebesgue integrable over \(E\) if \begin {multline*} \sup \biggl \{\int _E \varphi \biggm |\varphi :[a,b]\to \R \text { is a simple function and } \varphi (x)\leq f(x)\text { for all }x\in E\biggr \} \\= \inf \biggl \{\int _E \psi \biggm |\psi :[a,b]\to \R \text { is a simple function and } \psi (x)\geq f(x)\text { for all }x\in E\biggr \}. \end {multline*} If \(f\) is Lebesgue integrable we define \begin {equation*} \int _E f=\sup \biggl \{\int _E \varphi \biggm |\varphi :[a,b]\to \R \text { is a simple function and } \varphi (x)\leq f(x)\text { for all }x\in E\biggr \}. \end {equation*}
(Problem 1171) Let \(E\subset \R \) be measurable with \(m(E)<\infty \) and let \(\theta :E\to \R \) be simple. Show that \begin {align*} \int _E \theta &= \sup \biggl \{\int _E \varphi \biggm |\varphi :[a,b]\to \R \text { is a simple function and } \varphi (x)\leq \theta (x)\text { for all }x\in E\biggr \} \\ &=\inf \biggl \{\int _E \psi \biggm |\psi :[a,b]\to \R \text { is a simple function and } \psi (x)\geq \theta (x)\text { for all }x\in E\biggr \}. \end {align*}
Thus all simple functions with domains of bounded measure are integrable and there is no ambiguity in using \(\int _E \theta \) to denote both the integral of a simple function and of an arbitrary Lebesgue integrable function.
Theorem 4.3. If \(f\) is Riemann integrable on \([a,b]\), then \(f\) is Lebesgue integrable over \([a,b]\) and \(\int _a^b f=\int _{[a,b]}f\).
(Problem 1180) Prove Theorem 4.3 and give an example of a bounded measurable function defined on an interval \([a,b]\) that is Lebesgue integrable over \([a,b]\) but is not Riemann integrable.
Assume that \(f:[a,b]\to \R \) is bounded and Riemann integrable. By Problem 1140, if \(\varphi \) is a step function on \([a,b]\), then \(\varphi \) is a simple function and \(\int _a^b\varphi =\int _{[a,b]}\varphi \). Thus, \begin {multline*} \sup \biggl \{\int _a^b \varphi \biggm |\varphi :[a,b]\to \R \text { is a step function and } \varphi (x)\leq f(x)\text { for all }x\in [a,b]\biggr \} \\= \sup \biggl \{\int _{[a,b]} \varphi \biggm |\varphi :[a,b]\to \R \text { is a step function and } \varphi (x)\leq f(x)\text { for all }x\in [a,b]\biggr \} . \end {multline*} Because all step functions are simple functions, by definition of supremum \begin {multline*} \sup \biggl \{\int _{[a,b]} \varphi \biggm |\varphi :[a,b]\to \R \text { is a step function and } \varphi (x)\leq f(x)\text { for all }x\in [a,b]\biggr \} \\\leq \sup \biggl \{\int _{[a,b]} \varphi \biggm |\varphi :[a,b]\to \R \text { is a simple function and } \varphi (x)\leq f(x)\text { for all }x\in [a,b]\biggr \} . \end {multline*} Now, if \(\varphi \) and \(\psi \) are simple functions and \(\varphi \leq f\leq \psi \), then \(\varphi \leq \psi \) and so by Proposition 4.2 \(\int _{[a,b]}\varphi \leq \int _{[a,b]}\psi \). Thus, again by definition of supremum and infimum, \begin {multline*} \sup \biggl \{\int _{[a,b]} \varphi \biggm |\varphi :[a,b]\to \R \text { is a simple function and } \varphi (x)\leq f(x)\text { for all }x\in [a,b]\biggr \} \\\leq \inf \biggl \{\int _{[a,b]} \psi \biggm |\psi :[a,b]\to \R \text { is a simple function and } \psi (x)\geq f(x)\text { for all }x\in [a,b]\biggr \} . \end {multline*} Again by Problem 1140 and because all step functions are simple functions, \begin {multline*} \inf \biggl \{\int _{[a,b]} \psi \biggm |\psi :[a,b]\to \R \text { is a simple function and } \psi (x)\geq f(x)\text { for all }x\in [a,b]\biggr \} \\\leq \inf \biggl \{\int _a^b \psi \biggm |\psi :[a,b]\to \R \text { is a step function and } \psi (x)\geq f(x)\text { for all }x\in [a,b]\biggr \} . \end {multline*} But because \(f\) is Riemann integrable, we have that \begin {multline*} \inf \biggl \{\int _a^b \psi \biggm |\psi :[a,b]\to \R \text { is a step function and } \psi (x)\geq f(x)\text { for all }x\in [a,b]\biggr \} \\= \sup \biggl \{\int _a^b \varphi \biggm |\varphi :[a,b]\to \R \text { is a step function and } \varphi (x)\leq f(x)\text { for all }x\in [a,b]\biggr \} . \end {multline*} Thus the chain of inequalities collapses and we must have that all of the above quantities are equal. In particular, \begin {multline*} \sup \biggl \{\int _{[a,b]} \varphi \biggm |\varphi :[a,b]\to \R \text { is a simple function and } \varphi (x)\leq f(x)\text { for all }x\in [a,b]\biggr \} \\= \inf \biggl \{\int _{[a,b]} \psi \biggm |\psi :[a,b]\to \R \text { is a simple function and } \psi (x)\geq f(x)\text { for all }x\in [a,b]\biggr \} \end {multline*} This is the definition of Lebesgue integrability. This completes the proof.Now let \(f=\chi _{\Q \cap [0,1]}\) be the characteristic function of the rationals restricted to \([0,1]\). By Problem 370, \(f\) is not Riemann integrable. However, \(f\) is simple (the set \(\Q \) is measurable because it has outer measure zero), and so is Lebesgue integrable by Problem 1171.
Theorem 4.4. Let \(E\subset \R \) be measurable with \(m(E)<\infty \) and let \(f:E\to [-M,M]\) be bounded and measurable. Then \(f\) is Lebesgue integrable.
(Problem 1190) Prove Theorem 4.4.
Let \begin {align*} L&=\sup \biggl \{\int _{E} \varphi \biggm |\varphi :E\to \R \text { is a simple function and } \varphi (x)\leq f(x)\text { for all }x\in E\biggr \} ,\\ U&=\inf \biggl \{\int _{E} \psi \biggm |\psi :E\to \R \text { is a simple function and } \psi (x)\geq f(x)\text { for all }x\in E\biggr \} . \end {align*}As before, by Proposition 4.2, \(L\leq U\).
By the simple approximation lemma, if \(\varepsilon >0\), then there exist simple functions \(\varphi _\varepsilon \), \(\psi _\varepsilon :E\to \R \) such that \begin {equation*} \psi _\varepsilon (x)-\varepsilon \leq \varphi _\varepsilon (x)\leq f(x)\leq \psi _\varepsilon (x)\leq \varphi _\varepsilon (x)+\varepsilon \end {equation*} for all \(x\in E\).
Thus, if \(\varepsilon >0\), then \begin {equation*} L \geq \int _E \varphi _\varepsilon \end {equation*} and \begin {equation*} U\leq \int _E \psi _\varepsilon . \end {equation*} Thus \begin {equation*} 0\leq U-L\leq \int _E\psi _\varepsilon -\int _E \varphi _\varepsilon \end {equation*} and so by Proposition 4.2 \begin {equation*} 0\leq U-L\leq \int _E(\psi _\varepsilon -\varphi _\varepsilon ) \leq \int _E \varepsilon = \varepsilon \,m(E) \end {equation*} for all \(\varepsilon >0\). Thus \(U-L\leq 0\) and so \(U-L=0\). By definition of \(U\), \(L\), and Lebesgue integrablility, we have that \(f\) is Lebesgue integrable, as desired.
Theorem 5.7. Let \(E\subset \R \) be measurable with \(m(E)<\infty \) and let \(f:E\to [-M,M]\) be bounded. Then \(f\) is measurable if and only if it is Lebesgue integrable. (You may not use this result until we prove it in Chapter 5, but you may find it interesting at this point.)
Theorem 4.5. Let \(f\) and \(g\) be bounded measurable functions defined on a set of finite measure \(E\).
(Problem 1200) Prove Theorem 4.5, part (i).
(Problem 1210) Prove Theorem 4.5, part (ii).
Because \(f\) and \(g\) are measurable, we have that \(g-f\) is measurable by Theorem 3.6, and so \(g-f\) is integrable by Theorem 4.4. By part (i), \begin {equation*} \int _E g-\int _E f=\int _E (g-f). \end {equation*} But \(\varphi =0\) is a simple function with \(\varphi \leq g-f\) on \(E\), so \begin {equation*} 0=\int _E \varphi \leq \sup \biggl \{\int _{E} \varphi \biggm |\varphi :E\to \R \text { is a simple function and } \varphi (x)\leq g(x)-f(x)\text { for all }x\in E\biggr \} =\int _E (g-f)=\int _E g-\int _E f \end {equation*} and so \begin {equation*} \int _E f\leq \int _E g \end {equation*} as desired.
Corollary 4.6. If \(A\) and \(B\) are two disjoint measurable sets of finite measure and \(f:A\cup B\to \R \) is bounded and measurable, then \(\int _{A\cup B} f=\int _A f+\int _B f\).
(Problem 1220) Prove Corollary 4.6.
Corollary 4.7. If \(E\subset \R \) is measurable and has finite measure, and if \(f:E\to \R \) is bounded and measurable, then \begin {equation*} \biggl |\int _E f\biggr |\leq \int _E |f|. \end {equation*}
Proposition 4.8. If \(E\subset \R \) is measurable and has finite measure, if \(f_n:E\to \R \) is bounded and measurable for each \(n\), and if \(f_n\to f\) uniformly on \(E\), then \begin {equation*} \lim _{n\to \infty }\int _E f_n= \int _E f. \end {equation*}
(Problem 1230) Prove Proposition 4.8.
(Problem 1240) Give an example of a sequence of measurable functions \(\{f_n\}_{n=1}^\infty \), each of which is bounded, defined on a common measurable domain \(E\) of finite measure, such that \(f_n\to f\) pointwise on \(E\) for some bounded measurable function \(f:E\to \R \), but such that \begin {equation*} \int _E f_n\not \to \int _E f. \end {equation*} (The failure can be either because \(\lim _{n\to \infty }\int _E f_n\) does not exist, or because it exists but is not equal to \(\int _E f\).)
The bounded convergence theorem. If \(E\subset \R \) is measurable and has finite measure, if \(f_n:E\to \R \) is measurable for each \(n\), if there is a \(M\) such that \(|f_n(x)|<M\) for all \(x\in E\) and all \(n\in \N \), and if \(f_n\to f\) pointwise on \(E\), then \begin {equation*} \lim _{n\to \infty }\int _E f_n= \int _E f. \end {equation*}
(Problem 1250) Prove the Bounded Convergence Theorem. Hint: Use Egoroff’s theorem.
Choose some \(\varepsilon >0\).Let \(F\subseteq E\) be as in Egoroff’s theorem, so \(F\) is closed, \(m(E\setminus F)<\varepsilon \), and \(f_n\to f\) uniformly on \(F\). Thus, there is a \(N\in \N \) such that, if \(n\geq N\), then \(|f_n(x)-f(x)|<\varepsilon \) for all \(x\in F\).
If \(n\geq N\), then by Theorem 4.5(i), Corollary 4.7, and Corollary 4.6, \begin {equation*} \biggl |\int _E f_n-\int _E f\biggr |=\biggl |\int _E (f_n- f)\biggr | \leq \int _E |f_n-f| = \int _F |f_n-f|+\int _{E\setminus F} |f_n-f|. \end {equation*} Then by Theorem 4.5(ii), \begin {equation*} \int _F |f_n-f|+\int _{E\setminus F} |f_n-f| \leq \int _F \varepsilon + \int _{E\setminus F} 2M =m(F)\varepsilon + 2Mm(E\setminus F) \leq m(E)\varepsilon + 2M \varepsilon . \end {equation*} Thus, if \(n\geq N\) then \begin {equation*} \biggl |\int _E f_n-\int _E f\biggr |\leq m(E)\varepsilon + 2M \varepsilon . \end {equation*} This suffices to show that \(\int _E f_n\to \int _E f\).
[Definition: Finite support] Let \(E\subseteq \R \) be measurable and let \(h:E\to \R \). Suppose that there is a measurable set \(E_0\subseteq E\) with \(m(E_0)<\infty \) and such that \(h(x)=0\) for all \(x\in E\setminus E_0\). Then we say that \(h\) has finite support; if \(h\) is also bounded then we define \(\int _E h=\int _{E_0} h\).
[Definition: Integral of a nonnegative function] Let \(E\subseteq \R \) be measurable and let \(f:E\to [0,\infty ]\) be measurable. We define \begin {equation*} \int _E f=\sup \biggl \{\int _E h \biggm | h\text { is bounded, measurable, of finite support, and $0\leq h\leq f$ on }E\biggr \}. \end {equation*}
(Problem 1251) Show that if \(m(E)<\infty \) and \(f:E\to [0,M]\) is measurable, nonnegative, and bounded, then the above definition coincides with that in Section 4.2.
[Chapter 4, Problem 24] Let \(E\subseteq \R \) be measurable and let \(f:E\to [0,\infty ]\) be measurable. Then \begin {equation*} \int _E f=\sup \biggl \{\int _E \varphi \biggm | \varphi \text { is simple, of finite support, and $0\leq \varphi \leq f$ on }E\biggr \}. \end {equation*} If we define \(\int _E \varphi =\infty \) whenever \(\varphi \) is a nonnegative simple function that is not of finite support, then we also have that \begin {equation*} \int _E f=\sup \biggl \{\int _E \varphi \biggm | \varphi \text { is simple and $0\leq \varphi \leq f$ on }E\biggr \}. \end {equation*}
Chebychev’s inequality. Let \(f\) be a nonnegative measurable function on a measurable set \(E\). Let \(\lambda >0\). Then \begin {equation*} m(\{x\in E:f(x)\geq \lambda \}) \leq \frac {1}{\lambda } \int _E f. \end {equation*}
(Problem 1260) Prove Chebychev’s Inequality.
Let \(E_{\lambda ,N}=\{x\in E\cap [-N,N]:f(x)\geq \lambda \}\). Because \(f\) is measurable, so is \(E_{\lambda ,N}\). Let \(h_N=\lambda \chi _{E_{\lambda ,N}}\). Because \(f\) is nonnegative, \(h_N(x)=0\leq f(x)\) for all \(x\in E\setminus E_{\lambda ,N}\); by definition of \(E_{\lambda ,N}\), \(h_N(x)=\lambda \leq f(x)\) for all \(x\in E_{\lambda ,N}\).Furthermore, \(h_N\) is clearly a set of finite support.
Thus by definition of \(\int _E f\), \(\int _E f\geq \int _E h_N=\lambda m(E_{\lambda ,N})\) and so \(m(E_{\lambda ,N})\leq \frac {1}{\lambda }\int _E f\).
Observe that \(\{x\in E:f(x)\geq \lambda \} =\bigcup _{N=1}^\infty E_{\lambda ,N}\). Thus \begin {equation*} m(\{x\in E:f(x)\geq \lambda \}) =m\Bigl (\bigcup _{N=1}^\infty E_{\lambda ,N}\Bigr ). \end {equation*} The sets \(E_{\lambda ,N}\) are nondecreasing in \(N\), so by Theorem 2.15, \begin {equation*} m(\{x\in E:f(x)\geq \lambda \}) =m\Bigl (\bigcup _{N=1}^\infty E_{\lambda ,N}\Bigr ) =\lim _{N\to \infty } m(E_{\lambda ,N}). \end {equation*} Because the sets \(E_{\lambda ,N}\) are nondecreasing in \(N\), the right hand side is the limit of a nondecreasing sequence of real numbers, and so \begin {equation*} m(\{x\in E:f(x)\geq \lambda \}) =\sup _N m(E_{\lambda ,N}). \end {equation*} But \(m(E_{\lambda ,N})\leq \frac {1}{\lambda }\int _E f\) for each \(N\), and so \(\sup _N m(E_{\lambda ,N})\leq \frac {1}{\lambda }\int _E f\), as desired.
Proposition 4.9. Let \(f\) be a nonnegative measurable function on a measurable set \(E\). Then \(\int _E f=0\) if and only if \(f(x)=0\) for almost every \(x\in E\).
(Problem 1270) Prove Proposition 4.9.
Suppose that \(f\geq 0\) and \(\int _E f=0\). Let \(E_n=\{x\in E:f(x)\geq \frac {1}{n}\}\). Then \(\{x\in E:f(x)>0\}=\bigcup _{n\in \N } E_n\) by the Archimedean property of the real numebrs. By Chebychev’s inequality, \(m(E_n)\leq n\int _E f =0\) for each \(n\in \N \), and so by the subadditivity of Lebesgue measure (Proposition 2.3), \(m(\{x\in E:f(x)>0\})\leq \sum _{n=1}^\infty m(E_n)=0\), as desired.We now come to the converse. If \(\varphi \) is a simple function that is nonpositive almost everywhere, then \(\int _E \varphi \leq 0\) by definition of integral of a simple function.
If \(h\) is a bounded measurable function of finite support, then \(h\) is Lebesgue integrable by Theorem 4.4 and so \begin {equation*} \int _E h=\sup \biggl \{\int _E \varphi \biggm |\varphi :E\to \R \text { is simple, }\varphi \leq h\}. \end {equation*} If \(h\leq 0\) almost everywhere, then \(\varphi \leq 0\) almost everywhere for all such \(\varphi \), and so \(\int _E h\leq 0\).
Finally, if \(f\geq 0\) is measurable and \(f=0\) almost everywhere, then \begin {equation*} \int _E f=\sup \biggl \{\int _E h\biggm |h:E\to \R \text { is bounded measurable and finitely supported, }h\leq f\}. \end {equation*} All of the terms on the right hand side are nonnegative, so \(\int _E f\leq 0\). But \(h\equiv 0\) is a bounded measurable finitely supported function, and so \(\int _E f\geq \int _E 0=0\); thus \(\int _E f=0\), as desired.
Theorem 4.10. Let \(f\) and \(g\) be nonnegative measurable functions defined on a measurable set \(E\).
(Problem 1280) Prove Theorem 4.10, part (i).
(Problem 1281) Prove Theorem 4.10, part (ii).
Theorem 4.11. If \(A\) and \(B\) are two disjoint measurable sets and \(f:A\cup B\to \R \) is measurable and nonnegative, then \(\int _{A\cup B} f=\int _A f+\int _B f\). In particular, if \(m(E_0)=0\) and \(E_0\subseteq E\) for a measurable set \(E\), then \(\int _E f=\int _{E\setminus E_0} f\) for every nonnegative measurable function \(f:E\to [0,\infty ]\).
(Problem 1290) Prove Theorem 4.11.
Fatou’s lemma. Let \(E\subseteq \R \) be measurable and let \(\{f_n\}_{n=1}^\infty \) be a sequence of nonnegative measurable functions \(f_n:E\to [0,\infty ]\). Then \begin {equation*} \int _E \liminf _{n\to \infty } f_n\leq \liminf _{n\to \infty } \int _E f_n. \end {equation*}
(Problem 1300) Prove Fatou’s lemma.
Recall that \begin {equation*} \int _E \liminf _{n\to \infty } f_n=\sup \biggl \{\int _E h \biggm | h\text { is bounded, measurable, of finite support, and $0\leq h\leq \liminf _{n\to \infty } f_n$ on }E\biggr \}. \end {equation*} Let \(h\) be nonnegative, bounded, measurable, and have finite support, and let \(h\) satisfy \(0\leq h\leq \liminf _{n\to \infty } f_n\). It suffices to show that \(\int _E h \leq \liminf _{n\to \infty } \int _E f_n\) for all such \(h\).Let \(h_n=\min (h,f_n)\). Then \(h_n\) is also nonnegative, bounded, measurable, and of finite support.
I claim that \(h_n\to h\) pointwise. Recall \(h(x)\leq \liminf _{n\to \infty } f_n(x)=\lim _{n\to \infty } \inf _{k\geq n} f_k(x)=\sup _{n\in \N } \inf _{k\geq n} f_k(x)\). Thus, if \(\varepsilon >0\) and \(x\in E\), then there is some \(n\in \N \) such that \(h(x)<\inf _{k\geq n} f_k(x)+\varepsilon \).
Thus, if \(k\geq n\) then \(h(x)\geq h_k(x)=\min (h(x),f_k(x))>h(x)-\varepsilon \), and so \(|h(x)-h_k(x)|<\varepsilon \) for all \(k\geq n\). The claim is proven.
But \(h\) is bounded, and because \(h_n\leq h\) for all \(n\) the \(h_n\)s are uniformly bounded. Thus by the bounded convergence theorem, \(\lim _{n\to \infty } \int _E h_n=\int _E h\).
But \(h_n\leq f_n\) and so \(\int _E h_n\leq \int _E f_n\), and so \(\lim _{n\to \infty } \int _E h_n\leq \liminf _{n\to \infty } \int _E f_n\). Thus \begin {equation*} \int _E h=\lim _{n\to \infty } \int _E h_n\leq \liminf _{n\to \infty } \int _E f_n, \end {equation*} as desired.
The monotone convergence theorem. Let \(E\subseteq \R \) be measurable and let \(\{f_n\}_{n=1}^\infty \) be a sequence of nonnegative measurable functions \(f_n:E\to [0,\infty ]\). Suppose in addition that \(f_n(x)\leq f_{n+1}(x)\) for all \(x\in E\). Then \begin {equation*} \int _E \lim _{n\to \infty } f_n= \lim _{n\to \infty } \int _E f_n. \end {equation*}
(Problem 1310) Prove the monotone convergence theorem.
Corollary 4.12. Let \(E\subseteq \R \) be measurable and let \(\{f_n\}_{n=1}^\infty \) be a sequence of nonnegative measurable functions \(f_n:E\to [0,\infty ]\). Then \begin {equation*} \int _E \sum _{n=1}^\infty f_n= \sum _{n=1}^\infty \int _E f_n. \end {equation*}
(Problem 1311) Prove Corollary 4.12.
[Definition: Integrable function] A nonnegative measurable function \(f\) on a measurable set \(E\) is said to be integrable, integrable over \(E\), or in \(L^1(E)\), if \begin {equation*} \int _E f<\infty . \end {equation*}
Proposition 4.14. Let \(E\subseteq \R \) be measurable and let \(f:E\to [-\infty ,\infty ]\) be measurable. Then \(|f|\) is integrable (that is, \(\int _E |f|<\infty \)) if and only if both \(f^+\) and \(f^-\) are integrable.
[Definition: General Lebesgue integral] Suppose that \(f:E\to [-\infty ,\infty ]\) is measurable and that \(|f|\) is integrable. Then we say that \(f\) is integrable and that \begin {equation*} \int _E f=\int _E f^+-\int _E f^-. \end {equation*}
(Problem 1312) Show that if \(f\) is integrable over \(E\) and nonnegative, then the above definition of \(\int _E f\) coincides with that in Section 4.3.
Proposition 4.13. If \(f\) is integrable over \(E\), then \(f\) is finite almost everywhere on \(E\).
(Problem 1320) Prove Proposition 4.13.
[Chapter 4, Problem 28] Let \(f\) be integrable over \(E\) and let \(C\) be a measurable subset of \(E\). Show that \(\int _C f =\int _E f\,\chi _C\).
Proposition 4.15. If \(f\) is integrable over \(E\), then \(\int _E f=\int _{E\setminus E_0} f\) whenever \(m(E_0)=0\).
(Problem 1321) Prove Proposition 4.15.
By Problem 4.24, \begin {equation*} \int _E f=\sup \biggl \{\int _E \varphi \biggm | \varphi \text { is simple, of finite support, and $0\leq \varphi \leq f$ on }E\biggr \} \end {equation*} and \begin {equation*} \int _{E\setminus E_0} f=\sup \biggl \{\int _{E\setminus E_0} \varphi \biggm | \varphi \text { is simple, of finite support, and $0\leq \varphi \leq f$ on }E\biggr \} \end {equation*} Recall from Problem 1041 that simple functions may be extended to functions defined on all of \(\R \); it is clear that we may extend by zero and so the extension is also of finite support. Let \(\varphi :\R \to [0,\infty )\) be simple, of finite support and satisfy \(0\leq \varphi \leq f\). If \(\varphi =\sum _{k=1}^n c_k\chi _{E_k}\), then by Problem 1133 \begin {equation*} \int _E \varphi = \sum _{k=1}^n c_k \,m(E_k\cap E) = \sum _{k=1}^n c_k \,m(E_k\cap E\setminus E_0) =\int _{E\setminus E_0}\varphi \end {equation*}
Proposition 4.16. (The integral comparison test.) Suppose that \(g\) is nonnegative and integrable over \(E\) and that \(|f|\leq g\) on \(E\). If \(f\) is measurable, then \(f\) is also integrable and \(\left |\int _E f\right |\leq \int _E |f|\leq \int _E g\).
(Problem 1330) Prove Proposition 4.16.
Theorem 4.17. Let \(f\) and \(g\) be functions integrable over a measurable set \(E\).
(Problem 1340) Prove Theorem 4.17, part (i).
If \(\alpha >0\), then \((\alpha f)^+=\alpha f^+\) and \((\alpha f)^-=\alpha f^-\). By Theorem 4.10, \(\int _E \alpha (f^+)=\alpha \int _E f^+<\infty \) and \(\int _E \alpha (f^-)=\alpha \int _E f^-<\infty \), and so \(\alpha f\) is integrable. Furthermore, \begin {equation*} \int _E (\alpha f)=\int _E (\alpha f)^+-\int _E (\alpha f)^-=\int _E \alpha (f^+)-\int _E \alpha (f^-)=\alpha \int _E f^+-\alpha \int _E f^-=\alpha \int _E f \end {equation*} as desired.If \(\alpha =0\) then \(\alpha f=0\). The zero function is clearly integrable and satisfies \begin {equation*} \int _E (\alpha f)=\int _E 0 = 0 =0\int _E f=\alpha \int _E f \end {equation*} because \(\int _E f\) is finite.
If \(\alpha <0\), then \((\alpha f)^+=-\alpha f^-\) and \((\alpha f)^-=-\alpha f^+\). By Theorem 4.10, \(\int _E -\alpha (f^+)=-\alpha \int _E f^+<\infty \) and \(\int _E -\alpha (f^-)=-\alpha \int _E f^-<\infty \), and so \(\alpha f\) is integrable. Furthermore, \begin {align*} \int _E (\alpha f) &=\int _E (\alpha f)^+-\int _E (\alpha f)^- =\int _E (-\alpha ) (f^-)-\int _E (-\alpha ) (f^+) \\&=(-\alpha ) \int _E (f^-)-(-\alpha ) \int _E (f^+) =\alpha \int _E f^+-\alpha \int _E f^-=\alpha \int _E f \end {align*}
as desired.
Thus \(\alpha f\) and \(\beta g\) are integrable and have the expected integrals. To complete the proof, it suffices to show that \(f+g\) is integrable and has the expected integral.
But \(|f+g|\leq |f|+|g|\) by the triangle inequality in the real numbers, and \(\int _E(|f|+|g|)=\int _E |f|+\int _E |g|<\infty \) by Theorem 4.10 and because \(f\) and \(g\) are integrable.
By Proposition 4.13, \(E_0=\{x\in E:|f(x)|=\infty \}\cup \{x\in E:|g(x)|=\infty \}\) has measure zero. Thus, because \(f\), \(g\), and \(f+g\) are integrable, we have that \begin {equation*} \int _E f=\int _{E\setminus E_0} f,\qquad \int _E g=\int _{E\setminus E_0} g, \qquad \int _E(f+g) = \int _{E\setminus E_0} (f+g). \end {equation*}
Furthermore, in \(E\setminus E_0\) (where \(f+g\) is necessarily defined), \begin {equation*} (f+g)^+-(f+g)^-=f+g=f^+-f^-+g^+-g^-=(f^++g^+)-(f^-+g^-) \end {equation*} and so \begin {equation*} (f+g)^++(f^-+g^-)=(f^++g^+)+(f+g)^- \end {equation*} and so \begin {align*} \int _E (f+g)^++(f^-+g^-) &=\int _{E\setminus E_0} (f+g)^++(f^-+g^-) \\&=\int _{E\setminus E_0} (f^++g^+)+(f+g)^- =\int _{E} (f^++g^+)+(f+g)^-. \end {align*}
Applying Theorem 4.10 yields that \begin {equation*} \int _E (f+g)^++\int _E f^-+\int _E g^-=\int _{E} f^++\int _E g^++\int _E (f+g)^-. \end {equation*} Observe that all integrals are finite. We may thus rearrange the equation to see that \begin {equation*} \int _E (f+g)=\int _E (f+g)^+-\int _E (f+g)^- =\int _{E} f^+-\int _E f^-+\int _E g^+-\int _E g^-=\int _E f+\int _E g \end {equation*} as desired.
(Problem 1350) Prove Theorem 4.17, part (ii).
Suppose that \(f\leq g\). If \(x\in E\), then either:
\(0\leq g(x)\leq f(x)\). In this case \(g^-(x)=0=f^-(x)\) and \(g^+(x)=g(x)\leq f(x)=f^+(x)\).
\(g(x)\leq 0\leq f(x)\). In this case \(g^-(x)=-g(x)\geq 0=f^-(x)\) and \(g^+(x)=0\leq f(x)=f^+(x)\).
\(g(x)\leq f(x)\leq 0\). In this case \(g^-(x)=-g(x)\geq -f(x)=f^-(x)\) and \(g^+(x)=0=f^+(x)\).
In any case \(g^-(x)\geq f^-(x)\) and \(g^+(x)\leq f^+(x)\), and so by Theorem 4.10 \begin {equation*} \int _E g^-\geq \int _E f^-,\qquad \int _E g^+\leq \int _E f^+. \end {equation*} This immediately yields \begin {equation*} \int _E f=\int _E f^+-\int _E f^-\geq \int _E g^+-\int _E g^-=\int _E g \end {equation*} as desired.
Corollary 4.18. If \(A\) and \(B\) are two disjoint measurable sets and \(f:A\cup B\to \R \) is integrable over \(A\cup B\), then \(\int _{A\cup B} f=\int _A f+\int _B f\).
(Problem 1360) Prove Corollary 4.18.
The Lebesgue dominated convergence theorem. Let \(E\subseteq \R \) be measurable and let \(f\), \(f_n\), and \(g\) be measurable functions with domain \(E\). Suppose that \(g\) is nonnegative and integrable, that \(|f_n(x)|\leq g(x)\) for all \(n\in \N \) and almost every \(x\in E\), and that \(f_n\to f\) pointwise almost everywhere on \(E\). Then \(\int _E f_n\to \int _E f\).
(Problem 1370) Prove the Lebesgue dominated convergence theorem.
Let \begin {gather*} E_0=\{x\in E:|g(x)|=\infty \} \cup \{x\in E:f_n(x)\not \to f(x)\}\cup \gatherbreak \bigcup _{n=1}^\infty \{x\in E: |f_n(x)|>|g(x)|\}. \end {gather*} By Proposition 4.13, definition of almost everywhere, and countable subadditivity of measure (Proposition 2.3), we have that \(m(E_0)=0\). By Proposition 4.15, \(\int _E f=\int _{E\setminus E_0} f\) and \(\int _E f_n=\int _{E\setminus E_0} f_n\), so it suffices to show that \(\int _{E\setminus E_0} f_n\to \int _{E\setminus E_0} f\).Observe that \(f_n\leq |f_n|\leq g\) on \(E\setminus E_0\) and so \(g-f_n\geq 0\) on this set for all \(n\). Similarly, \(g+f_n\geq 0\). By linearity of limits, \(g\pm f_n\to g\pm f\) and so \(\liminf _{n\to \infty } (g-f_n)=g-f\) and \(\liminf _{n\to \infty } (g+f_n)=g+f\). By Fatou’s lemma, \begin {gather*} \int _{E\setminus E_0} (g-f)\leq \liminf _{n\to \infty } \int _{E\setminus E_0} (g-f_n), \gatherbreak \int _{E\setminus E_0} (g+f)\leq \liminf _{n\to \infty } \int _{E\setminus E_0} (g+f_n). \end {gather*} If \(G\) is a constant and \(a_n\) is any sequence of real numbers, then \begin {equation*} \liminf _{n\to \infty } (G+a_n)=G+\liminf _{n\to \infty } a_n. \end {equation*} Thus \begin {align*} &\int _{E\setminus E_0} (g-f)\leq \int _{E\setminus E_0} g+ \liminf _{n\to \infty } \int _{E\setminus E_0} (-f_n), \alignbreak \int _{E\setminus E_0} (g+f)\leq \int _{E\setminus E_0} g+\liminf _{n\to \infty } \int _{E\setminus E_0} f_n. \end {align*}
The function \(f\) is measurable by Proposition 3.9, and \(|f(x)|=\lim _{n\to \infty } |f_n(x)|\leq g(x)\) for all \(x\in E\setminus E_0\), so \(\int _{E\setminus E_0} |f|\leq \int _{E\setminus E_0} g=\int _E g<\infty \) by Theorem 4.17 and assumption on \(g\). Similarly \(\int _{E\setminus E_0} |f_n|\leq \int _E g<\infty \). Thus \(f_n\), \(f\), and \(g\) are integrable, and so \begin {gather*} \int _{E\setminus E_0} g-\int _{E\setminus E_0} f\leq \int _{E\setminus E_0} g+\liminf _{n\to \infty } \biggl (-\int _{E\setminus E_0} f_n\biggr ), \\ \int _{E\setminus E_0} g+\int _{E\setminus E_0} f\leq \int _{E\setminus E_0} g+\liminf _{n\to \infty } \int _{E\setminus E_0} f_n. \end {gather*} Canceling the integrals of \(g\), and recalling that \(\liminf _{n\to \infty } (-a_n)=-\limsup _{n\to \infty } a_n\) for any sequence \(\{a_n\}_{n=1}^\infty \), we see that \begin {gather*} \int _{E\setminus E_0} f\geq \limsup _{n\to \infty } \int _{E\setminus E_0} f_n, \\ \int _{E\setminus E_0} f\leq \liminf _{n\to \infty } \int _{E\setminus E_0} f_n. \end {gather*} Recalling also that \(\liminf _{n\to \infty } a_n\leq \limsup _{n\to \infty } a_n\), we see that \begin {equation*} \int _{E\setminus E_0} f\leq \liminf _{n\to \infty } \int _{E\setminus E_0} f_n \leq \limsup _{n\to \infty } \int _{E\setminus E_0} f_n \leq \int _{E\setminus E_0} f. \end {equation*} Thus we must have that \begin {equation*} \int _{E\setminus E_0} f= \liminf _{n\to \infty } \int _{E\setminus E_0} f_n =\limsup _{n\to \infty } \int _{E\setminus E_0} f_n =\int _{E\setminus E_0} f. \end {equation*} Recalling finally that if \(\liminf _{n\to \infty } a_n=\limsup _{n\to \infty } a_n\), then \(\lim _{n\to \infty } a_n\) exists and is equal to both, we complete the proof.
Theorem 4.20. Let \(\{E_n\}_{n=1}^\infty \) be a countable sequence of pairwise disjoint measurable sets. Let \(E=\bigcup _{n=1}^\infty E_n\). If \(f:E\to [-\infty ,\infty ]\) is integrable (that is, measurable and \(\int _E |f|<\infty \)), then \begin {equation*} \int _E f=\sum _{n=1}^\infty \int _{E_n} f. \end {equation*}
(Problem 1380) Prove Theorem 4.20.
Theorem 4.21. Let \(\{E_n\}_{n=1}^\infty \) be a countable sequence of measurable sets, let \(E=\bigcup _{n=1}^\infty E_n\), and suppose that \(f:E\to [-\infty ,\infty ]\) is integrable (that is, measurable and \(\int _E |f|<\infty \)).
Suppose that either:
\(E_n\subseteq E_{n+1}\) for all \(n\) and \(D=E=\bigcup _{n=1}^\infty E_n\).
\(E_n\supseteq E_{n+1}\) for all \(n\) and \(D=\bigcap _{n=1}^\infty E_n\).
Then \begin {equation*} \int _D f=\lim _{n\to \infty } \int _{E_n} f. \end {equation*}
[Chapter 4, Problem 39] Prove Theorem 4.21.
Lemma 4.22. Let \(E\subset \R \) be measurable and suppose \(m(E)<\infty \). Let \(\delta >0\). Then there is a \(n\in \N \) and a list of pairwise disjoint sets \(E_1\), \(E_2,\dots ,E_n\) such that \(m(E_k)<\delta \) for all \(k\) and such that \(E=\bigcup _{k=1}^n E_k\).
(Problem 1390) Prove Lemma 4.22.
Proposition 4.23. Let \(E\subseteq \R \) be measurable and let \(f\) be a measurable function on \(E\).
(Problem 1400) Prove Proposition 4.23, part (a).
(Problem 1410) Prove Proposition 4.23, part (b).
[Definition: Uniformly integrable] Let \(E\subseteq \R \) be measurable and let \(\mathcal {F}\) be a family of measurable functions on \(E\). We say that \(\mathcal {F}\) is uniformly integrable over \(E\) if, for each \(\varepsilon >0\), there is a \(\delta >0\) such that, for all \(f\in \mathcal {F}\), we have that if \(A\subseteq E\) is measurable and \(m(A)<\delta \), then \(\int _A |f|<\varepsilon \).
(Problem 1411) Let \(E\subseteq \R \) be measurable and assume that \(m(E)<\infty \). Let \(f:E\to [-\infty ,\infty ]\). Then \(f\) is integrable over \(E\) if and only if \(\{f\}\) is uniformly integrable over \(E\).
This follows immediately from the definition of uniformly integrable and from Proposition 4.23.
(Problem 1420) Let \(E\subseteq \R \) be measurable and let \(g\) be integrable over \(E\). Show that \(\mathcal {F}=\{f|f:E\to [-\infty ,\infty ]\) is measurable and \(|f(x)|\leq |g(x)|\) for all \(x\in E\}\) is a uniformly integrable family.
Proposition 4.24. Any finite collection of integrable functions over a common domain \(E\) is uniformly integrable.
(Problem 1430) Let \(E\subseteq \R \) be measurable and let \(\mathcal {F}_1\), \(\mathcal {F}_2,\dots \mathcal {F}_n\) be a finite collection of families, each of which is uniformly integrable over \(E\). Show that \(\mathcal {F}=\bigcup _{k=1}^n \mathcal {F}_k\) is uniformly integrable over \(E\).
Let \(\varepsilon >0\).For each \(k\in \{1,2,\dots ,n\}\), by definition of uniformly integrable there is a \(\delta _k>0\) such that, for all \(f\in \mathcal {F}_k\), we have that if \(A\subseteq E\) is measurable and \(m(A)<\delta _k\), then \(\int _A |f|<\varepsilon \).
Let \(\delta =\min \{\delta _1,\delta _2,\dots ,\delta _n\}\); because \(n\) is finite, \(\delta \) exists and is positive.
If \(A\subseteq E\) is measurable and \(m(A)<\delta \), and if \(f\in \mathcal {F}\), then \(f\in \mathcal {F}_k\) for some \(1\leq k\leq n\). By definition \(\delta \leq \delta _k\). Thus, \(m(A)<\delta _k\) and so \(\int _A |f|<\varepsilon \). This completes the proof.
(Problem 1431) Prove Proposition 4.24.
This follows immediately from Problem 1411 and Problem 1430.
Proposition 4.25. Let \(E\subset \R \) be measurable and assume \(m(E)<\infty \). Let \(\{f_n\}_{n=1}^\infty \) be a sequence of measurable functions on \(E\) and suppose that \(\mathcal {F}=\{f_n:n\in \N \}\) is uniformly integrable. Suppose that \(f_n\to f\) pointwise almost everywhere on \(E\) for some \(f\). Then \(f\) is integrable.
(Problem 1440) Prove Proposition 4.25.
By Proposition 3.9, \(f\) is measurable. We need only show \(\int _E |f|<\infty \).Let \(\delta >0\) be such that, if \(A\subseteq E\) is measurable and \(m(A)<\delta \), then \(\int _A |f_n|<1\) for all \(n\in \N \); such a \(\delta \) must exist by definition of uniform integrability.
By Lemma 4.22, \(E=\bigcup _{k=1}^\ell E_k\), where each \(E_k\) is measurable, the \(E_k\)s are pairwise-disjoint, and \(m(E_k)<\delta \).
Thus \begin {equation*} \int _E |f_n|= \sum _{k=1}^\ell \int _{E_k} |f_n| \end {equation*} by Corollary 4.18. But by definition of \(\delta \), \begin {equation*} \int _{E_k} |f_n|<1 \end {equation*} for all \(n\in \N \), so \begin {equation*} \int _E |f_n|= \sum _{k=1}^\ell \int _{E_k} |f_n|< \sum _{k=1}^\ell 1 =\ell . \end {equation*} Thus \(\liminf _{n\to \infty } \int _E |f_n|\leq \ell \).
By Fatou’s lemma, and because \(\liminf _{n\to \infty } |f_n(x)|=\lim _{n\to \infty } |f_n(x)|\) whenever the limit exists, we have that \begin {equation*} \int _E |f|=\int _E \lim _{n\to \infty } |f_n| =\int _E \liminf _{n\to \infty } |f_n| \leq \liminf _{n\to \infty } \int _E |f_n|\leq \ell . \end {equation*} Thus \(\int _E |f|<\infty \), as desired.
(Problem 1460) Let \(E\), \(f_n\), and \(f\) be as in Proposition 4.25. Show that the family \(\{f\}\cup \{f_n:n\in \N \}\) is also uniformly integrable.
This follows immediately from Problem 1411, Problem 1430, and the assumption that \(\{f_n:n\in \N \}\) is uniformly integrable.
The Vitali convergence theorem. Let \(E\subset \R \) be measurable and assume \(m(E)<\infty \). Let \(\{f_n\}_{n=1}^\infty \) be a sequence of measurable functions on \(E\) and suppose that \(\mathcal {F}=\{f_n:n\in \N \}\) is uniformly integrable. Suppose that \(f_n\to f\) pointwise almost everywhere on \(E\) for some \(f\). Then \(\int _E f_n\to \int _E f\).
(Problem 1470) Prove the Vitali convergence theorem.
Theorem 4.26. Let \(E\subset \R \) be measurable and assume \(m(E)<\infty \). Suppose that \(\{h_n\}_{n=1}^\infty \) is a sequence of nonnegative integrable functions on \(E\) that converges pointwise almost everywhere to zero. Then \(\lim _{n\to \infty } \int _E h_n=0\) if and only if \(\{h_n:n\in \N \}\) is uniformly integrable over \(E\).
(Problem 1480) Prove Theorem 4.26.
Proposition 5.1. Let \(E\subseteq \R \) be measurable and let \(f\) be integrable over \(E\). Then for every \(\varepsilon >0\), there is an \(E_0\subseteq E\) with \(m(E_0)<\infty \) such that \(\int _{E\setminus E_0} |f|<\varepsilon \).
(Problem 1490) Prove Proposition 5.1.
[Definition: Tight] Let \(E\subseteq \R \) be measurable and let \(\mathcal {F}\) be a family of measurable functions on \(E\). We say that \(\mathcal {F}\) is tight if for each \(\varepsilon >0\), there is an \(E_0\subseteq E\) with \(m(E_0)<\infty \) and such that \begin {equation*} \sup _{f\in \mathcal {F}} \int _{E\setminus E_0} |f|<\varepsilon . \end {equation*}
(Problem 1491) Show that if \(m(E)<\infty \), then every family of measurable functions on \(E\) is tight.
(Problem 1500) By Proposition 4.23b, if \(m(E)<\infty \) and \(\mathcal {F}\) is a family of uniformly integrable functions over \(E\), then each \(f\in \mathcal {F}\) is in fact integrable. Give an example to show that this is not true if \(m(E)=\infty \) and then prove that if \(\mathcal {F}\) is both uniformly integrable and tight, then every element of \(\mathcal {F}\) is integrable.
The Vitali convergence theorem (infinite measure case). Let \(E\subseteq \R \) be measurable and let \(\mathcal {F}\) be a family of functions that is uniformly integrable and tight over \(E\). Suppose \(f_n\to f\) pointwise almost everywhere on \(E\). Then \(f\) is integrable over \(E\) and \(\lim _{n\to \infty }\int _E f_n=\int _E f\).
(Problem 1510) Prove the Vitali convergence theorem (infinite measure case).
Corollary 5.2. Let \(E\subseteq \R \) be measurable. Suppose that \(\{h_n\}_{n=1}^\infty \) is a sequence of nonnegative integrable functions on \(E\) that converges pointwise almost everywhere to zero. Then \(\lim _{n\to \infty } \int _E h_n=0\) if and only if \(\{h_n:n\in \N \}\) is uniformly integrable and tight over \(E\).
[Chapter 5, Problem 2] Prove Corollary 5.2.
[Definition: Convergence in measure. Let \(E\subseteq \R \) be measurable and let \(f_n\), \(f\) be measurable functions defined on \(E\)] Assume that \(f_n\) and \(f\) are finite almost everywhere. We say that the sequence \(\{f_n\}_{n=1}^\infty \) converges to \(f\) in measure if, for every \(\eta >0\), we have that \begin {equation*} \lim _{n\to \infty } m(\{x\in E:|f_n(x)-f(x)|>\eta \})=0. \end {equation*}
Proposition 5.3. Let \(E\subset \R \) be measurable and assume \(m(E)<\infty \). Let \(f_n\), \(f\) be measurable functions on \(E\) and assume that \(f_n\) and \(f\) are finite almost everywhere on \(E\). If \(f_n\to f\) pointwise almost everywhere on \(E\), then \(f_n\to f\) in measure on \(E\).
(Problem 1520) Prove Proposition 5.3.
(Problem 1530) If \(n\in \N \), then there is a unique \(k\in \N \cup \{0\}\) with \(2^k\leq n<2^{k+1}\). For each \(n\in N\), let \(I_n=[\frac {n}{2^k}-1,\frac {n+1}{2^k}-1]\). Let \(f_n=\chi _{I_n}\). Show that \(f_n\) converges to the zero function on \(E=[0,1]\) but that \(f_n(x)\not \to 0\) for any \(x\in [0,1]\).
(Problem 1540) Let \(\{f_n\}_{n=1}^\infty \) be as in the previous problem. Find a subsequence \(\{f_{n_k}\}_{k=1}^\infty \) that converges pointwise to zero everywhere.
Theorem 5.4. (Riesz) Let \(E\subseteq \R \) be measurable and let \(f_n\), \(f\) be measurable functions defined on \(E\). Suppose that \(f_n\to f\) in measure on \(E\). Show that there is a subsequence \(\{f_{n_k}\}_{k=1}^\infty \) that converges pointwise to \(f\) almost everywhere.
(Problem 1550) Prove Theorem 5.4.
Corollary 5.5. Let \(E\subset \R \) be measurable. Suppose that \(\{h_n\}_{n=1}^\infty \) is a sequence of nonnegative integrable functions on \(E\). Then \(\lim _{n\to \infty } \int _E h_n=0\) if and only if the following three conditions hold:
\(\{h_n:n\in \N \}\) is uniformly integrable over \(E\).
\(\{h_n:n\in \N \}\) is tight over \(E\).
\(h_n\to 0\) in measure on \(E\).
(Problem 1560) Begin the proof of Corollary 5.5 by assuming that \(\{h_n:n\in \N \}\) is uniformly integrable and tight and that \(h_n\to 0\) in measure, and showing that \(\int _E h_n\to 0\).
(Problem 1570) Complete the proof by showing that if \(\int _E h_n\to 0\) then \(h_n\to 0\) in measure on \(E\).
(Memory 1571) Let \((X,d)\) be a metric space and let \(f_1\), \(f_2,\dots ,f_n:X\to \R \) be continuous. Then \(f=\max \{f_1,f_2,\dots ,f_n\}\) is also continous on \(X\).
(Problem 1580) Let \((X,d)\) be a metric space and let \(\{g_n\}_{n=1}^\infty \) be a sequence of continuous functions \(g_n:X\to \R \). Let \(g:X\to (-\infty ,\infty ]\) be given by \(g(x)=\sup _{n\in \N } g_n(x)\). If \(g(x)<\infty \), show that \(g\) is lower semicontinuous at \(x\), that is, if \(\varepsilon >0\), then there is a \(\delta >0\) such that, if \(d(x,y)<\delta \), then \(g(y)>g(x)-\varepsilon \).
(Problem 1581) Let \((X,d)\) be a metric space and let \(\{h_n\}_{n=1}^\infty \) be a sequence of continuous functions \(h_n:X\to \R \). Let \(h:X\to (-\infty ,\infty ]\) be given by \(h(x)=\inf _{n\in \N } h_n(x)\). If \(h(x)>-\infty \), show that \(h\) is upper semicontinuous at \(x\), that is, if \(\varepsilon >0\), then there is a \(\delta >0\) such that, if \(d(x,y)<\delta \), then \(h(y)<h(x)+\varepsilon \).
Lemma 5.6. Let \(E\subseteq \R \) be measurable. For each \(n\in \N \), let \(\varphi _n:E\to [-\infty ,\infty ]\) and \(\psi _n:E\to [-\infty ,\infty ]\) be integrable.
Suppose that for each \(x\in E\) and each \(n\), \(k\in \N \), we have that \(\varphi _n(x)\leq \psi _k(x)\).
Suppose further that \(\lim _{n\to \infty } \int _E(\psi _n-\varphi _n)=0\).
Then \(\limsup _{n\to \infty } \varphi _n(x)=\liminf _{n\to \infty } \psi _n(x)\) for almost every \(x\in E\). Furthermore, if we define \(f(x)=\lim _{n\to \infty } \varphi _n(x)=\lim _{n\to \infty } \psi _n(x)\) for all such \(x\), then \(f\) is integrable and satisfies \begin {equation*} \lim _{n\to \infty }\int _E\varphi _n=\int _E f=\lim _{n\to \infty } \psi _n. \end {equation*}
(Problem 1590) Begin the proof of Lemma 5.6 by showing that \(\limsup _{n\to \infty } \varphi _n(x)=\liminf _{n\to \infty } \psi _n(x)\) for almost every \(x\in E\).
(Problem 1600) Complete the proof of Lemma 5.6 by showing that \(f\) is integrable and satisfies \begin {equation*} \lim _{n\to \infty }\int _E\varphi _n=\int _E f=\lim _{n\to \infty } \psi _n. \end {equation*}
Recall [Theorem 4.4]: . Let \(E\subset \R \) be measurable and suppose that \(m(E)<\infty \). Let \(f:E\to [-M,M]\) be a bounded function. If \(f\) is measurable, then \(f\) is Lebesgue integrable in the sense of Section 4.2, that is, \begin {multline*} \sup \biggl \{\int _E \varphi \biggm |\varphi :E\to \R \text { is a simple function and } \varphi (x)\leq f(x)\text { for all }x\in E\biggr \} \\= \inf \biggl \{\int _E \psi \biggm |\psi :E\to \R \text { is a simple function and } \psi (x)\geq f(x)\text { for all }x\in E\biggr \}. \end {multline*}
Theorem 5.7. Let \(E\subset \R \) be measurable and suppose that \(m(E)<\infty \). Let \(f:E\to [-M,M]\) be a bounded function.
Suppose that \(f\) is Lebesgue integrable in the sense of Section 4.2. Then \(f\) is measurable.
(Problem 1610) Prove Theorem 5.7.
By definition of supremum, if \(n\in \N \) then there is a \(\varphi _n:E\to \R \) that is simple, satisfies \(\varphi _n\leq f\), and such that \begin {equation*} \int _E\varphi _n\geq \sup \biggl \{\int _E \varphi \biggm |\varphi :E\to \R \text { is a simple function and } \varphi (x)\leq f(x)\text { for all }x\in E\biggr \}-\frac {1}{n}. \end {equation*} Similarly, if \(n\in \N \) then there is a \(\psi _n:E\to \R \) that is simple, satisfies \(\psi _n\leq f\), and such that \begin {equation*} \int _E\psi _n\leq \inf \biggl \{\int _E \psi \biggm |\psi :E\to \R \text { is a simple function and } \psi (x)\geq f(x)\text { for all }x\in E\biggr \}+\frac {1}{n}. \end {equation*}Then each \(\varphi _n\) is measurable because simple functions are measurable by definition. Furthermore, we have that \(\varphi _n(x)\leq f(x)\leq \psi _n(x)\) for all \(x\in E\), and so by Proposition 4.2 we have that \begin {align*} 0\leq \int _E (\psi _n-\varphi _n)&=\int _E \psi _n-\int _E\varphi _n \\&\leq \frac {1}{n}+\inf \biggl \{\int _E \psi \biggm |\psi :E\to \R \text { is a simple function and } \psi (x)\geq f(x)\text { for all }x\in E\biggr \} \\&\qquad +\frac {1}{n}-\sup \biggl \{\int _E \varphi \biggm |\varphi :E\to \R \text { is a simple function and } \varphi (x)\leq f(x)\text { for all }x\in E\biggr \} . \end {align*}
By assumption on \(f\), the supremum and infimum are equal. Thus \begin {align*} 0\leq \int _E (\psi _n-\varphi _n)&\leq \frac {2}{n} \end {align*}
and so by the squeeze theorem \(\lim _{n\to \infty } \int _E (\psi _n-\varphi _n)=0\). Thus Lemma 5.6 applies and we have that \begin {equation*} \limsup _{n\to \infty }\varphi _n=\liminf _{n\to \infty }\psi _n \end {equation*} almost everywhere in \(E\).
But \(\varphi _n\leq f\leq \psi _n\) for all \(n\), and so \(\limsup _{n\to \infty }\varphi _n\leq f\leq \liminf _{n\to \infty }\psi _n\). Thus if \(\limsup _{n\to \infty }\varphi _n(x)=\liminf _{n\to \infty }\psi _n(x)\) then \(f(x)=\limsup _{n\to \infty }\varphi _n(x)\), and so we have that \(f=\limsup _{n\to \infty }\varphi _n\) almost everywhere in \(E\). But \(\limsup _{n\to \infty }\varphi _n\) is measurable by Proposition 3.9, and so \(f\) is measurable as well.
Theorem 5.8. (Lebesgue) Let \([a,b]\subset \R \) be a closed and bounded interval and let \(f:[a,b]\to [-M,M]\) be a bounded function. Then \(f\) is Riemann integrable over \([a,b]\) if and only if \begin {equation*} m(\{x\in [a,b]:f\text { is discontinuous at }x\})=0. \end {equation*}
(Problem 1620) Give an example of a bounded function \(f:[0,1]\to \R \) that is not Riemann integrable, but such that there is a set \(E\subset [0,1]\) with \(m(E)=0\) and such that \(f\) is continuous on \([0,1]\setminus E\).
(The point of this problem is that the statement \(m(\{x\in [a,b]:f\text { is discontinuous at }x\})=0\) is much stronger than the statement that \(f\) is continuous on \([a,b]\setminus E\) for some \(E\) with \(m(E)=0\).)
Let \(f(x)=\chi _{\Q }(x)=\begin {cases}1,&x\in \Q ,\\0,&x\notin \Q .\end {cases}\) Then \(f\) is not Riemann integrable, but \(f\big \vert _{[0,1]\setminus \Q }\) is the constant function \(0\) and so is continuous.
(Problem 1630) In this problem we begin the proof of Theorem 5.8. Let \([a,b]\subset \R \) be a closed and bounded interval and let \(f:[a,b]\to [-M,M]\) be a bounded function. If \(n\in \N \), define \(a_{n,k}=a+k(b-a)2^{-n}\). Let \(\varphi _n\) and \(\psi _n\) be step functions which satisfy \begin {equation*} \varphi _n\leq f\leq \psi _n\text { on }[a,b] \end {equation*} and \begin {equation*} \varphi _n=\inf _{[a_{n,{k-1}},a_{n,k}]} f\text { on }(a_{n,{k-1}},a_{n,k}), \qquad \psi _n=\sup _{[a_{n,{k-1}},a_{n,k}]} f\text { on }(a_{n,{k-1}},a_{n,k}) \end {equation*} for all \(1\leq k\leq 2^n\).
Suppose that \(x\in (a,b)\setminus \{a_{n,k}:n,k\in \N ,0\leq k\leq 2^n\}\). Suppose further that \(f\) is continuous at \(x\). Show that \(\varphi _n(x)\to f(x)\).
(Problem 1631) Show that \(\psi _n(x)\to f(x)\).
(Problem 1640) Let \([a,b]\subset \R \) be a closed and bounded interval and let \(f:[a,b]\to [-M,M]\) be a bounded function. Suppose that \(f\) is continuous at \(x\) for almost every \(x\in [a,b]\). Show that \(f\) is Riemann integrable.
(Problem 1650) Let \([a,b]\subset \R \) be a closed and bounded interval and let \(f:[a,b]\to [-M,M]\) be a bounded function. Show that \begin {multline*} \sup \biggl \{\int _E \varphi \biggm |\varphi :[a,b]\to \R \text { is a step function and } \varphi (x)\leq f(x)\text { for all }x\in E\biggr \} \\\geq \sup \biggl \{\int _E g\biggm |g:[a,b]\to \R \text { is a continuous function and } g(x)\leq f(x)\text { for all }x\in E\biggr \} . \end {multline*}
[I have changed the order of this and the following problem for the sake of clearer exposition.]Define \begin {gather*} L_{step}=\sup \biggl \{\int _E \varphi \biggm |\varphi :[a,b]\to \R \text { is a step function and } \varphi (x)\leq f(x)\text { for all }x\in E\biggr \} ,\\ L_{cts}=\sup \biggl \{\int _E g\biggm |g:[a,b]\to \R \text { is a continuous function and } g(x)\leq f(x)\text { for all }x\in E\biggr \} . \end {gather*}
Observe first that the constant function \(-M\) is both continuous and a step function and is always less than or equal to \(f\), and so \begin {equation*} -M(b-a)=\int _a^b(-M)\leq L_{step} \quad \text {and}\quad -M(b-a)=\int _a^b(-M)\leq L_{cts} . \end {equation*} Conversely, if \(\varphi \leq f\) then \(\varphi \leq M\) and so \(\int _a^b\varphi \leq \int _a^b M=M(b-a)\), and so we have that \begin {gather*} -M(b-a)\leq L_{step}\leq M(b-a) \quad \text {and}\quad -M(b-a)\leq L_{cts}\leq M(b-a) . \end {gather*} Thus both suprema are finite.
Choose some \(\varepsilon >0\). By definition of supremum, there is a continuous function \(g:[a,b]\to [-M,M]\) with \(g\leq f\) such that \begin {equation*} \int _a^b g\geq L_{cts}-\varepsilon . \end {equation*} We recall from undergraduate analysis that continuous functions on closed and bounded intervals are always Riemann integrable. Thus \begin {equation*} \int _a^b g=\sup \biggl \{\int _E \varphi \biggm |\varphi :[a,b]\to \R \text { is a step function and } \varphi (x)\leq g(x)\text { for all }x\in E\biggr \}. \end {equation*} Applying the definition of supremum again, we see that there is a step function \(\varphi :[a,b]\to [-M,M]\) with \(\varphi \leq g\) and with \begin {equation*} \int _a^b \varphi \geq \int _a^b g-\varepsilon \geq L_{cts}-2\varepsilon . \end {equation*} But then \(\varphi \leq g\leq f\) and so \(\varphi \leq f\), and so \begin {equation*} \int _a^b \varphi \in \biggl \{\int _E \varphi \biggm |\varphi :[a,b]\to \R \text { is a step function and } \varphi (x)\leq f(x)\text { for all }x\in E\biggr \}. \end {equation*} Thus \(L_{step}\geq \int _a^b \varphi \geq L_{cts}-2\varepsilon \). This is true for all \(\varepsilon >0\); thus we must have that \(L_{step}\geq L_{cts}\).
(Problem 1660) Let \([a,b]\subset \R \) be a closed and bounded interval and let \(f:[a,b]\to [-M,M]\) be a bounded function. Show that \begin {multline*} \sup \biggl \{\int _E g\biggm |g:[a,b]\to \R \text { is a continuous function and } g(x)\leq f(x)\text { for all }x\in E\biggr \} \\= \sup \biggl \{\int _E \varphi \biggm |\varphi :[a,b]\to \R \text { is a step function and } \varphi (x)\leq f(x)\text { for all }x\in E\biggr \} . \end {multline*}
Define \(L_{step}\) and \(L_{cts}\) as before and let \(\varepsilon >0\). Then there is a step function \(\varphi :[a,b]\to [-M,M]\) with \(\varphi \leq f\) such that \begin {equation*} \int _a^b \varphi \geq L_{step}-\varepsilon . \end {equation*} We seek to find a continuous function \(g\) with \(g\leq f\) and with \(\int _a^b g\geq \int _a^b \varphi -\varepsilon \); as before, this will show that \(L_{cts}\geq L_{step}-2\varepsilon \), and so will suffice to show \(L_{cts}\geq L_{step}\).Because \(\varphi \) is a step function, we may write \begin {equation*} \varphi =\sum _{k=1}^n c_k \chi _{I_k} \end {equation*} where each \(I_k\) is an interval.
We claim that if \(1\leq k\leq n\), then there is a continuous function \(g_k:[a,b]\to \R \) that satisfies \begin {equation*} g_k\leq c_k\chi _{I_k} \quad \text {and}\quad \int _a^b g_k\geq -\frac {\varepsilon }{n}+\int _a^b c_k\chi _{I_k}. \end {equation*}
Suppose momentarily that the claim is true. Let \(g=\sum _{k=1}^n g_k\). Then \(g\) is continuous, satisfies \(g\leq \varphi \), and satisfies \begin {equation*} \int _a^b g =\sum _{k=1}^n \int _a^b g_k \geq \sum _{k=1}^n -\frac {\varepsilon }{n}+\int _a^b c_k\chi _{I_k}=-\varepsilon +\int _a^b \varphi . \end {equation*} Thus, proving the claim suffices to complete the proof.
To prove the claim, define \(g_k\) as follows.
If \(I_k=\emptyset \) or \(c_k=0\), let \(g_k=0\). Then \(\int _a^b g_k=0=\int _a^b c_k\chi _{I_k}\) and \(g_k=0=c_k\chi _{I_k}\).
If \(I_k\) is a single point \(I_k=\{a_k\}\) and \(c_k\geq 0\), let \(g_k=0\). Then \(g_k\leq c_k\chi _{I_k}\) and \(\int _a^b g_k=0=\int _a^b c_k\chi _{I_k}\).
If \(I_k\) is a single point \(I_k=\{a_k\}\) and \(c_k<0\), let \(g_k\) be the piecewise-linear function with the following graph:
Observe that \(g_k\) is continuous, \(g_k\leq 0\) everywhere and \(g_k(a_k)=c_k=c_k\chi _{I_k}(a_k)\) and so \(g_k\leq c_k\chi _{I_k}\) everywhere, and \begin {equation*} \int _a^b g_k \geq \int _{a_k-{\varepsilon }/{n|c_k|}}^{a_k+{\varepsilon }/{n|c_k|}} g_k=\frac {\varepsilon }{n}. \end {equation*}Suppose that \(c_k>0\) and \(I_k\) is a nontrivial interval. Then there exist \(a_k\), \(b_k\) with \(a_k<b_k\) and with \((a_k,b_k)\subseteq I_k\subseteq [a_k,b_k]\). If \(c_k>0\), let \(g_k\) be the piecewise-linear function with the following graph:
where \(\widetilde a_k\) and \(\widetilde b_k\) satisfy \(a_k<\widetilde a_k\leq \widetilde b_k<b_k\) and \(\widetilde a_k<a_k+\frac {\varepsilon }{nc_k}\), \(\widetilde b_k>b_k-\frac {\varepsilon }{nc_k}\). Then \(g_k\leq c_k\chi _{I_k}\) and \(\int _a^b g_k\geq \int _a^b c_k\chi _{I_k}-\frac {\varepsilon }{n}\).Finally, suppose that \(c_k<0\) and \(I_k\) is a nontrivial interval. Then there exist \(a_k\), \(b_k\) with \(a_k<b_k\) and with \((a_k,b_k)\subseteq I_k\subseteq [a_k,b_k]\). If \(c_k>0\), let \(g_k\) be the piecewise-linear function with the following graph:
where \(\widetilde a_k\) and \(\widetilde b_k\) satisfy \(\widetilde a_k<a_k\leq b_k<\widetilde b_k\) and \(\widetilde a_k>a_k-\frac {\varepsilon }{nc_k}\), \(\widetilde b_k<b_k+\frac {\varepsilon }{nc_k}\). Then \(g_k\leq c_k\chi _{I_k}\) and \(\int _a^b g_k\geq \int _a^b c_k\chi _{I_k}-\frac {\varepsilon }{n}\).We see that the claim, and thus the result, is true.
(Problem 1661) Let \([a,b]\subset \R \) be a closed and bounded interval and let \(f:[a,b]\to [-M,M]\) be a bounded function. Show that \begin {multline*} \inf \biggl \{\int _E \psi \biggm |\varphi :[a,b]\to \R \text { is a step function and } \psi (x)\geq f(x)\text { for all }x\in E\biggr \} \\= \inf \biggl \{\int _E h\biggm |h:[a,b]\to \R \text { is a continuous function and } h(x)\geq f(x)\text { for all }x\in E\biggr \} . \end {multline*}
(Problem 1662) Recall from Section 4.1 that \(f\) is Riemann integrable if \begin {multline*} \sup \biggl \{\int _E \varphi \biggm |\varphi :[a,b]\to \R \text { is a step function and } \varphi (x)\leq f(x)\text { for all }x\in E\biggr \} \\= \inf \biggl \{\int _E \psi \biggm |\psi :[a,b]\to \R \text { is a step function and } \psi (x)\geq f(x)\text { for all }x\in E\biggr \}. \end {multline*} Show that \(f\) is Riemann integrable if and only if \begin {multline*} \sup \biggl \{\int _E g\biggm |g:[a,b]\to \R \text { is a continuous function and } g(x)\leq f(x)\text { for all }x\in E\biggr \} \\= \inf \biggl \{\int _E h\biggm |h:[a,b]\to \R \text { is a continuous function and } h(x)\geq f(x)\text { for all }x\in E\biggr \}. \end {multline*}
(Problem 1670) Let \([a,b]\subset \R \) be a closed and bounded interval and let \(f:[a,b]\to [-M,M]\) be a bounded function. Suppose that \(f\) is Riemann integrable. Show that \(f\) is continuous almost everywhere on \([a,b]\).
(Problem 1671) Let \(a\), \(b\in [-\infty ,\infty ]\) with \(a<b\) and let \(f:(a,b)\to \R \) be monotonic (either nonincreasing or nondecreasing). If \(a<x<b\), define \(f(x-)=\sup \{f(t):a<t<x\}\) and \(f(x+)=\inf \{f(t):x<t<b\}\). Show that:
\(f(x-)=\lim _{t\to x^-} f(t)\),
\(f(x+)=\lim _{t\to x^+} f(t)\),
If \(f(x-)<y<f(x+)\) then either \(f^{-1}(\{y\})=\emptyset \) or \(f^{-1}(\{y\})=\{x\}\).
If \(f\) is nondecreasing and \(a<x<y<b\), then \(f(x-)\leq f(x)\leq f(x+)\leq f(y-)\leq f(y)\leq f(y+)\).
If \(f\) is nonincreasing and \(a<x<y<b\), then \(f(x-)\geq f(x)\geq f(x+)\geq f(y-)\geq f(y)\geq f(y+)\).
\(f\) is continuous at \(x\) if and only if \(f(x-)=f(x+)\).
Theorem 6.1. Let \(a\), \(b\in [-\infty ,\infty ]\) with \(a<b\) and let \(f:(a,b)\to \R \) be monotonic. The set \(\{x\in (a,b):f\) is not continuous at \(x\}\) is at most countable.
(Problem 1680) Prove Theorem 6.1.
Proposition 6.2. Let \(a\), \(b\in \R \) with \(a<b\) and let \(C\subset (a,b)\). Then there exists a function \(f:(a,b)\to \R \) that is strictly increasing and such that \(C=\{x\in (a,b):f\) is not continuous at \(x\}\).
(Problem 1690) Prove Proposition 6.2.
Because \(C\) is countable, there is an enumeration; that is, there is a sequence \(\{c_n\}_{n=1}^\infty \) such that \(C=\{c_n:n\in \N \}\). (If we allow repetition, then such a sequence exists whether \(C\) is finite or countably infinite.)Define \(f\) by \begin {equation*} f(x)=x+\sum _{n=1}^\infty \frac {1}{2^n} \chi _{(-\infty ,x)}(c_n) =x+\sum _{\substack {n\in \N \\c_n<x}} \frac {1}{2^n} . \end {equation*}
If \(x<y\), then \begin {equation*} f(y)-f(x)=(y-x)+\sum _{n=1}^\infty \frac {1}{2^n} (\chi _{(-\infty ,y)}(c_n)-\chi _{(-\infty ,x)}(c_n)). \end {equation*} But \(y-x>0\), and if \(x<y\) then \((-\infty ,x)\subset (-\infty ,y)\) and so \(\chi _{(-\infty ,y)}\geq \chi _{(-\infty ,x)}\). Thus \(f(y)-f(x)\geq y-x>0\) if \(y>x\), and so \(f\) is strictly increasing.
Suppose that \(x\in C\). Then \(x=c_k\) for at least one value of \(k\). If \(y>x\), then \begin {equation*} f(y)-f(x)=(y-x)+\sum _{n=1}^\infty \frac {1}{2^n} (\chi _{(-\infty ,y)}(c_n)-\chi _{(-\infty ,x)}(c_n)) > \frac {1}{2^k} \end {equation*} because \(y>x\), \(\chi _{(-\infty ,y)}(c_k)-\chi _{(-\infty ,x)}(c_k) =\chi _{(-\infty ,y)}(x)-\chi _{(-\infty ,x)}(x)=1\), and all other terms in the sum are nonnegative. Thus, there is a \(\varepsilon \) (namely \(\varepsilon =\frac {1}{2^k}\)) such that, for all \(\delta >0\), there is a \(y\in (a,b)\) with \(|x-y|<\delta \) and with \(|f(x)-f(y)|>\varepsilon \) (any \(y\in (x,x+\delta )\cap (a,b)\) will do), and so \(f\) is not continuous at \(x\).
Now suppose that \(x\in (a,b)\setminus C\). Pick some \(\varepsilon >0\). Let \(N\) satisfy \(\frac {1}{2^N}< \frac {\varepsilon }{2}\). Such a (finite) \(N\) must exist. Observe that \(x\notin C\) and so \(x\neq c_n\) for any \(n\in \N \).
Let \(\delta =\min (\varepsilon /2,\min \{|x-c_n|:n\leq N\})\). Then \(\delta >0\). If \(|x-y|<\delta \), then \begin {equation*} f(y)-f(x)=(y-x)+\sum _{n=1}^\infty \frac {1}{2^n} (\chi _{(-\infty ,y)}(c_n)-\chi _{(-\infty ,x)}(c_n)). \end {equation*} But if \(\chi _{(-\infty ,y)}(c_n)-\chi _{(-\infty ,x)}(c_n)\neq 0\), then either \(x\leq c_n<y\) or \(y\leq c_n<x\) and so \(|x-c_n|<\delta \). Thus we must have that \(n>N\). Thus \begin {equation*} |f(y)-f(x)|\leq |y-x|+\sum _{n=N+1}^\infty \frac {1}{2^n} <\frac {\varepsilon }{2} + \frac {1}{2^N}<\varepsilon \end {equation*} as desired.
[Definition: Open ball] If \((X,d)\) is a metric space, \(x\in X\), and \(r>0\), then the open ball centered at \(x\) of radius \(r\) is \( B(x,r)=\{y\in X:d(x,y)< r\}\).
[Definition: Closed ball] If \((X,d)\) is a metric space, \(x\in X\), and \(r>0\), then the closed ball centered at \(x\) of radius \(r\) is \(\overline B(x,r)=\{y\in X:d(x,y)\leq r\}\).
[Definition: Diameter] If \((X,d)\) is a metric space and \(Y\subseteq X\), then \(\diam Y=\sup \{d(x,y):x,y\in Y\}\).
(Problem 1691) If \((X,d)\) is a metric space, \(x\in X\), and \(r>0\), then \(r\geq \frac {1}{2}\diam \overline B(x,r)\).
[Definition: Separable] A metric space is separable if it contains a countable dense subset.
[Definition: Scaled ball] Let \((X,d)\) be a metric space. If \(x\in X\) and \(r\), \(s>0\), then \(sB(x,r)=B(x,r)\) and \(s\overline {B}(x,r)=\overline {B}(x,sr)\).
(Problem 1700) (The Vitali covering lemma, finite version.) Let \((X,d)\) be a metric space. Let \(\mathcal {F}\) be a finite collection of closed balls in \(X\). Let \(E\subseteq X\) satisfy \(E\subseteq \bigcup _{B\in \mathcal {F}} B\).
Show that there is a subset \(\mathcal {S}\subseteq \mathcal {F}\) such that \begin {equation*} E\subseteq \bigcup _{B\in \mathcal {S}} 3B \end {equation*} and such that, if \(B\), \(\beta \in \mathcal {S}\), then \(B=\beta \) or \(B\cap \beta =\emptyset \). (For partial credit, you may prove this under the assumption that \(X=\R \) and \(d(x,y)=|x-y|\). Hint: Begin with the largest ball.)
(Problem 1701) Let \((X,d)\) be a separable metric space, let \(B\) and \(\beta \) be two closed balls in \(X\), and suppose that \(B\cap \beta \neq \emptyset \) and there is a \(r>0\) such that \(\diam \beta \leq 2r\) and \( \diam B\geq r\). Show that \(\beta \subseteq 5B\).
(Problem 1710) Let \((X,d)\) be a separable metric space. Let \(\mathcal {C}\) be a (possibly infinite) collection of closed balls in \(X\). Suppose that there is an \(r>0\) such that \(2r\leq \diam B\leq 3r\) for all \(B\in \mathcal {C}\). Let \(E\subseteq X\) satisfy \(E\subseteq \bigcup _{B\in \mathcal {C}} B\).
Show that there is a countable subset \(\mathcal {S}\subseteq \mathcal {C}\) such that \begin {equation*} E\subseteq \bigcup _{B\in \mathcal {S}} 5B \end {equation*} and such that, if \(B\), \(\beta \in \mathcal {S}\), then \(B\cap \beta =\emptyset \). (For partial credit, you may prove this under the assumption that \(X=\R \), \(d(x,y)=|x-y|\), and \(E\) is a measurable set of finite measure.)
(Bonus Problem 1711) Let \((X,d)\) be a metric space. Let \(\mathcal {C}\) be a (possibly infinite) collection of closed balls in \(X\). Suppose that there is an \(r>0\) such that \(r\leq \diam B\leq 2r\) for all \(B\in \mathcal {C}\). Let \(E\subseteq X\) satisfy \(E\subseteq \bigcup _{B\in \mathcal {C}} B\).
Use Zorn’s lemma to show that there is a subset \(\mathcal {S}\subseteq \mathcal {C}\) such that \begin {equation*} E\subseteq \bigcup _{B\in \mathcal {S}} 5B \end {equation*} and such that, if \(B\), \(\beta \in \mathcal {S}\), then \(B=\beta \) or \(B\cap \beta =\emptyset \). If \(X\) is separable, show that \(\mathcal {S}\) must be countable.
Let \begin {equation*} P=\{\mathcal {S}\subseteq \mathcal {C}:\text {if $B$, $\beta \in \mathcal {S}$ then $B=\beta $ or $B\cap \beta =\emptyset $}\}. \end {equation*} Then \(\mathcal {P}\) is a set (it is a subset of the power set \(2^{\mathcal {C}}\) of \(\mathcal {C}\)), and so it is a partially ordered set under inclusion \(\subseteq \).Suppose that \(T\subseteq {P}\) is a chain, that is, a totally ordered subset. Let \(\mathcal {U}=\bigcup _{\mathcal {S}\in T} \mathcal {S}\). Then \(\mathcal {U}\subseteq \mathcal {C}\) and is clearly an upper bound on \(T\). We need only show that \(\mathcal {U}\in P\), that is, if \(B\), \(\beta \in \mathcal {U}\) then \(B=\beta \) or \(B\cap \beta =\emptyset \).
But if \(B\), \(\beta \in \mathcal {U}\) then \(B\in \mathcal {S}\) and \(\beta \in \mathcal {R}\) for some \(\mathcal {R}\), \(\mathcal {S}\in T\). Because \(T\) is totally ordered, either \(\mathcal {R}\subseteq \mathcal {S}\) or \(\mathcal {S}\subseteq \mathcal {R}\). Without loss of generality \(\mathcal {R}\subseteq \mathcal {S}\). Then \(B\), \(\beta \in \mathcal {S}\in T\subseteq P\), and so by definition of \(P\), we have that \(B=\beta \) or \(B\cap \beta =\emptyset \).
Thus \(\mathcal {U}\in P\) and so every totally ordered subset \(T\) of \(P\) has an upper bound. Thus by Zorn’s lemma, there is a maximal element \(\mathcal {S}\) of \(P\).
By definition of \(P\), we have that \(\mathcal {S}\subseteq \mathcal {C}\) and that if \(B\), \(\beta \in \mathcal {S}\), then \(B=\beta \) or \(B\cap \beta =\emptyset \). If \(\beta \in \mathcal {C}\), then by maximality of \(\mathcal {S}\) we have that \(B\cap \beta \neq \emptyset \) for some \(B\in \mathcal {S}\); thus \(\beta \subseteq 5B\) by Problem 1701, and so \begin {equation*} E\subseteq \bigcup _{\beta \in \mathcal {C}}\beta \subseteq \bigcup _{B\in \mathcal {S}} 5B \end {equation*} as desired.
If \((X,d)\) is separable, then there is a countable dense set \(Q\subseteq X\). If \(B=\overline {B}(x,r)\in \mathcal {S}\), then \(B(x,r)\subseteq \overline {B}(x,r)\) is open and so there is a \(q=q_B\in Q\cap B\). If \(B\), \(\beta \in \mathcal {S}\), then either \(B=\beta \) or \(B\cap \beta =\emptyset \), and so \(q_B\neq q_\beta \) because \(q_B\in B\), \(q_\beta \in \beta \). Thus there is an injection from \(\mathcal {S}\) to \(Q\), and so because \(Q\) is countable, so too must \(\mathcal {S}\) be.
(Problem 1720) Let \(\mathcal {S}\) be a collection of pairwise-disjoint closed balls (that is, closed bounded intervals) in \(\R \). Suppose that \(r=\inf \{\ell (I):I\in \mathcal {S}\}\) is positive. Suppose in addition that \(m^*(\bigcup _{I\in \mathcal {S}} I)<\infty \). Show that \(\mathcal {S}\) is finite.
(Problem 1721) Let \(\mathcal {S}\) be a (possibly infinite) collection of pairwise-disjoint closed balls (that is, closed bounded intervals) in \(\R \). Suppose that \(r=\inf \{\ell (I):I\in \mathcal {S}\}\) is positive. Show that \(\bigcup _{I\in \mathcal {S}} I\) is closed.
(Problem 1722) Let \((X,d)\) be a metric space. Let \(\mathcal {C}\) be a (possibly infinite) collection of closed balls in \(X\). Suppose that there is a \(R<\infty \) such that \(0<\diam (B)\leq R\) for all \(B\in \mathcal {C}\). Let \(E\subseteq X\) satisfy \(E\subseteq \bigcup _{B\in \mathcal {C}} B\).
For each \(k\geq 0\), let \begin {equation*} \mathcal {D}_k=\{B\in \mathcal {C}:2^{-k}R<\diam (B)\leq 2^{1-k}R\}. \end {equation*} Define the sets \(\mathcal {S}_k\), \(\mathcal {C}_k\) inductively as follows. Let \begin {equation*} \mathcal {C}_k=\{B\in \mathcal {D}_k:B\cap \beta =\emptyset \text { for all }\beta \in \bigcup _{n=1}^{k-1} \mathcal {S}_k\} \end {equation*} (where we take \(\bigcup _{n=1}^0\mathcal {S}_k\) to be the empty union, so \(\mathcal {C}_1=\mathcal {D}_1\)) and where \(\mathcal {S}_k\) is the subset of \(\mathcal {C}_k\) given by Problem 1711. Let \(\mathcal {S}=\bigcup _{k=1}^\infty \mathcal {S}_k\).
Show that \(\mathcal {S}\subseteq \mathcal {C}\) and that, if \(B\), \(\beta \in \mathcal {S}\), then \(B\cap \beta =\emptyset \). (For partial credit, you may prove this under the assumption that \(X=\R \), \(d(x,y)=|x-y|\), and \(E\) is a measurable set of finite measure.)
Observe that \(\mathcal {C}=\bigcup _{k=1}^\infty \mathcal {D}_k\).For each \(k\geq 0\), let \begin {equation*} \mathcal {D}_k=\{B\in \mathcal {C}:2^{-k}R<\diam (B)\leq 2^{1-k}R\}. \end {equation*} Then \(\mathcal {C}=\bigcup _{k=1}^\infty \mathcal {D}_k\).
Define the sets \(\mathcal {S}_k\), \(\mathcal {C}_k\) inductively as follows. Let \begin {equation*} \mathcal {C}_k=\{B\in \mathcal {D}_k:B\cap \beta =\emptyset \text { for all }\beta \in \bigcup _{n=1}^{k-1} \mathcal {S}_k\} \end {equation*} (where we take \(\bigcup _{n=1}^0\mathcal {S}_k\) to be the empty union, so \(\mathcal {C}_1=\mathcal {D}_1\)) and where \(\mathcal {S}_k\) is the subset of \(\mathcal {C}_k\) given by Problem 1711. Let \(\mathcal {S}=\bigcup _{k=1}^\infty \mathcal {S}_k\).
Then \(\mathcal {S}\subseteq \mathcal {C}\) because each \(\mathcal {S}_k\) satisfies \(\mathcal {S}_k\subseteq \mathcal {C}_k\subseteq \mathcal {D}_k\subseteq \mathcal {C}\).
If \(B\), \(\beta \in \mathcal {S}\), then \(B\in \mathcal {S}_j\) and \(\beta \in \mathcal {S}_k\) for some \(j\), \(k\in \N \); without loss of generality \(j\leq k\). If \(j=k\) then either \(B=\beta \) or \(B\cap \beta =\emptyset \) by the conditions imposed in Problem 1711. If \(j<k\) then \(\beta \mathcal {S}_k\subseteq \mathcal {C}_k\) and so \(\beta \cap B=\emptyset \) because \(B\in \mathcal {S}_j\).
(Problem 1730) (The Vitali covering lemma, infinite version.) Let \((X,d)\) be a metric space. Let \(\mathcal {C}\) be a (possibly infinite) collection of closed balls in \(X\). Suppose that \(\diam (B)>0\) for all \(B\in \mathcal {C}\) and that \(\sup _{B\in \mathcal {C}} \diam B<\infty \); however, we do not impose a lower bound on the diameters of the balls in \(\mathcal {C}\). Let \(E\subseteq X\) satisfy \(E\subseteq \bigcup _{B\in \mathcal {C}} B\).
Show that there is a subcollection \(\mathcal {S}\subseteq \mathcal {C}\) such that \begin {equation*} E\subseteq \bigcup _{B\in \mathcal {S}} 5B \end {equation*} and such that, if \(B\), \(\beta \in \mathcal {S}\), then \(B\cap \beta =\emptyset \). (For partial credit, you may prove this under the assumption that \(X=\R \), \(d(x,y)=|x-y|\), and \(E\) is a measurable set of finite measure.)
Let \(\mathcal {S}\) be as in Problem 1722. We need only show that \(E\subseteq \bigcup _{B\in \mathcal {S}} 5B\).If \(\beta \in \mathcal {C}\), then \(\beta \in \mathcal {D}_k\) for some \(k\). Either \(\beta \in \mathcal {C}_k\) and so \(\beta \subseteq \bigcup _{B\in \mathcal {S}_k} 5B \subseteq \bigcup _{B\in \mathcal {S}} 5B\) by construction of \(\mathcal {S}_k\), or \(\beta \in \mathcal {D}_k\setminus \mathcal {C}_k\). Then \(\beta \cap B\neq \emptyset \) for some \(B\in \mathcal {S}_j\) and some \(j<k\). But then \(\diam \beta \leq 2\times 2^{-k}R\) and \(\diam B>2^{-j}R>2^{-k}R\), and so by Problem 1701 we have that \(\beta \subseteq 5B\). In any case \begin {equation*} E\subseteq \bigcup _{\beta \in \mathcal {C}}\beta \subseteq \bigcup _{B\in \mathcal {S}} 5B \end {equation*} as desired.
(Problem 1740) (The Vitali covering lemma, small ball version.) Let \(E\subseteq \R \). Let \(\mathcal {C}\) be a (possibly infinite) collection of closed bounded intervals in \(\R \) such that \(\diam (I)>0\) for all \(I\in \mathcal {C}\) and such that, for all \(\delta >0\), we have that \begin {equation*} E\subseteq \bigcup _{\substack {I\in \mathcal {C}\\\diam (I)<\delta }} I. \end {equation*} Show that there is a countable subcollection \(\mathcal {S}\subseteq \mathcal {C}\) such that, if \(\varepsilon >0\), then \begin {equation*} E\subseteq {\Bigl (\bigcup _{\substack {I\in \mathcal {S}\\\diam (I)\geq \varepsilon }}I\Bigr )} \cup \Bigl (\bigcup _{\substack {I\in \mathcal {S}\\\diam (I)<\varepsilon }} 5I\Bigr ) \end {equation*} and such that, if \(I\), \(\beta \in \mathcal {S}\), then \(I\cap \beta =\emptyset \). (For partial credit, you may prove this under the assumption that \(X=\R \), \(d(x,y)=|x-y|\), and \(E\) is a measurable set of finite measure.)
Because \begin {equation*} E\subseteq \bigcup _{\substack {I\in \mathcal {C}\\\diam (I)<1}} I \end {equation*} we may replace \(\mathcal {C}\) by \(\{I\in \mathcal {C}:\diam (I)<1\}\). Let \(R=1\) and let \(\mathcal {S}\), \(\mathcal {S}_k\) be as in Problem 1722. Note that each \(\mathcal {S}_k\) is countable by Problem 1711 and so \(\mathcal {S}\) must also be countable.We need only establish the inclusion \begin {equation*} E\subseteq {\Bigl (\bigcup _{\substack {I\in \mathcal {S}\\\diam (I)\geq \varepsilon }}I\Bigr )} \cup \Bigl (\bigcup _{\substack {I\in \mathcal {S}\\\diam (I)<\varepsilon }} 5I\Bigr ) \end {equation*} for all \(\varepsilon >0\).
Choose some \(\varepsilon >0\). Observe that \begin {equation*} \bigcup _{{I\in \mathcal {S}}}I \subseteq {\Bigl (\bigcup _{\substack {I\in \mathcal {S}\\\diam (I)\geq \varepsilon }}I\Bigr )} \cup \Bigl (\bigcup _{\substack {I\in \mathcal {S}\\\diam (I)<\varepsilon }} 5I\Bigr ). \end {equation*} Thus, suppose that \(e\in E\) and \(e\notin \bigcup _{{I\in \mathcal {S}}}I\). It suffices to show that \(e\in \bigcup _{\substack {I\in \mathcal {S}\\\diam (I)<\varepsilon }} 5I\) for all such \(e\).
Let \(n\in \N \) satisfy \(\varepsilon > 2^{-n}\). Recall that \(\mathcal {S}_k=\{I\in \mathcal {S}:2^{-k}<\diam (I)\leq 2^{1-k}\}\). Let \(\widetilde {\mathcal {S}}_n=\bigcup _{k=1}^n \mathcal {S}_k=\{I\in \mathcal {S}:\diam (i)>2^{-n}\}\).
Because \(e\notin \bigcup _{{I\in \mathcal {S}}}I\), we have that \(e\notin \bigcup _{I\in \widetilde {\mathcal {S}}_n}I\). But by Problem 1721, \(\bigcup _{I\in \widetilde {\mathcal {S}}_n}I\) is closed and so \(\dist (e,\bigcup _{I\in \widetilde {\mathcal {S}}_n}I)=r>0\).
Let \(\varrho \) satisfy \(\varrho <\min (r,2^{-n})\). Then \(e\in J\) for some \(J\in \mathcal {C}\) with \(\diam (J)<\varrho <\dist (e,\bigcup _{I\in \widetilde {\mathcal {S}}_n}I)\). In particular \(J\cap I=\emptyset \) for all \(I\in \widetilde {\mathcal {S}}_n\).
There is a \(k\in \N \) with \(2^{-k}<\diam (J)\leq 2^{1-k}\) (that is, with \(J\in \mathcal {D}_k\)). Observe that \(2^{-k}<\diam (J)<\varrho \) and so \(k>n\). Then either \(J\in \mathcal {C}_k\) or \(J\cap I\neq \emptyset \) for some \(I\in S_\ell \) for some \(\ell <k\).
If \(J\in \mathcal {C}_k\) then \(e\in J\subseteq \bigcup _{I\in \mathcal {S}_k} 5I\subseteq \bigcup _{\substack {I\in \mathcal {S}\\\diam (I)<\varepsilon }} 5I\) because \(k>n\) and so \(\diam (I)\leq 2^{1-k}\leq 2^{-n}<\varepsilon \).
Otherwise, \(J\cap I\neq \emptyset \) for some \(I\in S_\ell \) for some \(\ell <k\). But \(J\cap I=\emptyset \) for all \(I\in \in \widetilde {\mathcal {S}}_n\), and so \(n<\ell <k\). Thus \(\diam (I)\leq 2^{1-\ell }\leq 2^{-n}<\varepsilon \). As before, this implies \(e\in \bigcup _{\substack {I\in \mathcal {S}\\\diam (I)<\varepsilon }} 5I\).
Thus, if \(e\notin \bigcup _{{I\in \mathcal {S}}}I\), then in either case \(e\in \bigcup _{\substack {I\in \mathcal {S}\\\diam (I)<\varepsilon }} 5I\). This completes the proof.
The Vitali covering lemma, book version. Let \(E\subset \R \) satisfy \(m^*(E)<\infty \). Let \(\mathcal {C}\) be a (possibly infinite) collection of closed bounded intervals such that \(\diam (I)>0\) for all \(I\in \mathcal {C}\) and such that, for all \(\delta >0\), we have that \begin {equation*} E\subseteq \bigcup _{\substack {I\in \mathcal {C}\\\ell (I)<\delta }} I. \end {equation*} If \(\varepsilon >0\), then there is a finite collection \(\{I_k\}_{k=1}^n\subseteq \mathcal {C}\) such that \begin {equation*} m^*\Bigl (E\setminus \bigcup _{k=1}^n I_k\Bigr )<\varepsilon . \end {equation*}
(Problem 1750) Prove the Vitali covering lemma, book version.
Let \(\mathcal {O}\subset \R \) be open and satisfy \(E\subseteq \mathcal {O}\) and \(m(\mathcal {O})\leq m^*(E)+\varepsilon /2\); by definition of outer measure such a \(\mathcal {O}\) must exist.Let \(\widetilde {\mathcal {C}}=\{I\in \mathcal {C}:I\subset \mathcal {O}\}\). Note that if \(e\in E\), then \(B(e,r)=(e-r,e+r)\subseteq \mathcal {O}\) for some \(r>0\) by the definition of an open set. For each \(\delta >0\), we have that \(e\in I\) for some \(I\in \mathcal {C}\) with \(\ell (I)<\min (\delta ,r)\); but then \(I\subset (e-r,e+r)\subseteq \mathcal {O}\) and so \(I\in \widetilde {\mathcal {C}}\). Thus \(e\in \bigcup _{\substack {I\in \widetilde {\mathcal {C}}\\\ell (I)<\delta }} I\) for all \(e\in E\), and so \(\widetilde {\mathcal {C}}\) satisfies the conditions of Problem 1740.
Let \(\mathcal {S}\) be as in Problem 1740. Recall that \(\mathcal {S}\) is countable.
Define \(E_k=\bigcup _{\substack {I\in \mathcal {S}\\\diam (I)\geq 1/k}}I\). Then \(E_k\) is measurable, \(E_k\subseteq E_{k+1}\), and each \(E_k\subseteq \mathcal {O}\), so by Theorem 2.15, \begin {equation*} \lim _{k\to \infty } m(E_k)=m\Bigl (\bigcup _{k=1}^\infty E_k\Bigr )\leq m(\mathcal {O})<\infty . \end {equation*} Thus there is a \(k\) with \(m(E_k)> m\Bigl (\bigcup _{k=1}^\infty E_k\Bigr ) -\frac {\varepsilon }{10}\).
By Problem 1740, \begin {equation*} E\subseteq {\Bigl (\bigcup _{\substack {I\in \mathcal {S}\\\diam (I)\geq 1/k}}I\Bigr )} \cup \Bigl (\bigcup _{\substack {I\in \mathcal {S}\\\diam (I)<1/k}} 5I\Bigr ). \end {equation*} Now, because the intervals \(I\in \mathcal {S}\) are pairwise-disjoint, \begin {equation*} m(E_k)=m\Bigl (\bigcup _{\substack {I\in \mathcal {S}\\\diam (I)\geq 1/k}} I\Bigr ) =\sum _{\substack {I\in \mathcal {S}\\\diam (I)\geq 1/k}} \diam (I) \geq \frac {1}{k} |\{I\in \mathcal {S}:\diam (I)\geq 1/k\}| \end {equation*} and so \(\{I\in \mathcal {S}:\diam (I)\geq 1/k\}\) is finite. Let \(\{I_j\}_{j=1}^n=\{I\in \mathcal {S}:\diam (I)\geq 1/k\}\).
Then \begin {equation*} E\setminus \bigcup _{j=1}^n I_j\subset \bigcup _{\substack {I\in \mathcal {S}\\\diam (I)<1/k}} 5I \end {equation*} and so \begin {equation*} m^*(E\setminus \bigcup _{j=1}^n I_j) \leq \sum _{\substack {I\in \mathcal {S}\\\diam (I)<1/k}} 5\diam (I) =5\sum _{\substack {I\in \mathcal {S}\\\diam (I)<1/k}} \diam (I) =5m \Bigl (\bigcup _{\substack {I\in \mathcal {S}\\\diam (I)<1/k}} I\Bigr ) \end {equation*} again by pairwise disjointness of the intervals \(I\). But \begin {equation*} \bigcup _{\substack {I\in \mathcal {S}\\\diam (I)<1/k}} I\subset E\setminus E_k \subset \Bigl (\bigcup _{j=1}^\infty E_j\Bigr )\setminus E_k \end {equation*} and so has measure at most \(\varepsilon /10\). Combining these estimates completes the proof.
[Definition: The derivative] Let \(E\subseteq \R \) and let \(f:E\to \R \). Suppose that \(x\in E\) is an interior point, that is, \((x-\delta ,x+\delta )\subset E\) for some \(\delta >0\). We define \begin {align*} \overline {D} f(x)=\lim _{h\to 0^+} \sup \biggl \{\frac {f(x)-f(y)}{x-y}:0<|x-y|<h\biggr \},\\ \underline {D} f(x)=\lim _{h\to 0^+} \inf \biggl \{\frac {f(x)-f(y)}{x-y}:0<|x-y|<h\biggr \} . \end {align*}
If \(\overline {D} f(x)=\underline {D} f(x)\) and is finite, we say that \(f\) is differentiable at \(x\) and define \(f'(x)=\overline {D} f(x)=\underline {D} f(x)\).
(Memory 1751) (The Mean Value Theorem). Suppose that \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\). Then there is an \(x\in (a,b)\) such that \(f(b)-f(a)=(b-a)f'(x)\). In particular, if \(f'(x)\geq \lambda \) for all \(x\in (a,b)\), then \(f(b)-f(a)\geq (b-a)\lambda \).
Lemma 6.3. Let \(f:[a,b]\to \R \) be nondecreasing for some \(a<b\), \(a\), \(b\in \R \). Let \(\lambda >0\). Then \begin {equation*} \lambda m^*\{x\in (a,b):\overline {D} f(x)\geq \lambda \}\leq f(b)-f(a) \end {equation*} and \begin {equation*} m\{x\in (a,b):\overline {D} f(x)=\infty \}=0. \end {equation*}
(Problem 1760) Prove Lemma 6.3.
Lebesgue’s theorem. If \(f:(a,b)\to \R \) is monotonic, then \(f\) is differentiable almost everywhere in \((a,b)\).
(Problem 1770) Prove Lebesgue’s theorem.
[Definition: Divided difference and averaged function] Suppose \(f\) is integrable over \([a,b]\). Define \begin {equation*} \widetilde f(x)=\begin {cases} f(a), & x\leq a,\\f(x),&a\leq x\leq b, \\f(b),&b\leq x.\end {cases} \end {equation*} If \(h\neq 0\), we define \begin {equation*} \Diff _h f(x)=\frac {\widetilde f(x+h)-\widetilde f(x)}{h}, \qquad \mathop {\mathrm {Av}}\nolimits _h f(x) = \frac {1}{h}\int _x^{x+h} \widetilde f. \end {equation*}
(Problem 1780) Show that \(\int _u^v \mathop {\mathrm {Diff}}\nolimits _h f = \mathop {\mathrm {Av}}\nolimits _h f(v)-\mathop {\mathrm {Av}}\nolimits _h f(u)\).
Corollary 6.4. If \(f\) is a nondecreasing real-valued function over \([a,b]\), then \(f'\) (which exists almost everywhere by Lebesgue’s theorem) is integrable over \([a,b]\) and \begin {equation*} \int _a^b f'\leq f(b)-f(a). \end {equation*}
(Problem 1790) Prove Corollary 6.4.
By definition, we have that \(f'(x)=\lim _{h\to 0} \mathop {\mathrm {Diff}}\nolimits _h f(x)\) for all \(x\) such that \(f'(x)\) exists, that is, almost everywhere.In particular, let \(f_k=\mathop {\mathrm {Diff}}\nolimits _{1/k} f\). Then \(f'=\lim _{k\to \infty } f_k\). Because \(f\) is monotonic, \(f\) is measurable by Problem 3.24, and so each \(f_k\) is measurable by Theorem 3.6; thus \(f'\) is measurable by Proposition 3.9. We need only show that \(\int _{[a,b]}|f'|<\infty \).
Because \(f\) is nondecreasing, \(\mathop {\mathrm {Diff}}\nolimits _h f(x)\geq 0\) for all \(x\) and all \(h\neq 0\). Thus each \(f_k\) is a nonnegative function, and so \(f'\geq 0\) almost everywhere and \(\int _{[a,b]}|f'|=\int _{a}^b f'\).
Furthermore, by Fatou’s lemma \begin {equation*} \int _a^b f'=\int _{[a,b]} f' = \int _{[a,b]} \lim _{k\to \infty } f_k \leq \liminf _{k\to \infty } \int _{[a,b]} f_k. \end {equation*} But by the previous problem \begin {equation*} \int _{[a,b]} f_k=\int _a^b \mathop {\mathrm {Diff}}\nolimits _{1/k} f =\mathop {\mathrm {Av}}\nolimits _{1/k} f(b)-\mathop {\mathrm {Av}}\nolimits _{1/k} f(a). \end {equation*} Because \(1/k>0\) we have that \begin {equation*} \mathop {\mathrm {Av}}\nolimits _{1/k} f(b) = k\int _b^{b+1/k} \widetilde f = f(b) \end {equation*} because \(\widetilde f\equiv f(b)\) on \([b,\infty )\). Furthermore, \begin {equation*} \mathop {\mathrm {Av}}\nolimits _{1/k} f(a) = k\int _a^{a+1/k} \widetilde f \geq k\int _a^{a+1/k} f(a)=f(a) \end {equation*} because \(\widetilde f\geq f(a)\) everywhere. Thus \begin {equation*} \int _a^b f' \leq \liminf _{k\to \infty } \int _{[a,b]} f_k= \liminf _{k\to \infty } \bigl (\mathop {\mathrm {Av}}\nolimits _{1/k} f(b)-\mathop {\mathrm {Av}}\nolimits _{1/k} f(a)\bigr ) \leq f(b)-f(a). \end {equation*} This completes the proof.
(Problem 1800) Give an example of a function \(f\) that is an increasing function over \([a,b]\) and such that \begin {equation*} \int _a^b f'< f(b)-f(a). \end {equation*}
Let \begin {equation*} f(x)=\begin {cases}x^2,&0\leq x<\frac 12,\\x^2+1,&\frac 12\leq x\leq 1.\end {cases} \end {equation*} Let \(g(x)=x^2\). Then \(f'(x)=g'(x)\) for all \(x\in (0,1)\setminus \{1/2\}\), that is, almost everywhere, and so \(\int _0^1 f'=\int _0^1 g'\). But by the fundamental theorem of calculus, because \(g\) is continous on \([0,1]\) and continuously differentiable on \((0,1)\), we have that \(\int _0^1 g'=g(1)-g(0)=1\). But \(f(1)-f(0)=2\), and so \(f(1)-f(0)>\int _0^1 f'\).
[Definition: Total variation] Let \([a,b]\subset \R \) be a closed bounded interval and let \(f:[a,b]\to \R \). The total variation of \(f\) over \([a,b]\) is \begin {equation*} TV(f)=\sup \Bigl \{\sum _{k=1}^n |f(x_k)-f(x_{k-1})|:a\leq x_{k-1}\leq x_k\leq b\text { for all }1\leq k\leq n\Bigr \}. \end {equation*}
(Problem 1801) Show that we obtain an equivalent definition if we additionally require that \(a=x_0\) and \(b=x_n\).
(Problem 1810) If \(f:[a,b]\to \R \) is nondecreasing, show that \(TV(f)=f(b)-f(a)\).
(Problem 1820) If \(f:[a,b]\to \R \) is Lipschitz with Lipschitz constant \(M\), show that \(TV(f)\leq M(b-a)\).
(Problem 1830) If \(f:[a,b]\to \R \), \(g:[a,b]\to \R \), show that \(TV(f+g)\leq TV(f)+TV(g)\).
(Problem 1840) Give an example of a bounded, continuous function \(f\) from \([0,1]\) to \(\R \) such that \(TV(f)=\infty \).
(Problem 1841) Suppose that \(f:[a,b]\to \R \), that \(TV(f)<\infty \), and that \(a<c<b\). Show that \(TV(f\big \vert _{[a,c]})+TV(f\big \vert _{[c,b]})\leq TV(f)\).
Let \(a\leq x_0\leq x_1\leq x_2\leq x_n\leq b\). There is some \(k\) with \(1\leq k\leq n\) and with \(x_{k-1}\leq c< x_{k}\). Then \begin {align*} \sum _{j=1}^n |f(x_j)-f(x_{j-1})| &= \sum _{j=1}^{k-1} |f(x_j)-f(x_{j-1})| \\&\qquad + |f(x_{k})-f(x_{k-1})| \\&\qquad + \sum _{j=k+1}^{n} |f(x_j)-f(x_{j-1})|. \end {align*}
By the triangle inequality in the real numbers, \begin {align*} \sum _{j=1}^n |f(x_j)-f(x_{j-1})| &= \sum _{j=1}^{k-1} |f(x_j)-f(x_{j-1})| +|f(c)-f(x_{k-1})| \\&\qquad +|f(x_{k})-f(c)|+ \sum _{j=k+1}^{n} |f(x_j)-f(x_{j-1})|. \end {align*}
Observe that \(a\leq x_0\leq x_1\leq x_2\leq x_k\leq c\leq c\), and so the terms on the first line are at most \(TV(f\big \vert _{[a,c]})\). Similarly, \(c\leq c\leq x_k\leq \dots \leq x_n\leq b\) and so the terms on the second line are at most \(TV(f\big \vert _{[c,b]})\). Thus \begin {equation*} \sum _{j=1}^n |f(x_j)-f(x_{j-1})| \leq TV(f\big \vert _{[a,c]})+TV(f\big \vert _{[c,b]}) \end {equation*} whenever \(a\leq x_0\leq x_1\leq x_2\leq x_n\leq b\), and so taking the supremum over all such \(\{x_j\}_{j=1}^n\) yields the inequality \(TV(f)\leq TV(f\big \vert _{[a,c]})+TV(f\big \vert _{[c,b]})\).
(Problem 1842) Suppose that \(f:[a,b]\to \R \), that \(TV(f)<\infty \), and that \(a<c<b\). Show that \(TV(f\big \vert _{[a,c]})+TV(f\big \vert _{[c,b]})= TV(f)\).
We have that \(TV(f\big \vert _{[a,c]})+TV(f\big \vert _{[c,b]})\leq TV(f)\) by the previous problem. We now turn to the reverse inequality. Let \(a= y_0\leq y_1\leq \dots \leq y_m=c\) and \(c=z_0\leq z_1\leq \dots \leq z_\ell =b\). Let \(x_j=y_j\) if \(0\leq j\leq m\) and let \(x_j=z_{j-m}\) if \(m\leq j\leq m+\ell =n\). Then \(a=x_0\leq x_1\leq \dots \leq x_n=b\), and so \begin {equation*} \sum _{j=1}^m |f(y_j)-f(y_{j-1})| + \sum _{k=1}^\ell |f(z_k)-f(z_{k-1})| = \sum _{j=1}^n |f(x_j)-f(x_{j-1})|\leq TV(f). \end {equation*} Both of the sums on the left hand side are nonnegative. Thus, \begin {equation*} TV(f\big \vert _{[a,c]})=\sup \{\sum _{j=1}^m |f(y_j)-f(y_{j-1})|:a= y_0\leq y_1\leq \dots \leq y_m=c\}\leq TV(f) \end {equation*} and \begin {equation*} TV(f\big \vert _{[c,b]})=\sup \{\sum _{k=1}^\ell |f(z_k)-f(z_{k-1})|:c=z_0\leq z_1\leq \dots \leq z_\ell =b\}\leq TV(f) \end {equation*} and so in particular both are finite. Choose \(\varepsilon >0\). We may now impose the additional requirement on \(\{y_j\}_{j=1}^m\) and \(\{z_k\}_{k=1}^\ell \) that \begin {equation*} TV(f\big \vert _{[a,c]})\leq \varepsilon +\sum _{j=1}^m |f(y_j)-f(y_{j-1})|, \qquad TV(f\big \vert _{[c,b]})\leq \varepsilon +\sum _{k=1}^\ell |f(z_k)-f(z_{k-1})|. \end {equation*} Thus \(TV(f\big \vert _{[a,c]})+TV(f\big \vert _{[c,b]})<2\varepsilon + TV(f)\) for all \(\varepsilon >0\); thus \(TV(f\big \vert _{[a,c]})+TV(f\big \vert _{[c,b]})\leq TV(f)\) and so \(TV(f\big \vert _{[a,c]})+TV(f\big \vert _{[c,b]})= TV(f)\), as desired.
(Problem 1850) Suppose that \(f:[a,b]\to \R \), that \(TV(f)<\infty \), and that \(a\leq c<d\leq b\). Show that \(TV(f\big \vert _{[c,d]})\leq TV(f)\). In particular, the function \(F(x)=TV(f\big \vert _{[a,x]})\) is nondecreasing.
This follows immediately from the previous problem and from the fact that \(TF(f\big \vert _{[a,c]})\geq 0\) and \(TF(f\big \vert _{[d,b]})\geq 0\).
Lemma 6.5. Let \(f:[a,b]\to \R \) and suppose that \(TV(f)<\infty \). Then \(f=g-h\), where \(g\) and \(h\) are both nondecreasing.
(Problem 1860) Prove Lemma 6.5.
Let \(h(x)=TV(f\big \vert _{[a,x]})\). If \(a\leq x<y\leq b\), then \begin {align*} h(y)&=TV(f\big \vert _{[a,y]})=TV(f\big \vert _{[a,x]})+TV(f\big \vert _{[x,y]}) \alignbreak \geq TV(f\big \vert _{[a,x]})=h(x) \end {align*}by Problem 1842 and the fact that total variation is never negative. Thus \(h\) is nondecreasing.
Now, let \(g(x)=f(x)+h(x)\). Then \(f(x)=g(x)-h(x)\). It remains only to show that \(g\) is nondecreasing.
If \(a\leq x<y\leq b\), then \begin {align*} g(y)-g(x)&= f(y)+h(y)-f(x)-h(x) \alignbreak =f(y)-f(x) +TV(f\big \vert _{[a,y]})-TV(f\big \vert _{[a,x]}) \alignbreak =f(y)-f(x)+TV(f\big \vert _{[x,y]}) \end {align*}
But \(\{x,y\}\) is a partition of \([x,y]\), and so we must have that \begin {equation*} |f(y)-f(x)|\leq TV(f\big \vert _{[x,y]}) \end {equation*} by definition of total variation. Thus \(g(y)-g(x)\geq 0\) for all \(a\leq x<y\leq b\), and so \(g\) is nondecreasing as well.
Jordan’s Theorem. If \(f:[a,b]\to \R \), then \(f\) is of bounded variation (that is, \(TV(f)<\infty \)) if and only if \(f=g+h\), where \(g\) is nondecreasing and \(h\) is nonincreasing.
Corollary 6.6. If \(f\) is of bounded variation on the closed bounded interval \([a,b]\), then \(f\) is differentiable almost everywhere on \([a,b]\), and \(f'\) is integrable over \([a,b]\).
[Definition: Absolutely continuous] Let \([a,b]\subset \R \) be a closed and bounded interval and let \(f:[a,b]\to \R \) be a function.
We say that \(f\) is absolutely continuous if, for every \(\varepsilon >0\), there is a \(\delta >0\) such that, if \(\{a_k\}_{k=1}^n\) and \(\{b_k\}_{k=1}^n\) satisfy
\(a\leq a_k\leq b_k\leq b\) for all \(1\leq k\leq n\),
\(b_{k-1}\leq a_{k}\) for all \(2\leq k\leq n\),
\(\sum _{k=1}^n |b_k-a_k|<\delta \),
then \begin {equation*} \sum _{k=1}^n|f(b_k)-f(a_k)|<\varepsilon . \end {equation*}
(Problem 1861) Show that every absolutely continuous function is uniformly continuous.
[Chapter 6, Problem 38a] We may replace \(n\) by \(\infty \) in the above definition and recover an equivalent definition; that is, \(f\) is absolutely continuous over \([a,b]\) if and only if, for every \(\varepsilon >0\), there is a \(\delta >0\) such that, if \(\{a_k\}_{k=1}^\infty \) and \(\{b_k\}_{k=1}^\infty \) satisfy
\(a\leq a_k\leq b_k\leq b\) for all \(1\leq k<\infty \),
\((a_k,b_k)\cap (a_j,b_j)\) for all \(j\), \(k\in \N \) with \(j\neq k\),
\(\sum _{k=1}^\infty |b_k-a_k|<\delta \),
then \begin {equation*} \sum _{k=1}^\infty |f(b_k)-f(a_k)|<\varepsilon . \end {equation*} (Note in particular that the intervals may not be ordered; that is, we need only require that the intervals \((a_k,b_k)\) are pairwise-disjoint, not that \(b_k\leq a_{k+1}\) for all \(k\).)
(Problem 1870) Let \([a,b]\subseteq \mathbb {R}\) be a closed and bounded interval and let \(f:[a,b]\to \mathbb {R}\). Suppose that \(f'\) is defined almost everywhere on \([a,b]\) and is integrable on \([a,b]\), and that \(f(x)=f(a)+\int _a^x f'\) for all \(x\in [a,b]\). Show that \(f\) is absolutely continuous.
[Chapter 6, Problem 38b] Suppose that \(f\) satisfies the following weaker version of absolute continuity: for every \(\varepsilon >0\), there is a \(\delta >0\) such that, if \(\{a_k\}_{k=1}^n\) and \(\{b_k\}_{k=1}^n\) satisfy
\(a\leq a_k\leq b_k\leq b\) for all \(1\leq k\leq n\),
\(b_{k-1} < a_{k}\) for all \(2\leq k\leq n\),
\(\sum _{k=1}^n |b_k-a_k|<\delta \),
then \begin {equation*} \sum _{k=1}^n|f(b_k)-f(a_k)|<\varepsilon . \end {equation*} Clearly every absolutely continuous function satisfies this condition. Show that the converse is true, that is, any function that satisfies this condition is absolutely continuous.
(Problem 1880) Suppose that \(f\) is absolutely continuous. Determine whether this implies that \(f\) also satisfies the following stronger version of absolute continuity: for every \(\varepsilon >0\), there is a \(\delta >0\) such that, if \(\{a_k\}_{k=1}^\infty \) and \(\{b_k\}_{k=1}^\infty \) satisfy
\(a\leq a_k\leq b_k\leq b\) for all \(1\leq k\leq n\),
\(\sum _{k=1}^n |b_k-a_k|<\delta \),
then \(\sum _{k=1}^\infty |f(b_k)-f(a_k)|<\varepsilon \).
It does not. The function \(f(x)=\sqrt {x}\) is absolutely continuous by Problem 1870 but does not satisfy the above condition: take \(a_k=0\), \(b_k=\delta /2n\) for all \(1\leq k\leq n\), and observe that \(\sum _{k=1}^n |b_k-a_k|=\delta /2<\delta \) but \(\sum _{k=1}^\infty |f(b_k)-f(a_k)|=\sqrt {n\delta }\) grows without limit as \(n\to \infty \).
Proposition 6.7. A Lipschitz continuous function on a closed bounded interval is absolutely continuous.
(Problem 1890) Prove Proposition 6.7.
Theorem 6.8. Let \([a,b]\subseteq \mathbb {R}\) be a closed and bounded interval and let \(f:[a,b]\to \mathbb {R}\). If \(f\) is absolutely continous, then \(f\) is of bounded variation and \(f=g-h\), where \(g\) and \(h\) are nondecreasing functions both of which are absolutely continuous on \([a,b]\).
(Problem 1900) Begin the proof of Theorem 6.8 by showing that, if \(f\) is absolutely continuous on the closed bounded interval \([a,b]\), then \(f\) is of bounded variation. Hint: Use Problem 1842.
Let \(\delta \) respond to the \(\varepsilon =1\) challenge in the definition of absolute continuity and let \(N\in \N \) satisfy \(\frac {b-a}{N}<\delta \); such an \(N\) exists by the Archimedean property of the real numbers. Let \(c_j=a+\frac {j}{N}(b-a)\).By Problem 1842, \begin {equation*} TV(f)=\sum _{j=1}^N TV(f\big \vert _{[c_{j-1},c_j]}). \end {equation*} By definition of the total variation, \begin {equation*} TV(f\big \vert _{[c_{j-1},c_jk]}) = \sup \Bigl \{\sum _{k=1}^n |f(x_k)-f(x_{k-1})|:c_{j-1}\leq x_{k-1}\leq x_k\leq c_j\text { for all }1\leq k\leq n\Bigr \} . \end {equation*} Select some such partition \(\{x_k\}_{k=0}^n\) and choose \(a_k=x_{k-1}\) and \(b_k=x_k\). Then
\(a\leq c_{j-1}\leq a_k\leq b_k\leq c_j\leq b\),
\(b_k=a_{k+1}\leq a_{k+1}\),
\(\sum _{k=1}^n (b_k-a_k)=\sum _{k=1}^n (x_k-x_{k-1})=x_n-x_0=c_j-c_{j-1}=\frac {b-a}{N}<\delta \).
Thus by the definition of absolute continuity, we must have that \begin {equation*} \sum _{k=1}^n |f(x_k)-f(x_{k-1})| =\sum _{k=1}^n |f(b_k)-f(a_{k})|<\varepsilon =1. \end {equation*} Thus \begin {equation*} TV(f\big \vert _{[c_{j-1},c_jk]})\leq 1 \end {equation*} and so \begin {equation*} TV(f)=\sum _{j=1}^N TV(f\big \vert _{[c_{j-1},c_j]})\leq N<\infty \end {equation*} as desired.
(Problem 1901) Give an example of a function of bounded variation that is not absolutely continuous.
(Problem 1910) By Jordan’s Theorem and the previous problem, if \(f\) is absolutely continuous on the closed bounded interval \([a,b]\), then \(f=g-h\) where \(g\) and \(h\) are both nondecreasing functions. In particular, we may require that \(h(x)=TV(f\big \vert _{[a,x]})\). Show that \(h\) is absolutely continuous in addition to being nondecreasing.
(Problem 1920) Complete the proof of Theorem 6.8 by showing that the sum of two absolutely continuous functions is absolutely continuous.
Theorem 6.9. Let \([a,b]\subseteq \mathbb {R}\) be a closed and bounded interval and let \(f:[a,b]\to \mathbb {R}\). Suppose that \(f\) is continuous. Then \(f\) is absolutely continuous if and only if \(\{\Diff _h f:0<h\leq 1\}\) is a uniformly integrable family, where \(\Diff _h\) is as in Problem 1780.
(Problem 1930) Begin the proof of Theorem 6.9 by showing that, if \(\{\Diff _h f:0<h\leq 1\}\) is a uniformly integrable family, then \(f\) is absolutely continuous.
Let \(\varepsilon >0\) and let \(\delta >0\) respond to the \(\varepsilon \) challenge in the definition of \(\{\Diff _h f:0<h\leq 1\}\) being uniformly integrable. Let \(\{a_k\}_{k=1}^n\) and \(\{b_k\}_{k=1}^n\) satisfy
\(a\leq a_k\leq b_k\leq b\) for all \(1\leq k\leq n\),
\(b_{k-1}\leq a_{k}\) for all \(2\leq k\leq n\),
\(\sum _{k=1}^n |b_k-a_k|<\delta \).
By uniform integrability, we have that \begin {equation*} \sum _{k=1}^n \int _{a_k}^{b_k} |\Diff _h f|<\varepsilon . \end {equation*} By Proposition 4.16, we have that \begin {equation*} \sum _{k=1}^n \biggl |\int _{a_k}^{b_k} \Diff _h f\biggr |<\varepsilon . \end {equation*} By Problem 1780, \(\int _{a_k}^{b_k} \Diff _h f=\Av _h f(b_k)-\Av _h f(a_k)\) and so \begin {equation*} \sum _{k=1}^n \bigl |\Av _h f(b_k)-\Av _h f(a_k)\bigr |<\varepsilon . \end {equation*} Recall that we assumed that \(f\) is continuous. This implies that \(\Av _h f(x)\to f(x)\) for all \(x\in [a,b]\). Because the above sum involves only finitely many points, we may take the limit as \(h\to 0^+\) to see that \begin {equation*} \sum _{k=1}^n \bigl |f(b_k)-f(a_k)\bigr | =\lim _{h\to 0^+}\sum _{k=1}^n \bigl |\Av _h f(b_k)-\Av _h f(a_k)\bigr |\leq \varepsilon \end {equation*} as desired.
(Problem 1940) Complete the proof of Theorem 6.9.
(Problem 1950) Suppose that \(f:[a,b]\to \R \) is integrable over \([a,b]\) and continuous at \(a\) and \(b\). Show that \begin {equation*} f(b)-f(a)=\lim _{h\to 0} \int _a^b \mathop {\mathrm {Diff}}\nolimits _h f. \end {equation*}
By Problem 1780, we know that \begin {equation*} \int _a^b \mathop {\mathrm {Diff}}\nolimits _h f = \mathop {\mathrm {Av}}\nolimits _h f(b)-\mathop {\mathrm {Av}}\nolimits _h f(a). \end {equation*} Because \(f\) is continuous at \(a\) and at \(b\), we have that \(\lim _{h\to 0} \mathop {\mathrm {Av}}\nolimits _h f(b)= f(b)\) and \(\lim _{h\to 0} \mathop {\mathrm {Av}}\nolimits _h f(a)= f(a)\). This completes the proof.
Theorem 6.10. Let \([a,b]\subseteq \mathbb {R}\) be a closed and bounded interval and let \(f:[a,b]\to \mathbb {R}\). Suppose that \(f\) is absolutely continuous.
Then \(f\) is differentiable almost everywhere on \((a,b)\), its derivative \(f'\) is integrable over \([a,b]\), and \begin {equation*} \int _a^b f'=f(b)-f(a). \end {equation*}
(Problem 1960) Prove Theorem 6.10.
By Theorem 6.8, there are increasing absolutely continuous functions \(g\) and \(h\) defined on \([a,b]\) such that \(f=g-h\).By Lebesgue’s theorem and Corollary 6.4, we have that \(g\) and \(h\) are differentiable almost everywhere and their derivatives \(g'\) and \(h'\) are integrable, and so \(f\) must be differentiable almost everywhere (at every point where \(g\) and \(h\) are both integrable) and moreover \(f'=g'-h'\) must also be integrable.
By the definition of derivative, \(\lim _{n\to \infty } \Diff _{1/n} f(x)=f'(x)\) for almost every \(x\in [a,b]\). Thus \begin {equation*} \int _a^b f' = \int _a^b \lim _{n\to \infty } \Diff _{1/n} f. \end {equation*} By Theorem 6.9, the family \(\{\Diff _{1/n} f:n\in \N \}\) is uniformly integrable, and so by the Vitali convergence theorem, \begin {equation*} \int _a^b \lim _{n\to \infty } \Diff _{1/n} f=\lim _{n\to \infty } \int _a^b \Diff _{1/n} f. \end {equation*} By the previous problem, \begin {equation*} \lim _{n\to \infty } \int _a^b \Diff _{1/n} f=f(b)-f(a). \end {equation*} Thus \begin {equation*} \int _a^b f' = f(b)-f(a) \end {equation*} as desired.
(Problem 1970) Let \([a,b]\subseteq \mathbb {R}\) be a closed and bounded interval and let \(f:[a,b]\to \mathbb {R}\). Show that \(f\) is absolutely continuous over \([a,b]\) if and only if \(f\) is differentiable almost everywhere in \((a,b)\), \(f'\) is integrable over \([a,b]\), and \begin {equation*} f(x)=f(a)+\int _a^x f' \end {equation*} for all \(x\in [a,b]\).
Theorem 6.11. Let \([a,b]\subseteq \mathbb {R}\) be a closed and bounded interval and let \(f:[a,b]\to \mathbb {R}\). Then \(f\) is absolutely continuous if and only if there is a function \(g\) that is integrable over \([a,b]\) such that \begin {equation*} f(x)=f(a)+\int _a^x g \end {equation*} for all \(x\in [a,b]\). Furthermore, if these equivalent conditions are true, then one possible such \(g\) is \(f'\).
(Problem 1980) Prove Theorem 6.11.
If \(f\) is absolutely continuous over \([a,b]\), then \(f'\) exists and is integrable over \([a,b]\) by Theorem 6.10. Furthermore, \(f\) is clearly absolutely continuous over \([a,x]\) for any \(x\in (a,b]\), and so we have that \begin {equation*} f(x)=f(a)+\int _a^x f' \end {equation*} for all such \(x\). Choosing \(g=f'\) completes the proof.Conversely, suppose that \begin {equation*} f(x)=f(a)+\int _a^x g \end {equation*} for some function \(g\) integrable over \([a,b]\). Choose \(\varepsilon >0\) and let
Corollary 6.12. Let \([a,b]\subseteq \mathbb {R}\) be a closed and bounded interval and let \(f:[a,b]\to \mathbb {R}\) be monotonic. Then \(f\) is absolutely continuous on \([a,b]\) if and only if \begin {equation*} \int _a^b f'=f(b)-f(a). \end {equation*}
(Problem 1990) Prove Corollary 6.12.
Lemma 6.13. Let \([a,b]\subseteq \mathbb {R}\) be a closed and bounded interval and let \(f:[a,b]\to \mathbb {R}\) be integrable over \([a,b]\). Then \(f=0\) almost everywhere in \([a,b]\) if and only if \(\int _x^y f=0\) for all \(x\), \(y\) such that \(a<x<y<b\).
(Problem 2000) Prove Lemma 6.13.
Theorem 6.14. Let \([a,b]\subseteq \mathbb {R}\) be a closed and bounded interval and let \(f:[a,b]\to \mathbb {R}\) be integrable over \([a,b]\). Define \begin {equation*} F(x)=\int _a^x f. \end {equation*} Then \(F\) is differentiable almost everywhere on \((a,b)\) and \(F'=f\) almost everywhere on \((a,b)\).
(Problem 2010) Prove Theorem 6.14.
(Problem 2020) Let \(I\subseteq \R \) be an open interval and let \(\varphi :I\to \R \) be a continuous function. Suppose that \(\varphi \) is twice continuously differentiable. If \(0<a<1\) and \(b=1-a\), show that \begin {equation*} \varphi ''(x)=\frac {2}{ab} \lim _{h\to 0}\frac {b\varphi (x+ah)+a\varphi (x-bh)-\varphi (x)}{h^2}. \end {equation*}
[Definition: Convex function] If \(I\subseteq \R \) is an open interval and \(\varphi :I\to \R \), then \(\varphi \) is said to be convex if \begin {equation*} \varphi (x)\leq \frac {b}{a+b} \varphi (x+a) + \frac {a}{a+b} \varphi (x-b) \end {equation*} whenever \(a>0\), \(b>0\), and \(x+a\), \(x-b\in I\).
Proposition 6.15. If \(\varphi \) is differentiable on the open interval \(I\subseteq \R \) and \(\varphi '\) is nondecreasing, then \(\varphi \) is convex.
(Problem 2030) Prove Proposition 6.15.
Let \(x\in I\) and let \(a\), \(b>0\) be such that \(x+a\in I\) and \(x-b\in I\). Then \begin {align*} \hskip 2em&\hskip -2em\frac {b}{a+b} \varphi (x+a) + \frac {a}{a+b} \varphi (x-b) \\&= \frac {b}{a+b} \bigl (\varphi (x+a)-\varphi (x)\bigr ) - \frac {a}{a+b} \bigl (\varphi (x)-\varphi (x-b)\bigr ) \alignskipbreak + \frac {b}{a+b}\varphi (x)+\frac {a}{a+b} \varphi (x). \end {align*}Because \(\varphi '\) is nondecreasing, it is Riemann integrable over any closed bounded interval \([c,d]\subset I\). Because \(\varphi '\) exists everywhere in \(I\) we must have that \(\varphi \) is continuous in \(I\). Therefore we may apply the Fundamental Theorem of Calculus to see that, if \(c\), \(d\in I\) with \(c<d\), then \begin {equation*} \varphi (d)-\varphi (c)=\int _c^d \varphi '. \end {equation*} Thus \begin {align*} \hskip 2em&\hskip -2em\frac {b}{a+b} \varphi (x+a) + \frac {a}{a+b} \varphi (x-b) \\&= \frac {b}{a+b} \int _x^{x+a}\varphi ' - \frac {a}{a+b} \int _{x-b}^x \varphi ' + \varphi (x). \end {align*}
Because \(\varphi '\) is nondecreasing, we have that \(\varphi '\geq \varphi '(x)\) in \([x,x+a]\) and \(\varphi '\leq \varphi (x)\) in \([x-b,x]\). Thus by Theorem 4.17, and because \(a\) and \(b\) are positive \begin {align*} \hskip 2em&\hskip -2em\frac {b}{a+b} \varphi (x+a) + \frac {a}{a+b} \varphi (x-b) \\&\geq \frac {b}{a+b} \int _x^{x+a}\varphi '(x) - \frac {a}{a+b} \int _{x-b}^x \varphi '(x) + \varphi (x) \\&= \frac {b}{a+b} a\varphi '(x) - \frac {a}{a+b} b \varphi '(x) + \varphi (x)=\varphi (x). \end {align*}
Thus \(\varphi \) is convex.
The Chordal Slope Lemma. Let \(\varphi \) be convex on the open interval \(I\). If \(x\), \(y\), \(z\in I\) with \(x<y<z\), then \begin {equation*} \frac {\varphi (y)-\varphi (x)}{y-x}\leq \frac {\varphi (z)-\varphi (x)}{z-x}\leq \frac {\varphi (z)-\varphi (y)}{z-y}. \end {equation*}
(Problem 2040) Prove the Chordal Slope Lemma. Do not assume that \(\varphi \) is differentiable.
[Definition: One-sided derivative] If \(E\subseteq \R \), and if \([x,x+\delta )\subseteq E\) for some \(x\in \R \) and some \(\delta >0\), and if \(\varphi :E\to \R \), we define the right derivative of \(\varphi \) at \(x\), denoted \(\varphi '(x^+)\), to be equal to the following limit (provided that the limit exists): \begin {equation*} \varphi '(x^+)=\lim _{h\to 0^+} \frac {\varphi (x+h)-\varphi (x)}{h}. \end {equation*} If \(E\subseteq \R \), if \((x-\delta ,x]\subseteq E\) for some \(x\in \R \) and some \(\delta >0\), and if \(\varphi :E\to \R \), we define the left derivative of \(\varphi \) at \(x\), denoted \(\varphi '(x^-)\), to be equal to the following limit (provided that the limit exists): \begin {equation*} \varphi '(x^-)=\lim _{h\to 0^+} \frac {\varphi (x-h)-\varphi (x)}{-h}=\lim _{h\to 0^-} \frac {\varphi (x+h)-\varphi (x)}{h}. \end {equation*}
Lemma 6.16. Let \(\varphi \) be convex on the open interval \(I\). Then \(\varphi '(x^+)\) and \(\varphi '(x^-)\) exist for all \(x\in I\). Moreover, if \(u\), \(v\in I\) with \(u<v\), then \begin {equation*} \varphi '(u^-)\leq \varphi '(u^+)\leq \frac {\varphi (v)-\varphi (u)}{v-u}\leq \varphi '(v^-)\leq \varphi '(v^+). \end {equation*}
(Problem 2050) Prove Lemma 6.16.
Corollary 6.17. Let \(\varphi \) be convex on the open interval \(I\), where \(I\subseteq \R \) is an open interval. If \(K\subset I\) is a closed bounded interval, then \(\varphi \) is Lipschitz on \(K\).
(Problem 2060) Prove Corollary 6.17.
Theorem 6.18. Suppose that \(\varphi \) is convex on the open interval \(I\). Then \(\varphi \) is differentiable at all but countably many points, and its derivative \(\varphi '\) is an increasing function.
(Problem 2070) Prove Theorem 6.18.
Jensen’s Inequality. Let \(\varphi :\R \to \R \) be convex. Let \(f:[0,1]\to \R \) be integrable. Suppose that \(\varphi \circ f\) is also integrable over \([0,1]\). Then \begin {equation*} \varphi \biggl (\int _0^1 f\biggr )\leq \int _0^1 (\varphi \circ f). \end {equation*}
(Problem 2080) Prove Jensen’s Inequality.
Observe that \(X=\int _0^1 f\in \R \). By Lemma 6.16, we have that \(\varphi '(X^+)\) and \(\varphi '(X^-)\) exist.By Lemma 6.16, if \(y>X\), then \begin {equation*} \varphi '(X^+)\leq \frac {\varphi (y)-\varphi (X)}{y-X} \end {equation*} while if \(y<X\) then \begin {equation*} \varphi '(X^+)\geq \varphi '(X^-)\geq \frac {\varphi (y)-\varphi (X)}{y-X}. \end {equation*} In the first case \(y-X>0\), while in the second case \(y-X<0\), and so multiplying on both sides by \(y-X\) will either preserve the inequality or leave it unchanged. In either case we have that \begin {equation*} (y-X)\varphi '(X^+)\leq \varphi (y)-\varphi (X). \end {equation*} Taking \(y=f(x)\) for some \(x\in [0,1]\), we have that \begin {equation*} (f(x)-X)\varphi '(X^+)\leq \varphi (f(x))-\varphi (X). \end {equation*} We integrate both sides of the inequality. By Theorem 4.17, \begin {equation*} \int _0^1(f(x)-X)\varphi '(X^+)\,dx \leq \int _0^1\varphi (f(x))-\varphi (X)\,dx. \end {equation*} The numbers \(\varphi (X)\) and \(\varphi '(X^+)\) are constants, and thus by Theorem 4.17 may be pulled out of the integrals. Thus \begin {equation*} \varphi '(X^+) \biggl (\int _0^1 f\biggr )-X\varphi '(X^+)\leq \biggl (\int _0^1\varphi \circ f\biggr )-\varphi (X). \end {equation*} Recalling the definition of \(X\), we see that the left hand side is zero, and thus \begin {equation*} \varphi \biggl (\int _0^1 f\biggr )\leq \biggl (\int _0^1\varphi \circ f\biggr ) \end {equation*} as desired.
The Fundamental Theorem of Calculus. Suppose that \([a,b]\subset \R \) is a closed bounded interval, that \(F:[a,b]\to \R \) is continuous, that \(F'\) exists everywhere on \([a,b]\), and that \(F'\) is Riemann integrable over \([a,b]\). Then \(F(b)-F(a)=\int _a^b F'\).
Recall [Theorem 6.10]: Let \([a,b]\subseteq \mathbb {R}\) be a closed and bounded interval and let \(f:[a,b]\to \mathbb {R}\). Suppose that \(f\) is absolutely continuous.
Then \(f\) is differentiable almost everywhere on \((a,b)\), its derivative \(f'\) is integrable over \([a,b]\), and \begin {equation*} \int _a^b f'=f(b)-f(a). \end {equation*}
(Problem 2090) Give an example of a closed and bounded interval \([a,b]\) and a continuous function \(f:[a,b]\to \R \) such that \(f'\) exists almost everywhere on \([a,b]\) and is Lebesgue integrable on \([a,b]\), but such that \begin {equation*} f(b)-f(a)\neq \int _{[a,b]}f'. \end {equation*}
One such function is the Cantor function \(\Lambda \) discussed in Section 2.7.
(Problem 2091) Let \(A\), \(B\subseteq [-\infty ,\infty ]\) be two nonempty sets of extended real numbers. Suppose that \(a\leq b\) for all \(a\in A\) and all \(b\in B\). Then \(\sup A\leq \inf B\).
[Definition: Metric space] A metric space is a set \(X\) together with a function (metric) \(d:X\times X\to [0,\infty )\) such that
\(d(x,x)=0\) for all \(x\in X\),
if \(d(x,y)=0\) then \(x=y\),
\(d(x,y)=d(y,x)\) for all \(x\), \(y\in X\),
if \(x\), \(y\), \(z\in X\), then \(d(x,y)\leq d(x,z)+d(y,z)\).
[Definition: Vector space over \(\R \)] A real vector space is a set \(V\) together with two binary operations \(+:V\times V\to V\) and \(\cdot :\R \times V\to V\) such that, if \(u\), \(v\), \(w\in V\) and \(r\), \(s\in \R \), then
\(u+v=v+u\),
\((u+v)+w=u+(v+w)\),
If \(u+v=w+v\) then \(u=w\),
\(r\cdot (s\cdot v)=(rs)\cdot v\),
\(r\cdot (u+v)=r\cdot u+r\cdot v\),
\((r+s)\cdot v=r\cdot v+s\cdot v\),
\(1v=v\),
It is easy to show that \(0v=0w\) and \(v+0v\) for all \(v\), \(w\in V\); we let \(0_V=0v\) for any (all) \(v\in V\).
[Definition: Normed vector space] A normed vector space is a real vector space together with a metric \(d\) such that
\(d(v,w)=d(v-w,0)\),
\(d(rv,rw)=|r|d(v,w)\).
(We often write \(\|v\|=d(v,0)\).) We call \(d(v,0)\) the norm of \(v\).
(Problem 2092) Suppose that \(V\) is a real vector space and that \(\|\cdot \|:V\to \R \) is a function. Then \(\|\cdot \|\) is a norm on \(V\) (that is, the metric \(d(v,w)=\|v-w\|\) satisfies the above definition) if and only if
\(\|v\|\geq 0\) for all \(v\in V\),
\(\|v\|=0\) if and only if \(v=0_V\),
\(\|rv\|=|r|\,\|v\|\) for all \(r\in \R \) and all \(v\in V\),
\(\|v+w\|\leq \|v\|+\|w\|\) for all \(v\), \(w\in V\).
[Definition: Measurable and finite almost everywhere] Let \(E\subseteq \R \) be measurable and let \(\mathcal {F}\) be the set of all measurable functions \(f:E\to [-\infty ,\infty ]\) such that \(m(f^{-1}(\{-\infty ,\infty \}))=0\) (that is, \(m(f^{-1}(\R ))=m\{x\in E:f(x)\in \R \}=m(E)\)).
[Definition: Equal almost everywhere] Let \(f\), \(g\in \mathcal {F}\). We say that \(f\cong g\), or \(f=g\) almost everywhere, if \(m\{x\in E:f(x)\neq g(x)\}=0\).
(Problem 2093) Show that \(\cong \) is an equivalence relation and that, if \(f_1\cong f_2\) and \(g_1\cong g_2\), then
(Problem 2100) Suppose that \(f\), \(g\in \mathcal {F}\) and \(0<p<\infty \).
[Definition: Lebesgue (semi, quasi)-norm] Let \(0<p<\infty \) and let \(E\subseteq \R \) be measurable. If \(f\in \mathcal {F}\), we define \begin {equation*} \|f\|_{p}=\|f\|_{L^p(E)} = \biggl (\int _E |f|^p\biggr )^{1/p}. \end {equation*}
(Problem 2110) Let \(f:E\to [-\infty ,\infty ]\). Show that \begin {align*} \alignbreak \min \{M\in [0,\infty ]:|f(x)|\leq M\text { for almost every }x\in E\} \alignbreaknoand &= \sup \{\lambda \in [0,\infty ]:m\{x\in E:|f(x)|>\lambda \}>0\} \\&= \sup \{\lambda \in [0,\infty ]:m\{x\in E:|f(x)|\geq \lambda \}>0\} . \end {align*}
In particular, show that the first set contains its infimum.
First, if \(m\{x\in E:|f(x)|>\lambda \}>0\) then \(m\{x\in E:|f(x)|\geq \lambda \}>0\), and so \begin {equation*} \sup \{\lambda \in [0,\infty ]:m\{x\in E:|f(x)|>\lambda \}>0\} \leq \sup \{\lambda \in [0,\infty ]:m\{x\in E:|f(x)|\geq \lambda \}>0\} . \end {equation*} Now, suppose that \(m\{x\in E:|f(x)|\geq \lambda \}>0\). If \(M<\lambda \), then \(|f(x)|\geq \lambda >M\) on a set of positive measure, and so \(M\notin \{M\in [0,\infty ]:|f(x)|\leq M\text { for almost every }x\in E\}\). Thus, if \(M\in \{M\in [0,\infty ]:|f(x)|\leq M\text { for almost every }x\in E\}\), then \(\lambda \leq M\). By Problem 2091, this implies that \begin {equation*} \sup \{\lambda \in [0,\infty ]:m\{x\in E:|f(x)|\geq \lambda \}>0\} \leq \inf \{M\in [0,\infty ]:|f(x)|\leq M\text { for almost every }x\in E\}. \end {equation*}Define \begin {equation*} \mu =\inf \{M\in [0,\infty ]:|f(x)|\leq M\text { for almost every }x\in E\}. \end {equation*} We now seek to show that \(\mu \in \{M\in [0,\infty ]:|f(x)|\leq M\text { for almost every }x\in E\}\) and so the infimum is a minimum. If \(\mu =\infty \), then \(\mu \) is clearly an element of \(\{M\in [0,\infty ]:|f(x)|\leq M\text { for almost every }x\in E\}\). Otherwise, suppose that \(\mu \) is finite; by definition \(\mu \geq 0\). If \(n\in \N \), then \(\mu +\frac {1}{n}>\inf \{M\in [0,\infty ]:|f(x)|\leq M\text { for almost every }x\in E\}\) and so there is a \(M\in \{M\in [0,\infty ]:|f(x)|\leq M\text { for almost every }x\in E\}\) with \(M<\mu +\frac {1}{n}\). Let \(E_n=\{x\in E:|f(x)|>\mu +\frac {1}{n}\}\). Then \(E_n\subset \{x\in E:|f(x)|\geq M\}\), which has measure zero. Thus \(m(E_n)=0\). But \begin {equation*} \{x\in E:|f(x)|>\mu \}=\bigcup _{n=1}^\infty E_n \end {equation*} and so has measure zero, and so \(|f|\leq \mu \) almost everywhere in \(E\). Thus \(\mu \in \{M\in [0,\infty ]:|f(x)|\leq M\text { for almost every }x\in E\}\) and so \begin {equation*} \min \{M\in [0,\infty ]:|f(x)|\leq M\text { for almost every }x\in E\}=\inf \{M\in [0,\infty ]:|f(x)|\leq M\text { for almost every }x\in E\} \end {equation*} and in particular exists.
Finally, let \(\mu _n=\mu -\frac {1}{n}\) (if \(\mu \) is finite) or \(\mu _n=n\) (if \(\mu \) is infinite). Then \(\sup \{\mu _n:n\in \N \}=\mu \). Our goal now is to show that each \(\mu _n\) is in \(\{\lambda \in [0,\infty ]:m\{x\in E:|f(x)|>\lambda \}>0\}\). To see this, observe that \(\mu _n<\mu =\inf \{M\in [0,\infty ]:|f(x)|\leq M\text { for almost every }x\in E\}\) and so \(\mu _n\notin \{M\in [0,\infty ]:|f(x)|\leq M\text { for almost every }x\in E\}\). Thus \(|f(x)|\leq \mu _n\) is not true almost everywhere, and so \(\{x\in E:|f(x)|>\mu _n\}\) has positive measure. Thus \(\mu _n\in \{\lambda \in [0,\infty ]:m\{x\in E:|f(x)|>\lambda \}>0\}\) for all \(n\in \N \), and so \(\mu =\sup \{\mu _n:n\in \N \}\leq \sup \{\lambda \in [0,\infty ]:m\{x\in E:|f(x)|>\lambda \}>0\}\), as desired.
Thus \begin {align*} \mu &\leq \sup \{\lambda \in [0,\infty ]:m\{x\in E:|f(x)|>\lambda \}>0\} \\&\leq \sup \{\lambda \in [0,\infty ]:m\{x\in E:|f(x)|\geq \lambda \}>0\} \\&\leq \hskip \dimen 2\inf \{M\in [0,\infty ]:|f(x)|\leq M\text { for almost every }x\in E\} \\&=\min \{M\in [0,\infty ]:|f(x)|\leq M\text { for almost every }x\in E\} \alignbreak =\mu \end {align*}
and so all of the above numbers must be equal.
[Definition: Essential supremum] Let \(E\subseteq \R \) be measurable and let \(f:E\to \R \) be measurable. We define \begin {equation*} \esssup _E f=\min \{M\in [0,\infty ]:|f(x)|\leq M\text { for almost every }x\in E\}. \end {equation*}
(Problem 2120) Let \(E\subset \R \) be measurable with \(m(E)<\infty \) and let \(f:E\to \R \) be bounded. Show that \begin {equation*} \lim _{p\to \infty } \|f\|_{L^p(E)} = \esssup _E |f|. \end {equation*}
(Problem 2130) Let \(E\subset \R \) be measurable with \(m(E)<\infty \) and let \(f:E\to \R \) be bounded. Show that \begin {equation*} \lim _{p\to 0^+} \|f\|_{L^p(E)}^p = m\{x\in E:f(x)\neq 0\}. \end {equation*}
[Definition: \(L^\infty \)] If \(f\in \mathcal {F}\), we let \begin {equation*} \|f\|_{L^\infty (E)}=\esssup _E |f|. \end {equation*}
[Definition: Lebesgue space] Let \(\mathcal {F}/\cong \) be the set of equivalent classes of \(\mathcal {F}\) modulo the equivalence relation \(\cong \). If \(0<p\leq \infty \), we let \(L^p(E)\) be the set of all elements \(f\) of \(\mathcal {F}/\cong \) such that some representative \(\tilde f:E\to [-\infty ,\infty ]\) satisfies \(\|\tilde f\|_{L^p(E)}<\infty \). We let \(\|f\|_{L^p(E)}=\|\tilde f\|_{L^p(E)}\).
(Problem 2131) If \(f\in L^p(E)\), show that \(\|f\|_{L^p(E)}\) is well defined.
[Definition: Sequence space] If \(0<p\leq \infty \), we let \(\ell ^p\) be the space of all sequences of real numbers \(\{a_n\}_{n=1}^\infty \) such that the \(\ell ^p\) norm \begin {equation*} \|\{a_n\}_{n=1}^\infty \|_{\ell ^p} = \begin {cases}\Bigl (\sum _{n=1}^\infty |a_n|^p\Bigr )^{1/p}, & 0<p<\infty ,\\ \sup _{n\in \N } |a_n|,&p=\infty \end {cases} \end {equation*} is finite.
(Problem 2132) Let \(0<p\leq \infty \) and let \(E\subseteq \R \) be measurable. Show that:
\(\|f\|_{L^p(E)}\geq 0\) for all \(f\in L^p(E)\),
\(\|f\|_{L^p(E)}=0\) if and only if \(f=0\) as an element of \(L^p(E)\),
\(\|rf\|_{L^p(E)}=|r|\,\|f\|_{L^p(E)}\) for all \(r\in \R \) and all \(f\in L^p(E)\).
(Problem 2133) Let \(0<p\leq \infty \). Show that:
\(\|\{a_n\}_{n=1}^\infty \|_{\ell ^p}\geq 0\) for all \(\{a_n\}_{n=1}^\infty \in \ell ^p\),
\(\|\{a_n\}_{n=1}^\infty \|_{\ell ^p}=0\) if and only if \(a_n=0\) for all \(n\in \N \),
\(\|\{ra_n\}_{n=1}^\infty \|_{\ell ^p}=|r|\,\|\{a_n\}_{n=1}^\infty \|_{\ell ^p}\) for all \(r\in \R \) and all \(\{a_n\}_{n=1}^\infty \in \ell ^p\).
(Problem 2134) Show that \(L^1(E)\) and \(L^\infty (E)\) are normed vector spaces.
(Problem 2135) Show that \(\ell ^1\) and \(\ell ^\infty \) are normed vector spaces.
[Definition: Conjugate] If \(p\in [1,\infty ]\), we define the conjugate \(q\) of \(p\) to be the unique number in \([1,\infty ]\) that satisfies \begin {equation*} \frac {1}{p}+\frac {1}{q}=1. \end {equation*}
(Problem 2136) Show that \(q=\frac {p}{p-1}\).
(Problem 2137) Show that \(pq=p+q\).
Young’s Inequality. If \(1<p<\infty \), \(0\leq a<\infty \), and \(0\leq b<\infty \), then \begin {equation*} ab\leq \frac {a^p}{p}+\frac {b^q}{q}. \end {equation*}
(Problem 2140) Prove Young’s Inequality.
The statement is clearly true if \(a=0\) or \(b=0\), so we assume that \(a>0\) and \(b>0\). Fix some such \(b\) and let \(g:(0,\infty )\to \R \) be given by \(g(x)=x^p/p+b^q/q-xb\). Then \(g'(x)=x^{p-1}-b\) for all \(x\in (0,\infty )\), and so the only critical point of \(g\) (the only value of \(x\) for which \(g'(x)=0\)) is \(x=b^{q/p}\). \(g''(x)=(p-1)x^{p-2}\), which is positive for all \(x\in (0,\infty )\) because \(p>1\), and so \(g\) must have a local minimum at \(x=b^{q/p}\); recall from calculus that functions defined on an interval with a single local extremum necessarily has a global extremum at that point. Thus \(g(x)\geq g(b^{q/p})=b^{q}/p+b^q/q-b^{q/p+1}=0\). In particular, \(g(a)\geq 0\) and so \(\frac {a^p}{p}+\frac {b^q}{q}\geq ab\).
Hölder’s inequality. Let \(E\subseteq \R \) be measurable, \(1\leq p\leq \infty \), and \(q\) be the conjugate of \(p\). If \(f\in L^p(E)\), \(g\in L^q(E)\), then \(fg\in L^1(E)\) and \begin {equation*} \int _E |fg|\leq \|f\|_p\,\|g\|_q. \end {equation*}
(Problem 2150) Prove Hölder’s inequality in the case \(p=1\) or \(p=\infty \).
(Problem 2160) Prove Hölder’s inequality if \(1<p<\infty \).
If \(\|f\|_p=0\) or \(\|g\|_q=0\), then \(f\) or \(g\), respectively, is zero almost everywhere; therefore \(fg\) is zero almost everywhere and so \(\int _E |fg|=0=\|f\|_p\,\|g\|_q\). We therefore assume that \(\|f\|_p>0\) and \(\|g\|_q>0\).Let \(x\in E\). Suppose that \(f(x)\) and \(g(x)\) are finite; this is true for almost every \(x\in E\) by definition of \(L^p(E)\).
Then by Young’s inequality \begin {align*} |f(x)g(x)|&=\|f\|_p\|g\|_q \biggl (\frac {|f(x)|}{\|f\|_p}\biggr ) \biggl (\frac {|g(x)|}{\|g\|_q}\biggr ) \alignbreak \leq \|f\|_p\|g\|_q \biggl (\frac {1}{p\|f\|_p^p}|f(x)|^p+\frac {1}{q\|g\|_q^q}|g(x)|^q\biggr ). \end {align*}
By definition of \(L^p(E)\), \(|f|^p\) and \(|g|^q\) are integrable, and so the right hand side is integrable. Furthermore, \(f\) and \(g\) are measurable, so \(|fg|\) is also measurable, and by Theorem 4.10 we have that \begin {equation*} \int _E |fg|\leq \|f\|_p\|g\|_q \biggl (\frac {1}{p\|f\|_p^p}\int _E|f(x)|^p\,dx+\frac {1}{q\|g\|_q^q} \int _E|g(x)|^q\,dx\biggr ). \end {equation*} Recalling the definitions of \(q\) and \(\|f\|_p\), \(\|g\|_q\) completes the proof.
Theorem 7.1. If \(E\subseteq \R \) is measurable, \(1\leq p<\infty \), and \(f\in L^p(E)\), then either \(f=0\) or the function \(f^*\) given by \begin {equation*} f^*(x)=\begin {cases} \frac {1}{\|f\|_p^{p-1}} |f(x)|^{p-2}\,f(x), & f(x)\neq 0,\\0,&f(x)=0\end {cases} \end {equation*} satisfies \(f^*\in L^q(E)\), \(\|f^*\|_q=1\) and \begin {equation*} \int _E f\,f^*=\int _E |f\,f^*|= \|f\|_p. \end {equation*}
(Problem 2170) Prove Theorem 7.1.
Minkowski’s Inequality. Let \(E\subseteq \R \) be measurable and \(1\leq p\leq \infty \). If \(f\), \(g\in L^p(E)\), then \(f+g\in L^p(E)\) and \begin {equation*} \|f+g\|_p\leq \|f\|_p+\|g\|_p. \end {equation*}
(Problem 2180) Prove Minkowski’s Inequality in the cases \(p=1\) and \(p=\infty \).
(Problem 2190) Prove Minkowski’s Inequality in the case \(1<p<\infty \).
First observe that \(|f(x)+g(x)|\leq 2\max (|f(x)|,|g(x)|)\) and so \(|f(x)+g(x)|^p\leq 2^p\max (|f(x)|^p,|g(x)|^p)\leq 2^p |f(x)|^p+2^p|g(x)|^p\). Thus \begin {equation*} \int _E |f+g|^p \leq 2^p\int _E f^p+\int _E g^p \end {equation*} and so \(f+g\in L^p(E)\) with \(\|f+g\|_p\leq 2(\|f\|_p^p+\|g\|_p^p)^{1/p}\). The inclusion \(f+g\in L^p(E)\) is useful but we will need a better bound on \(\|f+g\|_p\).If \(f+g=0\) then there is nothing to prove. Otherwise, let \(h=(f+g)^*\), where \((f+g)^*\) is as in Theorem 7.1. Then \(h\in L^q(E)\), \(\|h\|_q=1\), and \(\|f+g\|_p=\int _E (f+g)h\).
But \begin {equation*} \int _E (f+g)h=\int _E fh+\int _E gh \leq \|f\|_p\|h\|_q+\|g\|_p\|h\|_q=\|f\|_p+\|g\|_p \end {equation*} by Hölder’s inequality. This completes the proof.
Corollary 7.2. Let \(E\subseteq \R \) be measuable and let \(1<p<\infty \). Let \(\mathcal {F}\subset L^p(E)\) and suppose that there is a \(M\in \R \) such that \(\sup _{f\in \mathcal {F}}\|f\|_p\leq M\). Then \(\mathcal {F}\) is uniformly integrable.
(Problem 2200) Prove Corollary 7.2.
Corollary 7.3. Let \(E\subseteq \R \) be measuable with \(m(E)<\infty \) and let \(1\leq p_1<p_2\leq \infty \). Then \(L^{p_2}(E)\subseteq L^{p_1}(E)\). Furthermore, if \(f\in L^{p_2}(E)\), then \begin {equation*} \|f\|_{p_1}\leq m(E)^{1/p_1-1/p_2}\|f\|_{p_2}. \end {equation*}
(Problem 2210) Prove Corollary 7.3.
(Problem 2220) Let \(1\leq p_1<p_2\leq \infty \). Show that if \(\{a_n\}_{n=1}^\infty \in \ell ^{p_1}\), then \begin {equation*} \|\{a_n\}_{n=1}^\infty \|_{\ell ^{p_2}}\leq \|\{a_n\}_{n=1}^\infty \|_{\ell ^{p_1}} \end {equation*} and so \(\ell ^{p_1}\subseteq \ell ^{p_2}\).
[Definition: Convergent sequence] Let \((X,d)\) be a metric space. If \(x\in X\) and \(x_n\in X\) for each \(n\in \N \), we say that \(x_n\) converges to \(x\) if \(\lim _{n\to \infty } d(x,x_n)=0\). In particular, if \((V,\|\,\|)\) is a normed linear space, \(v\in V\), and \(v_n\in V\) for each \(n\in \N \), we say that \(v_n\) converges to \(v\) if \(\lim _{n\to \infty } \|v-v_n\|=0\). If such an \(x\) (or \(v\)) exists, we call \(\{x_n\}_{n=1}^\infty \) (or \(\{v_n\}_{n=1}^\infty \)) a convergent sequence.
[Definition: Cauchy sequence] Let \((X,d)\) be a metric space. If \(x_n\in X\) for each \(n\in \N \), we say that \(\{x_n\}_{n=1}^\infty \) is a Cauchy sequence if \(\lim _{n\to \infty } \sup _{k\geq n} d(x_k,x_n)=0\). In particular, if \((V,\|\,\|)\) is a normed linear space, and \(v_n\in V\) for each \(n\in \N \), we say that \(\{v_n\}_{n=1}^\infty \) is a Cauchy sequence if \(\lim _{n\to \infty } \sup _{k\geq n} \|v_k-v_n\|=0\).
Proposition 7.4. Every convergent sequence is Cauchy. Every Cauchy sequence with a convergent subsequence is convergent.
(Problem 2221) Give an example of a normed vector space in which not all Cauchy sequences are convergent.
(Problem 2230) Give an example of a sequence of functions \(\{f_n\}_{n=1}^\infty \) such that each \(f_n\) is in \(L^2(\R )\cap L^3(\R )\), such that \(\{f_n\}_{n=1}^\infty \) is convergent in the \(L^2(\R )\)-norm, but such that \(\{f_n\}_{n=1}^\infty \) is not convergent in the \(L^3(\R )\)-norm.
The Riesz-Fischer Theorem. Let \(E\subseteq \R \) be measurable and let \(1\leq p\leq \infty \). Then \(L^p(E)\) is complete in the sense that every Cauchy sequence in \(L^p(E)\) is convergent.
Moreover, if \(f_n\to f\) in \(L^p(E)\), then there is a subsequence \(\{f_{n_k}\}_{k=1}^\infty \) such that \(f_{n_k}\to f\) pointwise almost everywhere.
[Chapter 7, Problem 33] The Riesz-Fischer theorem is true if \(p=\infty \). (We will prove the Riesz-Fischer theorem in the case \(1\leq p<\infty \) in Problem 2260 below; the next few theorems are important steps in the proof.)
[Definition: Rapidly Cauchy] Let \((V,\|\,\|)\) be a normed linear space and let \(\{v_n\}_{n=1}^\infty \) be a sequence of elements of \(V\). We say that \(\{v_n\}_{n=1}^\infty \) is rapidly Cauchy if \begin {equation*} \sum _{k=1}^\infty \sqrt {\|v_{n+1}-v_n\|}<\infty . \end {equation*}
Proposition 7.5. Every rapidly Cauchy sequence (in any normed vector space) is Cauchy. Every Cauchy sequence (in any normed vector space) has a rapidly Cauchy subsequence.
(Problem 2240) Prove Proposition 7.5.
Theorem 7.6. Let \(E\subseteq \R \) be measurable and let \(1\leq p<\infty \). Let \(\{f_n\}_{n=1}^\infty \) be a rapidly Cauchy sequence in \(L^p(E)\).
Then there exists a \(f\in L^p(E)\) such that \(f_n\to f\) in \(L^p(E)\) and such that \(f_n(x)\to f(x)\) in \(\R \) for almost every \(x\in E\).
(Problem 2250) Begin the proof of Theorem 7.6 by showing that, if \(1\leq p<\infty \) and \(\{f_n\}_{n=1}^\infty \) is a rapidly Cauchy sequence in \(L^p(E)\), then \(\{f_n(x)\}_{n=1}^\infty \) is a Cauchy sequence of real numbers for almost every \(x\in E\). Hint: Use the Borel-Cantelli Lemma.
(Problem 2260) Prove Theorem 7.6. Then use Theorem 7.6 to prove the Riesz-Fischer Theorem.
(Problem 2261) Let \(f_n(x)=\frac {1}{1+(x-n)^2}\). Then \(f_n\to 0\) pointwise everywhere but \(f_n\not \to 0\) in \(L^p(\R )\).
(Problem 2262) Let \(f_n(x)=\frac {n^{3/p}}{1+n^6(x-1/n)^2}\). Then \(f_n\to 0\) pointwise everywhere but \(f_n\not \to 0\) in \(L^p([0,1])\).
Theorem 7.7. Let \(E\subseteq \R \) be measurable and let \(1\leq p<\infty \). Suppose that \(\{f_n\}_{n=1}^\infty \) is a sequence in \(L^p(E)\) that converges pointwise almost everywhere on \(E\) to some function \(f:E\to \R \). Then \(f_n\to f\) in \(L^p(E)\) if and only if \(\|f_n\|_p\to \|f\|_p\) in \(\R \).
(Problem 2270) Prove Theorem 7.7.
(Problem 2280) Let \(E\subseteq \R \) be measurable and let \(1\leq p<\infty \). Let \(\varphi \) be a simple function defined on \(E\). Show that \(\varphi \in L^p(E)\) if and only if \(\varphi \) is finitely supported, that is, if \(m(\{x\in E:\varphi (x)\neq 0\})<\infty \).
[Definition: Dense] A subset \(Y\) of a metric space \((X,d)\) is dense if, for all \(x\in X\) and all \(r>0\), there is a \(y\in Y\) with \(d(x,y)<r\). In particular, a subset \(A\) of a normed vector space \((V,\|\,\|)\) is dense if, for every \(v\in V\) and every \(r>0\), there is an \(a\in A\) with \(\|v-a\|<r\).
[Chapter 7, Problem 36] A subset \(Y\) of a metric space \((X,d)\) is dense if and only if, whenever \(x\in X\), there is a sequence \(\{y_n\}_{n=1}^\infty \) of points in \(y\) with \(y_n\to x\).
[Chapter 7, Problem 37] Let \((X,d)\) be a metric space and let \(Z\subseteq Y\subseteq X\). If \(Y\) is dense in \((X,d)\), and if \(Z\) is dense in the subspace \((Y,d)\), then \(Z\) is also dense in \((X,d)\).
[Definition: Step function] We say that \(\varphi :\R \to \R \) is a step function if there are finitely many points \(x_0<x_1<\dots <x_n\) such that \(\varphi \) is constant on \((-\infty ,x_0)\), on \((x_n,\infty )\), and on each of the intervals \((x_{k-1},x_k)\) for all \(1\leq k\leq n\). We say that \(\varphi \) is a compactly supported step function if in addition \(\varphi =0\) on \((-\infty ,x_0)\) and on \((x_n,\infty )\).
The set of finitely supported simple functions is dense in \(L^1(\R )\).
The set of compactly supported step functions is dense in \(L^1(\R )\).
The set of compactly supported continuous functions is dense in \(L^1(\R )\).
Proposition 7.9. Let \(E\subseteq \R \) be measurable and let \(1\leq p\leq \infty \).
If \(1\leq p<\infty \), then the set of finitely supported simple functions is dense in \(L^p(E)\).
If \(p=\infty \), then the set of simple functions is dense in \(E\).
(Problem 2290) Prove Proposition 7.9.
Proposition 7.10. The set of compactly supported step functions is dense in \(L^p(\R )\) for \(1\leq p<\infty \).
(Problem 2300) Prove Proposition 7.10.
Theorem 7.12. The set of compactly supported continuous functions is dense in \(L^p(\R )\) for all \(1\leq p<\infty \) (and therefore is dense in \(L^p(E)\) for any \(1\leq p<\infty \) and for any \(E\subseteq \R \) measurable).
(Problem 2310) Prove Theorem 7.12.
Theorem 7.11. The set of compactly supported step functions that take rational values and have discontinuities at rational points is dense \(L^p(\R )\) for any \(1\leq p<\infty \). Therefore, \(L^p(E)\) is separable for \(1\leq p<\infty \) and for any \(E\subseteq \R \) measurable.
(Ashley, Problem 2320) Prove Theorem 7.11.
Let \(\mathcal {P}\) be the set of all compactly supported step functions that take rational values and have discontinuities at rational points. We must show that \(\mathcal {P}\) is countable and that \(\mathcal {P}\) is dense in \(L^p(\R )\).
To see that \(\mathcal {P}\) is countable, recall (Problem 240) that \(\Q \) is countable. We recall from undergraduate analysis that the Cartesian product of finitely many countable sets is countable. Thus \(\mathcal {Q}^{3n}\) is countable for any \(n\in \N \). Recall (Problem 250) that the countable union of countable sets is countable, and so \(\bigcup _{n=1}^\infty \mathcal {Q}^{3n}\) is countable.
Define \(\Psi :\Q ^{3n}\to L^p(\R )\) by \(\Psi ((q_{1},q_2,\dots ,q_{3n}))=\sum _{k=1}^n q_{3k} \chi _{[q_{3k-1},q_{3k-2}]}\) (where we define \([p,r]=\emptyset \) if \(p>r\)). Then \(\Psi ((q_{1},q_2,\dots ,q_{3n}))\) is a step function which takes rational values and has discontinuities at rational points. Furthermore, every such step function \(\varphi \) is equal to \(\Psi ((q_{1},q_2,\dots ,q_{3n}))\) for some \((q_{1},q_2,\dots ,q_{3n})\in \bigcup _{n=1}^\infty \mathcal {Q}^{3n}\). Thus \(\Psi \) is surjective. By countability of \(\bigcup _{n=1}^\infty \mathcal {Q}^{3n}\) and Problem 221 (the countable analogue to Problem 200), this implies that \(\mathcal {P}\) is countable.
It remains to show density of \(\mathcal {P}\). Let \(\mathcal {S}\) denote the set of all compactly supported step functions; recall that \(\mathcal {S}\) is dense in \(L^p(\R )\) for all \(1\leq p<\infty \). Thus by Problem 7.37 it suffices to show that \(\mathcal {P}\) is dense in \(\mathcal {S}\).
Let \(f\in \mathcal {S}\) and let \(\varepsilon >0\). We then have that \begin {equation*} f=\sum _{k=1}^n c_k\chi _{[a_k,b_k]} \end {equation*} for some finite integer \(n\) and some real numbers \(a_k\), \(b_k\), \(c_k\) with \(a_k\leq b_k\). (Allowing \(a_k=b_k\) allows us to redefine \(f\) at finitely many points, in particular, at the finitely many points of discontinuity of \(f\).)
Because \(n\) is finite, we have that \(C=\max _{1\leq k\leq n} |c_k|\) and \(L=\max _{1\leq k\leq n} (b_k-a_k)\) are also finite. Because \(\mathcal {Q}\) is dense in \(\mathcal {R}\), there are numbers \(p_k\), \(q_k\), \(r_k\in \Q \) with
\(|q_k-c_k|<\min \bigl (C,\frac {\varepsilon }{1+3n L^{1/p}}\bigr )\),
\(a_k-\frac {\varepsilon ^p}{1+6^pC^p}<p_k< a_k\),
\(b_k< r_k<b_k+\frac {\varepsilon ^p}{1+6^pC^p}\).
Let \(g=\sum _{k=1}^n q_k\chi _{[p_k,r_k]}\). Because \(p_k< a_k\) and \(r_k< b_k\), we have that \([p_k,r_k]=[p_k,a_k)\cup [a_k,b_k]\cup (b_k,r_k]\) and the three intervals on the right hand side are pairwise disjoint, so \(\chi _{[p_k,r_k]}=\chi _{[p_k,a_k)}+\chi _{[a_k,b_k]}+\chi _ {(b_k,r_k]}\). Thus \begin {align*} |g-f| &=\Bigl |\sum _{k=1}^n (q_k\chi _{[p_k,r_k]}-c_k\chi _{[a_k,b_k]}) \Bigr | \\&=\Bigl |\sum _{k=1}^n ((q_k-c_k)\chi _{[a_k,b_k]}+q_k\chi _{(b_k,r_k]}+q_k\chi _{[p_k,a_k)}) \Bigr | \\&\leq \sum _{k=1}^n |q_k-c_k|\chi _{[a_k,b_k]} +\sum _{k=1}^n |q_k|\chi _{(b_k,r_k]} +\sum _{k=1}^n |q_k|\chi _{[p_k,a_k)}). \end {align*}
We compute \begin {align*} \Bigl \|\sum _{k=1}^n |q_k-c_k|\chi _{[a_k,b_k]}\Bigr \|_p &\leq \sum _{k=1}^n |q_k-c_k|\|\chi _{[a_k,b_k]}\|_p \alignbreak = \sum _{k=1}^n |q_k-c_k|(b_k-a_k)^{1/p} \\&\leq \sum _{k=1}^n \frac {\varepsilon }{1+3n L^{1/p}}L^{1/p} \\&<\frac {\varepsilon }{3} ,\\\noalign {\allowbreak } \Bigl \|\sum _{k=1}^n |q_k|\chi _{(b_k,r_k]}\Bigr \|_p &\leq \sum _{k=1}^n |q_k|\|\chi _{(b_k,r_k]}\|_p \alignbreak = \sum _{k=1}^n |q_k|(r_k-b_k)^{1/p} \\&\leq \sum _{k=1}^n 2C\frac {\varepsilon }{(1+6^pC^p)^{1/p}} \\&<\frac {\varepsilon }{3} ,\\\noalign {\allowbreak } \Bigl \|\sum _{k=1}^n |q_k|\chi _{(b_k,r_k]}\Bigr \|_p &\leq \sum _{k=1}^n |q_k|\|\chi _{[p_k,a_k)}\|_p \alignbreak = \sum _{k=1}^n |q_k|(a_k-p_k)^{1/p} \\&\leq \sum _{k=1}^n 2C\frac {\varepsilon }{(1+6^pC^p)^{1/p}} \\&<\frac {\varepsilon }{3} \end {align*}
and so by Minkowski’s Inequality, \(\|g-f\|<\varepsilon \). Thus \(\mathcal {P}\) is dense in \(\mathcal {S}\), as desired.
(Bashar, Problem 2330) Show that \(\ell ^\infty \) is not separable.
If \(S\subseteq \N \), let \(1_S\) denote the sequence \(\{a_{S,n}\}_{n=1}^\infty \) that satisfies \(a_{S,n}=1\) if \(n\in S\) and \(a_{S,n}=0\) if \(n\notin S\).Observe that if \(S\subseteq \N \), then \(\|1_S\|_{\ell ^\infty }\leq 1\) (it equals zero if \(S=\emptyset \) and equals \(1\) if \(S\neq \emptyset \)). Furthermore, if \(S\), \(T\subseteq \N \) and \(S\neq T\), then either \(S\setminus T\) or \(T\setminus S\) must be nonempty. Let \(n\in (S\setminus T)\cup (T\setminus S)\). Then \(\|1_S-1_T\|_{\ell ^\infty }\geq |a_{S,n}-a_{T,n}|\). But either \(n\in S\setminus T\) and so \(a_{S,n}-a_{T,n}=1-0=1\) or \(n\in T\setminus S\) and so \(a_{S,n}-a_{T,n}=0-1=-1\). In either case \(\|1_S-1_T\|_{\ell ^\infty }\geq 1\).
Let \(B\subseteq \ell ^\infty \) be dense. If \(b\in B\), then by the triangle inequality there is at most one \(S\subseteq \N \) with \(\|b-1_S\|<\frac {1}{2}\). Define the mapping \(f:B\to 2^{\N }\) (where \(2^\N \) is as in Problem 270) by \(f(b)=S\) if \(\|b-1_S\|<\frac {1}{2}\) and \(f(b)=\emptyset \) if \(\|b-1_S\|\geq \frac {1}{2}\) for all \(S\subseteq \N \). Then \(f\) is well defined. Furthermore, by definition of dense, for each \(S\subseteq \N \) there must be at least one \(b_S\in B\cap B(1_S,1/2)\) and so at least one \(b_S\in B\) with \(f(b_S)=S\). Thus \(f\) is surjective \(f:B\to 2^\N \). But by Problem 270 the set \(2^\N \) is uncountable, and so \(B\) must also be uncountable. Thus no dense subset of \(\ell ^\infty \) can be countable, as desired.
(Dibyendu, Problem 2340) Let \(E\subseteq \R \) be measurable and suppose that \(m(E)>0\). Show that \(L^\infty (E)\) is not separable.
[Definition: Linear functional] If \(X\) is a real vector space, a linear functional on \(X\) is a function \(T:X\to \R \) that satisfies \begin {equation*} T(\alpha x+\beta y) =\alpha T(x)+\beta T(y) \end {equation*} for all \(x\), \(y\in X\) and all \(\alpha \), \(\beta \in \R \).
(Problem 2341) Let \(X=\R ^n\). Show that, if \(v\in \R ^n\), then \(T_v(x)=\langle v,x\rangle \) is a linear functional on \(X\), where \(\langle \,,\,\rangle \) denotes the usual inner product.
(Irina, Problem 2350) Let \(X=\R ^n\). Let \(T:X\to \R \) be a linear functional. Show that there is a \(v\in \R ^n\) such that \(T(x)=\langle v,x\rangle \) for all \(x\in X\).
[Definition: Bounded linear functional; norm of a linear functional] Let \(X\) be a normed real vector space and let \(T:X\to \R \) be a linear functional. If \begin {equation*} \sup _{x\in X\setminus \{0\}}\frac {|T(x)|}{\|x\|}<\infty \end {equation*} then we say that \(T\) is a bounded linear functional. We call \begin {equation*} \|T\|_*=\sup _{x\in X\setminus \{0\}}\frac {|T(x)|}{\|x\|} \end {equation*} the norm of \(T\).
(Problem 2351) Let \(T\) be a linear functional. Show that \begin {align*} \sup _{x\in X\setminus \{0\}}\frac {|T(x)|}{\|x\|} &=\sup _{\substack {x\in X\\ \|x\|=1}} |T(x)| =\sup _{\substack {x\in X\\ \|x\|<1}} |T(x)| =\sup _{\substack {x\in X\\ \|x\|\leq 1}} |T(x)| \\&=\inf \{M\in \R :|T(x)|\leq M\|x\|\text { for all }x\in X\} \end {align*}
where in the last expression we have that the infimum of the empty set is infinity, that is, \(\{M\in \R :|T(x)|\leq M\|x\|\text { for all }x\in X\}\) is empty if and only if the four suprema are infinite.
(Problem 2352) Give an example of a real vector space and a linear functional that is not bounded.
(Micah, Problem 2360) Let \(X\) be a normed real vector space (and therefore a metric space). Let \(T:X\to \R \) be a linear functional. Show that \(T\) is continuous if and only if \(T\) is bounded.
Proposition 8.3. Let \(X\) be a normed real vector space and let \(S\), \(T:X\to \R \) be two bounded linear functionals. Suppose that \(\{x\in X:T(x)=S(x)\}\) is dense in \(X\). Then \(T(x)=S(x)\) for all \(x\in X\).
[Definition: Dual space] Let \(X\) be a normed real vector space. Then the dual space \(X^*\) is the set of all bounded linear functionals on \(X\).
[Definition: Linear combinations of linear functions] If \(r\in \R \) and \(S\), \(T\) are linear functionals on a common vector space \(X\), then \(rT\) and \(S+T\) are also linear functionals on \(X\) defined by \begin {equation*} (rT)(x)=r(T(x)),\quad (S+T)(x)=S(x)+T(x). \end {equation*}
Proposition 8.1. Let \(X\) be a normed real vector space. Then \(X^*\) is also a normed real vector space (with the norm in \(X^*\) given by \(\|\,\|_*\)).
Proposition 8.2. Let \(E\subseteq \R \) be measurable, \(1\leq p\leq \infty \), \(1/p+1/q=1\), and \(g\in L^q(E)\) (with \(q=\infty \) if \(p=1\)). Define \(T_g\) by \begin {equation*} T_g(f)=\int _E fg. \end {equation*} Then \(T_g\in (L^p(E))^*\) with \(\|T_g\|_*=\|g\|_q\).
(Muhammad, Problem 2370) Prove Proposition 8.2 in the case \(p>1\).
By Hölder’s inequality, if \(1\leq p\leq \infty \), \(f\in L^p(E)\), and \(g\in L^q(E)\), then \(fg\in L^1(E)\) and \begin {equation*} \int _E |fg|\leq \|f\|_p\|g\|_q \end {equation*} and so by Proposition 4.16, \begin {equation*} |T_g(f)|=\biggl |\int _E fg\biggr |\leq \|f\|_p\|g\|_q. \end {equation*} Thus \(T_g\) is a bounded linear operator with \(\|T_g\|_*\leq \|g\|_q\).Suppose that \(1<p\leq \infty \) so \(1\leq q<\infty \). If \(g^*\) is as in Theorem 7.1 with \(f\) replaced by \(g\) and with \(p\) and \(q\) interchanged, then \(g^*\in L^p(E)\), \(\|g^*\|_p=1\) and \begin {equation*} \|g\|_q=\int _E g^*\,g = T_g(g^*) \end {equation*} and so \(\|T_g\|_*=\sup _{f\in L^p(E),\>\|f\|_p=1} |T_g(f)|\geq |T_g(g^*)|=\|g\|_q\). Combining with the above estimate yields that \(\|T_g\|_*= \|g\|_q\) if \(p>1\).
(Problem 2371) Prove Proposition 8.2 in the case \(p=1\).
If \(p=1\) then \(q=\infty \). By the above argument \(\|T_g\|_*\leq \|g\|_q=\|g\|_\infty \).If \(\|g\|_\infty =0\) then clearly \(\|T_g\|_*\geq \|g\|_\infty \) and we are done.
Suppose \(\|g\|_\infty >0\). Let \(0<\delta <\|g\|_\infty \) and let \(F_\delta =\{x\in E: |g(x)|>\|g\|_\infty -\delta \}\). Because \(\|g\|_\infty -\delta <\|g\|_\infty =\esssup _E |g|\), we have that \(m(F_\delta )>0\). If \(m(F_\delta )<\infty \) let \(F_\delta =E_\delta \); otherwise, let \(E_\delta \subset F_\delta \) satisfy \(0<m(E_\delta )<\infty \).
Let \(f_\delta =\frac {1}{m(E_\delta )}\chi _{E_\delta }\sgn (g)\), where \(\sgn (g(x))=1\) if \(g(x)>0\) and \(\sgn (g(x))=-1\) if \(g(x)<0\). Then \begin {equation*} \int _E f_\delta g =\frac {1}{m(E_\delta )} \int _{E_\delta } |g| \geq \|g\|_\infty -\delta \end {equation*} and \begin {equation*} \|f_\delta \|_1=\int _E |f_\delta |=\int _{E_\delta }\frac {1}{m(E_\delta )} =1 \end {equation*} and so \begin {equation*} \|T_g\|_*=\sup _{f\in L^1(E),\>\|f\|_1=1} |T_g(f)|\geq |T_g(f_\delta )|\geq \|g\|_\infty -\delta . \end {equation*} Because this is true for all \(\delta >0\), we must have that \(\|T_g\|_*\geq \|g\|_\infty \), as desired.
Lemma 8.4. Let \(E\subseteq \R \) be measurable and let \(1 \leq p \leq \infty \). Suppose that \(g:E\to [-\infty ,\infty ]\) is measurable and that there is a \(M\in [0,\infty )\) such that if \(\varphi \) is finitely supported and simple, then \(\varphi g\) is integrable and \begin {equation*} \biggl |\int _E \varphi g\biggr |\leq M\|\varphi \|_p. \end {equation*}
Then \(g\in L^q(E)\) and \(\|g\|_q\leq M\), where \(1/p+1/q=1\).
(Ashley, Problem 2380) Prove Lemma 8.4 in the case \(p=1\).
Our goal is to show that \(\|g\|_\infty \leq M\), that is, that \(\{x\in E:|g(x)|>M\}\) is a set of measure zero.If \(n\in \N \), let \(E_{n}=\{x\in E\cap [-n,n]:|g(x)|\geq M+1/n\}\). If \(|g(x)|>M\), then \(|g(x)|\geq M+1/m\) and \(|g(x)|\leq k\) for some \(m\), \(k\in \N \); thus \(x\in E_{\max (k,m)}\) and so \(\{x\in E:|g(x)|>M\}=\bigcup _{n=1}^\infty E_n\). By the countable subadditivity of the Lebesgue measure, it suffices to show that \(m(E_n)=0\) for all \(n\in \N \).
Let \(E^+_n=\{x\in E_n:g(x)>0\}=[-n,n]\cap g^{-1}([M+1/n,\infty ])\) and let \(E^-_n=\{x\in E_n:g(x)<0\}=[-n,n]\cap g^{-1}([-\infty ,-M-1/n])\). These sets are bounded and measurable. Let \(\varphi =\chi _{E^+}-\chi ^{E^-}\). Then \(\varphi \) is a finitely (in fact compactly) supported simple function and so \(\varphi g\) is integrable. We have that \(\varphi g=|g|\chi _{E_n}\) and so \begin {equation*} \int _E \varphi g=\int _{E_n} |g|. \end {equation*} By the conditions of the lemma, \begin {equation*} \int _E \varphi g\leq \biggl |\int _E \varphi g\biggr |\leq M\|\varphi \|_1 = M \,m(E_n). \end {equation*} By definition of \(E_n\), \begin {equation*} \int _{E_n} |g| \geq \int _{E_n} (M+1/n) = (M+1/n)m(E_n) \end {equation*} and so \((M+1/n)m(E_n)\leq M\,m(E_n)\). Because \(1/n>0\), this implies that \(m(E_n)=0\), as desired.
(Bashar, Problem 2390) Prove Lemma 8.4 in the case \(p>1\).
By the simple approximation theorem, there is a sequence of simple functions \(\{\psi _n\}_{n=1}^\infty \) such that \(\psi _n(x)\to g(x)\) and \(\{|\psi _n(x)|\}_{n=1}^\infty \) is nondecreasing for all \(x\in E\).Fix some \(k\in \N \) and let \(\vartheta _n=\psi _n \chi _{[-k,k]}\). Then \(\vartheta _n\) is still simple with \(\vartheta _n\to g\) pointwise on \([-k,k]\).
Then \(|g(x)|^q=\lim _{n\to \infty } |\vartheta _n(x)|^q =\liminf _{n\to \infty } |\vartheta _n(x)|^q \) in \(E\cap [-k,k]\), and so by Fatou’s lemma, \begin {equation*} \int _{E\cap [-k,k]} |g|^q\leq \liminf _{n\to \infty } \int _E |\vartheta _n|^q \leq \sup _{n\in \N } \int _E |\vartheta _n|^q. \end {equation*} Fix some \(n\in \N \). Because \(|\vartheta _n|\leq |\psi _n|\leq |g|\), we have that \begin {equation*} \int _E |\vartheta _n|^q\leq \int _E |g|\,|\vartheta _n|^{q-1}. \end {equation*} Let \(\varphi _n=|\vartheta _n|^{q-1}\sgn (g)\), that is, \(\varphi _n(x)=|\vartheta _n(x)|^{q-1}\) if \(g(x)>0\), \(\varphi _n(x)=-|\vartheta _n(x)|^{q-1}\) if \(g(x)<0\), and \(\varphi _n(x)=0\) if \(g(x)\) (and therefore \(\psi _n(x)\) and \(\vartheta (x)\)) are zero. Then \(\varphi _n\) is a compactly supported simple function and so \(g\varphi _n\) is integrable. Furthermore, \(g\varphi _n=|g|\,|\vartheta _n|^{q-1}\), and so \begin {equation*} \int _E |\vartheta _n|^q\leq \int _E g\varphi _n \leq \biggl |\int _E g\varphi _n\biggr |\leq M\|\varphi _n\|_p. \end {equation*}
If \(p=\infty \) then \(\varphi _n=|\vartheta _n|^{q-1}\sgn (g)=\sgn (g)\) takes on values \(0\), \(1\), and \(-1\), and so \(\|\varphi _n\|_p=\|\varphi _n\|_\infty \leq 1\). Thus \begin {equation*} \int _E |\vartheta _n|^q\leq M=M^{1/q}. \end {equation*} If \(p<\infty \), then a straightforward computation yields that \begin {equation*} |\varphi _n|^p = |\vartheta _n|^{p(q-1)}=|\vartheta _n|^{qp(1-1/q)}=|\vartheta _n|^{qp(1/p)}=|\vartheta _n|^{q} \end {equation*} and so \begin {equation*} \|\varphi _n\|_p = \biggl (\int _E|\varphi _n|^p\biggr )^{1/p} = \biggl (\int _E|\vartheta _n|^{q}\biggr )^{1/p}. \end {equation*} Thus \begin {equation*} \int _E |\vartheta _n|^q\leq M\|\varphi _n\|_p=M \biggl (\int _E|\vartheta _n|^{q}\biggr )^{1/p}. \end {equation*} But \(\vartheta _n\) is a compactly supported simple function and so the integral is finite, so \begin {equation*} \biggl (\int _E |\vartheta _n|^q\biggr )^{1-1/p}\leq M. \end {equation*} Recalling that \(1-1/p=1/q\), we have that \begin {equation*} \int _E |\vartheta _n|^q\leq M^q. \end {equation*} Recall that the above equation is true for \(p=\infty \) (and thus \(q=1\)) as well.
Because \begin {equation*} \int _{E\cap [-k,k]} |g|^q\leq \sup _{n\in \N } \int _E |\vartheta _n|^q \end {equation*} we have that \begin {equation*} \int _{E\cap [-k,k]} |g|^q\leq M^q. \end {equation*} Another application of Fatou’s lemma (taking limits as \(k\to \infty \)) completes the proof.
Theorem 8.5. Let \(I\subset \R \) be a closed bounded interval and let \(1\leq p<\infty \). Let \(T\in (L^p(I))^*\). Then there is a \(g\in L^q(I)\), \(1/p+1/q=1\), such that \begin {equation*} T(f)=\int _I fg \end {equation*} for all \(f\in L^p(I)\). Furthermore, \(\|g\|_q=\|T\|_*\).
(Dibyendu, Problem 2400) In this problem we begin the proof of Theorem 8.5. Let \(I=[a,b]\), \(T\), and \(p\) be as in Theorem 8.5. Define \begin {equation*} \Phi (x)=T(\chi _{[a,x)}). \end {equation*} Show that \(\Phi \) is absolutely continuous.
Recall that \(\Phi \) is absolutely continuous if, for every \(\varepsilon >0\), there is a \(\delta >0\) such that, if \(\{a_k\}_{k=1}^n\) and \(\{b_k\}_{k=1}^n\) satisfy
\(a\leq a_k\leq b_k\leq b\) for all \(1\leq k\leq n\),
\(b_{k-1}\leq a_{k}\) for all \(2\leq k\leq n\),
\(\sum _{k=1}^n |b_k-a_k|<\delta \),
then \begin {equation*} \sum _{k=1}^n|\Phi (b_k)-\Phi (a_k)|< \varepsilon . \end {equation*}
Fix \(\varepsilon >0\) and let \(\delta =\bigl (\frac {\varepsilon }{1+\|T\|_*}\bigr )^p\). Suppose that \(a_k\) and \(b_k\) satisfy the given conditions. Then by definition of \(\Phi \) \begin {align*} \sum _{k=1}^n|\Phi (b_k)-\Phi (a_k)| =\sum _{k=1}^n|T(\chi _{[a,b_k)})-T(\chi _{[a,a_k)})| . \end {align*}
Because \(T\) is linear, \(|T(\chi _{[a,b_k)})-T(\chi _{[a,a_k)})| =|T(\chi _{[a_k,b_k)})|\). Because \(T\) is bounded and \(\chi _{[a_k,b_k)}\in L^p\), this quantity is finite. Let \(c_k=1\) if \(T(\chi _{[a_k,b_k)})\geq 0\) and \(c_k=-1\) if \(T(\chi _{[a_k,b_k)})<0\). Then \begin {align*} \sum _{k=1}^n|\Phi (b_k)-\Phi (a_k)| &=\sum _{k=1}^n|T(\chi _{[a_k,b_k)})| \alignbreak =\sum _{k=1}^nc_k T(\chi _{[a_k,b_k)}) . \end {align*}
Again by linearity of \(T\), \begin {align*} \sum _{k=1}^n|\Phi (b_k)-\Phi (a_k)| &=T\Bigl (\sum _{k=1}^nc_k \chi _{[a_k,b_k)}\Bigr ) . \end {align*}
Because \(T\) is a bounded operator, we have that \begin {align*} \sum _{k=1}^n|\Phi (b_k)-\Phi (a_k)| &\leq \|T\|_*\Bigl \|\sum _{k=1}^nc_k \chi _{[a_k,b_k)}\Bigr \|_p \alignbreak =\|T\|_* \Bigl (\sum _{k=1}^n (b_k-a_k)\Bigr )^{1/p} . \end {align*}
The sum is less than \(\delta \) by choice of \(a_k\) and \(b_k\). Thus \begin {align*} \sum _{k=1}^n|\Phi (b_k)-\Phi (a_k)| &\leq \|T\|_* \delta ^{1/p} <\varepsilon \end {align*}
as desired.
(Irina, Problem 2410) Prove Theorem 8.5.
(Problem 2411) Let \(1\leq p<\infty \). Let \(T\in (L^p(\R ))^*\). Show that there is a \(g\in L^q(\R )\), \(1/p+1/q=1\), such that \begin {equation*} T(f)=\int _\R fg \end {equation*} for all \(f\in L^p(\R )\). Furthermore, \(\|g\|_q=\|T\|_*\).
(Problem 2412) Let \(1\leq p<\infty \). Let \(E\subseteq \R \) be measurable and let \(T\in (L^p(E))^*\). Show that there is a \(g\in L^q(E)\), \(1/p+1/q=1\), such that \begin {equation*} T(f)=\int _E fg \end {equation*} for all \(f\in L^p(E)\). Furthermore, \(\|g\|_q=\|T\|_*\).
The Riesz representation theorem for the dual of \(L^p(E)\). Let \(E\subseteq \R \) be measurable, \(1\leq p<\infty \), and \(1/p+1/q=1\).
Then there is a canonical isomorphism between \((L^p(E))^*\) and \(L^q(E)\): if we define \(\mathcal {R}_g(f)=\int _E fg\), then for every \(g\in L^q(E)\) we have that \(\mathcal {R}_g\in (L^p(E))^*\) with \(\|\mathcal {R}_g\|_*=\|g\|_q\), and conversely if \(T\in (L^p(E))^*\) then there is a \(g\in L^q(E)\) such that \(T=\mathcal {R}_g\) and \(\|g\|_q=\|T\|_*\). Furthermore, if \(\mathcal {R}_g=\mathcal {R}_h\) (in the sense of functions from \(L^p(E)\) to \(\R \)) then \(g=h\) as elements of \(L^q(E)\), that is, \(g(x)=h(x)\) for almost every \(x\in E\).
(Micah, Problem 2420) Prove the Riesz representation theorem for the dual of \(L^p(E)\).
Recall [Bolzano-Weierstrauß theorem]: If \(\{x_n\}_{n=1}^\infty \) is a bounded sequence of points in \(\R \), then there is a subsequence \(\{x_{n_k}\}_{k=1}^\infty \) that converges.
(Problem 2421) Let \(\{a_n\}_{n=1}^\infty \) be a sequence in \(\R \) and let \(\{a_{n_k}\}_{k=1}^\infty \) be a convergent subsequence. Show that \begin {equation*} \liminf _{n\to \infty } a_{n}\leq \lim _{k\to \infty } a_{n_k}\leq \limsup _{n\to \infty } a_{n}. \end {equation*}
(Problem 2422) Let \(\{a_n\}_{n=1}^\infty \) be a sequence in \(\R \). Show that there is a subsequence \(\{a_{n_k}\}_{k=1}^\infty \) that converges to \(\liminf _{n\to \infty } a_{n}\).
[Definition: Weak convergence] Let \(X\) be a normed linear space, let \(x\in X\), and let \(\{x_n\}_{n=1}^\infty \) be a sequence in \(X\). We say that the sequence \(\{x_n\}_{n=1}^\infty \) converges weakly to \(x\), or \(x_n\rightharpoonup x\) in \(X\), if \begin {equation*} \lim _{n\to \infty } T(x_n)=T(x)\quad \text {for all }T\in X^*. \end {equation*}
Proposition 8.6. Let \(E\subseteq \R \) be measurable, let \(1\leq p<\infty \), and let\(1/p+1/q=1\). Let \(f\), \(f_n\in L^p(E)\). Then \(f_n\rightharpoonup f\) in \(L^p(E)\) if and only if \begin {equation*} \lim _{n\to \infty } \int _E f_n\, g=\int _E f\,g \end {equation*} for all \(g\in L^q(E)\).
(Muhammad, Problem 2430) Give an example of a bounded sequence in \(L^p(\R )\), \(1\leq p\leq \infty \), that does not have a convergent subsequence.
(Ashley, Problem 2440) Give an example of a measurable set \(E\subseteq \R \), a \(p\in (1,\infty )\), and functions \(f_n\), \(f\in L^p(E)\) such that \(f_n\rightharpoonup f\) weakly in \(L^p(E)\) but such that \(f_n\not \to f\) strongly in \(L^p(E)\).
(Problem 2441) Show that if \(f_n\to f\) (meaning \(\{f_n\}_{n=1}^\infty \) converges strongly to \(f\)) then \(\|f_n\|\to \|f\|\). Give an example to show that the condition \(f_n\rightharpoonup f\) (meaning \(\{f_n\}_{n=1}^\infty \) converges weakly to \(f\)) does not imply that \(\|f_n\|\to \|f\|\).
[Homework 16.1] Let \(1\leq p\leq \infty \) and let \(q\) satisfy \(1/p+1/q=1\). Let \(E\subseteq \mathbb {R}\) be measurable.
Suppose that \(f\in L^p(E)\) and that \(\int _E fg=0\) for all \(g\in L^q(E)\). Show that \(f=0\) almost everywhere in \(E\).
(Problem 2442) Let \(E\subseteq \R \) be measurable, let \(1\leq p<\infty \), and let \(1/p+1/q=1\). Let \(f\), \(F\), \(f_n\in L^p(E)\). Suppose that \(f_n\rightharpoonup f\) and \(f_n\rightharpoonup F\). Show that \(f=F\) in \(L^p(E)\).
Theorem 8.7. Let \(E\subseteq \R \) be measurable and let \(1\leq p<\infty \). Suppose that \(f_n\rightharpoonup f\) in \(L^p(E)\) (that is, \(\{f_n\}_{n=1}^\infty \) converges weakly to \(f\) in \(L^p(E)\)).
Then \(\{f_n\}_{n=1}^\infty \) is bounded in \(L^p(E)\) (that is, \(\sup _{n\in \N } \|f_n\|_p<\infty \)) and \(\|f\|_p\leq \liminf _{n\to \infty } \|f_n\|_p\).
(Problem 2443) This problem is a useful first step in the proof of Theorem 8.7. Let \(E\subseteq \R \) be measurable and let \(1\leq p<\infty \). Let \(\{f_n\}_{n=1}^\infty \) be a sequence in \(L^p(E)\) and let \(\{\varepsilon _n\}_{n=1}^\infty \) be a sequence of positive real numbers. Show that there exists a sequence \(\{g_n\}_{n=1}^\infty \) in \(L^q(E)\) such that \(\|g_{n+1}-g_n\|_q\leq \varepsilon _{n+1}\) for each \(n\in \N \) and such that \begin {equation*} \left |\int _E f_n \,g_n\right |\geq \varepsilon _n \|f_n\|_p. \end {equation*}
[Chapter 8, Problem 18] Let \(\{a_n\}_{n=1}^\infty \) be a sequence of positive numbers with \(a_n\to \infty \). Let \(X\) be a normed linear space. Suppose that \(X\) contains an unbounded weakly convergent sequence; that is, suppose that there is an \(x\in X\) and a sequence \(\{x_n\}_{n=1}^\infty \) such that \(x_n\rightharpoonup x\) in \(X\) and \(\{\|x_n\|\}_{n=1}^\infty \) is unbounded.
Show that there exists a \(y\in X\) and a sequence \(\{y_n\}_{n=1}^\infty \) in \(X\) with \(\|y_n\|=a_n\) and with \(y_n\rightharpoonup y\). Hint: The easiest choice of \(y\) is \(y=0\).
(Bashar, Problem 2450) Prove the bound \(\sup _{n\in \N } \|f_n\|_p<\infty \) in Theorem 8.7.
(Dibyendu, Problem 2460) Prove the bound \(\|f\|_p\leq \liminf _{n\to \infty } \|f_n\|_p\) in Theorem 8.7.
Let \(f^*\) be as in Theorem 7.1. Then \(f\in L^q(E)\), \(\|f^*\|_q=1\), and \begin {equation*} \|f\|_p=\int _E f^*\,f. \end {equation*} By the definition of weak convergence, we have that \begin {equation*} \|f\|_p=\int _E f^*\,f=\lim _{n\to \infty } \int _E f^*\,f_n. \end {equation*} By Hölder’s inequality and Proposition 4.16, \begin {equation*} \biggl |\int _E f^*\,f_n\biggr |\leq \int _E |f^*\,f_n|\leq \|f^*\|_q\|f_n\|_p=\|f_n\|_p. \end {equation*} By elementary real analysis arguments, this implies that \begin {equation*} \|f\|_p=\lim _{n\to \infty } \int _E f^*\,f_n\leq \liminf _{n\to \infty } \|f_n\|_p \end {equation*} as desired.
Corollary 8.8. Let \(E\subseteq \R \) be measurable and let \(1\leq p<\infty \). Suppose that \(f_n\rightharpoonup f\) weakly in \(L^p(E)\). Let \(1/p+1/q=1\) and let \(g_n\to g\) strongly in \(L^q(E)\). Then \begin {equation*} \lim _{n\to \infty } \int _E f_n\,g_n=\int _E fg. \end {equation*}
(Irina, Problem 2470) Prove Corollary 8.8.
Observe that \begin {equation*} \int _E f_n\,g_n-\int _E f\,g = \int _E f_n\,g_n-\int _E f_n\,g+\int _E f_n\,g-\int _E f\,g = \int _E f_n\,(g_n-g)+\int _E (f_n - f) g. \end {equation*} By definition of weak convergence and linearity of integration, \begin {equation*} \lim _{n\to \infty } \int _E (f_n - f) g=0. \end {equation*} By Theorem 8.7, there is an \(M<\infty \) such that \(\|f_n\|_p\leq M\) for all \(n\in \N \). Thus by Hölder’s inequality \begin {equation*} \biggl |\int _E f_n\,(g_n-g)\biggr |\leq M\|g_n-g\|_q. \end {equation*} Recall that \(g_n\to g\) strongly in \(L^q(E)\); this means that \(\lim _{n\to \infty }\|g_n-g\|_q=0\). By linearity of limits \(\lim _{n\to \infty }M\|g_n-g\|_q=0\), and so by the squeeze theorem \begin {equation*} \lim _{n\to \infty } \int _E f_n\,(g_n-g)=0. \end {equation*} Again by linearity of limits this implies that \begin {equation*} \lim _{n\to \infty } \biggl (\int _E f_n\,g_n-\int _E f\,g\biggr ) =\lim _{n\to \infty } \int _E (f_n - f) g+\lim _{n\to \infty } \int _E f_n\,(g_n-g)=0 \end {equation*} as desired.
(Micah, Problem 2480) Give an example of a measurable set \(E\subseteq \R \), a \(p\), \(q\in (1,\infty )\) with \(1/p+1/q=1\), and sequences \(\{f_n\}_{n=1}^\infty \) and \(\{g_n\}_{n=1}^\infty \) such that \(f_n\rightharpoonup f\) weakly in \(L^p(E)\) and \(g_n\rightharpoonup g\) weakly in \(L^q(E)\) but such that \(\int _E f_n\,g_n\) does not converge to \(\int _E f\,g\).
Take \(E=\R \), \(p\in (1,\infty )\) and \(f_n=g_n=\chi _{[n,n+1]}\). As observed above, if \(1<p<\infty \) then \(f_n\rightharpoonup 0\) weakly in \(L^p(\R )\). Thus \(f=g=0\) and so \(\int _E fg=0\). But \(\int _E f_ng_n=\int _n^{n+1} 1=1\) and \(\lim _{n\to \infty } 1\neq 0\). Thus \(\int _E f_n\,g_n\) does not converge to \(\int _E f\,g\), as desired.
(Muhammad, Problem 2490) Let \(E\subseteq \R \) be measurable and let \(1\leq p<\infty \). Let \(1/p+1/q=1\) and let \(S\subset L^q(E)\) be dense. Suppose that \(f_n\), \(f\in L^p(E)\), that \(\sup _{n\in \N } \|f_n\|_p<\infty \), and that \begin {equation*} \lim _{n\to \infty }\int _E f_n\,g= \int _E f\,g\quad \text {for all $g\in S$}. \end {equation*} Show that \(f_n\rightharpoonup f\) weakly in \(L^p(E)\), that is, that \begin {equation*} \lim _{n\to \infty }\int _E f_n\,g= \int _E f\,g\quad \text {for all $g\in L^q(E)$}. \end {equation*}
Choose some \(g\in L^q(E)\). It suffices to prove that \(\lim _{n\to \infty } \int _E f_n\,g= \int _E f\,g\).By definition of density, there is a sequence \(\{g_k\}_{k=1}^\infty \) of functions in \(S\) that converge strongly to \(g\), that is, such that \(\lim _{k\to \infty } \|g-g_k\|_q=0\).
Choose some \(\varepsilon >0\). Then by the triangle inequality and linearity of integration \begin {equation*} \biggl |\int _E f_n\,g-\int _E f\,g\biggr | \leq \biggl |\int _E (f_n-f) g_k\biggr | + \biggl |\int _E (f_n-f)(g-g_k)\biggr | . \end {equation*} Fix some \(k\) such that \(\|g_k-g\|_q<\varepsilon \). By Hölder’s inequality \begin {equation*} \biggl |\int _E (f_n-f)(g-g_k)\biggr | \leq \varepsilon \|f_n-f\|_p \end {equation*} and by the triangle inequality \begin {equation*} \biggl |\int _E (f_n-f)(g-g_k)\biggr | \leq \varepsilon \bigl (\|f\|_p+\sup _{m\in \N } \|f_m\|_p\bigr ). \end {equation*} Now, \(g_k\in L^q(E)\) and \(f_n\rightharpoonup f\), so there is some \(N\in \N \) such that, if \(n\geq N\), then \begin {equation*} \biggl |\int _E f_n\,g_k-\int _E f\,g_k\biggr |<\varepsilon . \end {equation*} Combining the above estimates, we see that if \(\varepsilon >0\) then there is a \(N\in \N \) such that, if \(n\geq N\), then \begin {equation*} \biggl |\int _E f_n\,g-\int _E f\,g\biggr | \leq \varepsilon \bigl (1+ \|f\|_p+\sup _{m\in \N } \|f_m\|_p\bigr ) . \end {equation*} The right hand side is a finite multiple of \(\varepsilon \) that is independent of \(n\), and so \begin {equation*} \lim _{n\to \infty }\int _E f_n\,g= \int _E f\,g \end {equation*} as desired.
Proposition 8.9. Let \(E\subseteq \R \) be measurable and let \(1\leq p<\infty \). Let \(1/p+1/q=1\) and let \(\mathcal {H}\subset L^q(E)\) be such that \begin {equation*} S=\Bigl \{\sum _{k=1}^\ell a_k h_k : \ell \in \N ,\>a_k\in \R ,\>h_k\in \mathcal {H}\Bigr \} \end {equation*} is dense in \(L^q(E)\). If \(f_n\), \(f\in L^p(E)\), \(\sup _{n\in \N } \|f_n\|_p<\infty \), and \begin {equation*} \lim _{n\to \infty }\int _E f_n\,h= \int _E f\,h\quad \text {for all $h\in \mathcal {H}$} \end {equation*} then \(f_n\rightharpoonup f\) weakly in \(L^p(E)\).
Proof: By the previous problem, it suffices to show that \begin {equation*} \lim _{n\to \infty }\int _E f_n\,g= \int _E f\,g \end {equation*} for all \(g\in S\).Choose some \(g\in S\). Then \(g=\sum _{k=1}^\ell a_k h_k\) for some \(a_k\in \R \) and some \(h_k\in \mathcal {H}\). We then have that \begin {equation*} \lim _{n\to \infty } \int _E f_n\, h_k=\int _E f\,h_k \end {equation*} by assumption. Because limits and integrals are linear, we then have that \begin {equation*} \int _E f\,g = \sum _{k=1}^\ell a_k\int _E f\,h_k = \sum _{k=1}^\ell a_k\lim _{n\to \infty }\int _E f_n\,h_k = \lim _{n\to \infty }\int _E f_n\,g \end {equation*} as desired.
Theorem 8.10. Let \(E\subseteq \R \) be measurable and let \(1\leq p<\infty \). Let \(f_n\), \(f\in L^p(E)\). Suppose that \(\sup _{n\in \N } \|f_n\|_p<\infty \) and that \begin {equation*} \lim _{n\to \infty }\int _A f_n= \int _A f\quad \text {for all $A\subseteq E$ measurable.} \end {equation*} Then \(f_n\rightharpoonup f\) weakly in \(L^p(E)\). If \(p>1\), it suffices to require this condition for all \(A\subseteq E\) measurable with \(m(A)<\infty \).
Define \(S\) by \begin {equation*} S=\{\varphi :E\to \R |\varphi \text { is simple and finitely supported}\}\quad \text {if }1<p<\infty , \end {equation*} \begin {equation*} S=\{\varphi :E\to \R |\varphi \text { is simple}\}\quad \text {if } p=1. \end {equation*} By Proposition 7.9, \(S\) is dense in \(L^q(E)\). But recall that \(\varphi \) is simple if and only if \(\varphi =\sum _{k=1}^\ell a_k \chi _{A_k}\) for some real numbers \(a_k\) and some measurable sets \(A_k\), and that if \(\varphi \) is finitely supported we may in addition require that \(m(A_k)<\infty \) for all \(1\leq k\leq \ell \). Thus, if we take \(\mathcal {H}=\{\chi _{A}:A\) measurable\(\}\) (if \(p=1\)) or \(\mathcal {H}=\{\chi _{A}:A\) measurable and \(m(A)<\infty \}\) (if \(1<p<\infty \)), then \(S\) is the set of all finite linear combinations of elements of \(\mathcal {H}\). Thus, by Proposition 8.9 it suffices to show that \begin {equation*} \lim _{n\to \infty }\int _E f_n\,\chi _A= \int _E f\,\chi _A \end {equation*} for all \(A\) measurable (and of finite measure if \(p>1\)).But \(\int _E f\,\chi _A =\int _A f\) by Problem 4.28, and so the above limit is valid by the assumptions in the theorem statement. This completes the proof.
Theorem 8.11. Let \(-\infty <a<b<\infty \) and let \(1< p<\infty \). Let \(f_n\), \(f\in L^p([a,b])\). Suppose that \(\sup _{n\in \N } \|f_n\|_p<\infty \) and that \begin {equation*} \lim _{n\to \infty }\int _a^x f_n= \int _a^x f\quad \text {for all $x\in [a,b]$.} \end {equation*} Then \(f_n\rightharpoonup f\) weakly in \(L^p([a,b])\).
The argument is similar to the proof of Theorem 8.10 with \(S\) equal to the set of step functions, \(\mathcal {H}\) equal to the set of characteristic functions of intervals, and with Proposition 7.10 in place of Proposition 7.9.
(Bashar, Problem 2500) Prove Proposition 8.9, Theorem 8.10, and Theorem 8.11.
The Riemann-Lebesgue lemma. If \(1\leq p<\infty \) and \(f_n(x)=\sin (nx)\), then \(f_n\rightharpoonup 0\) in \(L^p([-\pi ,\pi ])\).
(Ashley, Problem 2510) Prove the Riemann-Lebesgue lemma.
(Dibyendu, Problem 2520) Let \(1\leq p<\infty \) and let \(f_n(x)=n^{1/p}\,\chi _{[1/n,2/n]}\). Show that \(\|f_n\|_p=1\), that \(\{f_n\}_{n=1}^\infty \) converges to zero pointwise, that if \(1<p<\infty \) then \(\{f_n\}_{n=1}^\infty \) converges to zero weakly in \(L^p([0,1])\), but that if \(p=1\) then \(\{f_n\}_{n=1}^\infty \) does not converge to zero weakly in \(L^1([0,1])\).
Theorem 8.12. Let \(E\subseteq \R \) be measurable and let \(1<p<\infty \). Suppose that \(\{f_n\}_{n=1}^\infty \) is a bounded sequence in \(L^p(E)\) and that \(f_n\) converges pointwise almost everywhere in \(E\) to some function \(f\). Then \(f\in L^p(E)\) and \(\{f_n\}_{n=1}^\infty \) converges to \(f\) weakly in \(L^p(E)\).
(Irina, Problem 2530) Prove Theorem 8.12.
Observe that \(\{|f_n|^p\}_{n=1}^\infty \) is a sequence of nonnegative measurable functions. Thus, by Fatou’s lemma, \begin {equation*} \int _E |f|^p=\int _E \lim _{n\to \infty } |f_n|^p=\int _E \liminf _{n\to \infty } |f_n|^p \leq \liminf _{n\to \infty } \int _E |f_n|^p\leq \sup _{n\in \N }\|f_n\|_p^p. \end {equation*} By assumption the rightmost term is finite, so the leftmost term is finite and \(f\in L^p(E)\).By Corollary 7.2, the set \(\{f_n:n\in \N \}\) is uniformly integrable over \(E\). Therefore, if \(A\subseteq E\) is measurable with \(m(A)<\infty \), then \(\{f_n:n\in \N \}\) is uniformly integrable over \(A\). So by Vitali convergence theorem, for all such \(A\), \begin {equation*} \lim _{n\to \infty }\int _A f_n=\int _A f. \end {equation*}
Applying Theorem 8.10 completes the proof.
(Problem 2531) Give an example of a sequence of functions in \(L^p(E)\) that converge pointwise but not weakly.
The Radon-Riesz theorem. Let \(E\subseteq \R \) be measurable and let \(1<p<\infty \). Suppose that \(f_n\rightharpoonup f\) weakly in \(L^p(E)\). Then \(f_n\to f\) strongly in \(L^p(E)\) if and only if \(\lim _{n\to \infty }\|f_n\|_p= \|f\|_p\).
(Micah, Problem 2540) Prove the Radon-Riesz theorem in the case \(p=2\).
(Bonus Problem 2541) Prove the Radon-Riesz theorem in the cases \(1<p<2\) and/or \(2<p<\infty \).
(Problem 2542) Let \(\{a_n\}_{n=1}^\infty \) be a sequence in \(\R \). Show that there is a subsequence \(\{a_{n_k}\}_{k=1}^\infty \) that converges to \(\liminf _{n\to \infty } a_{n}\).
Corollary 8.13. Let \(E\subseteq \R \) be measurable and let \(1<p<\infty \). Suppose that \(f_n\rightharpoonup f\) weakly in \(L^p(E)\). Then \(\|f\|_p=\liminf _{n\to \infty } \|f_n\|_p\) if and only if a subsequence of \(\{f_n\}_{n=1}^\infty \) converges to \(f\) strongly in \(L^p(E)\).
(Muhammad, Problem 2550) Prove Corollary 8.13.
(Problem 2551) Let \(f_n(x)=1+\sin (nx)\). Show that \(\{f_n\}_{n=1}^\infty \) is a counterexample to the Radon-Riesz theorem in the case \(p=1\).
Helly’s Theorem. Let \(X\) be a separable normed linear space and let \(\{T_n\}_{n=1}^\infty \) be a sequence in the dual space \(X^*\). Suppose that \(\{T_n\}_{n=1}^\infty \) is bounded, that is, there is some \(M<\infty \) such that \begin {equation*} |T_n(x)|\leq M\|x\| \end {equation*} for all \(x\in X\) and all \(n\in \N \).
Then there is a subsequence \(\{T_{n_k}\}_{k=1}^\infty \) of \(\{T_n\}_{n=1}^\infty \) and a \(T\in X^*\) such that \begin {equation*} \lim _{k\to \infty } T_{n_k}(x)=T(x) \end {equation*} for all \(x\in X\).
(Ashley, Problem 2560) Prove Helly’s Theorem.
Let \(\{x_\ell \}_{\ell =1}^\infty \) be a countable dense subset of \(X\).Consider the sequence \(\{T_n(x_1)\}_{n=1}^\infty \). Then \(|T_n(x_1)|\leq \|T_n\|_*\|x_1\|\leq M\|x_1\|\). The right hand side is independent of \(n\), and so \(\{T_n(x_1)\}_{n=1}^\infty \) is a bounded sequence of real numbers. By the Bolzano-Weierstrauß theorem, there is a subsequence that converges. Let the numbers \(n(1,k)\) be such that \(\{T_{n(1,k)}(x_1)\}_{k=1}^\infty \) is a convergent subsequence. Let \(n_1=n(1,1)\).
We have now defined a number \(n_1\) and a strictly increasing sequence of natural numbers \(\{n(1,k)\}_{k=1}^\infty \) such that \(n(1,k)=n_k\) for all \(k\leq 1\) and such that \(\{T_{n(1,k)}(x_1)\}_{k=1}^\infty \) is a convergent sequence of real numbers.
Let \(\ell \in \N \). Suppose that we have defined the numbers \(n_1\), \(n_2,\dots ,n_\ell \) and the strictly increasing sequences of natural numbers \(\{n(j,k)\}_{k=1}^\infty \), for \(1\leq j\leq \ell \), such that
\(\{n(j+1,k)\}_{k=1}^\infty \) is a subsequence of \(\{n(j,k)\}_{k=1}^\infty \) for all \(1\leq j\leq j+1\leq \ell \),
\(n(j,k)=n_k\) whenever \(1\leq k\leq j\leq \ell \),
The sequence \(\{n(j,k)\}_{k=1}^\infty \) is strictly increasing for all \(1\leq j\leq \ell \),
\(\{T_{n(j,k)}(x_j)\}_{k=1}^\infty \) is a convergent sequence of real numbers for all \(1\leq j\leq \ell \).
We have that \(|T_n(x_{\ell +1})|\leq \|T_n\|_*\|x_{\ell +1}\|\leq M\|x_{\ell +1}\|\) for all \(n\in \N \), and so \(\{T_{n(\ell ,k)}(x_{\ell +1})\}_{k=\ell +1}^\infty \) is a bounded sequence of real numebrs. By the Bolzano-Weierstrauß theorem, there is a subsequence that converges. Let \(\{n(\ell +1,k)\}_{k=\ell +1}^\infty \) be a subsequence of \(\{n(\ell ,k)\}_{k=\ell +1}^\infty \) such that \(\{T_{n(\ell +1,k)}(x_{\ell +1})\}_{k=\ell +1}^\infty \) converges. Prepending finitely many terms to a convergent series does not change its convergence properties; thus, if we define \(n(\ell +1,k)=n_k\) for all \(k\leq \ell \), the sequence \(\{T_{n(\ell +1,k)}(x_{\ell +1})\}_{k=1}^\infty \) still converges. Choosing \(n_{\ell +1}=n(\ell +1,\ell +1)\), we see that we have now constructed numbers \(n_1\), \(n_2,\dots ,n_{\ell +1}\) and sequences of natural numbers \(\{n(j,k)\}_{k=1}^\infty \), for \(1\leq j\leq \ell +1\), such that
\(\{n(j+1,k)\}_{k=1}^\infty \) is a subsequence of \(\{n(j,k)\}_{k=1}^\infty \) for all \(1\leq j\leq j+1\leq \ell +1\),
\(n(j,k)=n_k\) whenever \(1\leq k\leq j\leq \ell +1\),
The sequence \(\{n(j,k)\}_{k=1}^\infty \) is strictly increasing for all \(1\leq j\leq \ell +1\),
\(\{T_{n(j,k)}(x_j)\}_{k=1}^\infty \) is a convergent sequence of real numbers for all \(1\leq j\leq \ell +1\).
By induction, we may construct the infinite sequence \(\{n_k\}_{k=1}^\infty \).
It is elementary to establish that, if \(\{a_n\}_{n=1}^\infty \) is a strictly increasing sequence, then \(\{b_n\}_{n=1}^\infty \) is a subsequence of \(\{a_n\}_{n=1}^\infty \) if \(\{b_n:n\in \N \}\subseteq \{a_n:n\in \N \}\) and \(\{b_n\}_{n=1}^\infty \) is strictly increasing. This implies that \(\{n_k\}_{k=1}^\infty \) is a subsequence of \(\{n(j,k)\}_{k=1}^\infty \) for all \(j\in \N \): clearly \(\{n_k\}_{k=1}^\infty \) is strictly increasing, and if \(k\leq j\) then \(n_k=n(j,k)\), while if \(k\geq j\) then \(n_k=n(k,k)\in \{n(k,\ell ):\ell \in \N \}\subseteq \{n(j,\ell ):\ell \in \N \}\). Thus \(\{n_k:k\in \N \}\subseteq \{n(j,\ell ):\ell \in \N \}\) and so \(\{n_k\}_{k=1}^\infty \) is a subsequence of \(\{n(j,k)\}_{k=1}^\infty \) for all \(j\in \N \).
But then \(\{T_{n_k}(x_j)\}_{k=1}^\infty \) is a subsequence of \(\{T_{n(j,k)}(x_j)\}_{k=1}^\infty \), and so is convergent, for all \(j\in \N \).
We now must show that \(\{T_{n_k}(x)\}_{k=1}^\infty \) is convergent for all \(x\in X\). Choose some such \(x\). Because \(\R \) is complete it suffices to show that \(\{T_{n_k}(x)\}_{k=1}^\infty \) is Cauchy.
Let \(\varepsilon >0\). Let \(j\in \N \) be such that \(\|x-x_j\|<\frac {\varepsilon }{M+1}\); by definition of density such a \(j\) must exist. Then \(\{T_{n_k}(x_j)\}_{k=1}^\infty \) converges to \(a_j\) for some \(a_j\in \R \). Let \(K\in \N \) be such that, if \(k\geq K\), then \(|T_{n_k}(x_j)-a_j|<\varepsilon \).
If \(k\), \(\ell \geq K\), then \begin {equation*} |T_{n_k}(x)-T_{n_\ell }(x)| \leq |T_{n_k}(x)-T_{n_k}(x_j)|+ |T_{n_k}(x_j)-T_{n_\ell }(x_j)|+ |T_{n_\ell }(x_j)-T_{n_\ell }(x)| \end {equation*} by the triangle inequality. By linearity uniform boundedness of the \(T_n\)s, the first term satisfies \begin {equation*} |T_{n_k}(x)-T_{n_k}(x_j)|=|T_{n_k}(x-x_j)|\leq \|T_{n_k}\|_* \|x-x_j\| <M\frac {\varepsilon }{M+1}<\varepsilon . \end {equation*} Similarly \(|T_{n_\ell }(x_j)-T_{n_\ell }(x)|<\varepsilon \). By choice of \(K\), \begin {equation*} |T_{n_k}(x_j)-T_{n_\ell }(x_j)| \leq |T_{n_k}(x_j)-a_j|+|a_j-T_{n_\ell }(x_j)|<2\varepsilon \end {equation*} and so if \(k\), \(\ell \geq K\) then \(|T_{n_k}(x)-T_{n_\ell }(x)|<4\varepsilon \). Thus \(\{T_{n_k}(x)\}_{k=1}^\infty \) is Cauchy, as desired.
Define \(T(x)=\lim _{k\to \infty } T_{n_k}(x)\); we have just seen that \(T(x)\) exists for all \(x\in X\). It is clear that \(T\) is linear because each \(T_{n_k}\) is. To show that \(T\) is bounded, observe that the norm is by the triangle inequality a continuous map from \(X\) to \(\R \), and so \begin {align*} \|T(x)\|&=\|\lim _{k\to \infty } T_{n_k}(x)\|=\lim _{k\to \infty } \|T_{n_k}(x)\| \\&=\limsup _{k\to \infty } \|T_{n_k}(x)\|\leq \sup _{k\in \N } \|T_{n_k}(x)\| \\&\leq \sup _{n\in \N } \|T_n\|_*\|x\|\leq M\|x\| \end {align*}
and so \(T\) is a bounded linear operator with operator norm at most \(M\), as desired.
Theorem 8.14. Let \(E\subseteq \R \) be measurable and let \(1<p<\infty \). Let \(\{f_n\}_{n=1}^\infty \) be a bounded sequence in \(L^p(E)\). Then there is a subsequence \(\{f_{n_k}\}_{k=1}^\infty \) that converges weakly to some \(f\in L^p(E)\).
(Bashar, Problem 2570) Prove Theorem 8.14.
Define \(T_n:L^q(E)\to \R \) by \(T_n(g)=\int _E f_n\,g\). Then \(T_n\) is clearly linear, and by Hölder’s inequality \(|T_n(g)|\leq \|f_n\|_p\|g\|_q\). Thus each \(T_n\) is bounded with \begin {equation*} \|T_n\|_*\leq \|f_n\|_p \leq \sup _{n\in \N }\|f_n\|_p, \end {equation*} which by assumption is finite.We may thus apply Helly’s Theorem to see that there is a subsequence \(\{T_{n_k}\}_{k=1}^\infty \) and a \(T\in (L^q(E))^*\) such that \begin {equation*} \lim _{k\to \infty } \int _E f_{n_k}\,g=\lim _{k\to \infty } T_{n_k}(g)=T(g) \end {equation*} for all \(g\in L^q(E)\).
But by the Riesz representation theorem for the dual of \(L^p(E)\), if \(1<q<\infty \) then there is a \(f\in L^p(E)\) such that \(T(g)=\int _E fg\) for all \(g\in L^q(E)\). Thus \begin {equation*} \lim _{k\to \infty } \int _E f_{n_k}\,g=\int _E f\,g \end {equation*} for all \(g\in L^q(E)\), and so \(f_{n_k}\rightharpoonup f\) weakly in \(L^p(E)\), as desired.
(Dibyendu, Problem 2580) Give an example that shows that Theorem 8.14 is not true for \(p=1\), that is, a bounded sequence in \(L^1(E)\) that has no weakly convergent subsequence.
[Definition: Weakly sequentially compact] A subset \(K\) of a normed linear space \(X\) is weakly sequentially compact in \(X\) if, for every sequence \(\{x_n\}_{n=1}^\infty \) in \(K\), there is a subsequence \(\{x_{n_k}\}_{k=1}^\infty \) that converges weakly to an element of \(K\).
Theorem 8.15. If \(E\subseteq \R \) is measurable and \(1<p<\infty \), then \(\{f\in L^p(E):\|f\|_p\leq 1\}\) is weakly sequentially compact in \(L^p(E)\).
(Problem 2581) Give an example of a measurable set \(E\subseteq \R \), a \(p\in (1,\infty )\), and a sequence \(\{f_n\}_{n=1}^\infty \) such that \(\{f_n\}_{n=1}^\infty \) converges weakly but such that no subsequence of \(\{f_n\}_{n=1}^\infty \) converges strongly.
The Banach-Saks theorem. Let \(E\subseteq \R \) be measurable and let \(1<p<\infty \). Let \(f_n\), \(f\in L^p(E)\) and suppose that \(f_n\rightharpoonup f\) weakly in \(L^p(E)\).
Then there is a subsequence \(\{f_{n_k}\}_{k=1}^\infty \) of \(\{f_n\}_{n=1}^\infty \) such that, if we define \begin {equation*} \varphi _k=\frac {1}{k}\sum _{\ell =1}^k f_{n_\ell }, \end {equation*} then \(\varphi _k\to f\) strongly in \(L^p(E)\).
(Irina, Problem 2590) Prove the Banach-Saks theorem in the case \(p=2\).
By Theorem 8.7, we have that \(M=\sup _{n\in \N } \|f_n\|_p<\infty \) and \(\|f\|_p\leq M\). If \(M=0\) then \(f_n=0\) for all \(n\) and so there is nothing to prove. Thus we assume \(M>0\).Let \(n_1=1\). Then \(\varphi _1=f_{n_1}=f_1\) and so by Minkowski’s Inequality \begin {equation*} \|\varphi _1-f\|_p\leq \|\varphi _1\|_p+\|f\|_p\leq 2M. \end {equation*}
Now suppose that \(n_1\), \(n_2,\dots ,n_k\) have been chosen such that \(\|\varphi _k-f\|_p^2\leq \frac {6M^2}{k}\).
The function \(\varphi _k-f=\frac {1}{k}\sum _{\ell =1}^k f_{n_\ell }-f\) is a sum of functions in \(L^2(E)\), and so lies in \(L^2(E)\). Because \(f_n\rightharpoonup f\), there is a \(n_{k+1}>n_k\) such that \begin {equation*} \biggl |\int _E (f_{n_{k+1}}-f) (\varphi _k-f)\biggr |\leq \frac {M^2}{k}. \end {equation*} We may then compute that \begin {align*} \|\varphi _{k+1}-f\|_p^2 &=\|\varphi _{k+1}-f\|_2^2 \alignbreak =\biggl \|\frac {1}{k+1} f_{n_{k+1}}+\frac {1}{k+1}\sum _{\ell =1}^k f_{n_\ell }-f\biggr \|_2^2 \\&=\biggl \|\frac {1}{k+1} f_{n_{k+1}}+\frac {k}{k+1}\varphi _k-f\biggr \|_2^2 \alignbreak = \biggl \|\frac {1}{k+1} (f_{n_{k+1}}-f) + \frac {k}{k+1}(\varphi _k-f)\biggr \|_2^2 . \end {align*}
Because \(p=2\), we have that \(\|g+h\|_p^2=\|g+h\|_2^2=\|g\|_2^2+\|h\|_2^2 +2\int _E g\,h\), and so \begin {align*} \|\varphi _{k+1}-f\|_p^2 &= \frac {1}{(k+1)^2}\|f_{n_{k+1}}-f\|_2^2 +\frac {k^2}{(k+1)^2} \|\varphi _k-f\|_2^2 \alignbreak +2\frac {k}{(k+1)^2}\int _E (f_{n_{k+1}}-f)\varphi _k . \end {align*}
Because \(\|f\|_2\leq M\) and \(\|f_{n_{k+1}}\|_2\leq \sup _{n\in \N }\|f_n\|_2\leq M\), the first term is at most \(\frac {4M^2}{(k+1)^2}\). By the induction hypothesis the second term is at most \( k\frac {6M^2}{(k+1)^2}\). Finally, by choice of \(n_{k+1}\) the final term is at most \(\frac {2M^2}{(k+1)^2}\). Summing, we see that we have chosen \(n_{k+1}\) such that \begin {equation*} \|\varphi _{k+1}-f\|_p^2\leq \frac {6M^2}{k+1}. \end {equation*} Thus, we have chosen \(n_k\) such that \(\varphi _k\to f\) strongly in \(L^2(E)\), as desired.
[Definition: Continuous function] If \((X,d)\) is a metric space, then a function \(T:X\to \R \) is continuous if, whenever \(\{x_n\}_{n=1}^\infty \) is a sequence in \(X\) that satisfies \(x_n\to x\) for some \(x\in X\), we have that \(T(x_n)\to T(x)\).
(Memory 2591) A linear functional on a normed linear space is continuous if and only if it is bounded.
[Definition: Closed set] If \((X,d)\) is a metric space, then a subset \(C\subseteq X\) is closed if, whenever \(\{x_n\}_{n=1}^\infty \) is a sequence in \(C\) that satisfies \(x_n\to x\) for some \(x\in X\), we have that \(x\in C\).
(Memory 2592) If \((K,d)\) is a compact metric space and \(f:K\to \R \) is continuous, then there is an \(x\in K\) such that \(f(x)\geq f(y)\) for all \(y\in K\).
The Heine-Borel theorem. If \(K\subset \R ^n\) is closed and bounded, then \(K\) is compact.
(Memory 2593) If \((K,d)\) is a compact metric space and \(\{x_n\}_{n=1}^\infty \) is a sequence in \(K\), then \(\{x_n\}_{n=1}^\infty \) has a convergent subsequence. (If \(K\subset \R ^n\) then this is the Bolzano-Weierstrauß theorem.)
(Micah, Problem 2600) Let \(\{a_n\}_{n=1}^\infty \) be a sequence in \(\R \). Define \(b_n\) by \begin {equation*} b_n=\frac {1}{n}\sum _{k=1}^n a_n. \end {equation*} Show that \begin {equation*} \liminf _{n\to \infty } a_n \leq \liminf _{n\to \infty }b_n\leq \limsup _{n\to \infty } b_n\leq \limsup _{n\to \infty } a_n. \end {equation*} Give examples to show that all three inequalities may be strict.
We provide only the example, as the proof was given in class. Let \(a_n=(-1)^n (2n-1)\).Then \(b_1=a_1=1\), and if \(n\in \N \) then \begin {equation*} b_{n+1}=\frac {1}{n+1}\sum _{k=1}^{n+1} (-1)^k (2k-1) =\frac {1}{n+1}\bigl (nb_{n} - (-1)^n (2n+1)\bigr ) . \end {equation*} A straightforward induction argument establishes that \(b_n=(-1)^n\). Thus \begin {equation*} \liminf _{n\to \infty } a_n =-\infty , \qquad \liminf _{n\to \infty }b_n=-1, \qquad \limsup _{n\to \infty } b_n=+1, \qquad \limsup _{n\to \infty } a_n=\infty \end {equation*} and so \begin {equation*} \liminf _{n\to \infty } a_n<\liminf _{n\to \infty }b_n<\limsup _{n\to \infty } b_n<\limsup _{n\to \infty } a_n \end {equation*} as desired.
(Muhammad, Problem 2610) Let \(E\subseteq \R \) be measurable and let \(1\leq p\leq \infty \). If we regard \(L^p(E)\) as a metric space with the metric \(d(f,g)=\|f-g\|_p\), show that \begin {equation*} \{f\in L^p(E):\|f\|_p\leq 1\} \end {equation*} is both closed and bounded.
(Ashley, Problem 2620) With \(E\) and \(p\) as before, show that \(\{f\in L^p(E):\|f\|_p\leq 1\}\) is not compact.
[Definition: Convex set] Let \(X\) be a linear space and let \(C\subseteq X\). We say that \(C\) is convex if, whenever \(f\), \(g\in C\), we have that \(\{\lambda f+(1-\lambda g):0\leq \lambda \leq 1\}\subseteq C\).
(Problem 2621) If \(r\in (0,\infty )\), show that \(\{x\in X:\|x\|\leq r\}\) is convex.
(Problem 2622) Show that the intersection of a collection of convex sets is convex.
Lemma 8.16a. Let \(E\subseteq \R \) be measurable and let \(1<p<\infty \). Suppose that \(C\subset L^p(E)\) is closed and convex. Suppose further that \(\{f_n\}_{n=1}^\infty \) is a sequence in \(C\) and that \(f_n\rightharpoonup f\) weakly in \(L^p(E)\) for some \(f\in L^p(E)\). Then in addition \(f\in C\).
(Bashar, Problem 2630) Prove Lemma 8.16a.
By Banach-Saks theorem, there is a subsequence \(\{f_{n_k}\}_{k=1}^\infty \) such that, if \begin {equation*} \phi _k=\frac {1}{k}\sum _{\ell =1}^k f_{n_\ell }, \end {equation*} then \(\phi _k\to f\) strongly in \(L^p(E)\).We observe that \(\phi _1=f_{n_1}\in C\) by assumption. Now suppose that \(\phi _k\in C\). Then \begin {equation*} \phi _{k+1}=\frac {1}{k+1} f_{n_{k+1}} + \frac {k}{k+1}\phi _k. \end {equation*} The number \(\lambda =\frac {1}{k+1}\) is in \([0,1]\), and the functions \(f_{n_{k+1}}\) and \(\phi _k\) are in \(C\) by assumption. Thus, because \(C\) is convex, we have that \(\phi _{k+1}\in C\).
Thus, by induction, \(\phi _k\in C\) for all \(k\in \N \).
But \(C\) is closed. By definition of closed set, if \(\phi _k\to f\) strongly in \(L^p(E)\), and \(\phi _k\in C\), then \(f\in C\), as desired.
(Problem 2631) Give an example to show that the assumption that \(C\) is closed in Lemma 8.16a is necessary.
(Dibyendu, Problem 2640) Give an example to show that the assumption that \(C\) is convex in Lemma 8.16a is necessary.
[Definition: Convex functional] Let \(X\) be a linear space, let \(C\subseteq X\) be convex, and let \(T:C\to \R \). We do not assume that \(T\) is either linear or continuous. We say that \(T\) is convex if \begin {equation*} T(\lambda f+(1-\lambda ) g) \leq \lambda T(f)+(1-\lambda )\,T(g) \end {equation*} for all \(f\), \(g\in C\) and all \(\lambda \in [0,1]\).
(Problem 2641) Show that every linear functional is convex.
(Problem 2642) If \(X\) is a normed linear space, show that the functional \(T:X\to \R \) given by \(T(x)=\|x\|\) is convex but not linear.
Theorem 8.17. Let \(E\subseteq \R \) be measurable and let \(1<p<\infty \). Suppose that \(C\subset L^p(E)\) is closed, bounded, and convex. Suppose further that \(T:C\to \R \) is continuous and convex. Then there is a \(f\in C\) such that \begin {equation*} T(f)\leq T(g)\quad \text {for all }g\in C. \end {equation*}
Lemma 8.16b. Let \(E\subseteq \R \) be measurable and let \(1<p<\infty \). Suppose that \(C\subset L^p(E)\) is closed, bounded, and convex. Suppose further that \(T:C\to \R \) is continuous and convex. Suppose further that \(\{f_n\}_{n=1}^\infty \) is a sequence in \(C\) and that \(f_n\rightharpoonup f\) weakly in \(L^p(E)\) for some \(f\in L^p(E)\). Then \(f\in C\), and \begin {equation*} T(f)\leq \liminf _{n\to \infty } T(f_n). \end {equation*}
(Irina, Problem 2650) Prove Lemma 8.16b. (Note that we will use Lemma 8.16b in the proof of Theorem 8.17, so you may not use Theorem 8.17 to prove Lemma 8.16b.)
By Problem 2542, there is a subsequence \(\{T(f_{n_\ell })\}_{\ell =1}^\infty \) of \(\{T(f_n)\}_{n=1}^\infty \) such that \begin {equation*} \lim _{\ell \to \infty } T(f_{n_\ell })=\liminf _{n\to \infty } T(f_n). \end {equation*} We furthermore have that \(f_{n_\ell }\rightharpoonup f\) weakly in \(L^p(E)\). Thus, there is a sequence of functions \(\{f_{n_\ell }\}_{\ell =1}^\infty \) in \(C\) that converges weakly to \(f\) and such that \begin {equation*} \lim _{\ell \to \infty } T(f_{n_\ell })=\liminf _{\ell \to \infty } T(f_{n_\ell })=\liminf _{n\to \infty } T(f_n). \end {equation*} It is inconvenient to work with subsequences, so at this point we will redefine \(f_\ell \) so that \(\{f_{\ell }\}_{\ell =1}^\infty \) now refers to the subsequence \(\{f_{n_\ell }\}_{\ell =1}^\infty \) of the original subsequence. This procedure happens fairly often at the beginning of complicated proofs and is traditionally described as follows:By passing to a subsequence (see Problem 2542), we may assume that \begin {equation*} \lim _{n\to \infty } T(f_{n})=\liminf _{n\to \infty } T(f_n). \end {equation*}
That \(f\in C\) is Lemma 8.16a. As in the proof of Lemma 8.16a, by the Banach-Saks theorem, there is a subsequence \(\{f_{\ell _k}\}_{k=1}^\infty \) of \(\{f_\ell \}_{\ell =1}^\infty \) such that if \begin {equation*} \phi _k=\frac {1}{k}\sum _{j=1}^k f_{\ell _j} \end {equation*} then \(\phi _k\to f\) strongly in \(L^p(E)\). Because \(T\) is continuous \(T:C\to \R \) and \(\phi _k\), \(f\in C\), we have that \begin {equation*} T(f)=\lim _{k\to \infty } T(\phi _k). \end {equation*} A straightforward induction argument (similar to that in Lemma 8.16a) establishes that \begin {equation*} T(\phi _k)=T\Bigl (\frac {1}{k}\sum _{j=1}^k f_{\ell _j}\Bigr ) \leq \frac {1}{k}\sum _{j=1}^k T(f_{\ell _j}). \end {equation*} Because \(\{T(f_{\ell _j})\}_{j=1}^\infty \) is a subsequence of \(\{T(f_{\ell })\}_{\ell =1}^\infty \), and \begin {equation*} \lim _{\ell \to \infty } T(f_\ell )=\liminf _{\ell \to \infty } T(f_\ell ), \end {equation*} we may apply Problem 2600 to see that \begin {align*} T(f)&=\lim _{k\to \infty } T(\phi _k) \\&\leq \lim _{k\to \infty }\frac {1}{k}\sum _{j=1}^k T(f_{\ell _j}) = \lim _{k\to \infty } T(f_{\ell _k}) = \lim _{\ell \to \infty } T(f_{\ell }) \alignbreak = \liminf _{\ell \to \infty } T(f_{\ell }). \end {align*}
(The second limit exists by the squeeze theorem.) This completes the proof.
(Micah, Problem 2660) Let \(T\) and \(C\) be as in Theorem 8.17. Prove that \(\{T(f):f\in C\}\) is bounded below.
(Muhammad, Problem 2670) Prove Theorem 8.17.
[Definition: Measure space] Recall the definition of a \(\sigma \)-algebra: let \(X\) be a set and let \(\M \subseteq 2^X\). We say that \(\M \) is a \(\sigma \)-algebra of subsets of \(X\), or a \(\sigma \)-algebra over \(X\), if
Recall the definition of a measure: if \(X\) is a set and \(\M \) is a \(\sigma \)-algebra of subsets of \(X\), then we call \((X,\M )\) a measurable space. A measure on a measurable space \((X,\M )\) is a function \(\mu \) such that:
A measure space is a triple \((X,\M ,\mu )\), where \(X\) is a set, \(\M \) is a \(\sigma \)-algebra over \(X\), and \(\mu \) is a measure on \((X,\M )\).
(Problem 2671) The following are measure spaces:
\((\R ,\mathcal {L},m)\), where \(\mathcal {L}\) is the set of Lebesgue measurable sets and where \(m\) denotes Lebesgue measure.
\((\R ,\mathcal {B},m)\), where \(\mathcal {B}\) is the set of Borel sets.
\((X,2^X,\eta )\), where \(X\) is any set and where \(\eta \) is given by \(\eta (S)=|S|\) if \(S\) is finite and \(\eta (S)=\infty \) if \(S\) is not finite.
\((X,\M ,\delta _{x_0})\), where \(X\) is any nonempty set, \(\M \) is any \(\sigma \)-algebra over \(X\), \(x_0\in X\), and \(\delta _{x_0}(S)=1\) if \(x_0\in S\) and \(\delta _{x_0}(S)=0\) if \(x_0\notin S\).
\((X,\mathcal {C},\mu )\), where \(X\) is any uncountable set, \begin {equation*} \mathcal {C}=\{S\subseteq X:S\text { is countable}\}\cup \{S\subseteq X:X\setminus S\text { is countable}\}, \end {equation*} and \(\mu (S)=0\) if \(S\) is countable and \(\mu (S)=1\) if \(S\) is uncountable.
[Chapter 17, Problem 6] Let \((X,\M ,\mu )\) be a measure space. If \(E\in \M \), then \((E,\M _E,\mu _E)\) is a measure space, where \(\M _E=\{S\cap E:S\in \M \}=\{S\in \M :S\subseteq E\}\) and \(\mu _E=\mu \big \vert _{\M _E}\).
(Problem 2672) Let \(f:\R \to [0,\infty ]\) be measurable. Define \(\mu (A)=\int _A f\) for all measurable sets \(A\subseteq \R \). Show that \(\mu \) is a measure on the \(\sigma \)-algebra \(\M \) of Lebesgue measurable subsets of \(\R \).
Proposition 17.1. Let \((X,\M ,\mu )\) be a measure space.
Finite additivity:
If \(\{E_k\}_{k=1}^n\subseteq \M \) is a finite collection of sets in \(\M \) and \(E_k\cap E_j=\emptyset \) for all \(j\neq k\), then \begin {equation*} \mu \Bigl (\bigcup _{k=1}^n E_k\Bigr )=\sum _{k=1}^n \mu (E_k). \end {equation*}
Monotonicity:
If \(A\), \(B\in \M \) with \(A\subseteq B\), then \begin {equation*} \mu (A)\leq \mu (B). \end {equation*}
Excision:
If \(A\), \(B\in \M \) with \(A\subseteq B\) and \(\mu (A)<\infty \), then \begin {equation*} \mu (B\setminus A)=\mu (B)-\mu (A). \end {equation*}
Countable monotonicity:
If \(\{E_k\}_{k=1}^\infty \subseteq \M \), then \begin {equation*} \mu \Bigl (\bigcup _{k=1}^\infty E_k\Bigr )\leq \sum _{k=1}^\infty \mu (E_k). \end {equation*}
(Problem 2673) Prove Proposition 17.1.
Proposition 17.2. Let \((X,\M ,\mu )\) be a measure space.
(Problem 2674) Prove Proposition 17.2.
The proof is identical to the proof of Theorem 2.15.
The Borel-Cantelli Lemma in general measure spaces. Let \((X,\M ,\mu )\) be a measure space. Let \(\{E_k\}_{k=1}^\infty \) be a sequence of sets in \(\M \). Suppose that \(\sum _{k=1}^\infty \mu (E_k)<\infty \). Then \(|\{k\in \N :x\in E_k\}|<\infty \) for almost every \(x\in X\).
(Problem 2675) Prove the Borel-Cantelli Lemma in general measure spaces.
The proof is identical to the proof of the version of the Borel-Cantelli Lemma in \((\R ,\mathcal {L},m)\).
[Definition: Finite measure] Let \((X,\M ,\mu )\) be a measure space. We say that \(\mu \) is finite if \(\mu (X)<\infty \). (Note that \(X\) need not be finite!)
[Definition: \(\sigma \)-finite measure]Let \((X,\M ,\mu )\) be a measure space. We say that \(\mu \) is finite if \begin {equation*} X=\bigcup _{n=1}^\infty X_n \end {equation*} where \(\mu (X_n)<\infty \) for all \(n\in \N \).
(Ashley, Problem 2680) Give an example of a measure space that is not \(\sigma \)-finite.
Let \(X=\{1,2,3\}\). This is a set. Let \(\M =\{\emptyset ,X\}\); then \(\M \) is a \(\sigma \)-algebra over \(X\). Define \(\mu \) by \(\mu (\emptyset )=0\) and \(\mu (X)=\infty \). Then \(\mu \) is a measure, but is not \(\sigma \)-finite.
[Definition: Complete measure space] Let \((X,\M ,\mu )\) be a measure space. We say that \((X,\M ,\mu )\) is complete if, whenever \(E\in \M \), \(\mu (E)=0\), and \(D\subseteq E\), we have that \(D\in \M \).
(Bashar, Problem 2690) Show that \((\R ,\mathcal {B},m)\) is not complete.
(Dibyendu, Problem 2700) Let \((X,\M ,\mu )\) be a measure space that is not complete. How should we define the completion of \((X,\M ,\mu )\) and why should we define it that way?
Let \[\M _0=\{S\cup Z:S\in \M ,Z\subseteq E\text { for some }E\in \M \text { with }\mu (E)=0\}.\] Clearly \(\emptyset \in \M _0\) and \(\M _0\) is closed under countable unions. To show that \(\M _0\) is a \(\sigma \)-algebra we need only show that it is closed under complements.Suppose \(S\cup Z\in \M _0\). Then \(Z\subseteq E\) for some \(E\in \M \) with \(\mu (E)=0\). We observe that \[X\setminus (S\cup Z)= (X\setminus (S\cup E)) \cup (E\setminus Z).\] But \(E\setminus Z\subseteq E\) and \(X\setminus (S\cup E)\in \M \), so \(\M _0\) is closed under complements. Thus \(\M _0\) is a \(\sigma \)-algebra over \(X\).
Define \(\mu _0:\M _0\to [0,\infty ]\) by \[\mu _0(S\cup Z)=\mu (S)\] whenever \(S\in \M \) and \(Z\subseteq E\) for some \(E\in \M \) with \(\mu (E)=0\). We must first show that \(\mu \) is well defined. Suppose that \(S\cup Z=A\cup B\) where \(S\), \(A\) are measurable and \(Z\subseteq E\), \(B\subseteq D\) where \(\mu (E)=0=\mu (D)\). We want to show that \(\mu (S)=\mu (A)\).
Then \(\mu (S)=\mu (S\cup E)\) because \(\mu (E)=0\). But \[A\subseteq A\cup B=S\cup Z\subseteq S\cup E\] and so \(\mu (A)\leq \mu (S\cup E)=\mu (S)\). By symmetry \(\mu (S)\leq \mu (A)\) and so \(\mu (S)=\mu (A)\) and \(\mu _0\) is well defined.
Clearly \(\mu _0\) is a complete measure. We call \((X,\M _0,\mu _0)\) the completion of \((X,\M ,\mu )\).
[Definition: Outer measure] Let \(X\) be a set. An outer measure on \(2^X\) is a function \(\mu ^*\) such that
(Problem 2701) Show that if \(\mu \) is a measure on the measure space \((X,2^X)\), then \(\mu \) is an outer measure on \(2^X\).
(Problem 2702) Give an example of an outer measure that is not a measure.
[Chapter 17, Problem 20] If \(X\) is a nonempty set and \(\alpha \in [0,\infty ]\), then the function \(\eta :2^X\to [0,\infty ]\) given by \(\eta (\emptyset )=0\) and \(\eta (S)=\alpha \) for all nonempty subsets \(S\) of \(X\) is an outer measure. In particular, the constant function zero is an outer measure.
[Definition: Measurable] If \(\mu ^*\) is an outer measure on \(2^X\) and \(E\subseteq X\), we call \(E\) measurable with respect to \(\mu ^*\) if \[\mu ^*(A)=\mu ^*(A\cap E)+\mu ^*(A\setminus E)\] for all \(A\subseteq X\).
(Problem 2703) A set \(E\subseteq X\) is measurable with respect to \(\mu ^*\) if and only if \[\mu ^*(A\cap E)+\mu ^*(A\setminus E)\leq \mu ^*(A)\] for all \(A\subseteq X\) that satisfy \(\mu ^*(A)<\infty \).
Proposition 17.5. The union of a finite collection of measurable sets is measurable.
(Irina, Problem 2710) Prove Proposition 17.5.
The proof is identical to that of Proposition 2.5.
Proposition 17.6. Let \(\mu ^*\) be an outer measure on \(2^X\), let \(A\subseteq X\), and let \(\{E_k\}_{k=1}^n\) be a finite collection such that each \(E_k\) is measurable with respect to \(\mu ^*\) and such that \(E_k\cap E_\ell =\emptyset \) if \(k\neq \ell \). Show that \[\mu ^*\Bigl (A\cap \Bigl [\bigcup _{k=1}^n E_k\Bigr ]\Bigr ) = \sum _{k=1}^n \mu ^*(A\cap E_k).\]
(Micah, Problem 2720) Prove Proposition 17.6.
The proof is identical to that of Proposition 2.6
Proposition 17.7. The union of a countable collection of measurable sets is measurable.
(Problem 2721) Prove Proposition 17.6.
The proof is identical to that of Proposition 2.7
Theorem 17.8. Let \(\mu ^*\) be an outer measure on \(2^X\). Let \(\M =\{E\subseteq X:E\) is measurable with respect to \(\mu ^*\}\). Let \(\bar \mu =\mu ^*\big \vert _{\M }\).
Then \(\M \) is a \(\sigma \)-algebra over \(X\), and \((X,\M ,\bar \mu )\) is a complete measure space.
(Muhammad, Problem 2730) Let \(\mu ^*\) and \(\M \) be as in Theorem 17.8. Show that \(\M \) is a \(\sigma \)-algebra over \(X\) and that \((X,\M ,\bar \mu )\) is a measure space.
(Ashley, Problem 2740) Complete the proof of Theorem 17.8 by showing that \((X,\M ,\bar \mu )\) is complete.
By the definition of a complete measure space, we need only show that if \(E\in \M \), \(D\subseteq E\), and \(\mu ^*(E)=0\), then \(D\in \M \).Let \(A\in 2^X\). We need only show that \[\mu ^*(A)=\mu ^*(A\cap D)+\mu ^*(A\setminus D)\] for all such \(A\).
First, \(A\cap D\subseteq D\subseteq E\), and so \(\mu ^*(A\cap D)\leq \mu ^*(E)\) by monotonicity of outer measures. But \(\mu ^*:2^X\to [0,\infty ]\) so \(\mu ^*(A\cap D)\geq 0\), and \(\mu ^*(E)=0\) by assumption so \(\mu ^*(A\cap D)\leq 0\). Thus \(\mu ^*(A\cap D)=0\).
Again by monotonicity \(\mu ^*(A\setminus D)\leq \mu ^*(A)\). But by countable subadditivity \[\mu ^*(A)=\mu ^*((A\cap D)\cup (A\setminus D))\leq \mu ^*(A\cap D)+\mu ^*(A\setminus D).\] Combining the above estimates, we have that \[ \mu ^*(A)\leq \mu ^*(A\cap D)+\mu ^*(A\setminus D)=\mu ^*(A\setminus D)\leq \mu ^*(A)\] and so we must have that \[\mu ^*(A)=\mu ^*(A\cap D)+\mu ^*(A\setminus D)\] as desired.
[Definition: Empty sum] If \(\S \) is a set and \(f:\S \to [-\infty ,\infty ]\) is a function, then \(\emptyset \) is a finite (therefore countable) subset of \(\S \) and \[\sum _{x\in \emptyset } f(x)=0.\]
[Definition: Empty union] If \(\S \) is a collection of sets, then \(\emptyset \) is a finite (therefore countable) subset of \(\S \) and \[\bigcup _{I\in \emptyset } I=\emptyset .\]
[Definition: Supremum and infimum of the empty set] Recall that if \(A\subseteq [-\infty ,\infty ]\), then \(\sup A\) is the smallest element of \([-\infty ,\infty ]\) such that \(\sup A\geq x\) for all \(x\in A\), and \(\inf A\) is the largest element of \([-\infty ,\infty ]\) such that \(\inf A\leq x\) for all \(x\in A\). Applying these definitions with \(A=\emptyset \), we see that \[\sup \emptyset = -\infty ,\qquad \inf \emptyset = \infty .\]
[Definition: Addition and subtraction on the extended real numbers]
If \(r\), \(s\in \R \), then \(r+s\) and \(r-s\) take their usual values.
If \(r\in \R \), then \(\infty +r=\infty -r=\infty \).
If \(r\in \R \), then \(r-\infty =-\infty \).
\(\infty +\infty =\infty \).
\(-\infty -\infty =-\infty \).
\(\infty -\infty \) is undefined.
In particular, if \(r+s=t+s\), you can only conclude that \(r=t\) if \(s\) is finite.
[Definition: Outer measure induced by a set function] Let \(X\) be a set, let \(\mathcal {S}\subseteq 2^X\), and let \(\mu :\mathcal {S}\to [0,\infty ]\) be a function. Then the outer measure \(\mu ^*\) induced by \(\mu \) is defined by \[\mu ^*(\emptyset )=0,\quad \mu ^*(A)=\inf \Bigl \{\sum _{I\in \mathcal {C}}\mu (I):\mathcal {C}\subseteq \mathcal {S} \text { is countable and }A\subseteq \bigcup _{I\in \mathcal {C}} I\Bigr \}.\] (Here a set is countable if it is either finite or countably infinite, and we take the infimum of the empty set to be \(\infty \). Note that the definition \(\mu ^*(\emptyset )=0\) is redundant because \(0=\sum _{I\in \emptyset } \mu (I)\) is in the set on the right.)
Theorem 17.9. The outer measure induced by a set function is an outer measure.
(Bashar, Problem 2750) Prove Theorem 17.9.
By the definition of outer measure, we must prove that
- (a)
- \(\mu ^*:2^X\to [0,\infty ]\).
- (b)
- \(\mu ^*(\emptyset )=0\),
- (c)
- If \(D\subseteq E\subseteq X\), then \(\mu ^*(D)\leq \mu ^*(E)\).
- (d)
- If \(\{E_k\}_{k=1}^\infty \subseteq 2^X\), then \begin {equation*} \mu ^*\Bigl (\bigcup _{k=1}^\infty E_k\Bigr )\leq \sum _{k=1}^\infty \mu ^*(E_k). \end {equation*}
The first two conditions are true by definition.
Suppose \(D\subseteq E\subseteq X\). If there are no countable covers of \(E\) by elements of \(\mathcal {S}\), then \(\mu ^*(E)=\infty \) and so \(\mu ^*(D)\leq \mu ^*(E)\) for any value of \(\mu ^*(D)\). Otherwise, let \(\mathcal {C}\) be a countable cover of \(E\) by elements of \(\mathcal {S}\). Then \(\mathcal {C}\) is a cover of \(D\) as well. Thus, \[\Bigl \{\sum _{I\in \mathcal {C}}\mu (I):\mathcal {C}\subseteq \mathcal {S} \text { is countable and }E\subseteq \bigcup _{I\in \mathcal {C}} I\Bigr \} \subseteq \Bigl \{\sum _{I\in \mathcal {C}}\mu (I):\mathcal {C}\subseteq \mathcal {S} \text { is countable and }D\subseteq \bigcup _{I\in \mathcal {C}} I\Bigr \}\] and so their respective infima must satisfy the desired inequality, that is, \(\mu ^*(E)\geq \mu ^*(D)\).
Finally, let \(\{E_k\}_{k=1}^\infty \subseteq 2^X\). We seek to show that \begin {equation*} \mu ^*\Bigl (\bigcup _{k=1}^\infty E_k\Bigr )\leq \sum _{k=1}^\infty \mu ^*(E_k). \end {equation*} The proof is identical to the proof of Proposition 2.3 and so we are done.
[Definition: Induced Carathéodory measure] If \(\mu ^*\) is the outer measure on \(2^X\) induced by the set function \(\mu \), and if \((X,\M ,\bar \mu )\) is defined from \(\mu ^*\) as in Theorem 17.8, we call \(\bar \mu \) the Carathéodory measure induced by \(\mu \).
(Problem 2751) The Lebesgue measure is the Carathéodory measure induced by \(\ell \), where \(\mathcal {I}\) is the collection of all bounded open intervals in \(\R \) and \(\ell :\mathcal {I}\to [0,\infty ]\) maps \(I\) to the length of \(I\).
[Definition: \(\mathcal {S}_{\sigma \delta }\)] Let \(\mathcal {S}\subseteq 2^X\) for some set \(X\). Then \(\mathcal {S}_\sigma \) is the collection of countable unions of elements of \(\mathcal {S}\) and \(\mathcal {S}_{\sigma \delta }\) is the collection of countable intersections of elements of \(\mathcal {S}_\sigma \).
(Dibyendu, Problem 2760) Let \(X\) be a set and let \(\mathcal {S}\subseteq 2^X\). Let \(\mu :\mathcal {S}\to [0,\infty ]\) be a set function and let \(\mu ^*\) and \((X,\M ,\bar \mu )\) be the outer measure and Carathéodory measure space induced by \(\mu \).
If \(E\in X\) and \(\varepsilon >0\), show that there is an \(A\in \mathcal {S}_\sigma \) with \(E\subseteq A\) and with \(\mu ^*(A)<\mu ^*(E)+\varepsilon \).
Proposition 17.10. Let \(X\) be a set and let \(\mathcal {S}\subseteq 2^X\). Let \(\mu :\mathcal {S}\to [0,\infty ]\) be a set function and let \((X,\M ,\bar \mu )\) be the Carathéodory measure space induced by \(\mu \).
If \(E\in X\) and \(\mu ^*(E)<\infty \), then there is an \(A\) with \[A\in \mathcal {S}_{\sigma \delta },\quad E\subseteq A,\quad \text {and}\quad \mu ^*(E)=\mu ^*(A).\]
(Irina, Problem 2770) Prove Proposition 17.10.
[Definition: Arc length] If \((X,d)\) is a metric space and \(\gamma :[0,1]\to X\) is a continuous function, we define its arc length by \[\ell (\gamma )=\sup \Bigl \{\sum _{i=1}^n d(\gamma (x_i),\gamma (x_{i-1})) : a=x_0<x_1<x_2<\dots <x_n=b\Bigr \}.\]
(Memory 2771) If \(\gamma :[0,1]\to \R ^n\) is continuously differentiable with bounded derivative, then \[\ell (\gamma )=\int _a^b\|\gamma '\|.\] In particular, if \(\gamma :[a,b]\to \R ^2\) is given by \(\gamma (t)=(t,g(t))\) for a continuously differentiable function \(g\) with bounded derivative, then \[\ell (\gamma )=\int _a^b \sqrt {1+|g'(x)|^2}\,dx.\]
(Micah, Problem 2780) Let \(X\) be a set. Let \(\mathcal {I}\) be an index set and let \(\mu ^{(i)}\) be an outer measure on \(2^X\) for each \(i\in \mathcal {I}\). Show that \(\mu ^*\) given by \(\mu ^*(S)=\sup _{i\in \mathcal {I}}\mu ^{(i)}(S)\) is also an outer measure.
[Definition: Hausdorff measure] Let \((X,d)\) be a metric space. Let \(\alpha \geq 0\). If \(\delta >0\), define \[\mathcal {S}_\delta = \{E\subseteq X:0\leq \diam (E)<\delta \}\] and let \(h_\alpha ^{(\delta )}:\mathcal {S}_\delta \to [0,\infty ]\) be given by \[h_\alpha ^{(\delta )}(E)=\diam (E)^\alpha .\] (Observe that \(\diam (\emptyset )=-\infty \) and so \(\emptyset \notin \mathcal {S}_\delta \). We take \(h_0^{(\delta )}(E)=1\) if \(E\neq \emptyset \), even if \(\diam (E)=0.\))
We let \(H_\alpha ^{(\delta )}\) be the outer measure induced by the set function \(h_\alpha ^{(\delta )}\) and let the Hausdorff outer measure of dimension \(\alpha \) on \((X,d)\) be given by \(H^*_\alpha =\sup _{\delta >0} H^{(\delta )}_\alpha \).
(Muhammad, Problem 2790) Show that \(H_\alpha ^{(\delta )}\) is decreasing in \(\delta \), that is, if \(0<\delta _1<\delta _2\) and \(Y\subseteq X\) then \(H_\alpha ^{(\delta _1)}(Y)\geq H_\alpha ^{(\delta _2)}(Y)\). Conclude that \(H^*_\alpha (Y)=\lim _{\delta \to 0^+} H^{(\delta )}_\alpha (Y)\).
Proposition 20.30. Let \(0<\alpha <\beta \) and let \(Y\subseteq X\) for some metric space \((X,d)\). If \(H^*_\alpha (Y)<\infty \), then \(H^*_\beta (Y)=0\).
(Ashley, Problem 2800) Prove Proposition 20.30.
(Bashar, Problem 2810) If \(X=\R \) and \(d(x,y)=|x-y|\), show that \(H_1^{(\delta )}(Y)=m^*(Y)\) for all \(Y\subseteq \R \) and all \(\delta >0\). Conclude that \(H_1^*=m^*\).
(Dibyendu, Problem 2820) Define \(\mathcal {O}_\delta = \{E\subseteq X:E\) is open and \(0\leq \diam (E)<\delta \}\). Let \(\tilde H_\alpha ^{(\delta )}\) be the outer measure induced by the set function \(h_\alpha ^{(\delta )}\big \vert _{\mathcal {O}_\delta }\) (that is, by \(h_\alpha ^{(\delta )}\) restricted to \(\mathcal {O}_\delta \)). Show that \(\tilde H_\alpha ^{(\delta )}=H_\alpha ^{(\delta )}\).
By the definition of outer measure induced by a set function, if \(Y\subseteq X\) then \begin {align*} H^{(\delta )}_\alpha (Y)&= \inf \Bigl \{\sum _{I\in \mathcal {C}}h_{\alpha }^{(\delta )}(I):\mathcal {C}\subseteq \mathcal {S}_\delta \text { is countable and }Y\subseteq \bigcup _{I\in \mathcal {C}} I\Bigr \} ,\\ \widetilde H^{(\delta )}_\alpha (Y)&= \inf \Bigl \{\sum _{I\in \mathcal {C}}h_{\alpha }^{(\delta )}(I):\mathcal {C}\subseteq \mathcal {O}_\delta \text { is countable and }Y\subseteq \bigcup _{I\in \mathcal {C}} I\Bigr \}. \end {align*}But \(\mathcal {O}_\delta \subset \mathcal {S}_\delta \), and so \[\Bigl \{\sum _{I\in \mathcal {C}}h_{\alpha }^{(\delta )}(I):\mathcal {C}\subseteq \mathcal {O}_\delta \text { is countable and }Y\subseteq \bigcup _{I\in \mathcal {C}} I\Bigr \} \subseteq \Bigl \{\sum _{I\in \mathcal {C}}h_{\alpha }^{(\delta )}(I):\mathcal {C}\subseteq \mathcal {S}_\delta \text { is countable and }Y\subseteq \bigcup _{I\in \mathcal {C}} I\Bigr \}. \] Therefore, because \(\inf \) is decreasing, we must have that \(H^{(\delta )}_\alpha (Y)\leq \widetilde H^{(\delta )}_\alpha (Y)\).
Conversely, suppose that \(H^{(\delta )}_\alpha (Y)<\infty \); if not, then \(H^{(\delta )}_\alpha (Y)=\infty \geq \widetilde H^{(\delta )}_\alpha (Y)\) and so the two must be equal.
Let \(\varepsilon >0\) and let \(\mathcal {C}\subseteq \mathcal {S}_\delta \) be countable and such that \(Y\subseteq \bigcup _{I\in \mathcal {C}} I\) and \(\sum _{I\in \mathcal {C}}h_{\alpha }^{(\delta )}(I)<H^{(\delta )}_\alpha (Y)+\varepsilon \). Such a \(\mathcal {C}\) must exist by definition of infimum; observe that \(h_{\alpha }^{(\delta )}(I)<\infty \) for all \(I\in \mathcal {C}\).
We may write \(\mathcal {C}=\{I_k:1\leq k< \infty \}\) or \(\mathcal {C}=\{I_k:1\leq k\leq N\}\) for some \(N\in \N \). If \(I=I_k\in \mathcal {C}\), then \(\diam (I)<\delta \) by definition of \(\mathcal {S}_\delta \). Let \(r_k>0\) be such that \(\diam (I)+2r_k<\delta \) and such that \((\diam I+2r_k)^\alpha < \diam (I)^\alpha + 2^{-k}\varepsilon \); such a \(r_k\) must exist because \(\delta >\diam (I)\) and because the function \(f(x)=x^\alpha \) is continuous.
Let \(U_k=\bigcup _{x\in I_k} B(x,r_k)\). An elementary real analysis argument yields that \(I_k\subseteq U_k\), \(U_k\) is open, and \(\diam (U_k)\leq \diam (I_k)+2r_k<\delta \).
Then \(\widetilde {\mathcal {C}}=\{U_k:I_k\in \mathcal {C}\}\) is a countable collection of elements of \(\mathcal {O}_\delta \) that cover \(Y\). Therefore \[\sum _{U\in \widetilde {\mathcal {C}}} h_\alpha ^{(\delta )}\big \vert _{\mathcal {O}_\delta }(U) \geq \widetilde H_\alpha ^{(\delta )}(Y)\] by definition of \(\widetilde H_\alpha ^{(\delta )}\). But \[\sum _{U\in \widetilde {\mathcal {C}}} h_\alpha ^{(\delta )}\big \vert _{\mathcal {O}_\delta }(U) =\sum _{U\in \widetilde {\mathcal {C}}} \diam (U)^\alpha <\sum _{k=1}^N \diam (I_k)^\alpha +2^{-k}\varepsilon < H_\alpha ^{(\delta )}(Y)+2\varepsilon . \] Thus, \(\widetilde H_\alpha ^{(\delta )}(Y)<H_\alpha ^{(\delta )}(Y)+2\varepsilon \) for all \(\varepsilon >0\), and so \(\widetilde H_\alpha ^{(\delta )}(Y)\leq H_\alpha ^{(\delta )}(Y)\), as desired.
[Chapter 20, Problem 54] If \(\gamma :[0,1]\to X\) is a continuous curve that is one-to-one, show that \(\ell (\gamma )=H_1^*(\gamma ([0,1]))\).
(Irina, Problem 2830) If \((X,d)\) is a metric space, show that \(H_0^*\) is the counting measure.
Let \(Y\subseteq X\). We must show that \(H_0^*(Y)=|Y|\).We first observe that \(\diam (I)^0=1\) by definition whenever \(I\) is nonempty, and so if \(\mathcal {C}\) is a collection of elements of \(\mathcal {S}_\delta \), then \(\sum _{I\in \mathcal {C}} h_\alpha ^{(\delta )}(I)=|\mathcal {C}|\).
There are two cases: the case where \(|Y|<\infty \) and where \(|Y|=\infty \).
If \(Y\) is finite, then so is \(\Delta =\{d(x,y):x,y\in Y,\>x\neq y\}\). Furthermore, every element of this set is positive. Let \(\delta <\min (\Delta )\). If \(I\subset X\) with \(\diam (Y)<\delta \) and \(x\in Y\cap I\), then \(y\notin I\) for any other \(y\in Y\) because \(d(x,y)>\delta >\diam (Y)\). Thus, any cover of \(Y\) by elements of \(\mathcal {S}_\delta \) must contain at least \(|Y|\) elements, and so \(H_\alpha ^{(\delta )}(Y)\geq |Y|\). Conversely, we can always cover \(Y\) by \(\mathcal {C}=\{\{y\}:y\in Y\}\), a finite collection of nonempty sets with \(|\mathcal {C}|=|Y|\). Then \(H^{(\delta )}_0(Y)\leq |\mathcal {C}|=|Y|\) and so the two must be equal.
Now suppose that \(Y\) is an infinite set. Then there is a sequence \(\{y_n\}_{n=1}^\infty \) of distinct elements of \(Y\) (that is, such that \(y_n\neq y_k\) if \(n\neq k\)). For each \(N\in \N \), by the previous argument there is a \(\delta _N>0\) such that \(H_0^{(\delta )}(\{y_1,y_2,\dots ,y_N\})=N\) for all \(\delta <\delta _N\). Because \(\{y_1,y_2,\dots ,y_N\}\subseteq Y\), we have that for each \(N\in \N \) there is a \(\delta _N>0\) such that, if \(0<\delta <\delta _N\), then \(H_0^{(\delta )}(Y)\geq N\). This is the definition of limit; thus, \(\lim _{\delta \to 0^+}H_0^{(\delta )}(Y)=\infty \) and so \(H_0^*(Y)=\infty \) by Problem 2790.
(Bonus Problem 2831) Let \(X=\R ^n\) and let \[\S =\{I_1\times I_2\times \dots \times I_n:I_k\subseteq \R \text { is a bounded interval}\}.\] Define \(\vol _n:\S \to [0,\infty ]\) by \[\vol _n(I_1\times I_2\times \dots \times I_n)=\prod _{k=1}^n \ell (I_k)\] where \(\ell \) denotes the length of the interval (that is, its Lebesgue measure). Let \(\vol ^*_n\) be the outer measure induced by the set function \(\vol _n\). Show that \(\vol ^*_n(R)=\vol _n(R)\) for all \(R\in \S \).
(Micah, Problem 2840) Show that this is not true for the Hausdorff measure \(H_\alpha ^{(\delta )}\); that is, give an example of a set \(Y\) of diameter less than \(\delta \) such that \(h_\alpha ^{(\delta )}(Y)=\diam (Y)^\alpha \) is not equal to \(H_\alpha ^{(\delta )}(Y)\).
[Chapter 17, Problem 27] Let \(\mathcal {S}\) be a collection of subsets of a set \(X\), and let \(\mu :\mathcal {S}\to [0,\infty ]\) be a set function. Let \(\mu ^*\) be the induced outer measure on \(2^X\). Show that \(\mu ^*(E)=\mu (E)\) for all \(E\in \mathcal {S}\) if and only if \(\mu \) is countably monotone, that is, if and only if \[\mu (E)\leq \sum _{A\in \mathcal {C}} \mu (A)\text { whenever } \mathcal {C}\subseteq \mathcal {S}\text { is countable and }E\subseteq \bigcup _{A\in \mathcal {C}} \mu (A).\]
[Definition: Premeasure]Let \(X\) be a set and let \(\S \subseteq 2^X\). Let \(\mu :S\to [0,\infty ]\). We call \(\mu \) a premeasure if:
(Problem 2841) Let \(\S =\{\{1,2\},\{3,4\},\{2,3\}\}\) and define \(\mu :\S \to [0,\infty ]\) by
\(\mu (\{1,2\})=\mu (\{3,4\})=1\),
\(\mu (\{2,3\})=5\).
Show that \(\mu \) is a premeasure but that \(\mu (\{2,3\})\neq \mu ^*(\{2,3\})\).
Proposition 17.11. Let \(\mu :\S \to [0,\infty ]\) be a set function, let \(\mu ^*\) be the outer measure induced by \(\mu \), let \(\bar \mu \) be the measure induced by \(\mu ^*\) and let \(\M \) be the \(\sigma \)-algebra of sets measurable with respect to \(\mu ^*\). Suppose that \(\S \subseteq \M \) and that \(\mu (Y)=\bar \mu (Y)\) for all \(Y\in \S \). Then \(\mu \) is a premeasure.
(Muhammad, Problem 2850) Prove Proposition 17.11.
[Definition: Closed with respect to the formation of relative complements] Let \(X\) be a set and let \(\S \subseteq 2^X\). We say that \(\S \) is closed with respect to the formation of relative complements if, whenever \(A\), \(B\in \S \), we have that \(A\setminus B\in \S \).
[Definition: Closed under finite intersections] Let \(X\) be a set and let \(\S \subseteq 2^X\). We say that \(\S \) is closed under finite intersections if, whenever \(A\), \(B\in \S \), we have that \(A\cap B\in \S \).
(Ashley, Problem 2860) Show that, if \(\S \) is closed with respect to the formation of relative complements, then \(\S \) is closed under finite intersection. Give an example to show that the reverse is not true.
Theorem 17.12. Let \(X\) be a set and let \(\S \subseteq 2^X\). Let \(\mu :\S \to [0,\infty ]\). Suppose that
\(\S \) is closed with respect to the formation of relative complements.
\(\mu \) is a premeasure.
Let \(\mu ^*\) be the outer measure induced by \(\mu \), let \(\bar \mu \) be the measure induced by \(\mu ^*\) and let \(\M \) be the \(\sigma \)-algebra of sets measurable with respect to \(\mu ^*\). Then \(\S \subseteq \M \) and \(\mu (Y)=\bar \mu (Y)\) for all \(Y\in \S \).
(Bashar, Problem 2870) Begin the proof of Theorem 17.12 by showing that all elements of \(\S \) are measurable.
(Dibyendu, Problem 2880) Complete the proof of Theorem 17.12.
(Problem 2881) Let \(\S =\{\{1,2\},\{3,4\},\{2,3\}\}\) and define \(\mu :\S \to [0,\infty ]\) by
\(\mu (\{1,2\})=\mu (\{3,4\})=1\),
\(\mu (\{2,3\})=5\).
Show that \(\mu \) is a premeasure but that \(\mu (\{2,3\})\neq \mu ^*(\{2,3\})\).
[Definition: Ring] Let \(X\) be a set and let \(\S \subseteq 2^X\) with \(\S \neq \emptyset \). We say that \(\S \) is a ring if
\(\S \) is closed under finite unions.
\(\S \) is closed under the formation of relative complements: if \(A\), \(B\in \S \), then \(A\setminus B\in \S \). (By Problem 2860, this implies that \(\S \) is closed under finite intersections.)
If \(X\in \S \) we say that \(\S \) is an algebra.
(Bonus Problem 2882) Let \(\S \) be a collection of sets. Equip \(\S \) with the operations \(\oplus \), \(\otimes \) defined by \(A\oplus B=(A\setminus B)\cup (B\setminus A)\) and \(A\otimes B=A\cap B\). Show that \(\S \) is a ring in the sense defined above if and only if \((\S ,\oplus ,\otimes )\) is a ring (possibly without multiplicative identity) in the sense of MATH 51203. (\(\S \) is an algebra in the sense given above if and only if \((\S ,\oplus ,\otimes )\) is a ring with multiplicative identity.)
[Definition: Semiring] Let \(X\) be a set and let \(\S \subseteq 2^X\) with \(\S \neq \emptyset \). We say that \(\S \) is a semiring if
\(\S \) is closed under finite intersections.
If \(A\), \(B\in \S \), then \(A\setminus B=\bigcup _{k=1}^n C_k\), where the \(C_k\)s are pairwise disjoint elements of \(\S \).
If \(X\in \S \) we say that \(\S \) is a semialgebra.
(Problem 2883) Give an example of a semiring that is not a ring.
Proposition 17.13a. Let \(\S \) be a semiring of subsets of \(X\). Let \[\S '=\Bigl \{\bigcup _{k=1}^n C_k:C_k\in \S , C_k\cap C_j=\emptyset \text { if }j\neq k\Bigr \}.\] Then \(\S '\) is a ring.
(Irina, Problem 2890) Prove Proposition 17.13a.
We claim that, if \(A\in \S '\) and \(R\in \S \), then \(A\setminus R\in \S '\).To prove the claim, observe that \(A=\bigsqcup _{j=1}^n A_j\) for some sets \(A_j\) with \(A_j\in \S \) and with \(A_j\cap A_J=\emptyset \) whenever \(j\neq J\). Then \[A\setminus R=\bigcup _{j=1}^n (A_j\setminus R).\] Because \(\S \) is a semiring, we may write \[A_j\setminus R=\bigsqcup _{k=1}^{m_j} C_{j,k}\] where \(C_{j,k}\cap C_{j,K}=\emptyset \) if \(k\neq K\); because \(C_{j,k}\subseteq A_j\) and \(A_j\cap A_J=\emptyset \) whenever \(j\neq J\), we have that \(C_{j,k}\cap C_{J,K}=\emptyset \) if either \(j\neq J\) or \(k\neq K\). Also each \(C_{j,k}\subset (A_j\setminus R)\) is disjoint from \(R\). Thus \[A\setminus R = \bigcup _{j=1}^n \bigcup _{k=1}^{m_j} C_{j,k}\] is the union of finitely many disjoint elements of \(\S \), and so \(A\setminus R\in \S '\). This proves the claim.
Furthermore, if \(A\in \S '\) and \(R\in \S \), then \(A\cup R=(A\setminus R)\cup R\). Since \(R\) and \(A\setminus R\) are disjoint, because \(A\setminus R\in \S '\), \(A\setminus R\) must be a union of elements of \(\S \) that are disjoint from \(R\), and so \((A\setminus R)\cup R\in \S '\) as well.
Thus, if \(A\in \S '\) and \(R\in \S \), then \(A\cup R\) and \(A\setminus R\) are in \(\S '\).
Let \(B\in \S '\). Then \(B=\bigsqcup _{k=1}^m B_k\) for some sets \(B_k\) with \(B_k\in \S \) and with \(B_k\cap B_K=\emptyset \) whenever \(k\neq K\), respectively.
Observe that \(A\cup B = (A\cup \bigsqcup _{k=1}^{m-1} B_k)\cup B_m\) and \(A\setminus B = A\setminus (\bigsqcup _{k=1}^m B_k) = (A\setminus \bigsqcup _{k=1}^{m-1} B_k)\setminus B_m\). Thus, inducting on the value of \(m\) yields that \(A\cup B\in \S '\) and \(A\setminus B\in \S '\). This yields that \(\S '\) is closed under the formation of relative complements, while another induction argument yields that \(\S '\) is closed under finite unions.
Proposition 17.13b. Let \(\S \), \(\S '\), and \(X\) be as in Proposition 17.13a. Let \(\mu :\S \to [0,\infty ]\) be a premeasure. Then there is a unique \(\mu ':\S '\to [0,\infty ]\) that is also a premeasure and satisfies \(\mu '(E)=\mu (E)\) for all \(E\in \S \).
(Micah, Problem 2900) Prove Proposition 17.13b.
(Problem 2901) Please carefully review the definitions in the Undergraduate Analysis presection of Section 17.4.
The Carathéodory-Hahn Theorem. Let \(\S \) be a semiring of subsets of \(X\) and let \(\mu :\S \to [0,\infty ]\) be a premeasure. Then the Carathéodory measure \(\bar \mu \) induced by \(\mu \) is an extension of \(\mu \) in the sense that every \(E\in \S \) is measurable with respect to \(\mu ^*\) and \(\mu (E)=\bar \mu (E)\) for all \(E\in \S \).
Furthermore, if \(\mu \) is \(\sigma \)-finite, (where we use the obvious definition of a \(\sigma \)-finite premeasure), then \(\bar \mu \) is also \(\sigma \)-finite, and is also the only measure on the \(\sigma \)-algebra of \(\mu ^*\)-measurable sets that extends \(\mu \).
(Problem 2902) Begin the proof of the Carathéodory-Hahn Theorem by showing that every \(E\in \S \) is measurable with respect to \(\mu ^*\) and that \(\mu (E)=\bar \mu (E)\) for all \(E\in \S \).
(Problem 2903) Let \(\M \) be the collection of subsets of \(X\) that are measurable with respect to \(\mu ^*\). Suppose that \(\nu :\M \to [0,\infty ]\) is a measure that satisfies \(\nu (E)=\mu (E)\) for all \(E\in \S \). Show that \(\nu (E)=\bar \mu (E)\) for all \(E\in \S _\sigma \), where \(\S _\sigma \) is the collection of countable unions of elements of \(\mathcal {S}\).
(Problem 2904) Let \(Y\in \S _\sigma \) satisfy \(\bar \mu (Y)<\infty \). Suppose that \(E\in \S _{\sigma \delta }\) and that \(E\subseteq Y\). Show that \(\bar \mu (E)=\nu (E)\). (Here \(\S _{\sigma \delta }\) is the collection of countable intersections of elements of \(\mathcal {S}_\sigma \).)
(Problem 2905) Let \(Y\in \S _\sigma \) satisfy \(\bar \mu (Y)<\infty \). Suppose that \(E\in \M \) and that \(E\subseteq Y\). Show that \(\bar \mu (E)=\nu (E)\).
(Muhammad, Problem 2910) Complete the proof of the Carathéodory-Hahn Theorem.
Corollary 17.14. Let \(\S \) be a semiring of subsets of \(X\) and let \(\B \) be the smallest \(\sigma \)-algebra of subsets of \(X\) that contains \(\S \). If \(\nu \), \(\lambda :\B \to [0,\infty ]\) are \(\sigma \)-finite measures, then \(\nu =\lambda \) if and only if \(\nu \big \vert _S=\lambda \big \vert _\S \).
[Homework 21.2] Let \(X\) be a set and let \(f:X\to [0,\infty ]\). Define \(\mu :2^X\to [0,\infty ]\) by \[\mu (A)=\sup \Bigl \{\sum _{x\in \mathcal {C}} f(x):\mathcal {C}\subseteq A\text { is countable}\Bigr \}.\] Show that \(\mu \) is a measure on \(2^X\).
Proposition 20.9. Let \[\S =\{I_1\times I_2\times \dots \times I_n:I_k\subseteq \R \text { is a bounded interval}\}.\] A bounded interval is a bounded subset \(I\) of \(\mathbb {R}\) such that, if \(x\), \(z\in I\) and \(x<y<z\), then \(y\in I\). We call \(\S \) the set of bounded intervals, or bounded rectangles, in \(\R ^n\). Then \(\S \) is a semiring of subsets of \(\R ^n\).
(Problem 2911) Show that \(I\) is a bounded interval if and only if \(I\) satisfies one of the following six conditions
\(I=\emptyset \) is empty,
\(I\) is a singleton set \(I=\{r\}\) for some \(r\in \R \),
\(I=(a,b)=\{c\in \R :a<c<b\}\) for some \(a\), \(b\in \R \) with \(a<b\),
\(I=(a,b]=\{c\in \R :a<c\leq b\}\) for some \(a\), \(b\in \R \) with \(a<b\),
\(I=[a,b)=\{c\in \R :a\leq c<b\}\) for some \(a\), \(b\in \R \) with \(a<b\), or
\(I=[a,b]=\{c\in \R :a\leq c\leq b\}\) for some \(a\), \(b\in \R \) with \(a<b\).
Proposition 20.10. Define \(\vol _n:\S \to [0,\infty ]\) by \[\vol _n(I_1\times I_2\times \dots \times I_n)=\prod _{k=1}^n \ell (I_k)\] where \(\ell \) denotes the length of the interval (that is, its Lebesgue measure). Then \(\vol _n\) is a premeasure on \(\S \).
(Ashley, Problem 2920) Prove Proposition 20.10.
Because \(\emptyset \) is an interval of length zero, we have that \(\emptyset =\emptyset \times \emptyset \times \dots \times \emptyset \) is in \(\S \) and \(\vol _n(\emptyset )=0^n=0\). We need only establish finite additivity and countable subadditivity.Let \(\varepsilon >0\). Define \(f_\varepsilon :\R \to \R \) by \[f_\varepsilon (r)=\begin {cases}\varepsilon , & r/\varepsilon \in \Z ,\\0, &\text {otherwise}.\end {cases}\] Define \(f_{n,\varepsilon }:\R ^n\to \R \) by \(f_{n,\varepsilon }(x_1,x_2,\dots ,x_n)=\prod _{k=1}^n f_\varepsilon (x_k)\), so \[f_{n,\varepsilon }(x_1,x_2,\dots ,x_n)=\begin {cases} \varepsilon ^n, & x_k/\varepsilon \in \Z \text { for all }k\text { with }1\leq k\leq n,\\ 0, & x_k/\varepsilon \notin \Z \text { for at least one }k\text { with }1\leq k\leq n.\end {cases}\] Let \(\mu _{n,\varepsilon }\) be the measure given by taking \(f=f_{n,\varepsilon }\) in Homework 21.2. Because \(\mu _{n,\varepsilon }\) is a measure, we have that \(\mu _{n,\varepsilon }\) is finitely additive and countably subadditive (in fact countably additive).
It is straightforward to establish that \[\mu _{n,\varepsilon }(I_1\times \dots \times I_n)=\mu _{1,\varepsilon }(I_1)\times \dots \times \mu _{1,\varepsilon }(I_n)\] and that if \(I\subset \R \) is a bounded interval then \(\ell (I)-\varepsilon \leq \mu _{1,\varepsilon }(I)\leq \ell (I)+\varepsilon \). We then have that \[\vol _n(R)=\lim _{\varepsilon \to 0^+} \mu _{n,\varepsilon }(R).\] By taking limits and using the fact that each \(\mu _{n,\varepsilon }\) is a measure and thus finitely additive, we may establish that \(\vol _n\) is also finitely additive.
We finally must establish that \(\vol _n\) is countably subadditive. Let \(R\in \mathcal {S}\) and suppose that \(R\subseteq \bigcup _{\ell =1}^\infty R_\ell \) for some rectangles \(R_\ell \in \mathcal {S}\).
Fix \(\delta >0\). Let \(K\) be a closed rectangle with \(K\subseteq R\) and \(\vol _n(K)>\vol _n(R)-\delta \), and for each \(\ell \) let \(G_\ell \) be an open rectangle with \(G_\ell \supseteq R_\ell \) and with \(\vol _n(G_\ell )\leq \vol _n(R_\ell )+\frac {\delta }{2^\ell }\). It is an elementary real analysis argument to show that such \(K\) and \(G_\ell \) must exist.
Then \[K\subseteq R\subseteq \bigcup _{\ell =1}^\infty R_\ell \subseteq \bigcup _{\ell =1}^\infty G_\ell .\] Because \(K\) is compact and each \(G_\ell \) is open, we must have that \[K\subseteq \bigcup _{\ell =1}^M G_\ell \] for some finite integer \(M\). Then \[\vol _n(K)=\lim _{\varepsilon \to 0^+}\mu _{n,\varepsilon }(K) \leq \lim _{\varepsilon \to 0^+}\sum _{\ell =1}^M\mu _{n,\varepsilon }(G_\ell )\] by subadditivity of the measure \(\mu _{n,\varepsilon }\), and so \[\vol _n(K)\leq \sum _{\ell =1}^M\lim _{\varepsilon \to 0^+}\mu _{n,\varepsilon }(G_\ell ) \sum _{\ell =1}^M \vol _n(G_\ell )\] because \(M\) is finite. We thus have that \[\vol _n(R)-\delta <\vol _n(K)\leq \sum _{\ell =1}^M \vol _n(G_\ell ) \leq \sum _{\ell =1}^\infty \vol _n(G_\ell ) \leq \sum _{\ell =1}^\infty (\vol _n(R_\ell )+2^{-k}\delta ) \leq \Bigl (\sum _{\ell =1}^\infty \vol _n(R_\ell )\Bigr )+\delta \] and so \[\vol _n(R)<\Bigl (\sum _{\ell =1}^\infty \vol _n(R_\ell )\Bigr )+2\delta \] for all \(\delta >0\). Letting \(\delta \to 0^+\) yields that \[\vol _n(R)\leq \Bigl (\sum _{\ell =1}^\infty \vol _n(R_\ell )\Bigr )\] as desired.
[Definition: Lebesgue measure on \(\R ^n\)] The outer measure \(\mu ^*_n\) induced by the premeasure \(\vol \) on the semiring \(\S \) of bounded rectangles in \(\R ^n\) is called the Lebesgue outer measure on \(\R ^n\), and the induced Carathéodory measure \(\mu _n\) is called the Lebesgue measure on \(\R ^n\), or \(n\)-dimensional Lebesgue measure.
Theorem 20.11. The \(\sigma \)-algebra \(\mathcal {L}^n\) of Lebesgue measurable subsets of \(\R ^n\) contains the Borel sets in \(\R ^n\) (the smallest \(\sigma \)-algebra of subsets of \(\R ^n\) that contains all the open sets). The Lebesgue measure is \(\sigma \)-finite, complete, and \(\mu _n(R)=\vol (R)\) for all bounded rectangles \(R\).
(Bashar, Problem 2930) Prove Theorem 20.11.
By Theorem 17.8, \((\R ^n,\mathcal {L}^n,\mu _n)\) is a complete maesure space. By the Carathéodory-Hahn Theorem, every element of \(\S \) is measurable and \(\mu _n(R)=\vol (R)\) for all \(R\in \S \). We have that \[\R ^n=\bigcup _{k=1}^\infty [-k,k]^n\] and each cube \([-k,k]^n\) is an element of \(\S \), and therefore measurable with finite measure, and so \(\mu _n\) is \(\sigma \)-finite.\(\mathcal {L}^n\) is a \(\sigma \)-algebra by Theorem 17.8, so to show that \(\mathcal {L}^n\) contains the Borel sets, it suffices to show that \(\mathcal {L}^n\) contains all nonempty open sets.
Recall that the rational numbers \(\Q \) are countable, and thus so is \(\Q ^{2n}\) for any finite integer \(n\). Let \[\mathcal {G}=\Bigl \{\frac {r}{\sqrt {n}}:p_i,q_i\in \Q ,p_i<q_i\Bigr \}\] be the set of all nonempty open rectangles in \(\R ^n\) whose vertices have rational coordinates. There is clearly a bijection from \(\mathcal {G}\) to a subset of \(\Q ^{2n}\) given by \[\prod _{i=1}^n (p_i,q_i)\mapsto (p_1,q_1,p_2,q_2,\dots ,p_n,q_n)\] and so \(\mathcal {G}\) is countable. But each element of \(\mathcal {G}\) is a rectangle, so each element of \(\mathcal {G}\) is measurable.
Because the collection of measurable sets is a \(\sigma \)-algebra, the union of countably many elements of \(\mathcal {G}\) is measurable. But \(\mathcal {G}\) is countable; thus all subsets of \(\mathcal {G}\) are countable, and so the union of an arbitrary collection of elements of \(\mathcal {G}\) is measurable.
Let \(\mathcal {O}\subseteq \R ^n\) be open and nonempty. It suffices to show that every such \(\mathcal {O}\) may be written as a union of elements of \(\mathcal {G}\). Let \(x\in \mathcal {O}\). By definition of open set, there is a \(r>0\) such that \(B(x,r)\subseteq \mathcal {O}\).
Let \(x=(x_1,x_2,\dots ,x_n)\). For each \(1\leq i\leq n\), choose rational numbers \(p_i\), \(q_i\) such that \[x_i-\frac {r}{\sqrt {n}}<p_i<x<q_i<x_i+\frac {r}{\sqrt {n}}.\] Let \(R_x=\frac {r}{\sqrt {n}}\). Clearly \(x\in R_x\in \mathcal {G}\). Furthermore, if \(y=(y_1,y_2,y_3,\dots ,y_n)\in R_x\), then \(x_i-\frac {r}{\sqrt {n}}<p_i<y_i<q_i<x_i+\frac {r}{\sqrt {n}}\) and so \(-\frac {r}{\sqrt {n}}<y_i-x_i<\frac {r}{\sqrt {n}}\) for all \(n\), and so \(|y_i-x_i|<\frac {r}{\sqrt {n}}\). Thus \[\|x-y\|=\sqrt {\sum _{i=1}^n |x_i-y_i|^2} < \sqrt {\sum _{i=1}^n \biggl (\frac {r}{\sqrt {n}}\biggr )^2}=r\] and so \(y\in B(x,r)\subseteq \mathcal {O}\). Thus \(x\in R_x\subset \mathcal {O}\), and so \[\mathcal {O}=\bigcup _{x\in \mathcal {O}} R_x\] is a union of elements of \(\mathcal {G}\), as desired.
[Definition: Measurable function] Let \((X,\M ,\mu )\) be a measure space and let \(E\in \mathcal {M}\) (that is, let \(E\) be measurable). Let \(f:E\to [-\infty ,\infty ]\). Suppose that for every \(c\in \R \) the set \begin {equation*} \{x\in E:f(x)>c\}=f^{-1}((c,\infty ]) \end {equation*} is measurable, that is, lies in \(\mathcal {M}\). Then we say that \(f\) is a \(\M \)-measurable function (or that \(f\) is measurable on \(E\)).
Corollary 20.12. If \(E\subseteq \R ^n\) is Lebesgue measurable and \(f:E\to \R \) is continuous, then \(f\) is a Lebesgue measurable function.
(Problem 2931) Let \(X\) be a set. Let \(\mu ^*:2^X\to [0,\infty ]\) be an outer measure on \(2^X\), and let \(\bar \mu \) and \(\M \) be the induced measure and \(\sigma \)-algebra of measurable sets given by Theorem 17.8.
If \(\varphi :X\to [-\infty ,\infty ]\), show that \(\varphi \) is \(\M \)-measurable if and only if, for every \(c\in \R \) and every \(A\), \(B\subseteq X\) that satisfy \[\varphi (x)\leq c\text { for every }x\in A,\qquad \varphi (x)>c\text { for every }x\in B,\] we have that \(\mu ^*(A\cup B)=\mu ^*(A)+\mu ^*(B)\).
By definition of measurable function, \(\varphi \) is \(\M \)-measurable if and only if \[\{x\in X:\varphi (x)>c\}\in \M \] for all \(c\in \R \).By the definition of a measurable set, \(\{x\in X:\varphi (x)>c\}\) is measurable if and only if, for all \(T\subseteq X\), we have that \[\mu ^*(T)=\mu ^*(T\cap \{x\in X:\varphi (x)>c\})+\mu ^*(T\setminus \{x\in X:\varphi (x)>c\}).\] We may write \(T\cap \{x\in X:\varphi (x)>c\}=\{x\in T:\varphi (x)>c\}\) and \(T\setminus \{x\in X:\varphi (x)>c\}=\{x\in T:\varphi (x)\leq c\}\). Thus \(\varphi \) is measurable if and only if, for all \(T\subseteq X\), we have that \[\mu ^*(T)=\mu ^*(\{x\in T:\varphi (x)>c\})+\mu ^*(\{x\in T:\varphi (x)\leq c\}).\]
Suppose that \(\varphi \) is measurable. Let \(c\in \R \) and let \(A\), \(B\subseteq X\) satisfy \[\varphi (x)\leq c\text { for every }x\in A,\qquad \varphi (x)>c\text { for every }x\in B.\] Define \(T=A\cup B\). Then \(B=\{x\in T:\varphi (x)>c\}\) and \(A=\{x\in T:\varphi (x)\leq c\}\). Thus \[\mu ^*(A\cup B)=\mu ^*(T)=\mu ^*(\{x\in T:\varphi (x)>c\})+\mu ^*(\{x\in T:\varphi (x)\leq c\}) =\mu ^*(B)+\mu ^*(A)\] and the given condition holds.
Conversely, suppose that the given condition holds. We want to show \(\varphi \) is measurable; thus, we want to show that if \(c\in \R \), then \(\{x\in X:\varphi (x)>c\}\) is measurable. Let \(T\subseteq X\). Let \(B=\{x\in T:\varphi (x)>c\}\) and let \(A=\{x\in T:\varphi (x)\leq c\}\). Then \(T=A\cup B\) and \[\varphi (x)\leq c\text { for every }x\in A,\qquad \varphi (x)>c\text { for every }x\in B,\] and so by assumption we have that \(\mu ^*(A\cup B)=\mu ^*(A)+\mu ^*(B)\). Collecting definitions, we see that \[\mu ^*(T)=\mu ^*(A\cup B)=\mu ^*(A)+\mu ^*(B)= \mu ^*(\{x\in T:\varphi (x)>c\})+\mu ^*(\{x\in T:\varphi (x)\leq c\}) =\mu ^*(T\cap \{x\in X:\varphi (x)>c\})+\mu ^*(T\setminus \{x\in X:\varphi (x)\leq c\}).\] Thus \(\{x\in X:\varphi (x)>c\}\) is measurable for all \(c\in \R \), and so \(\varphi \) is measurable, as desired.
[Homework 23.1] Give an example of an outer measure \(\mu ^*:2^{\mathbb {R}}\to [0,\infty )\) such that \(\mu ^*(\mathbb {R})\lt \infty \) but such that \[\mu ^*(\{3\})\neq \lim _{\delta \to 0^+} \mu ^*([3,3+\delta )).\]
Proposition 20.27. Let \(X\) be a set. Let \(\mu ^*:2^X\to [0,\infty ]\) be an outer measure on \(2^X\), and let \(\bar \mu \) and \(\M \) be the induced measure and \(\sigma \)-algebra of measurable sets given by Theorem 17.8.
Let \(\varphi :X\to [-\infty ,\infty ]\). Then \(\varphi \) is \(\M \)-measurable if and only if, for all \(A\), \(B\subseteq X\) that satisfy \[\sup _{x\in A} \varphi (x)<\inf _{y\in B} \varphi (y)\] we have that \[\mu ^*(A\cup B)=\mu ^*(A)+\mu ^*(B).\]
(Dibyendu, Problem 2940) Prove Proposition 20.27.
[Definition: Carathéodory outer measure] Let \((X,d)\) be a metric space. An outer measure \(\mu ^*:2^X\to [0,\infty ]\) is a Carathéodory outer measure if \[\dist (A,B)>0\text { implies }\mu ^*(A\cup B)=\mu ^*(A)+\mu ^*(B).\]
Theorem 20.28. If \(\mu ^*\) is a Carathéodory outer measure on a metric space \((X,d)\), then every Borel subset of \(X\) is measurable with respect to \(\mu ^*\).
(Irina, Problem 2950) Prove Theorem 20.28.
Proposition 20.29. If \((X,d)\) is a metric space and \(\alpha \geq 0\), then the Hausdorff outer measure \(H_\alpha ^*\) on \(X\) is a Carathéodory outer measure. Thus every Borel subset of \(X\) is Hausdorff measurable.
(Micah, Problem 2960) Prove Proposition 20.29.
Theorem 20.13. Let \(E\subseteq \R ^n\) be Lebesgue measurable. Then \begin {align*} \mu _n(E)&=\inf \{\mu _n(\mathcal {O}):E\subseteq \mathcal {O},\>\mathcal {O}\text { is open}\} \\&=\sup \{\mu _n(K):K\subseteq E,\>K\text { is compact}\}. \end {align*}
(Muhammad, Problem 2970) Prove that \(\mu _n(E)=\inf \{\mu _n(\mathcal {O}):E\subseteq \mathcal {O},\>\mathcal {O}\text { is open}\}\).
(Ashley, Problem 2980) Prove that \(\mu _n(E)=\sup \{\mu _n(K):K\subseteq E,\>K\text { is compact}\}\) in the case where \(E\) is bounded.
Corollary 20.14. Let \(E\subseteq \R ^n\). The following statements are equivalent.
[Definition: Infinite sum] Let \(\{a_k\}_{k=1}^\infty \) be a sequence of extended real numbers. If \(\sum _{k=1}^n a_k\) exists for all \(n\) (that is, if it is not the case that \(a_\ell =\infty \) and \(a_m=-\infty \) for some \(\ell \) and \(m\)), and if \(\lim _{n\to \infty } \sum _{k=1}^n a_k\) exists (in the sense of either finite or infinite limits), then we say that the series \(\sum _{k=1}^\infty a_k\) converges and write \[\sum _{k=1}^\infty a_k=\lim _{n\to \infty } \sum _{k=1}^n a_k.\] If the limit does not exist (including the case where some \(\sum _{k=1}^n a_k\) does not exist), we say that the series \(\sum _{k=1}^\infty a_k\) diverges.
[Definition: Absolute convergence] Let \(\{a_k\}_{k=1}^\infty \) be a sequence of real numbers. We say that \(\sum _{k=1}^\infty a_k\) converges absolutely if \(\sum _{k=1}^\infty |a_k|\) converges to a finite real number.
(Problem 2981) Show that \(\sum _{k=1}^\infty a_k\) converges absolutely if and only if \(\sup _{n\in \N } \sum _{k=1}^n | a_k|\) is finite.
[Definition: Unconditional convergence] Let \(\{a_n\}_{n=1}^\infty \) be a sequence of extended real numbers. We say that \(\sum _{n=1}^\infty a_n\) converges unconditionally if, for every bijection \(\sigma :\N \to \N \), we have that \(\sum _{n=1}^\infty a_{\sigma (n)}\) is convergent and \[\sum _{n=1}^\infty a_{\sigma (n)}=\sum _{n=1}^\infty a_n.\]
The Riemann rearrangement theorem. All absolutely convergent series converge unconditionally. All unconditionally convergent series of real numbers that converge to a finite value converge absolutely.
(Problem 2982) Suppose that \(a_n\geq 0\) for all \(n\). Show that \(\sum _{n=1}^\infty a_n\) is unconditionally convergent.
(Problem 2983) Let \(\sum _{n=1}^\infty a_n\) be an unconditionally convergent series that converges to \(\infty \). Suppose that \(a_n\neq \infty \) for all \(n\). Show that \(\sum _{n=1}^\infty \max (a_n,0)\) converges absolutely.
[Definition: Signed measure] If \((X,\M )\) is a measurable space, then a signed measure on \((X,\M )\) is a function \(\nu \) such that
\(\nu :\M \to [-\infty ,\infty ]\).
\(\nu (\emptyset )=0\).
If \(\{E_k\}_{k=1}^\infty \subseteq \M \) and \(E_k\cap E_j=\emptyset \) whenever \(k\neq j\), then \(\sum _{k=1}^\infty \nu (E_k)\) converges and \[\nu \Bigl (\bigcup _{k=1}^\infty E_k\Bigr )=\sum _{k=1}^\infty \nu (E_k).\]
(Problem 2984) Suppose that \(\{E_k\}_{k=1}^\infty \subseteq \M \) and that \(E_k\cap E_j=\emptyset \) whenever \(k\neq j\). Show that \(\sum _{k=1}^\infty \nu (E_k)\) converges unconditionally.
Let \(\sigma :\N \to \N \) be a bijection. If \(E_{\sigma (m)}\cap E_{\sigma (n)}\neq \emptyset \), then \(\sigma (m)=\sigma (n)\) because the sets \(E_k\) are pairwise disjoint, and so \(m=n\) because \(\sigma \) is a bijection. Thus \(\{E_\sigma (k)\}_{k=1}^\infty \) is a sequence of pairwise-disjoint measurable sets, and so \(\sum _{k=1}^\infty \nu (E_{\sigma (k)})\) converges by assumption.Clearly \[\bigcup _{k=1}^\infty E_k=\bigcup _{k=1}^\infty E_{\sigma (k)}.\] We thus must have that \[\sum _{k=1}^\infty \nu (E_k)=\nu \Bigl (\bigcup _{k=1}^\infty E_k\Bigr )=\sum _{k=1}^\infty \nu (E_{\sigma (k)}).\] This is the definition of unconditional convergence.
(Problem 2985) Suppose that \(\{E_k\}_{k=1}^\infty \subseteq \M \) and that \(E_k\cap E_j=\emptyset \) whenever \(k\neq j\), and that \[-\infty <\nu \Bigl (\bigcup _{k=1}^\infty E_k\Bigr )<\infty .\] Show that the series \(\sum _{k=1}^\infty \nu (E_k)\) converges absolutely.
This follows from the previous problem and the Riemann rearrangement theorem.
(Irina, Problem 3000) Let \((X,\M )\) be a measurable space and \(\nu \) a signed measure on \((X,\M )\). Suppose that \(B\in \M \) and that \(\nu (B)=\pm \infty \). Show that \(\nu (E)=\pm \infty \) for all \(E\in \M \) with \(B\subseteq E\).
Suppose that \(\nu (B)=\infty \) (the case \(\nu (B)=-\infty \) is similar). Because \(\M \) is a \(\sigma \)-algebra, we have that \(E\setminus B\in \M \). Because \(B\) and \(E\setminus B\) are disjoint, we have that \[\infty =\nu (E)=\nu (B)+\nu (E\setminus B).\] Recall that \(\nu (B)=\infty \). Because \(\nu (B)+\nu (E\setminus B)\) exists, we cannot have \(\nu (E\setminus B)=-\infty \); we must have \(\nu (E\setminus B)=\infty \) or \(\nu (E\setminus B)\in \R \). In either case \(\nu (B)+\nu (E\setminus B)=\infty \), and so \(\nu (E)=\infty \).
(Problem 2987) Let \(\nu \) be a signed measure. Show that \(\nu \) assumes at most one of the two values \(-\infty \) and \(\infty \); that is, show that either \(\nu :\M \to [-\infty ,\infty )\) or \(\nu :\M \to (-\infty ,\infty ]\) (or both).
If \(\nu (A)<\infty \) for all \(A\in \M \) we are done, so let \(A\in \M \) with \(\nu (A)=\infty \). We want to show that \(\nu (B)>-\infty \) for all \(B\in \M \).First, let \(B\in \M \) with \(A\cap B=\emptyset \). Let \(E_1=A\), \(E_2=B\), and \(E_k=\emptyset \) for all \(k\geq 3\). Then \(E_j\cap E_k=\emptyset \) if \(j\neq k\). Thus by assumption \(\sum _{k=1}^\infty \nu (E_k)=\nu (A)+\nu (B)\) must exist, and so \(\nu (B)\neq -\infty \), as \(\infty +(-\infty )\) does not exist. Thus if \(B\in \M \) and \(A\cap B=\emptyset \), then \(\nu (B)\neq -\infty \).
Next, let \(B\in \M \) with \(B\subseteq A\). By Problem 3000, if \(\nu (B)=-\infty \) then \(\nu (A)=-\infty \), contradicting the assumed value of \(\nu (A)\). Thus if \(B\in \M \) and \(B\subseteq A\), then \(\nu (B)\neq -\infty \).
Finally, choose an arbitrary \(B\in \M \). Then \(B=(A\cap B)\cup (B\setminus A)\), and both of the sets \(A\cap B\) and \(B\setminus A\) are elements of \(\M \), that is, are measurable. Because \(A\cap B\subseteq A\), we have that \(\nu (A\cap B)\neq -\infty \); because \(B\setminus A\) is disjoint from \(A\), we have that \(\nu (B\setminus A)\neq -\infty \). Thus \(\nu (B)=\nu (B\cap A)+\nu (B\setminus A)>-\infty \), as desired.
(Problem 2988) Let \(f:\R \to [0,\infty ]\) be measurable. Define \(\mu (A)=\int _A f\) for all measurable sets \(A\subseteq \R \). Show that \(\mu \) is a measure on the \(\sigma \)-algebra \(\M \) of Lebesgue measurable subsets of \(\R \).
(Dibyendu, Problem 2990) Let \((X,\M )\) be a measurable space, let \(\mu :\M \to [0,\infty ]\) be a measure, and let \(\lambda :\M \to [0,\infty )\) be a finite measure (that is, a measure that satisfies \(\lambda (X)<\infty \)). Let \(\alpha \), \(\beta \in \R \) and let \(\nu =\alpha \mu +\beta \lambda \). Show that \(\nu \) is a signed measure on \((X,\M )\).
(Bashar, Problem 3010) Let \((X,\M )\) be a measurable space and \(\nu \) a signed measure on \((X,\M )\). Suppose that \(E\in \M \) and that \(\nu (E)>-\infty \). Show that \(\inf \{\nu (B):B\in \M ,B\subseteq E\}>-\infty \).
(Problem 3011) Let \((X,\M )\) be a measurable space and \(\nu \) a signed measure on \((X,\M )\). Suppose that \(E\in \M \) and that \(\nu (E)<\infty \). Show that \(\sup \{\nu (B):B\in \M ,B\subseteq E\}<\infty \).
[Definition: Positive with respect to a signed measure] Let \((X,\M )\) be a measurable space and \(\nu \) a signed measure on \((X,\M )\). If \(A\in \M \), we say that \(A\) is positive with respect to \(\nu \) if \(\nu (E)\geq 0\) for every \(E\in \M \) with \(E\subseteq A\).
(Problem 3012) Let \(A\) be positive with respect to \(\nu \) and let \(\M _A=\{E\in \M :E\subseteq A\}\). Then \((A,\M _A,\nu \big \vert _{\M _A})\) is a measure space.
[Definition: Null with respect to a signed measure] Let \((X,\M )\) be a measurable space and let \(\nu \) be a signed measure on \((X,\M )\). If \(A\in \M \), we say that \(A\) is null with respect to \(\nu \) if \(\nu (E)= 0\) for every \(E\in \M \) with \(E\subseteq A\).
(Problem 3013) Give an example of a set of (signed) measure zero that is not a null set.
Proposition 17.4. Let \((X,\M )\) be a measurable space and \(\nu \) a signed measure on \((X,\M )\). Then every measurable subset of a positive set is positive and the union of countably many positive sets is positive.
(Micah, Problem 3020) Prove Proposition 17.4.
Hahn’s Lemma. Let \((X,\M )\) be a measurable space and \(\nu \) a signed measure on \((X,\M )\). Let \(E\in \M \) satisfy \(0<\nu (E)<\infty \). Then there is an \(A\in \M \) such that \(A\subseteq E\), \(A\) is positive, and \(\nu (A)>0\).
(Muhammad, Problem 3030) Prove Hahn’s Lemma.
(Problem 3031) Is the condition \(\nu (E)<\infty \) in Hahn’s Lemma a necessary condition?
The Hahn Decomposition Theorem. Let \((X,\M )\) be a measurable space and let \(\nu \) be a signed measure on \(\M \). Then there is a positive set \(A\) and a negative set \(B\) such that \[X=A\cup B,\qquad A\cap B=\emptyset .\]
(Ashley, Problem 3040) Prove the Hahn Decomposition Theorem.
Observe that if \(\nu \) is a signed measure then so is \(-\nu \). Then \(A\in \M \) is positive for \(\nu \) if and only if \(A\) is negative for \(-\nu \), and so if a Hahn decomposition exists for \(-\nu \) then it exists for \(\nu \). We may thus prove the Hahn Decomposition Theorem for either \(\nu \) or \(-\nu \). Observe that at most one of the measures \(\nu \) and \(-\nu \) attain the value \(+\infty \). We may at this point change the names of \(\nu \) and \(-\nu \), without loss of generality, so that \(\nu \neq \infty \) and \(-\nu \neq -\infty \). We will prove the Hahn decomposition theorem for \(\nu \) after this renaming.Let \(\alpha =\sup \{\nu (A):A\in \M ,A\) is positive\(\}\). Let \(\{A_n\}_{n=1}^\infty \) be a sequence of positive sets such that \(\nu (A_n)\to \alpha \).
Define \(A=\bigcup _{n=1}^\infty \) and let \(B=X\setminus A\). Then \(A\) is positive by Proposition 17.4 and clearly \(X=A\cup B\), \(\emptyset =A\cap B\). We need only show that \(B\) is negative.
It will be helpful to compute \(\nu (A)\). Because \(A\) is positie, we have that \(\nu (A)\leq \alpha \). Conversely, \(A_n\subseteq A\) for all \(n\), and so \(\nu (A)=\nu (A_n)+\nu (A\setminus A_n)\geq \nu (A_n)\) because \(A\setminus A_n\subseteq A\) and \(A\) is positive. Thus \(\nu (A)\geq \lim _{n\to \infty } \nu (A_n)=\alpha \). Thus \(\nu (A)=\alpha \). In particular, because \(\nu \) does not attain the value \(\infty \), we have that \(\alpha \) must be finite.
We now turn to the proof that \(B\) is negative. Let \(E\subseteq B\) be positive. Then \(E\cup A\) is positive and so \(\nu (E\cup A)\leq \alpha \). Conversely, \(E\cap A=\emptyset \) because \(E\subseteq B\), and so \(\nu (E\cup A)=\nu (E)+\nu (A)\geq \nu (A)=\alpha \) because \(E\) is positive and so \(\nu (E)\geq 0\). Thus \(\alpha \leq \nu (E)+\alpha \leq \alpha \). Because \(\alpha \) is finite, this implies that \(\nu (E)=0\). Thus every positive subset of \(B\) has measure zero and so in fact is a null set. Combined with the contrapositive of Hahn’s Lemma, we must have that \(\nu (D)\leq 0\) for all \(D\subseteq B\), and thus \(B\) is negative, as desired.
[Definition: Mutually singular measure] Let \((X,\M )\) be a measurable space and let \(\mu \), \(\lambda \) be two (nonnegative) measures on \(\M \). We say that \(\mu \perp \lambda \), or \(\mu \) and \(\lambda \) are mutually singular, if there are sets \(A\) and \(B\) in \(\M \) with \(X=A\cup B\) and with \(\mu (A)=\lambda (B)=0\).
(Problem 3041) Let \(m\) denote the Lebesgue measure on \(\R \) and let \(\delta _3\) denote the Dirac measure (that is, \(\delta _3(E)=1\) if \(3\in E\) and \(\delta _3(E)=0\) if \(3\notin E\)). Show that \(m\perp \delta _3\).
The Jordan Decomposition Theorem. Let \((X,\M )\) be a measurable space and let \(\nu \) be a signed measure on \(\M \). Then there is a unique pair of nonnegative measures \(\nu ^+\) and \(\nu ^-\) on \(\M \) that are mutually singular and such that \(\nu =\nu ^+-\nu ^-\).
(Bashar, Problem 3050) Show that \(\nu ^+\) and \(\nu ^-\) exist.
[Chapter 17, Problem 13] Show that the ordered pair \((\nu ^+,\nu ^-)\) is unique.
[Definition: Measurable function] Let \((X,\M )\) be a measurable space and let \(f:X\to [-\infty ,\infty ]\). Suppose that for every \(c\in \R \) the set \begin {equation*} \{x\in E:f(x)>c\}=f^{-1}((c,\infty ]) \end {equation*} is an element of \(\M \). Then we say that \(f\) is a measurable function (or that \(f\) is measurable with respect to \(\M \)).
Proposition 18.1. Let \((X,\M )\) be a measurable space and let \(f:X\to [-\infty ,\infty ]\). The following statements are equivalent.
Furthermore, if any of these conditions is true, then \(f^{-1}(\{c\})\) is an element of \(\M \) for all \(c\in [-\infty ,\infty ]\).
The proof is identical to the proof of Proposition 3.1.
Proposition 18.2. If \(f\) is measurable, then \(f^{-1}(\mathcal {O})\) is measurable for all open sets \(\mathcal {O}\subseteq \R \).
The proof is identical to the proof of Proposition 3.2.
[Chapter 18, Problem 1] If \(f\) is measurable, then \(f^{-1}(B)\) is measurable for all Borel sets \(B\subseteq \R \).
Proposition 18.3. Let \((X,\M ,\mu )\) be a complete measure space and let \(f:X\to [-\infty ,\infty ]\). Suppose that \(g:X\to [-\infty ,\infty ]\) satisfies \(f=g\) almost everywhere on \(X\) and that \(g\) is measurable. Then \(f\) is also measurable.
(Dibyendu, Problem 3060) Prove Proposition 18.3.
[Chapter 18, Problem 2] If \((X,\M ,\mu )\) is not complete, then Proposition 18.3 is false.
Theorem 18.4. If \(f\) and \(g\) are both real-valued measurable functions on the measurable space \((X,\M )\), then so are \(\alpha f+\beta g\), \(fg\), \(\max (f,g)\), and \(\min (f,g)\) for any \(\alpha \), \(\beta \in \R \).
The proof is identical to that of Theorem 3.6 and Proposition 3.8.
Theorem 2.41. (Axler) Let \(\B \) be the \(\sigma \)-algebra of Borel sets of real numbers. Suppose that \(B\subseteq \R \) is a Borel set and that \(\varphi :B\to \R \) is continuous. Then \(\varphi \) is measurable with respect to \(\B \). (We call such functions Borel measurable.)
(Problem 3061) Prove Theorem 2.41 (Axler).
Theorem 2.44. (Axler) Let \((X,\M )\) be a measurable space, let \(B\subseteq \R \) be a Borel set, let \(f:X\to B\) be measurable with respect to \(\M \), and let \(\varphi :B\to \R \) be Borel measurable. Show that \(\varphi \circ f:X\to \R \) is measurable with respect to \(\M \).
(Irina, Problem 3070) Prove Theorem 2.44 (Axler).
We need to show that \((\varphi \circ f)^{-1}((a,\infty ])\) is in \(\M \) for all \(a\in \R \). But \[(\varphi \circ f)^{-1}((a,\infty ])=f^{-1}(\varphi ^{-1}((a,\infty ])).\] Because \(\varphi \) is Borel measurable, we have that \(\varphi ^{-1}((a,\infty ])\) is a Borel set. By Problem 18.1, \(f^{-1}(\varphi ^{-1}((a,\infty ]))\) is measurable, that is, is in \(\M \). This completes the proof.
Theorem 2.48. (Axler) Let \((X,\M )\) be a measurable space. Let \(f_n:X\to [-\infty ,\infty ]\) be \(\M \)-measurable. Suppose that \(f:X\to [-\infty ,\infty ]\) and that \(f_n(x)\to f(x)\) for every \(x\in X\). Then \(f\) is measurable.
(Micah, Problem 3080) Prove Theorem 2.48 (Axler).
We need to show that \(f^{-1}((a,\infty ])\) is measurable for all \(a\in \R \).Suppose \(x\in f^{-1}((c,\infty ])\) for some \(c\in \R \). Then \(f(x)> c\), so \(\lim _{n\to \infty } f_n(x)=c\). If \(\varepsilon >0\), there exists a \(k\in \N \) such that, for all \(n\geq k\), we have that \(|f_n(x)-f(x)|<\varepsilon \). Thus, for all \(n\geq k\), we have that \(f_n(x)>f(x)-\varepsilon >c-\varepsilon \) and so \(x\in \bigcap _{n=k}^\infty f_n^{-1}((c-\varepsilon ,\infty ])\) for at least one \(k\); thus \(x\in \bigcup _{k=1}^\infty \bigcap _{n=k}^\infty f_n^{-1}((c-\varepsilon ,\infty ])\) and so \[f^{-1}((c,\infty ])\subseteq \bigcup _{k=1}^\infty \bigcap _{n=k}^\infty f_n^{-1}((c-\varepsilon ,\infty ])\] for all \(\varepsilon >0\).
Conversely, if \(x\in \bigcup _{k=1}^\infty \bigcap _{n=k}^\infty f_n^{-1}((c+\varepsilon ,\infty ])\), then there exists a \(k\) such that, for all \(n\geq k\), we have that \(f_n(x)>c+\varepsilon \) and so \(f(x)=\lim _{n\to \infty } f_n(x)\geq c+\varepsilon \). In particular \(x\in f^{-1}([c+\varepsilon ,\infty ])\subseteq f^{-1}((c,\infty ])\). Thus \[\bigcup _{k=1}^\infty \bigcap _{n=k}^\infty f_n^{-1}((c+\varepsilon ,\infty ]) \subseteq f^{-1}((c,\infty ])\subseteq \bigcup _{k=1}^\infty \bigcap _{n=k}^\infty f_n^{-1}((c-\varepsilon ,\infty ]).\] We now observe that \(f^{-1}((a,\infty ])=\bigcup _{j=1}^\infty f^{-1}((a+2/j,\infty ])\). Choosing \(\varepsilon =1/j\), we see that \[f^{-1}((a,\infty ]) =\bigcup _{j=1}^\infty f^{-1}((a+2/j,\infty ]) \subseteq \bigcup _{j=1}^\infty \bigcup _{k=1}^\infty \bigcap _{n=k}^\infty f_n^{-1}((a+1/j,\infty ]).\] Similarly (with \(c=a+1/2j\) and \(\varepsilon =1/2j\)) we see that \[f^{-1}((a,\infty ]) =\bigcup _{j=1}^\infty f^{-1}((a+1/2j,\infty ]) \supseteq \bigcup _{j=1}^\infty \bigcup _{k=1}^\infty \bigcap _{n=k}^\infty f_n^{-1}((a+1/j,\infty ]).\] Thus \[f^{-1}((a,\infty ])=\bigcup _{j=1}^\infty \bigcup _{k=1}^\infty \bigcap _{n=k}^\infty f_n^{-1}((a+1/j,\infty ]).\] Because \(f_n\) is measurable, each \(f_n^{-1}((a+1/j,\infty ])\) is measurable, that is, is an element of \(\M \). Because \(\M \) is a \(\sigma \)-algebra, it is closed under countable unions and intersections, and so \(f^{-1}((a,\infty ])\in \M \), as desired.
Theorem 18.6. Let \((X,\M ,\mu )\) be a complete measure space. Let \(f_n:X\to [-\infty ,\infty ]\) be measurable. Suppose that \(f:X\to [-\infty ,\infty ]\) and that \(f_n(x)\to f(x)\) for \(\mu \)-almost every \(x\in X\). Then \(f\) is measurable.
[Chapter 18, Problem 3] If \((X,\M ,\mu )\) is a measure space that is not complete, then there is a sequence \(\{f_n\}_{n=1}^\infty \) of \(\M \)-measurable functions that converge pointwise almost everywhere to a function that is not \(\mu \)-measurable.
(Muhammad, Problem 3090) Prove Theorem 18.6.
Corollary 18.7. Let \((X,\M )\) be a measurable space. Let \(f_n:X\to [-\infty ,\infty ]\) be \(\M \)-measurable. Then \(\sup _n f_n\), \(\inf _n f_n\), \(\liminf _n f_n\), and \(\limsup _n f_n\) are all measurable.
The simple approximation lemma. Let \((X,\M )\) be a measurable space. Let \(f:X\to \R \) be measurable and bounded. Let \(\varepsilon >0\). Then there are two simple functions \(\varphi _\varepsilon \) and \(\psi _\varepsilon \) with \begin {equation*} \psi _\varepsilon (x)-\varepsilon \leq \varphi _\varepsilon (x)\leq f(x)\leq \psi _\varepsilon (x)\leq \varphi _\varepsilon (x)+\varepsilon \end {equation*} for all \(x\in X\).
The simple approximation theorem. Let \((X,\M )\) be a measurable space. Let \(f:X\to [-\infty ,\infty ]\). Then \(f\) is measurable if and only if there is a sequence \(\{\varphi _n\}_{n=1}^\infty \) such that
If \(\mu :\M \to [0,\infty ]\) is a measure and \((X,\M ,\mu )\) is \(\sigma \)-finite, then we may in addition require that \(\mu \{x\in X:\varphi _n(x)\neq 0\}<\infty \).
Egoroff’s theorem. Let \((X,\M ,\mu )\) be a measure space with \(\mu (X)<\infty \). Let \(\{f_n\}_{n=1}^\infty \) be a sequence of measurable functions on \(X\) that converges pointwise almost everywhere to some function \(f\) that is finite almost everywhere.
Then for every \(\varepsilon >0\) there is a measurable set \(F\subseteq X\) with \(m(X\setminus F)<\varepsilon \) and such that \(f_n\to f\) uniformly on \(F\).
[Axler, Definition 3.1: \(\M \)-partition] Let \(\M \) be a \(\sigma \)-algebra on a set \(X\). An \(\M \)-partition of \(X\) is a collection \(\P \) such that
\(\P \subseteq \M \),
\(\P \) is finite (that is, \(\P \) is a collection of finitely many sets),
\(X=\bigcup _{A\in \P } A\),
if \(A\), \(B\in \P \) then either \(A=B\) or \(A\cap B=\emptyset \).
[Axler, Definition 3.2: Lower Lebesgue sum] Let \((X,\M ,\mu )\) be a measure space and let \(f:X\to [0,\infty ]\) be measurable with respect to \(\M \). If \(P\subset \M \) is a \(\M \)-partition of \(X\), then \[\mathcal {L}(f,P)=\sum _{A\in P} \mu (A)\cdot \inf _A f\] where we take \(0\cdot \infty =0=\infty \cdot 0\).
[Axler, Definition 3.3: Integral of a nonnegative measurable function] Let \((X,\M ,\mu )\) be a measure space and let \(f:X\to [0,\infty ]\) be measurable. We define \[\int _X f\,d\mu =\sup \{\mathcal {L}(f,P):P\text { is a $\M $-partition of~$X$}\}.\]
(Ashley, Problem 3100) Let \((X,\M ,\mu )\) be a measure space and let \(f\), \(g:X\to [0,\infty ]\) be measurable. Suppose that \(g=f\) \(\mu \)-almost everywhere on \(X\). Show that \(\int _X f\,d\mu =\int _X g\,d\mu \).
Let \(E\in \M \) be such that \(\mu (X\setminus E)=0\) and such that \(f=g\) on \(E\). (It is true, but annoying to prove, that \(\tilde E=\{x\in X:f(x)=g(x)\}\) is measurable, and so we simply work with a set of full measure on which \(f\) and \(g\) are equal.)Let \(\P \) be a \(\M \)-partition of \(X\). Then \(\L (f,\P )=\sum _{A\in \P }\mu (A)\cdot \inf _A f\).
If \(A\in \P \), then \(A\in \M \). Then \(A\cap E\) and \(A\setminus E\) are also in \(\M \) (measurable) because \(\M \) is a \(\sigma \)-algebra. Because \(\mu \) is a measure, we have that \[0\leq \mu (A\setminus E)\leq \mu (X\setminus E)=0\] so \(\mu (A\setminus E)=0\), and \[\mu (A)=\mu (A\cap E)+\mu (A\setminus E)=\mu (A\cap E).\] Let \[\mathcal {Q}=\{A\cap E:A\in \P \}\cup \{A\setminus E:A\in \P \}.\] Then \(\mathcal {Q}\) is also a \(\M \)-partition of \(X\). We have that \begin {align*} \L (f,\P )&=\sum _{A\in \P }\mu (A)\cdot \inf _A f&&\text {by definition of }\L (f,P) \\&=\sum _{A\in \P }\mu (A\cap E)\cdot \inf _A f&&\text {because }\mu (A\cap E)=\mu (A) \\&\leq \sum _{A\in \P }\mu (A\cap E)\cdot \inf _{A\cap E} f&&\text {by monotonicity of the infimum} \\&=\sum _{A\in \P }\mu (A\cap E)\cdot \inf _{A\cap E} g&&\text {because $f=g$ in $E$} \\&=\sum _{A\in \P }\mu (A\cap E)\cdot \inf _{A\cap E} g \alignbreak \ifdim \hsize <4in\qquad \fi +\sum _{A\in \P }\mu (A\setminus E)\cdot \inf _{A\setminus E} g&&\text {because $\mu (A\setminus E)=0$} \\&=\sum _{B\in \mathcal {Q}}\mu (B)\cdot \inf _{B} g&&\text {by definition of $\mathcal {Q}$} \\&=\L (g,\mathcal {Q})\leq \int _X g\,d\mu &&\text {by definition of $\L $ and $\int _X$}. \end {align*}
Thus \[\int _X f\,d\mu =\sup \{\L (f,\P ):\P \text { is a $\M $-partition}\}\leq \int _X g\,d\mu .\] Similarly \(\int _X g\,d\mu \leq \int _X f\,d\mu \) and so the two must be equal.
Theorem 3.8. (Axler) Let \((X,\M ,\mu )\) be a measure space and let \(f\), \(g:X\to [0,\infty ]\) be measurable functions. Suppose that \(f(x)\leq g(x)\) for all \(x\in X\). Then \(\int _X f\,d\mu \leq \int _X g\,d\mu \).
(Bashar, Problem 3110) Prove Theorem 3.8 (Axler).
Let \(\mathcal {P}\) be any \(\M \)-partition of \(X\). Then \[\L (f,\P )=\sum _{A\in \P }\mu (A)\cdot \inf _A f\leq \sum _{A\in \P }\mu (A)\cdot \inf _A g=\L (g,P)\leq \int _X g\,d\mu \] because \(f\leq g\) in \(A\) for each \(A\in \mathcal {P}\) and by definition of the integral. Again by definition of the integral, \[\int _X f\,d\mu =\sup \{\L (f,\P ):\P \text { is a $\M $-partition}\}\leq \int _X g\,d\mu \] as desired.
Theorem 3.7. (Axler) Suppose that \(\varphi \) is a nonnegative simple function, that is, \(\varphi =\sum _{k=1}^n c_k \chi _{E_k}\) for some real numbers \(c_k\), where the \(E_k\)s are pairwise disjoint measurable sets. Then \[\int _X \varphi \,d\mu =\sum _{k=1}^n c_k \cdot \mu (E_k).\]
Theorem 3.15. (Axler) Suppose that \(\varphi =\sum _{k=1}^n c_k \chi _{E_k}\) for some real numbers \(c_k\) where the \(E_k\)s are measurable sets. Then \[\int _X \varphi \,d\mu =\sum _{k=1}^n c_k \cdot \mu (E_k)\] even if the \(E_k\)s are not pairwise disjoint.
The proof of Theorem 3.15 (Axler) (given Theorem 3.7 (Axler)) is identical to the proof of Lemma 4.1.
(Dibyendu, Problem 3120) Prove Theorem 3.7 (Axler).
Theorem 3.9. (Axler) Let \((X,\M ,\mu )\) be a measure space and let \(f:X\to [0,\infty ]\) be measurable. Then \[\int _X f\,d\mu =\sup \biggl \{\int _X \varphi \,d\mu :\varphi \text { is simple and } 0\leq \varphi (x)\leq f(x)\text { for all }x\in X\biggr \}.\]
(Irina, Problem 3130) Prove Theorem 3.9 (Axler).
If \(0\leq \varphi \leq f\) and \(\varphi \) is measurable, then \(\int _X\varphi \,d\mu \leq \int _X f\,d\mu \) by Theorem 3.8 (Axler), and so \[\int _X f\,d\mu \geq \sup \biggl \{\int _X \varphi \,d\mu :\varphi \text { is simple and } 0\leq \varphi (x)\leq f(x)\text { for all }x\in X\biggr \}.\] We now must establish the reverse inequality. Let \(\P \) be a \(\M \)-partition of \(X\). Then \[\L (f,P)=\sum _{A\in \P } \mu (A)\cdot \inf _A f.\] Let \(n\in \N \) and define \[c_A^n=\begin {cases} n,&\inf _A f=\infty ,\\ \inf _A f,& \inf _A f<\infty .\end {cases}\] Then \(c_A^n\) is finite for all \(n\in \N \) and all \(A\in \P \). Therefore \[\varphi _n = \sum _{A\in \P } c_A^n \,\chi _{A}\] is simple (recalling that, if \(\P \) is a \(\M \)-partition, then there are finitely many \(A\in \P \) and each \(A\in \P \) is measurable). Furthermore, if \(x\in X\), then \(x\in B\) for exactly one \(B\in \P \) by definition of partition; thus \[\varphi _n(x)=\sum _{A\in \P } c_A^n \,\chi _{A}(x)=c_B^n \leq \inf _B f\leq f(x).\] Thus \(\varphi _n\) is simple and \(0\leq \varphi (x)\leq f(x)\) for all \(x\in X\).We compute \begin {align*} \L (f,\P )&=\sum _{A\in \P } \mu (A)\cdot \inf _A f=\sum _{\substack {A\in \P \\\mu (A)>0}} \mu (A)\cdot \inf _A f, \\ \int _X \varphi _n \,d\mu &= \sum _{A\in \P } \mu (A)\cdot c_A^n = \sum _{\substack {A\in \P \\\inf _A f<\infty \\\mu (A)>0}} \mu (A)\cdot \inf _A f + n\sum _{\substack {A\in \P \\\inf _A f=\infty \\\mu (A)>0}}\mu (A). \end {align*}
If \(\inf _A f<\infty \) for all \(A\in \P \) with \(\mu (A)>0\), then the two terms are equal and so \[\L (f,\P )\leq \sup \biggl \{\int _X \varphi \,d\mu :\varphi \text { is simple and } 0\leq \varphi (x)\leq f(x)\text { for all }x\in X\biggr \}.\] If \(\inf _A f=\infty \) for at least one \(A\in \P \) with \(\mu (A)>0\), then \(\L (f,\P )=\infty \) and \(\int _X \varphi _n\,d\mu \geq n\mu (A)\), and so \[\sup \biggl \{\int _X \varphi \,d\mu :\varphi \text { is simple and } 0\leq \varphi (x)\leq f(x)\text { for all }x\in X\biggr \} \geq \sup _{n\in \N } n\mu (A)=\infty =\L (f,\P ).\] Thus, in either case, \[\L (f,\P )\leq \sup \biggl \{\int _X \varphi \,d\mu :\varphi \text { is simple and } 0\leq \varphi (x)\leq f(x)\text { for all }x\in X\biggr \}\] which completes the proof.
Theorem 3.11. (Axler) (The monotone convergence theorem.) Let \((X,\M ,\mu )\) be a measure space, and let \(\{f_n\}_{n=1}^\infty \) be a sequence of nonnegative measurable functions \(f_n:X\to [0,\infty ]\). Suppose in addition that \(f_n(x)\leq f_{n+1}(x)\) for all \(x\in X\). Then \begin {equation*} \int _X \lim _{n\to \infty } f_n\,d\mu = \lim _{n\to \infty } \int _X f_n\,d\mu . \end {equation*}
(Micah, Problem 3140) Prove Theorem 3.11 (Axler).
Theorem 18.11. If \((X,\M ,\mu )\) is a measure space, \(f\), \(g:X\to [0,\infty ]\) are measurable, and \(\alpha \), \(\beta >0\), then \[\int _X (\alpha f+\beta g)\,d\mu =\alpha \int _X f\,d\mu +\beta \int _X g\,d\mu .\]
(Muhammad, Problem 3150) Use Theorem 3.15 (Axler), Theorem 3.11 (Axler) (the monotone convergence theorem), and the simple approximation theorem to prove Theorem 18.11.
By the simple approximation theorem, and because \(f\) and \(g\) are nonnegative, there exist sequences of functions \(\{\varphi _n\}_{n=1}^\infty \) and \(\{\psi _n\}_{n=1}^\infty \) such that
Each \(\varphi _n\), \(\psi _n\) is simple
\(\varphi _n\to f\) and \(\psi _n\to g\) pointwise
Each \(\varphi _n\) and \(\psi _n\) is nonnegative
The sequences \(\{\varphi _n(x)\}_{n=1}^\infty \) and \(\{\psi _n(x)\}_{n=1}^\infty \) are nondecreasing for every \(x\in X\).
Then \(\{\alpha \varphi _n+\beta \psi _n\}_{n=1}^\infty \) is also a sequence of nonnegative, nondecreasing simple functions, and this sequence converges pointwise to \(\alpha f+\beta g\).
By the monotone convergence theorem, \begin {align*} \int _X (\alpha f+\beta g)\,d\mu &= \int _X \lim _{n\to \infty } \alpha \varphi _n+\beta \psi _n\,d\mu = \lim _{n\to \infty } \int _X \alpha \varphi _n+\beta \psi _n\,d\mu ,\\ \int _X f\,d\mu &= \int _X \lim _{n\to \infty } \varphi _n\,d\mu = \lim _{n\to \infty } \int _X \varphi _n\,d\mu ,\\ \int _X g\,d\mu &= \int _X \lim _{n\to \infty } \psi _n\,d\mu = \lim _{n\to \infty } \int _X \psi _n\,d\mu . \end {align*}
Because \(\varphi _n\) and \(\psi _n\) are simple, we may write \[\varphi _n=\sum _{k=1}^{m_n} a_{n,k} \chi _{A_{n,k}},\qquad \psi _n=\sum _{k=1}^{\ell _n} b_{n,k} \chi _{B_{n,k}}\] for some integers \(m_n\) and \(\ell _n\), some real numbers \(a_{n,k}\) and \(b_{n,k}\), and some measurable sets \(A_{n,k}\) and \(B_{n,k}\). Let \[e_{n,k}=\begin {cases}\alpha \, a_{n,k}, & 1\leq k\leq m_n, \\\beta \, b_{n,k-m_n}, & m_n+1\leq k\leq m_n+\ell _n,\end {cases} \qquad E_{n,k}=\begin {cases}A_{n,k}, & 1\leq k\leq m_n, \\B_{n,k-m_n}, & m_n+1\leq k\leq m_n+\ell _n.\end {cases}\] Then \[\alpha \varphi _n+\beta \psi _n=\sum _{k=1}^{m_n+\ell _n} e_{n,k}\chi _{E_{n,k}}.\] By Theorem 3.15 (Axler), \begin {align*} \int _X \varphi _n\,d\mu &= \sum _{k=1}^{m_n} a_{n,k} \mu (A_{n,k}) ,\\ \int _X \psi _n\,d\mu &= \sum _{k=1}^{\ell _n} b_{n,k} \mu (B_{n,k}) ,\\ \int _X \alpha \varphi _n+\beta \psi _n\,d\mu &= \sum _{k=1}^{m_n+\ell _n} e_{n,k}\chi _{E_{n,k}} \end {align*}
We compute \begin {align*} \int _X \alpha \varphi _n+\beta \psi _n\,d\mu &= \sum _{k=1}^{m_n+\ell _n} e_{n,k}\chi _{E_{n,k}} \\&= \Bigl (\sum _{k=1}^{m_n}e_{n,k}\chi _{E_{n,k}}\Bigr )+ \Bigl (\sum _{k=m_n+1}^{m_n+\ell _n}e_{n,k}\chi _{E_{n,k}}\Bigr ) \\&= \Bigl (\sum _{k=1}^{m_n}\alpha a_{n,k}\chi _{A_{n,k}}\Bigr )+ \Bigl (\sum _{k=1}^{\ell _n}\beta b_{n,k}\chi _{B_{n,k}}\Bigr ) \\&= \alpha \Bigl (\sum _{k=1}^{m_n}a_{n,k}\chi _{A_{n,k}}\Bigr )+ \beta \Bigl (\sum _{k=1}^{\ell _n}b_{n,k}\chi _{B_{n,k}}\Bigr ) \\&= \alpha \int _X \varphi _n\,d\mu + \beta \int _X \psi _n\,d\mu . \end {align*}
Thus by linearity of limits, \begin {align*} \int _X (\alpha f+\beta g)\,d\mu &=\lim _{n\to \infty } \int _X \alpha \varphi _n+\beta \psi _n\,d\mu \\&=\lim _{n\to \infty } \biggl (\alpha \int _X \varphi _n\,d\mu + \beta \int _X \psi _n\,d\mu \biggr ) \\&=\alpha \biggl (\lim _{n\to \infty } \int _X \varphi _n\,d\mu \biggr )+ \beta \biggl (\lim _{n\to \infty } \int _X \psi _n\,d\mu \biggr ) \\&=\alpha \int _X f\,d\mu + \beta \int _X g\,d\mu \end {align*}
as desired.
Fatou’s lemma. Let \((X,\M ,\mu )\) be a measure space, and let \(\{f_n\}_{n=1}^\infty \) be a sequence of nonnegative measurable functions \(f_n:X\to [0,\infty ]\). Then \begin {equation*} \int _X \liminf _{n\to \infty } f_n\,d\mu \leq \liminf _{n\to \infty } \int _X f_n\,d\mu . \end {equation*}
[Chapter 3A, Problem 17 (Axler)] Use Theorem 3.11 (Axler) (the monotone convergence theorem) to prove Fatou’s lemma.
[Axler, Definition 3.18: Integral of a real valued function] Let \((X,\M ,\mu )\) be a measure space and let \(f:X\to [-\infty ,\infty ]\) be \(\M \)-measurable. Suppose that \(\int _X f^+\,d\mu \) and \(\int _X f^-\,d\mu \) are not both infinite. Then we define \[\int _X f\,d\mu =\int _X f^+\,d\mu -\int _X f^-\,d\mu .\] (Here \(f^\pm \) are as defined in Problem 1022.)
Theorem 3.20. (Axler) Let \((X,\M ,\mu )\) be a measure space and let \(f\), \(g:X\to [-\infty ,\infty ]\) be \(\M \)-measurable. If \(\int _X f\,d\mu \) exists and \(c\in \R \), then \(\int _X cf\,d\mu =c\int _X f\,d\mu \).
Theorem 3.21. (Axler) Let \((X,\M ,\mu )\) be a measure space and let \(f\), \(g:X\to [-\infty ,\infty ]\) be \(\M \)-measurable. If \(\int _X |f|\,d\mu +\int _X |g|\,d\mu <\infty \), then \(\int _X (f+g)\,d\mu \) exists and \[\int _X (f+g)\,d\mu =\int _X f\,d\mu +\int _X g\,d\mu .\]
(Ashley, Problem 3160) Prove Theorem 3.21 (Axler).
Theorem 3.22. (Axler) Let \((X,\M ,\mu )\) be a measure space and let \(f\), \(g:X\to [-\infty ,\infty ]\) be \(\M \)-measurable. Suppose that \(f(x)\leq g(x)\) for all \(x\in X\) and that \(\int _X f\,d\mu \), \(\int _X g\,d\mu \) exist. Then \(\int _X f\,d\mu \leq \int _X g\,d\mu \).
(Bashar, Problem 3170) Prove Theorem 3.22 (Axler) in the case where \(\int _X |f|\,d\mu \) and \(\int _X |g|\,d\mu \) are both finite.
If \(\int _X |f|\,d\mu <\infty \) and \(\int _X |g|\,d\mu <\infty \), then \(|f|\) and \(|g|\) must be finite \(\mu \)-almost everywhere, and so \(g-f\) is defined and finite \(\mu \)-almost everywhere. But \(g-f\geq 0\) everywhere it is defined, and so we must have that \[\int _X g-f\,d\mu \geq 0\] by the definition of the integral of a nonnegative function. Applying Theorem 3.20 (Axler) and Theorem 3.21 (Axler) yields that \[\int _X g\,d\mu -\int _X f\,d\mu \geq 0\] which immediately yields the desired result.
(Problem 3171) Prove Theorem 3.22 (Axler) in the case where \(\int _X |f|\,d\mu \) or \(\int _X |g|\,d\mu \) is infinite.
If \(\int _X |f|\,d\mu =\int _X f^+\,d\mu +\int _X f^-\,d\mu =\infty \), then at least one of \(\int _X f^+\,d\mu \) and \(\int _X f^-\,d\mu \) must be infinite. Recall that \(\int _X f\,d\mu \) is assumed to exist, so \(\int _X f^+\,d\mu \) and \(\int _X f^-\,d\mu \) cannot both be infinite. Thus if \(\int _X |f|\,d\mu \) is infinite, then either \(\int _X f\,d\mu =\infty \) or \(\int _X f\,d\mu =-\infty \). Similarly if \(\int _X |g|\,d\mu \) is infinite, then either \(\int _X g\,d\mu =\infty \) or \(\int _X g\,d\mu =-\infty \).If either \(\int _X g\,d\mu =\infty \) or \(\int _X f\,d\mu =-\infty \) then Theorem 3.22 (Axler) is trivially true, so we need only consider the case \(\int _X f\,d\mu =\infty \) or \(\int _X g\,d\mu =-\infty \). We will consider the case \(\int _X g\,d\mu =-\infty \); the other case is similar.
If \(\int _X g\,d\mu =-\infty \), then \(\int _X g^+\,d\mu \) is finite and \(\int _X g^-\,d\mu =\infty \). If \(g\geq f\), then \(g^-\leq f^-\), and so by Theorem 3.8 (Axler), we have that \(\int _X f^-\,d\mu =\infty \) and so \(\int _X f\,d\mu =-\infty \). Thus \(\int _X f\,d\mu =-\infty \leq -\infty =\int _X g\,d\mu \), as desired.
Theorem 3.23. (Axler) Let \((X,\M ,\mu )\) be a measure space and let \(f:X\to [-\infty ,\infty ]\) be \(\M \)-measurable. If \(\int _X f\,d\mu \) exists then \[\biggl |\int _X f\,d\mu \biggr |\leq \int _X |f|\,d\mu \] and if \(\int _X f\,d\mu \) does not exist then \[\int _X |f|\,d\mu =\infty .\]
(Dibyendu, Problem 3180) Prove Theorem 3.23 (Axler).
This follows immediately from the fact that \[\int _X f\,d\mu =\int _X f^+\,d\mu -\int _X f^-\,d\mu \] and \[\int _X|f|\,d\mu =\int _X f^+\,d\mu +\int _X f^-\,d\mu .\]
(Irina, Problem 3190) Let \((X,\M ,\mu )\) be a measure space, let \(E\in \M \), and let \(f:X\to [-\infty ,\infty ]\) be \(\M \)-measurable. Show that \[\int _E (f\big \vert _E)\,d\mu =\int _X (f\chi _E)\,d\mu \] if either is defined.
[Axler, Definition 3.24: Integral over a subset] Let \((X,\M ,\mu )\) be a measure space, let \(E\in \M \), and let \(f:X\to [-\infty ,\infty ]\) be \(\M \)-measurable. We define \[\int _E f\,d\mu =\int _X (f\chi _E)\,d\mu .\]
Theorem 3.25. (Axler) Let \((X,\M ,\mu )\) be a measure space, let \(E\in \M \), and let \(f:X\to [-\infty ,\infty ]\) be \(\M \)-measurable and such that \(\int _E f\,d\mu \) is defined. Then \[\biggl |\int _E f\,d\mu \biggr |\leq \mu (E)\cdot \sup _E |f|.\]
(Micah, Problem 3200) Prove Theorem 3.25 (Axler).
Theorem 3.28. (Axler) Let \((X,\M ,\mu )\) be a measure space, let \(g:X\to [0,\infty ]\) be \(\M \)-measurable, and suppose that \(\int _X g\,d\mu <\infty \). Then for every \(\varepsilon >0\), there exists a \(\delta >0\) such that, if \(B\in \M \) and \(\mu (B)<\delta \), then \[\int _B g\,d\mu <\varepsilon .\]
The proof of Theorem 3.28 (Axler) is identical to the proof of Proposition 4.23, part (a).
Theorem 3.29. (Axler) Let \((X,\M ,\mu )\) be a measure space, let \(g:X\to [0,\infty ]\) be \(\M \)-measurable, and suppose that \(\int _X g\,d\mu <\infty \). Then for every \(\varepsilon >0\), there exists an \(E\in \M \) with \(\mu (E)<\infty \) and with \[\int _{X\setminus E} g\,d\mu <\varepsilon .\]
The proof of Theorem 3.28 (Axler) is identical to the proof of Proposition 5.1.
Theorem 2.85. (Axler) (Egorov’s theorem.) Let \((X,\M ,\mu )\) be a measure space with \(\mu (X)<\infty \). Let \(\{f_n\}_{n=1}^\infty \) be a sequence of \(\M \)-measurable functions on \(X\) that converges pointwise almost everywhere to some function \(f\) that is finite almost everywhere.
Then for every \(\varepsilon >0\) there is a set \(E\in \M \) with \(m(X\setminus E)<\varepsilon \) and such that \(f_n\to f\) uniformly on \(E\).
The proof of the general case of Egorov’s theorem is the same as in the case of one-dimensional Lebesgue measure.
Theorem 3.31. (Axler) (The Dominated Convergence Theorem). Let \((X,\M ,\mu )\) be a measure space, let \(g:X\to [0,\infty ]\) be \(\M \)-measurable, and suppose that \(\int _X g\,d\mu <\infty \).
If \(f_k:X\to [-\infty ,\infty ]\) is \(\M \)-measurable, \(|f_k(x)|\leq g(x)\) for all \(k\in \N \) and all \(x\in X\), and if \(f(x)=\lim _{k\to \infty } f_k(x)\) exists for almost every \(x\in X\), then \[\lim _{k\to \infty } \int _X f_k\,d\mu =\int _X f\,d\mu .\]
(Muhammad, Problem 3210) Prove Theorem 3.31 (Axler) in the case where \(\mu (X)<\infty \).
(Ashley, Problem 3220) Prove Theorem 3.31 (Axler) in the case where \(\mu (X)=\infty \).
(Bashar, Problem 3230) Let \((X,\M ,\mu )\) be a measure space, let \(f:X\to [0,\infty ]\) be \(\M \)-measurable, and let \(\{A_n\}_{n=1}^\infty \) be a sequence of pairwise-disjoint sets in \(\M \).
If \(A=\bigcup _{n=1}^\infty A_n\), show that \[\int _A f\,d\mu =\sum _{n=1}^\infty \int _{A_n} f\,d\mu .\]
Theorem 18.13. Let \((X,\M ,\mu )\) be a measure space, let \(f:X\to [-\infty ,\infty ]\) be \(\M \)-measurable, and let \(\{A_n\}_{n=1}^\infty \) be a sequence of pairwise-disjoint sets in \(\M \). If at least one of \(\int _X f^+\,d\mu \) and \(\int _X f^-\,d\mu \) is finite, then \[\int _A f\,d\mu =\sum _{n=1}^\infty \int _{A_n} f\,d\mu \] where \(A=\bigcup _{n=1}^\infty A_n\).
(Dibyendu, Problem 3240) Prove Theorem 18.13.
(Problem 3241) Let \((X,\M ,\mu )\) be a measure space and let \(f:X\to [0,\infty ]\). Show that \(\int _\emptyset f\,d\mu =0\).
(Problem 3242) Let \((X,\M ,\mu )\) be a measure space and let \(f:X\to [0,\infty ]\). Define \(\nu :\M \to [0,\infty ]\) by \[\nu (E)=\int _E f\,d\mu .\] Show that \(\nu \) is a measure on \(\M \).
[Definition: Absolutely continuous] If \((X,\M )\) is a measurable space and \(\mu \), \(\nu \) are two measures on \((X,\M )\), we say that \(\nu \) is absolutely continuous with respect to \(\mu \), or \(\nu \ll \mu \), if for all \(E\in \M \) such that \(\mu (E)=0\) we have that \(\nu (E)=0\).
(Problem 3243) Let \((X,\M ,\mu )\) be a measure space and let \(f:X\to [0,\infty ]\). Define \(\nu :\M \to [0,\infty ]\) by \[\nu (E)=\int _E f\,d\mu .\] Recall that \(\nu \) is a measure on \(\M \). Show that \(\nu \ll \mu \).
The Radon-Nikodym Theorem. Let \((X,\M )\) be a measurable space and let \(\mu \), \(\nu \) be two measures on \((X,\M )\) such that \(\nu \ll \mu \) and such that \((X,\M ,\mu )\) is \(\sigma \)-finite.
Then there is a \(\M \)-measurable function \(f:\M \to [0,\infty ]\) such that \[\nu (E)=\int _E f\,d\mu \] for all \(E\in \M \).
[Chapter 18, Problem 60] Let \(X=[0,1]\) and let \(\M \) be the set of Lebesgue measurable subsets of \(X\). Let \(\mu \) denote the counting measure and let \(\nu \) denote the Lebesgue measure. Then \(\nu \ll \mu \) but there is no \(\M \)-measurable function \(f:X\to [0,\infty ]\) that satisfies the conclusions of the Radon-Nikodym theorem.
Proposition 18.19. (This is the first step in the proof of the Radon-Nikodym theorem.) Let \((X,\M )\) be a measurable space and let \(\mu \), \(\nu \) be two measures on \((X,\M )\) such that \(\nu (X)<\infty \). Then \(\nu \ll \mu \) if and only if, for every \(\varepsilon >0\), there is a \(\delta >0\) such that, if \(E\in \M \) and \(\mu (E)<\delta \), then \(\nu (E)<\varepsilon \).
(Problem 3244) Assume that for every \(\varepsilon >0\), there is a \(\delta >0\) such that, if \(E\in \M \) and \(\mu (E)<\delta \), then \(\nu (E)<\varepsilon \). Show that \(\nu \ll \mu \).
(Irina, Problem 3250) Complete the proof of Proposition 18.19.
(Micah, Problem 3260) Let \((X,\M )\) be a measurable space and let \(\mu \), \(\nu \) be two measures on \((X,\M )\) such that \(\nu \ll \mu \). Define \begin {gather*} \mathcal {F}=\Bigl \{f:X\to [0,\infty ]:f\text { is $\M $-measurable and } \gatherbreak \int _E f\,d\mu \leq \nu (E)\text { for all }E\in \M \Bigr \}. \end {gather*} Show that if \(f\), \(g\in \mathcal {F}\), then so is \(h=\max (f,g)\).
(Muhammad, Problem 3270) Let \(f\in \mathcal {F}\). Suppose that \(\int _X f\,d\mu =\nu (X)<\infty \). Show that \(\int _E f\,d\mu =\nu (E)\) for all \(E\in \M \).
(Ashley, Problem 3280) Let \((X,\M )\) be a measurable space and let \(\mu \), \(\theta \) be two measures on \((X,\M )\) such that \(\theta \ll \mu \) and such that \(0<\theta (X)\) and \(\mu (X)<\infty \). Show that there exists a function \(\varphi \) such that
\(\varphi :X\to [0,\infty )\),
\(\varphi \) is \(\M \)-measurable,
\(\int _E \varphi \,d\mu \leq \theta (E)\) for all \(E\in \M \),
\(\int _X \varphi \,d\mu >0\).
(Bashar, Problem 3290) Prove the Radon-Nikodym Theorem in the case where \(\mu (X)<\infty \) and \(\nu (X)<\infty \).
(Bonus Problem 3291) Show that the Radon-Nikodym theorem is true if \((X,\M ,\mu )\) is \(\sigma \)-finite, whether \(\nu \) is finite, \(\sigma \)-finite, or not \(\sigma \)-finite.
[Chapter 18, Problem 49] If the Radon-Nikodym theorem is true whenever \(\mu (X)<\infty \) and \(\nu (X)<\infty \), then it is true whenever \((X,\M ,\mu )\) and \((X,\M ,\nu )\) are \(\sigma \)-finite.
(Bonus Problem 3292) Show that the Radon-Nikodym theorem is true if \(\mu (X)<\infty \), whether \(\nu \) is finite, \(\sigma \)-finite, or not \(\sigma \)-finite.
(Problem 3293) Show that the function \(f\) in the Radon-Nikodym theorem is unique up to redefinition on sets of measure zero.
[Definition: Radon-Nikodym derivative] The function \(f\) in the Radon-Nikodym theorem is called the Radon-Nikodym derivative of \(\nu \) with respect to \(\mu \) and is often denoted \(\frac {d\nu }{d\mu }\).
