(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 \(a\), \(b\), \(c\in F\),
(Elliott, 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 \(\R \cup \{-\infty ,\infty \}=[-\infty ,\infty ]\).
(Irina, 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 \(F\).
(Zach, 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.
(Juan, 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 \(n>a\).
(Micah, 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 \(\R \).
(Muhammad, 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 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.
[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 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.
(Anjuman, 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 bijection \(f:S\to 2^S\).
[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 \(X\).
(Ashley, 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 \(\mathcal {F}\).
(Bashar, Problem 330) Prove Proposition 1.13.
(Dibyendu, 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.
(Elliott, 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 \(\mathcal {J}(E)=\int _{-\infty }^\infty \chi _E\).
(Irina, 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*}
(Zach, 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 \((X,\M )\).
(Juan, 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 \(m^*(S)=0\).
(Micah, 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.
(Muhammad, Problem 410) Let \(I=[a,b]\) be a closed bounded interval (that is, \(a\) and \(b\) are both real numbers). Show that \(m^*(I)=\ell (I)=b-a\).
(Anjuman, 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*}
(Ashley, 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*}
(Bashar, 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 \(r-s\in \Q \).
(Dibyendu, 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 \(V\).
(Elliott, 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.
(Irina, Problem 470) Show that \begin {equation*} [-1,1]\subseteq \bigcup _{q\in [-2,2]\cap {\Q }} V_q\subseteq [-3,3]. \end {equation*}
(Zach, 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 \(m^*(V)>0\).
(Juan, 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 \(\sum _{k=1}^\infty m^*(V_{q_k}) \neq m^*\Bigl (\bigcup _{k=1}^\infty V_{q_k}\Bigr )\).
(Micah, 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.
(Muhammad, 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.
(Anjuman, Problem 540) Prove Proposition 2.4.
Proposition 2.5. The union of finitely many measurable sets is measurable.
(Ashley, 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*}
(Bashar, 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.
(Dibyendu, 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.
(Elliott, 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 \(\R \).
(Irina, 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.
(Zach, 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.
(Juan, 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 \(m^*(E)<\infty \).
(Micah, Problem 660) Show that (v) implies (i) in general.
(Muhammad, Problem 670) Show that (iii) implies (iv).
(Anjuman, 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*}
(Ashley, 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?
(Bashar, 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 \(E\).
(Dibyendu, 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.
(Elliott, 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 \(j\neq k\).
(Irina, 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 \(3^{-n}\).
(Zach, 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 \(\Lambda _2\).
(Juan, Problem 760) Suppose that \(x\notin F_n\). Show that \(\Lambda (x)=2^{-n}|\{k\in \{1,2,\dots ,2^n\}:b_{k,n}<x\}|\).
(Micah, 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 \(x\notin F_n\).
(Muhammad, Problem 780) Show that \(\{\Lambda _k\}_{k=1}^\infty \) is uniformly Cauchy.
[Definition: The Cantor function] Let \(\Lambda (x)=\lim _{k\to \infty } \Lambda _k(x)\).
(Ashley, 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 \(E\).
(Bashar, Problem 800) Show that \(\Lambda '(x)=0\) for every \(x\in \R \setminus C\), and thus for almost every \(x\in \R \).
(Dibyendu, 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.
(Elliott, 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.
(Irina, 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 \(m^*(F\cap V_{q_k})=0\).
(Zach, 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.
(Juan, 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.
(Micah, Problem 860) Show that \(f(C)\) has positive measure.
(Muhammad, 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.
(Ashley, Problem 880) Prove part (ii) of Theorem 2.15 without using part (i).
(Bashar, 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 \(x\in \R \).
(Dibyendu, 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.
(Elliott, Problem 910) Prove Proposition 3.3.
(Irina, 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 \(c\in [-\infty ,\infty ]\).
(Zach, 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 \(\{x\in E:f(x)>c\}=f^{-1}((c,\infty ))\) 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 \(Y\).
(Juan, 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 \(\mathcal {O}\).
(Micah, Problem 950) Prove that in fact, if \(f\) is measurable, then then \(f^{-1}(B)\) is measurable for all Borel sets \(B\).
(Muhammad, Problem 960) If \(g\) is measurable, is it true that \(g^{-1}(E)\) is measurable for all measurable sets \(E\)?
(Ashley, 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).
(Bashar, 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.
(Dibyendu, 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.
(Elliott, 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.
(Irina, 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.
(Zach, 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.
(Juan, 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.
(Micah, 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.
(Muhammad, 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 \(x\in E\).
(Ashley, 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.
(Bashar, 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.
(Dibyendu, 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 \(n\geq k\).
(Elliott, 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.)
(Zach, 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.
(Irina, 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 \(m^*(\{x\in \R :g(x)\neq \varphi (x)\})<\varepsilon \).
(Juan, 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 \(m(E\setminus F)<\varepsilon \).
(Micah, 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 \(1\leq k\leq n\).
[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 1132) 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 1133) 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.
(Muhammad, 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.
(Ashley, 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 1133, \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 \(E\).
(Bashar, 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.
(Dibyendu, 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 \(\int _a^b f=\int _{[a,b]}f\).
(Elliott, 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.
(Irina, 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 \(E\).
(Zach, Problem 1200) Prove Theorem 4.5, part (i).
(Juan, 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 \(\int _{A\cup B} f=\int _A f+\int _B f\).
(Micah, 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*}
(Muhammad, Problem 1230) Prove Proposition 4.8.
(Ashley, 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*}
(Bashar, 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*}
(Dibyendu, 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 \(x\in E\).
(Elliott, 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 \(E\).
(Irina, 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 \(f:E\to [0,\infty ]\).
(Juan, 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*}
(Zach, 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*}
(Micah, 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 \(E\).
(Muhammad, 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 1132 \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 \(\left |\int _E f\right |\leq \int _E |f|\leq \int _E g\).
(Ashley, Problem 1330) Prove Proposition 4.16.
Theorem 4.17. Let \(f\) and \(g\) be functions integrable over a measurable set \(E\).
(Bashar, 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.
(Elliott, 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 \(\int _{A\cup B} f=\int _A f+\int _B f\).
(Dibyendu, 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 \(\int _E f_n\to \int _E f\).
(Zach, 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*}
(Irina, 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 \(E=\bigcup _{k=1}^n E_k\).
(Micah, Problem 1390) Prove Lemma 4.22.
Proposition 4.23. Let \(E\subseteq \R \) be measurable and let \(f\) be a measurable function on \(E\).
(Muhammad, Problem 1400) Prove Proposition 4.23, part (a).
(Juan, 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.
(Ashley, 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.
(Bashar, 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.
(Elliott, 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.
(Irina, 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 \(\int _E f_n\to \int _E f\).
(Zach, 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 \(E\).
(Juan, 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 \(\int _{E\setminus E_0} |f|<\varepsilon \).
(Micah, 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.
(Muhammad, 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 \(\lim _{n\to \infty }\int _E f_n=\int _E f\).
(Ashley, 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 \(E\).
(Bashar, Problem 1520) Prove Proposition 5.3.
(Dibyendu, 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 \(x\in [0,1]\).
(Irina, 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.
(Elliott, 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 \(E\).
(Zach, 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 \(\int _E h_n\to 0\).
(Juan, 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 \(X\).
(Micah, 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*}
(Muhammad, 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 \(x\in E\).
(Ashley, 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.
(Dibyendu, 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*}
(Bashar, 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.
(Elliott, 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 \(\psi _n(x)\to f(x)\).
(Irina, 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.
(Zach, 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 \(L_{step}\geq L_{cts}\).
(Juan, 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*}
(Micah, 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.
(Muhammad, 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 \(x\}\).
(Ashley, 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 \(s\overline {B}(x,r)=\overline {B}(x,sr)\).
(Bashar, 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 \(\beta \subseteq 5B\).
(Elliott, 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.
(Dibyendu, 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 \(B\in \mathcal {S}_j\).
(Zach, 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.
(Irina, 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*}
(Micah, 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*}
(Muhammad, Problem 1760) Prove Lemma 6.3.
Lebesgue’s theorem. If \(f:(a,b)\to \R \) is monotonic, then \(f\) is differentiable almost everywhere in \((a,b)\).
(Ashley, 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*}
(Juan, 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*}
(Bashar, 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.
(Dibyendu, 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 \(b=x_n\).
(Elliott, Problem 1810) If \(f:[a,b]\to \R \) is nondecreasing, show that \(TV(f)=f(b)-f(a)\).
(Irina, Problem 1820) If \(f:[a,b]\to \R \) is Lipschitz with Lipschitz constant \(M\), show that \(TV(f)\leq M(b-a)\).
(Zach, Problem 1830) If \(f:[a,b]\to \R \), \(g:[a,b]\to \R \), show that \(TV(f+g)\leq TV(f)+TV(g)\).
(Micah, 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.
(Ashley, 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.
(Dibyendu, 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 \(k\).)
(Muhammad, 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.
(Bashar, 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.
(Elliott, 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 \([a,b]\).
(Irina, 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.
(Zach, 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.
(Micah, 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.
(Ashley, 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.
(Bashar, Problem 1940) Complete the proof of Theorem 6.9.
(Dibyendu, 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*}
(Elliott, 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.
(Muhammad, 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 \(f'\).
(Irina, 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*}
(Zach, 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 \(a<x<y<b\).
(Micah, 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 \((a,b)\).
(Ashley, Problem 2010) Prove Theorem 6.14.
(Dibyendu, 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.
(Bashar, 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*}
(Elliott, 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*}
(Irina, 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 \(K\).
(Muhammad, 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.
(Zach, 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*}
(Ashley, 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*}
(Dibyendu, 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
(Micah, 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*}
(Bashar, 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*}
(Elliott, 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*}
(Irina, 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*}
(Zach, 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*}
(Micah, Problem 2150) Prove Hölder’s inequality in the case \(p=1\) or \(p=\infty \).
(Ashley, 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*}
(Dibyendu, 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*}
(Muhammad, Problem 2180) Prove Minkowski’s Inequality in the cases \(p=1\) and \(p=\infty \).
(Bashar, 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 \[\int _E |f+g|^p \leq 2^p\int _E f^p+\int _E g^p\] 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 \[\int _E (f+g)h=\int _E fh+\int _E gh \leq \|f\|_p\|h\|_q+\|g\|_p\|h\|_q=\|f\|_p+\|g\|_p\] 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.
(Elliott, 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*}
(Irina, Problem 2210) Prove Corollary 7.3.
(Zach, 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.
(Ashley, 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 \[\sum _{k=1}^\infty \sqrt {\|v_{n+1}-v_n\|}<\infty .\]
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.
(Micah, 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 \(x\in E\).
(Dibyendu, 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.
(Muhammad, 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 \(\R \).
(Bashar, Problem 2270) Prove Theorem 7.7.
(Elliott, 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 \(\|v-a\|<r\).
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 \(E\).
(Bonus Problem 2281) Prove Proposition 7.9.
Proposition 7.10. The set of compactly supported step functions is dense in \(L^p(\R )\) for \(1\leq p<\infty \).
(Bonus Problem 2282) 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).
(Bonus Problem 2283) 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.
(Bonus Problem 2284) Prove Theorem 7.11.
(Bonus Problem 2285) Show that \(\ell ^\infty \) is not separable.
(Bonus Problem 2286) Show that \(L^\infty (\R )\) is not separable.
