# Measure Theory: Sets

## Fall 2021.3 #real_analysis/qual/work

Recall that a set $$E \subset \mathbb{R}^{d}$$ is measurable if for every $$c>0$$ there is an open set $$U \subseteq {\mathbf{R}}^d$$ such that $$m^{*}(U \setminus E)<\epsilon$$.

• Prove that if $$E$$ is measurable then for all $$\epsilon>0$$ there exists an elementary $$\operatorname{set} F$$, such that $$m(E \Delta F)<\epsilon$$.

Here $$m(E)$$ denotes the Lebesgue measure of $$E$$, a set $$F$$ is called elementary if it is a finite union of rectangles and $$E \Delta F$$ denotes the symmetric difference of the sets $$E$$ and $$F$$.

• Let $$E \subset \mathbb{R}$$ be a measurable set, such that $$0<m(E)<\infty$$. Use part (a) to show that \begin{align*} \lim _{n \rightarrow \infty} \int_{E} \sin (n t) d t=0 \end{align*}

## Spring 2020.2 #real_analysis/qual/completed

Let $$m_*$$ denote the Lebesgue outer measure on $${\mathbf{R}}$$.

a.. Prove that for every $$E\subseteq {\mathbf{R}}$$ there exists a Borel set $$B$$ containing $$E$$ such that \begin{align*} m_*(B) = m_*(E) .\end{align*}

b.. Prove that if $$E\subseteq {\mathbf{R}}$$ has the property that \begin{align*} m_*(A) = m_*(A\bigcap E) + m_*(A\bigcap E^c) \end{align*} for every set $$A\subseteq {\mathbf{R}}$$, then there exists a Borel set $$B\subseteq {\mathbf{R}}$$ such that $$E = B\setminus N$$ with $$m_*(N) = 0$$.

Be sure to address the case when $$m_*(E) = \infty$$.

• Definition of outer measure: \begin{align*} m_*(E) = \inf_{\left\{{Q_j}\right\} \rightrightarrows E} \sum {\left\lvert {Q_j} \right\rvert} \end{align*} where $$\left\{{Q_j}\right\}$$ is a countable collection of closed cubes.
• Break $${\mathbf{R}}$$ into $${\textstyle\coprod}_{n\in {\mathbf{Z}}} [n, n+1)$$, each with finite measure.
• Theorem: $$m_*(Q) = {\left\lvert {Q} \right\rvert}$$ for $$Q$$ a closed cube (i.e. the outer measure equals the volume).

• $$m_*(Q) \leq {\left\lvert {Q} \right\rvert}$$:

• Since $$Q\subseteq Q$$, $$Q\rightrightarrows Q$$ and $$m_*(Q) \leq {\left\lvert {Q} \right\rvert}$$ since $$m_*$$ is an infimum over such coverings.

• $${\left\lvert {Q} \right\rvert} \leq m_*(Q)$$:

• Fix $${\varepsilon}> 0$$.

• Let $$\left\{{Q_i}\right\}_{i=1}^\infty \rightrightarrows Q$$ be arbitrary, it suffices to show that \begin{align*}{\left\lvert {Q} \right\rvert} \leq \qty{\sum_{i=1}^\infty {\left\lvert {Q_i} \right\rvert}} + {\varepsilon}.\end{align*}

• Pick open cubes $$S_i$$ such that $$Q_i\subseteq S_i$$ and $${\left\lvert {Q_i} \right\rvert} \leq {\left\lvert {S_i} \right\rvert} \leq (1+{\varepsilon}){\left\lvert {Q_i} \right\rvert}$$.

• Then $$\left\{{S_i}\right\} \rightrightarrows Q$$, so by compactness of $$Q$$ pick a finite subcover with $$N$$ elements.

• Note \begin{align*} Q \subseteq \bigcup_{i=1}^N S_i \implies {\left\lvert {Q} \right\rvert} \leq \sum_{i=1}^N {\left\lvert {S_i} \right\rvert} \leq \sum_{i=1}^N (1+{\varepsilon}) {\left\lvert {Q_j} \right\rvert} \leq (1+{\varepsilon})\sum_{i=1}^\infty {\left\lvert {Q_i } \right\rvert} .\end{align*}

• Taking an infimum over coverings on the RHS preserves the inequality, so \begin{align*}{\left\lvert {Q} \right\rvert} \leq (1+{\varepsilon}) m_*(Q)\end{align*}

• Take $${\varepsilon}\to 0$$ to obtain final inequality.

• If $$m_*(E) = \infty$$, then take $$B = {\mathbf{R}}^n$$ since $$m({\mathbf{R}}^n) = \infty$$.

• Suppose $$N \coloneqq m_*(E) < \infty$$.

• Since $$m_*(E)$$ is an infimum, by definition, for every $${\varepsilon}> 0$$ there exists a covering by closed cubes $$\left\{{Q_i({\varepsilon})}\right\}_{i=1}^\infty \rightrightarrows E$$ depending on $${\varepsilon}$$ such that \begin{align*} \sum_{i=1}^\infty {\left\lvert {Q_i({\varepsilon})} \right\rvert} < N + {\varepsilon} .\end{align*}

• For each fixed $$n$$, set $${\varepsilon}_n = {1\over n}$$ to produce such a covering $$\left\{{Q_i({\varepsilon}_n)}\right\}_{i=1}^\infty$$ and set $$B_n \coloneqq\bigcup_{i=1}^\infty Q_i({\varepsilon}_n)$$.

• The outer measure of cubes is equal to the sum of their volumes, so \begin{align*} m_*(B_n) = \sum_{i=1}^\infty {\left\lvert {Q_i({\varepsilon}_n)} \right\rvert} < N + {\varepsilon}_n = N + {1\over n} .\end{align*}

• Now set $$B \coloneqq\bigcap_{n=1}^\infty B_n$$.

• Since $$E\subseteq B_n$$ for every $$n$$, $$E\subseteq B$$
• Since $$B$$ is a countable intersection of countable unions of closed sets, $$B$$ is Borel.
• Since $$B_n \subseteq B$$ for every $$n$$, we can apply subadditivity to obtain the inequality \begin{align*} E \subseteq B \subseteq B_n \implies N \leq m_*(B) \leq m_*(B_n) < N + {1\over n} {\quad \operatorname{for all} \quad} n\in {\mathbf{Z}}^{\geq 1} .\end{align*}
• This forces $$m_*(E) = m_*(B)$$.

Suppose $$m_*(E) < \infty$$.

• By (a), find a Borel set $$B\supseteq E$$ such that $$m_*(B) = m_*(E)$$
• Note that $$E\subseteq B \implies B\bigcap E = E$$ and $$B\bigcap E^c = B\setminus E$$.
• By assumption, \begin{align*} m_*(B) &= m_*(B\bigcap E) + m_*(B\bigcap E^c) \\ m_*(E) &= m_*(E) + m_*(B\setminus E) \\ m_*(E) - m_*(E) &= m_*(B\setminus E) \qquad\qquad\text{since } m_*(E) < \infty \\ \implies m_*(B\setminus E) &= 0 .\end{align*}
• So take $$N = B\setminus E$$; this shows $$m_*(N) = 0$$ and $$E = B\setminus (B\setminus E) = B\setminus N$$.

If $$m_*(E) = \infty$$:

• Apply result to $$E_R\coloneqq E \bigcap[R, R+1)^n \subset {\mathbf{R}}^n$$ for $$R\in {\mathbf{Z}}$$, so $$E = {\textstyle\coprod}_R E_R$$
• Obtain $$B_R, N_R$$ such that $$E_R = B_R \setminus N_R$$, $$m_*(E_R) = m_*(B_R)$$, and $$m_*(N_R) = 0$$.
• Note that
• $$B\coloneqq\bigcup_R B_R$$ is a union of Borel sets and thus still Borel
• $$E = \bigcup_R E_R$$
• $$N\coloneqq B\setminus E$$
• $$N' \coloneqq\bigcup_R N_R$$ is a union of null sets and thus still null
• Since $$E_R \subset B_R$$ for every $$R$$, we have $$E\subset B$$
• We can compute \begin{align*} N = B\setminus E = \qty{ \bigcup_R B_R } \setminus \qty{\bigcup_R E_R } \subseteq \bigcup_R \qty{B_R\setminus E_R} = \bigcup_R N_R \coloneqq N' \end{align*} where $$m_*(N') = 0$$ since $$N'$$ is null, and thus subadditivity forces $$m_*(N) = 0$$.

## Fall 2019.3. #real_analysis/qual/completed

Let $$(X, \mathcal B, \mu)$$ be a measure space with $$\mu(X) = 1$$ and $$\{B_n\}_{n=1}^\infty$$ be a sequence of $$\mathcal B$$-measurable subsets of $$X$$, and \begin{align*} B \coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}x\in B_n \text{ for infinitely many } n}\right\}. \end{align*}

• Argue that $$B$$ is also a $$\mathcal{B} {\hbox{-}}$$measurable subset of $$X$$.

• Prove that if $$\sum_{n=1}^\infty \mu(B_n) < \infty$$ then $$\mu(B)= 0$$.

• Prove that if $$\sum_{n=1}^\infty \mu(B_n) = \infty$$ and the sequence of set complements $$\left\{{B_n^c}\right\}_{n=1}^\infty$$ satisfies \begin{align*} \mu\left(\bigcap_{n=k}^{K} B_{n}^{c}\right)=\prod_{n=k}^{K}\left(1-\mu\left(B_{n}\right)\right) \end{align*} for all positive integers $$k$$ and $$K$$ with $$k < K$$, then $$\mu(B) = 1$$.

Hint: Use the fact that $$1 - x ≤ e^{-x}$$ for all $$x$$.

• Borel-Cantelli: for a sequence of sets $$X_n$$, \begin{align*} \left\{{x {~\mathrel{\Big\vert}~}x\in X_n \text{ for infinitely many $n$} }\right\} &= \bigcap_{N\geq 1} \bigcup_{n\geq N} X_n = \limsup_n X_n \\ \left\{{x {~\mathrel{\Big\vert}~}x\in X_n \text{ for all but finitely many $n$} }\right\} &= \bigcup_{N\geq 1} \bigcap_{n\geq N} X_n = \liminf X_n .\end{align*}

• Properties of logs and exponentials: \begin{align*} \prod_n e^{x_n} = e^{\Sigma_n x_n} \quad\text{and} \quad \sum_n \log(x_n) = \log\left(\prod_n x_n\right) .\end{align*}

• Tails of convergent sums vanish.

• Continuity of measure: $$B_n \searrow B$$ and $$\mu(B_0)<\infty$$ implies $$\lim_n \mu(B_n) = \mu(B)$$, and $$B_n\nearrow B \implies \lim_n \mu(B_n) = \mu(B)$$.

• The Borel $$\sigma{\hbox{-}}$$algebra is closed under countable unions/intersections/complements,
• $$B = \limsup_n B_n = \cap_{N\geq 1} \cup_{n\geq N} B_n$$ is an intersection of unions of measurable sets.

• Tails of convergent sums vanish, so \begin{align*} \sum_{n\geq N} \mu(B_n) \xrightarrow{N\to\infty} 0 .\end{align*}
• Also, \begin{align*} B_M \coloneqq\bigcap_{N = 1}^M \bigcup_{n\geq N} B_n \searrow B .\end{align*}
• A computation: \begin{align*} \mu(B) &\coloneqq\mu\left(\bigcap_{N\geq 1} \bigcup_{n\geq N} B_n\right) \\ &\leq \mu\left( \bigcup_{n\geq N} B_n \right) && \forall N \\ &\leq \sum_{n\geq N} \mu(B_n) && \forall N \\ &\overset{N\to\infty}\longrightarrow 0 ,\end{align*} where we’ve used that we’re intersecting over fewer sets and this can only increase measure.

• Since $$\mu(X) = 1$$, in order to show $$\mu(B) = 1$$ it suffices to show $$\mu(X\setminus B) = 0$$.

• A computation: \begin{align*} \mu(B^c) &= \mu\qty{ \qty{ \bigcap_{N=1}^\infty \bigcup_{n=N}^\infty B_n }^c }\\ &= \mu\qty{ \bigcup_{N=1}^\infty \bigcap_{n=N}^\infty B_n^c } \\ &\leq \sum_{N=1}^\infty \mu\qty{ \bigcap_{n=N}^\infty B_n^c } \\ &= \sum_{N=1}^\infty \lim_{K\to\infty} \mu\qty{ \bigcap_{n=N}^K B_n^c } && \text{continuity of measure from above} \\ &= \sum_{N=1}^\infty \lim_{K\to\infty} \prod_{n=N}^K \qty{1 - \mu(B_n)} && \text{by assumption} \\ &\leq \sum_{N=1}^\infty \lim_{K\to\infty} \prod_{n=N}^K e^{-\mu(B_n)} && \text{by hint} \\ &= \sum_{N=1}^\infty \lim_{K\to\infty} e^{-\sum_{n=N}^K \mu(B_n)} \\ &= \sum_{N=1}^\infty e^{-\lim_{K\to\infty} \sum_{n=N}^K \mu(B_n)} && \text{by continuity of } f(x) = e^x \\ &= \sum_{N=1}^\infty e^{-\sum_{n=N}^\infty \mu(B_n)} \\ &= \sum_{N=1}^\infty 0 \\ &= 0 .\end{align*}

• Here we’ve used that every tail of a divergent sum is divergent: if $$\sum_{n=1}^\infty a_n \to \infty$$ then for every $$N$$, the tail $$\sum_{n=N}^\infty a_n \to \infty$$ as well.

• We’ve also use that if $$b_n\to \infty$$ then $$e^{-b_n} \to 0$$.

## Spring 2019.2 #real_analysis/qual/completed

Let $$\mathcal B$$ denote the set of all Borel subsets of $${\mathbf{R}}$$ and $$\mu : \mathcal B \to [0, \infty)$$ denote a finite Borel measure on $${\mathbf{R}}$$.

• Prove that if $$\{F_k\}$$ is a sequence of Borel sets for which $$F_k \supseteq F_{k+1}$$ for all $$k$$, then \begin{align*} \lim _{k \rightarrow \infty} \mu\left(F_{k}\right)=\mu\left(\bigcap_{k=1}^{\infty} F_{k}\right) \end{align*}

• Suppose $$\mu$$ has the property that $$\mu (E) = 0$$ for every $$E \in \mathcal B$$ with Lebesgue measure $$m(E) = 0$$. Prove that for every $$\epsilon > 0$$ there exists $$\delta > 0$$ so that if $$E \in \mathcal B$$ with $$m(E) < δ$$, then $$\mu(E) < ε$$.

• Proof of continuity of measure.
• Using limsup/liminf sets (intersections of unions and vice-versa) and (sub)additivity to bound measures.
• Control over lower bound: use tails of convergent sums
• Control over upper bound: use rapidly converging coefficients like $$\sum 1/2^n$$
• Convergent sums have vanishing tails.
• Intersecting over more sets can only lose measure, taking a union over more can only gain measure.
• Similarly intersecting over fewer sets can only gain measure, and taking a union over fewer sets can only lose measure.

Use a limsup or liminf of sets and continuity of measure. Note that choosing a limsup vs a liminf is fiddly – for one choice, you can only get one of the bounds you need, for the other choice you can get both.

• Observation: $$\mu$$ finite means $$\mu(E) < \infty$$ for all $$E \in\mathcal{B}$$, which we’ll need in several places.

• Prove a more general statement: for any measure $$\mu$$, \begin{align*} \mu(F_1) < \infty,\, F_k \searrow F \implies \lim_{k\to\infty}\mu(F_k) = \mu(F) ,\end{align*} where $$F_k \searrow F$$ means $$F_1 \supseteq F_2 \supseteq \cdots$$ with $$\bigcap_{k=1}^\infty F_k = F$$.

• Note that $$\mu(F)$$ makes sense: each $$F_k \in \mathcal{B}$$, which is a $$\sigma{\hbox{-}}$$algebra and closed under countable intersections.
• Take disjoint annuli by setting $$E_k \coloneqq F_k \setminus F_{k+1}$$

• Funny step: write \begin{align*} F_1 = F {\textstyle\coprod}\coprod_{k=1}^{\infty} E_k .\end{align*}

• This is because $$x\in F_1$$ iff $$x$$ is in every $$F_k$$, so in $$F$$, or
• $$x\not \in F_1$$ but $$x\in F_2$$, noting incidentally $$x\in F_3, F_4,\cdots$$, or,
• $$x\not\in F_2$$ but $$x\in F_3$$, and so on.
• Now take measures, and note that we get a telescoping sum: \begin{align*} \mu(F_1) &= \mu(F) + \sum_{k=1}^\infty \mu(E_k) \\ &= \mu(F) + \lim_{N\to\infty} \sum_{k=1}^N \mu(E_k) \\ &\coloneqq\mu(F) + \lim_{N\to\infty} \sum_{k=1}^N \mu(F_k \setminus F_{k+1} ) \\ &\coloneqq\mu(F) + \lim_{N\to\infty} \sum_{k=1}^N \mu(F_k) - \mu(F_{k+1} ) \hspace{5em}\text{to be justified}\\ &= \mu(F) + \lim_{N\to\infty} [ (\mu(F_1) - \mu(F_2)) + (\mu(F_2) - \mu(F_3)) + \cdots \\ & \hspace{8em} + (\mu(F_{N-1}) - \mu(F_N)) + (\mu(F_N) - \mu(F_{N+1})) ] \\ \\ &= \mu(F) + \lim_{N\to\infty} \mu(F_1) - \mu(F_{N+1}) \\ &= \mu(F) + \mu(F_1) - \lim_{N\to\infty} \mu(F_{N+1}) .\end{align*}

• Justifying the measure subtraction: the general statement is that for any pair of sets $$A\subseteq X$$, $$\mu(X\setminus A) = \mu(X) - \mu(A)$$ when $$\mu(A) < \infty$$: \begin{align*} X &= A {\textstyle\coprod}(X\setminus A) \\ \implies \mu(X) &= \mu(A) + \mu(X\setminus A) && \text{countable additivity} \\ \implies \mu(X) -\mu(A) &= \mu(X\setminus A) && \text{if } \mu(A) < \infty .\end{align*}

• Now use that $$\mu(F_1)<\infty$$ to justify subtracting it from both sides: \begin{align*} \mu(F_1) &= \mu(F) + \mu(F_1) - \lim_{N\to\infty} \mu(F_{N+1}) \\ \implies 0 &= \mu(F_1) - \lim_{N\to\infty} \mu(F_{N+1}) \\ \lim_{N\to\infty} \mu(F_{N+1}) &= \mu(F_1) .\end{align*}

• Now use that $$\lim_{N\to\infty}\mu(F_{N+1}) = \lim_{N\to\infty} \mu(F_N)$$ to conclude.

• Toward a contradiction, negate the implication: there exists an $${\varepsilon}>0$$ such that for all $$\delta$$, there exists an $$E\in \mathcal{B}$$ \begin{align*} m(E) < \delta && \text{but} \hspace{4em} \mu(E) > {\varepsilon} .\end{align*}

• Goal: produce a set $$A$$ with $$m(A)= 0$$ but $$\mu(A)\neq 0$$.
• Take a sequence $$\delta_n = \alpha(n)$$, some function to be determined later, produce sets $$E_n$$ with \begin{align*} m(E_n) < \delta_n && \text{but} \hspace{4em} \mu(E_n) > {\varepsilon}\quad \forall n .\end{align*}

• Set \begin{align*} A_M \coloneqq\bigcap_{N=1}^M \bigcup_{n=N}^\infty E_n \coloneqq\bigcap_{N=1}^M F_N \hspace{4em} F_N \coloneqq\bigcup_{n=N}^\infty E_n .\end{align*}

• Observation: $$F_N \supseteq F_{N+1}$$ for all $$N$$, since the right-hand side involves taking a union over fewer sets.
• Notation: define \begin{align*} A_\infty \coloneqq\bigcap_{N=1}^\infty \bigcup_{n=N}^\infty E_n .\end{align*}
• Bounding the Lebesgue measure $$m$$ from above: \begin{align*} m(A_\infty) &\coloneqq m\qty{ \bigcap_{N=1}^\infty \bigcup_{n=N}^\infty E_n } \\ &\leq m\qty{ \bigcup_{n=N}^\infty E_n } && \forall N \\ &\leq \sum_{n=N}^\infty m(E_n) && \forall N \quad \text{by countable subadditivity} \\ &\leq \sum_{n=N}^\infty \alpha(n) \\ \\ &\overset{N\to\infty}\longrightarrow 0 ,\end{align*} where we’ve used that intersecting over fewer sets (i.e. none) can only increase measure in the first bound.

• We have control over the sequence $$\alpha(n)$$, so we can choose it to be summable so that the tails converge to zero as rapidly as we’d like.
• So e.g. for any $${\varepsilon}_1 >0$$, we can choose $$\alpha(n) \coloneqq{\varepsilon}_1/2^n$$, then \begin{align*} \sum_{n=N}^\infty \alpha(n) &\leq \sum_{n=1}^\infty {{\varepsilon}_1 \over 2^n} = {\varepsilon}_1 \to 0 .\end{align*}
• Bounding the $$\mu$$ measure from below: \begin{align*} \mu(A_\infty) &\coloneqq \mu\qty{\bigcap_{N=1}^\infty F_N} \\ &= \lim_{N\to\infty} \mu(F_N) && \text{by part (1) }\\ &= \lim_{N\to\infty} \mu\qty{ \bigcup_{n=N}^\infty E_n } \\ &\geq \lim_{N\to\infty} \mu(E_N ) \\ &\geq \lim_{N\to\infty} {\varepsilon}\\ &= {\varepsilon}\\ &>0 ,\end{align*} where we’ve used that taking a union over fewer sets can only make the measure smaller.

## Fall 2018.2 #real_analysis/qual/work

Let $$E\subset {\mathbf{R}}$$ be a Lebesgue measurable set. Show that there is a Borel set $$B \subset E$$ such that $$m(E\setminus B) = 0$$.

\todo[inline]{Move this to review notes to clean things up.}

\todo[inline]{What a mess, redo!!}


• Definition of measurability: there exists an open $$O\supset E$$ such that $$m_*(O\setminus E) < {\varepsilon}$$ for all $${\varepsilon}> 0$$.
• Theorem: $$E$$ is Lebesgue measurable iff there exists a closed set $$F\subseteq E$$ such that $$m_*(E\setminus F) < {\varepsilon}$$ for all $${\varepsilon}>0$$.
• Every $$F_\sigma, G_\delta$$ is Borel.
• Claim: $$E$$ is measurable $$\iff$$ for every $$\varepsilon$$ there exist $$F_\varepsilon \subset E \subset G_\varepsilon$$ with $$F_\varepsilon$$ closed and $$G_\varepsilon$$ open and $$m(G_\varepsilon \setminus E)< \varepsilon$$ and $$m(E\setminus F_\varepsilon) < \varepsilon$$.
• Proof: existence of $$G_{\varepsilon}$$ is the definition of measurability.
• Existence of $$F_{\varepsilon}$$: ?
• Claim: $$E$$ is measurable $$\implies$$ there exists an open $$O\supseteq E$$ such that $$m(O\setminus E) = 0$$.
• Since $$E$$ is measurable, for each $$n\in {\mathbb{N}}$$ choose $$G_n \supseteq E$$ such that $$m_*(G_n\setminus E) < {1\over n}$$.
• Set $$O_N \coloneqq\bigcap_{n=1}^N G_n$$ and $$O\coloneqq\bigcap_{n=1}^\infty G_n$$.
• Suppose $$E$$ is bounded.
• Note $$O_N \searrow O$$ and $$m_*(O_1) < \infty$$ if $$E$$ is bounded, since in this case \begin{align*} m_*(G_n\setminus E) = m_*(G_1) - m_*(E) < 1 \iff m_*(G_1) < m_*(E) + {1\over n} < \infty .\end{align*}
• Note $$O_N \setminus E \searrow O \setminus E$$ since $$O_N\setminus E \coloneqq O_N \bigcap E^c \supseteq O_{N+1} \bigcap E^c$$ for all $$N$$, and again $$m_*(O_1 \setminus E) < \infty$$.
• So it’s valid to apply continuity of measure from above: \begin{align*} m_*(O\setminus E) &= \lim_{N\to\infty} m_*(O_N\setminus E) \\ &\leq \lim_{N\to \infty} m_*(G_N\setminus E) \\ &= \lim_{N\to\infty} {1\over N} = 0 ,\end{align*} where the inequality uses subadditivity on $$\bigcap_{n=1}^N G_n \subseteq G_N$$
• Suppose $$E$$ is unbounded.
• Write $$E^k = E \bigcap[k, k+1]^d \subset {\mathbf{R}}^d$$ as the intersection of $$E$$ with an annulus, and note that $$E = {\textstyle\coprod}_{k\in {\mathbb{N}}} E_k$$.
• Each $$E_k$$ is bounded, so apply the previous case to obtain $$O_k \supseteq E_k$$ with $$m(O_k\setminus E_k) = 0$$.
• So write $$O_k = E_k {\textstyle\coprod}N_k$$ where $$N_k \coloneqq O_k \setminus E_k$$ is a null set.
• Define $$O = \bigcup_{k\in {\mathbb{N}}} O_k$$, note that $$E\subseteq O$$.
• Now note \begin{align*} O\setminus E &= \qty{{\textstyle\coprod}_k O_k}\setminus \qty{{\textstyle\coprod}_K E_k} \\ &\subseteq {\textstyle\coprod}_k \qty{O_k \setminus E_k} \\ \implies m_*(O\setminus E) &\leq m_*\qty{{\textstyle\coprod}\qty{O_k \setminus E_k} } = 0 ,\end{align*} since any countable union of null sets is again null.
• So $$O\supseteq E$$ with $$m(O\setminus E) = 0$$.
• Theorem: since $$E$$ is measurable, $$E^c$$ is measurable
• Proof: It suffices to write $$E^c$$ as the union of two measurable sets, $$E^c = S \bigcup(E^c - S)$$, where $$S$$ is to be determined.
• We’ll produce an $$S$$ such that $$m_*(E^c - S) = 0$$ and use the fact that any subset of a null set is measurable.
• Since $$E$$ is measurable, for every $${\varepsilon}> 0$$ there exists an open $${\mathcal{O}}_{\varepsilon}\supseteq E$$ such that $$m_*({\mathcal{O}}_{\varepsilon}\setminus E) < {\varepsilon}$$.
• Take the sequence $$\left\{{{\varepsilon}_n \coloneqq{1\over n}}\right\}$$ to produce a sequence of sets $${\mathcal{O}}_n$$.
• Note that each $${\mathcal{O}}_n^c$$ is closed and \begin{align*} {\mathcal{O}}_n \supseteq E \iff {\mathcal{O}}_n^c \subseteq E^c .\end{align*}
• Set $$S \coloneqq\bigcup_n {\mathcal{O}}_n^c$$, which is a union of closed sets, thus an $$F_\sigma$$ set, thus Borel, thus measurable.
• Note that $$S\subseteq E^c$$ since each $${\mathcal{O}}_n \subseteq E^c$$.
• Note that \begin{align*} E^c\setminus S &\coloneqq E^c \setminus \qty{\bigcup_{n=1}^\infty {\mathcal{O}}_n^c} \\ &\coloneqq E^c \bigcap\qty{\bigcup_{n=1}^\infty {\mathcal{O}}_n^c}^c \quad\text{definition of set minus} \\ &= E^c \bigcap\qty{\bigcap_{n=1}^\infty {\mathcal{O}}_n}^c \quad \text{De Morgan's law}\\ &= E^c \bigcup\qty{\bigcap_{n=1}^\infty {\mathcal{O}}_n} \\ &\coloneqq\qty{ \bigcap_{n=1}^\infty {\mathcal{O}}_n} \setminus E \\ & \subseteq {\mathcal{O}}_N \setminus E \quad \text{for every } N\in {\mathbb{N}} .\end{align*}
• Then by subadditivity, \begin{align*} m_*(E^c\setminus S) \leq m_*({\mathcal{O}}_N \setminus E) \leq {1\over N} \quad \forall N \implies m_*(E^c\setminus S) = 0 .\end{align*}
• Thus $$E^c\setminus S$$ is measurable.

• Since $$E$$ is measurable, $$E^c$$ is measurable.
• Since $$E^c$$ is measurable exists an open $$O\supseteq E^c$$ such that $$m(O\setminus E^c) = 0$$.
• Set $$B \coloneqq O^c$$, then $$O\supseteq E^c \iff {\mathcal{O}}^c \subseteq E \iff B\subseteq E$$.
• Computing measures yields \begin{align*} E\setminus B \coloneqq E\setminus {\mathcal{O}}^c \coloneqq E\bigcap({\mathcal{O}}^c)^c = E\bigcap{\mathcal{O}}= {\mathcal{O}}\bigcap(E^c)^c \coloneqq{\mathcal{O}}\setminus E^c ,\end{align*} thus $$m(E\setminus B) = m({\mathcal{O}}\setminus E^c) = 0$$.
• Since $${\mathcal{O}}$$ is open, $$B$$ is closed and thus Borel.

d.irect Proof (Todo)

\todo[inline]{Try to construct the set.}


## Spring 2018.1 #real_analysis/qual/completed

Define \begin{align*} E:=\left\{x \in \mathbb{R}:\left|x-\frac{p}{q}\right|

Prove that $$m(E) = 0$$.

• Borel-Cantelli: If $$\left\{{E_k}\right\}_{k\in{\mathbf{Z}}}\subset 2^{\mathbf{R}}$$ is a countable collection of Lebesgue measurable sets with $$\sum_{k\in {\mathbf{Z}}} m(E_k) < \infty$$, then almost every $$x\in {\mathbf{R}}$$ is in at most finitely many $$E_k$$.
• Equivalently (?), $$m(\limsup_{k\to\infty} E_k) = 0$$, where $$\limsup_{k\to\infty} E_k = \bigcap_{k=1}^\infty \bigcup_{j\geq k} E_j$$, the elements which are in $$E_k$$ for infinitely many $$k$$.

• Strategy: Borel-Cantelli.

• We’ll show that $$m(E) \bigcap[n, n+1] = 0$$ for all $$n\in {\mathbf{Z}}$$; then the result follows from \begin{align*} m(E) = m \qty{\bigcup_{n\in {\mathbf{Z}}} E \bigcap[n, n+1]} \leq \sum_{n=1}^\infty m(E \bigcap[n, n+1]) = 0 .\end{align*}

• By translation invariance of measure, it suffices to show $$m(E \bigcap[0, 1]) = 0$$.

• So WLOG, replace $$E$$ with $$E\bigcap[0, 1]$$.
• Define \begin{align*} E_j \coloneqq\left\{{x\in [0, 1] {~\mathrel{\Big\vert}~}\ \exists p\in {\mathbf{Z}}^{\geq 0} \text{ s.t. } {\left\lvert {x - \frac{p}{j} } \right\rvert} < \frac 1 {j^3}}\right\} .\end{align*}

• Note that $$E_j \subseteq {\textstyle\coprod}_{p\in {\mathbf{Z}}^{\geq 0}} B_{j^{-3}}\qty{p\over j}$$, i.e. a union over integers $$p$$ of intervals of radius $$1/j^3$$ around the points $$p/j$$. Since $$1/j^3 < 1/j$$, this union is in fact disjoint.
• Importantly, note that \begin{align*} \limsup_{j\to\infty} E_j \coloneqq\bigcap_{n=1}^\infty \bigcup_{j=n}^\infty E_j = E \end{align*}

since

\begin{align*} x \in \limsup_j E_j &\iff x \in E_j \text{ for infinitely many } j \\ &\iff \text{ there are infinitely many $j$ for which there exist a $p$ such that } {\left\lvert {x - {p\over j}} \right\rvert} < j^{-3} \\ &\iff \text{ there are infinitely many such pairs $p, j$} \\ &\iff x\in E .\end{align*}

• Intersecting with $$[0, 1]$$, we can write $$E_j$$ as a union of intervals: \begin{align*} E_j =& \qty{0, {j^{-3}}} \quad {\textstyle\coprod}\quad B_{j^{-3}}\qty{1\over j} {\textstyle\coprod} B_{j^{-3}}\qty{2\over j} {\textstyle\coprod} \cdots {\textstyle\coprod} B_{j^{-3}}\qty{j-1\over j} \quad {\textstyle\coprod}\quad (1 - {j^{-3}}, 1) ,\end{align*} where we’ve separated out the “boundary” terms to emphasize that they are balls about $$0$$ and $$1$$ intersected with $$[0, 1]$$.

• Since $$E_j$$ is a union of open sets, it is Borel and thus Lebesgue measurable.

• Computing the measure of $$E_j$$:

• For a fixed $$j$$, there are exactly $$j+1$$ possible choices for a numerator ($$0, 1, \cdots, j$$), thus there are exactly $$j+1$$ sets appearing in the above decomposition.

• The first and last intervals are length $$1 \over j^3$$

• The remaining $$(j+1)-2 = j-1$$ intervals are twice this length, $$2 \over j^3$$

• Thus \begin{align*} m(E_j) = 2 \qty{1 \over j^3} + (j-1) \qty{2 \over j^3} = {2 \over j^2} \end{align*}

• Note that \begin{align*} \sum_{j\in {\mathbb{N}}} m(E_j) = 2\sum_{j\in {\mathbb{N}}} \frac 1 {j^2} < \infty ,\end{align*} which converges by the $$p{\hbox{-}}$$test for sums.

• But then \begin{align*} m(E) &= m(\limsup_j E_j) \\ &= m(\bigcap_{n\in {\mathbb{N}}} \bigcup_{j\geq n} E_j) \\ &\leq m(\bigcup_{j\geq N} E_j) \quad\text{for every } N \\ &\leq \sum_{j\geq N} m(E_j) \\ &\overset{N\to\infty}\to 0 \quad\text{} .\end{align*}

• Thus $$E$$ is measurable as a subset of a null set and $$m(E) = 0$$.

## Fall 2017.2 #real_analysis/qual/completed

Let $$f(x) = x^2$$ and $$E \subset [0, \infty) \coloneqq{\mathbf{R}}^+$$.

• Show that \begin{align*} m^*(E) = 0 \iff m^*(f(E)) = 0. \end{align*}

• Deduce that the map

\begin{align*} \phi: \mathcal{L}({\mathbf{R}}^+) &\to \mathcal{L}({\mathbf{R}}^+) \\ E &\mapsto f(E) \end{align*} is a bijection from the class of Lebesgue measurable sets of $$[0, \infty)$$ to itself.

\todo[inline]{Walk through.}


It suffices to consider the bounded case, i.e. $$E \subseteq B_M(0)$$ for some $$M$$. Then write $$E_n = B_n(0) \bigcap E$$ and apply the theorem to $$E_n$$, and by subadditivity, $$m^*(E) = m^*(\bigcup_n E_n) \leq \sum_n m^*(E_n) = 0$$.

Lemma: $$f(x) = x^2, f^{-1}(x) = \sqrt{x}$$ are Lipschitz on any compact subset of $$[0, \infty)$$.

Proof: Let $$g = f$$ or $$f^{-1}$$. Then $$g\in C^1([0, M])$$ for any $$M$$, so $$g$$ is differentiable and $$g'$$ is continuous. Since $$g'$$ is continuous on a compact interval, it is bounded, so $${\left\lvert {g'(x)} \right\rvert} \leq L$$ for all $$x$$. Applying the MVT, \begin{align*} {\left\lvert {f(x) - f(y)} \right\rvert} = f'(c) {\left\lvert {x-y} \right\rvert} \leq L {\left\lvert {x-y} \right\rvert} .\end{align*}

Lemma: If $$g$$ is Lipschitz on $${\mathbf{R}}^n$$, then $$m(E) = 0 \implies m(g(E)) = 0$$.

Proof: If $$g$$ is Lipschitz, then \begin{align*} g(B_r(x)) \subseteq B_{Lr}(x) ,\end{align*} which is a dilated ball/cube, and so \begin{align*} m^*(B_{Lr}(x)) \leq L^n \cdot m^*(B_{r}(x)) .\end{align*}

Now choose $$\left\{{Q_j}\right\} \rightrightarrows E$$; then $$\left\{{g(Q_j)}\right\} \rightrightarrows g(E)$$.

By the above observation, \begin{align*} {\left\lvert {g(Q_j)} \right\rvert} \leq L^n {\left\lvert {Q_j} \right\rvert} ,\end{align*}

and so \begin{align*} m^*(g(E)) \leq \sum_j {\left\lvert {g(Q_j)} \right\rvert} \leq \sum_j L^n {\left\lvert {Q_j} \right\rvert} = L^n \sum_j {\left\lvert {Q_j} \right\rvert} \to 0 .\end{align*}

Now just take $$g(x) = x^2$$ for one direction, and $$g(x) = f^{-1}(x) = \sqrt{x}$$ for the other.

Lemma: $$E$$ is measurable iff $$E = K {\textstyle\coprod}N$$ for some $$K$$ compact, $$N$$ null.

Write $$E = K {\textstyle\coprod}N$$ where $$K$$ is compact and $$N$$ is null.

Then $$\phi^{-1}(E) = \phi^{-1}(K {\textstyle\coprod}N) = \phi^{-1}(K) {\textstyle\coprod}\phi^{-1}(N)$$.

Since $$\phi^{-1}(N)$$ is null by part (a) and $$\phi^{-1}(K)$$ is the preimage of a compact set under a continuous map and thus compact, $$\phi^{-1}(E) = K' {\textstyle\coprod}N'$$ where $$K'$$ is compact and $$N'$$ is null, so $$\phi^{-1}(E)$$ is measurable.

So $$\phi$$ is a measurable function, and thus yields a well-defined map $$\mathcal L({\mathbf{R}}) \to \mathcal L({\mathbf{R}})$$ since it preserves measurable sets. Restricting to $$[0, \infty)$$, $$f$$ is bijection, and thus so is $$\phi$$.

## Spring 2017.1 #real_analysis/qual/completed

Let $$K$$ be the set of numbers in $$[0, 1]$$ whose decimal expansions do not use the digit $$4$$.

We use the convention that when a decimal number ends with 4 but all other digits are different from 4, we replace the digit $$4$$ with $$399\cdots$$. For example, $$0.8754 = 0.8753999\cdots$$.

Show that $$K$$ is a compact, nowhere dense set without isolated points, and find the Lebesgue measure $$m(K)$$.

• Definition: $$A$$ is nowhere dense $$\iff$$ every interval $$I$$ contains a subinterval $$S \subseteq A^c$$.
• Equivalently, the interior of the closure is empty, $$\qty{\overline{K}}^\circ = \emptyset$$.

Claim: $$K$$ is compact.

• It suffices to show that $$K^c \coloneqq[0, 1]\setminus K$$ is open; Then $$K$$ will be a closed and bounded subset of $${\mathbf{R}}$$ and thus compact by Heine-Borel.

• Strategy: write $$K^c$$ as the union of open balls (since these form a basis for the Euclidean topology on $${\mathbf{R}}$$).

• Do this by showing every point $$x\in K^c$$ is an interior point, i.e. $$x$$ admits a neighborhood $$N_x$$ such that $$N_x \subseteq K^c$$.
• Identify $$K^c$$ as the set of real numbers in $$[0, 1]$$ whose decimal expansion does contain a 4.

• We will show that there exists a neighborhood small enough such that all points in it contain a $$4$$ in their decimal expansions.
• Let $$x\in K^c$$, suppose a 4 occurs as the $$k$$th digit, and write \begin{align*} x = 0.d_1 d_2 \cdots d_{k-1}~ 4 ~d_{k+1}\cdots = \qty{\sum_{j=1}^k d_j 10^{-j}} + \qty{4\cdot 10^{-k}} + \qty{\sum_{j=k+1}^\infty d_j 10^{-j}} .\end{align*}

• Set $$r_x < 10^{-k}$$ and let $$y \in [0, 1] \bigcap B_{r_x}(x)$$ be arbitrary and write \begin{align*} y = \sum_{j=1}^\infty c_j 10^{-j} .\end{align*}

• Thus $${\left\lvert {x-y} \right\rvert} < r_x < 10^{-k}$$, and the first $$k$$ digits of $$x$$ and $$y$$ must agree:

• We first compute the difference: \begin{align*} x - y &= \sum_{i=1}^\infty d_j 10^{-j} - \sum_{i=1}^\infty c_j 10^{-j} = \sum_{i=1}^\infty \qty{d_j - c_j} 10^{-j} \\ \end{align*}
• Thus (claim) \begin{align*} {\left\lvert {x-y} \right\rvert} &\leq \sum_{j=1}^\infty {\left\lvert {d_j - c_j} \right\rvert} 10^j < 10^{-k} \iff {\left\lvert {d_j - c_j} \right\rvert} = 0 \quad \forall j\leq k .\end{align*}
• Otherwise we can note that any term $${\left\lvert {d_j - c_j} \right\rvert}\geq 1$$ and there is a contribution to $${\left\lvert {x-y} \right\rvert}$$ of at least $$1\cdot 10^{-j}$$ for some $$j < k$$, whereas \begin{align*} j < k \iff 10^{-j} > 10^{-k} ,\end{align*} a contradiction.
• This means that for all $$j \leq k$$ we have $$d_j = c_j$$, and in particular $$d_k = 4 = c_k$$, so $$y$$ has a 4 in its decimal expansion.

• But then $$K^c = \bigcup_x B_{r_x}(x)$$ is a union of open sets and thus open.

Claim: $$K$$ is nowhere dense and $$m(K) = 0$$:

• Strategy: Show $$\qty{\overline{K}}^\circ = \emptyset$$.

• Since $$K$$ is closed, $$\overline{K} = K$$, so it suffices to show that $$K$$ does not properly contain any interval.

• It suffices to show $$m(K^c) = 1$$, since this implies $$m(K) = 0$$ and since any interval has strictly positive measure, this will mean $$K$$ can not contain an interval.

• As in the construction of the Cantor set, let

• $$K_0$$ denote $$[0, 1]$$ with 1 interval $$\left({4 \over 10}, {5 \over 10} \right)$$ of length $$1 \over 10$$ deleted, so \begin{align*}m(K_0^c) = {1\over 10}.\end{align*}
• $$K_1$$ denote $$K_0$$ with 9 intervals $$\left({1 \over 100}, {5\over 100}\right), ~\left({14 \over 100}, {15 \over 100}\right), \cdots \left({94\over 100}, {95 \over 100}\right)$$ of length $${1 \over 100}$$ deleted, so \begin{align*}m(K_1^c) = {1\over 10} + {9 \over 100}.\end{align*}
• $$K_n$$ denote $$K_{n-1}$$ with $$9^{n}$$ such intervals of length $$1 \over 10^{n+1}$$ deleted, so \begin{align*}m(K_n^c) = {1\over 10} + {9 \over 100} + \cdots + {9^{n} \over 10^{n+1}}.\end{align*}
• Then compute \begin{align*} m(K^c) = \sum_{j=0}^\infty {9^n \over 10^{n+1} } = {1\over 10} \sum_{j=0}^\infty \qty{9\over 10}^n = {1 \over 10} \qty{ {1 \over 1 - {9 \over 10 } } } = 1. \end{align*}

Claim: $$K$$ has no isolated points:

• A point $$x\in K$$ is isolated iff there there is an open ball $$B_r(x)$$ containing $$x$$ such that $$B_r(x) \subsetneq K^c$$.

• So every point in this ball should have a 4 in its decimal expansion.
• Strategy: show that if $$x\in K$$, every neighborhood of $$x$$ intersects $$K$$.

• Note that $$m(K_n) = \left( \frac 9 {10} \right)^n \overset{n\to\infty}\to 0$$

• Also note that we deleted open intervals, and the endpoints of these intervals are never deleted.

• Thus endpoints of deleted intervals are elements of $$K$$.
• Fix $$x$$. Then for every $$\varepsilon$$, by the Archimedean property of $${\mathbf{R}}$$, choose $$n$$ such that $$\left( \frac 9 {10} \right)^n < \varepsilon$$.

• Then there is an endpoint $$x_n$$ of some deleted interval $$I_n$$ satisfying \begin{align*}{\left\lvert {x - x_n} \right\rvert} \leq \left( \frac 9 {10} \right)^n < {\varepsilon}.\end{align*}

• So every ball containing $$x$$ contains some endpoint of a removed interval, and thus an element of $$K$$.

## Spring 2017.2 #real_analysis/qual/completed

• Let $$\mu$$ be a measure on a measurable space $$(X, \mathcal M)$$ and $$f$$ a positive measurable function.

Define a measure $$\lambda$$ by \begin{align*} \lambda(E):=\int_{E} f ~d \mu, \quad E \in \mathcal{M} \end{align*}

Show that for $$g$$ any positive measurable function, \begin{align*} \int_{X} g ~d \lambda=\int_{X} f g ~d \mu \end{align*}

• Let $$E \subset {\mathbf{R}}$$ be a measurable set such that \begin{align*} \int_{E} x^{2} ~d m=0. \end{align*} Show that $$m(E) = 0$$.

• Absolute continuity of measures: $$\lambda \ll \mu \iff E\in\mathcal{M}, \mu(E) = 0 \implies \lambda(E) = 0$$.
• Radon-Nikodym: if $$\lambda \ll \mu$$, then there exists a measurable function $${\frac{\partial \lambda}{\partial \mu}\,} \coloneqq f$$ where $$\lambda(E) = \int_E f \,d\mu$$.
• Chebyshev’s inequality: \begin{align*} A_c \coloneqq\left\{{ x\in X {~\mathrel{\Big\vert}~}{\left\lvert {f(x)} \right\rvert} \geq c }\right\} \implies \mu(A_c) \leq c^{-p} \int_{A_c} {\left\lvert {f} \right\rvert}^p \,d\mu \quad \forall 0 < p < \infty .\end{align*}

• Strategy: use approximation by simple functions to show absolute continuity and apply Radon-Nikodym

• Claim: $$\lambda \ll \mu$$, i.e. $$\mu(E) = 0 \implies \lambda(E) = 0$$.

• Note that if this holds, by Radon-Nikodym, $$f = {\frac{\partial \lambda}{\partial \mu}\,} \implies d\lambda = f d\mu$$, which would yield \begin{align*} \int g ~d\lambda = \int g f ~d\mu .\end{align*}
• So let $$E$$ be measurable and suppose $$\mu(E) = 0$$.

• Then \begin{align*} \lambda(E) \coloneqq\int_E f ~d\mu &= \lim_{n\to\infty} \left\{{\int_E s_n \,d\mu {~\mathrel{\Big\vert}~}s_n \coloneqq\sum_{j=1}^\infty c_j \mu(E_j),\, s_n \nearrow f}\right\} \end{align*} where we take a sequence of simple functions increasing to $$f$$.

• But since each $$E_j \subseteq E$$, we must have $$\mu(E_j) = 0$$ for any such $$E_j$$, so every such $$s_n$$ must be zero and thus $$\lambda(E) = 0$$.

\todo[inline]{What is the final step in this approximation?}

• Set $$g(x) = x^2$$, note that $$g$$ is positive and measurable.

• By part (a), there exists a positive $$f$$ such that for any $$E\subseteq {\mathbf{R}}$$, \begin{align*} \int_E g ~dm = \int_E gf ~d\mu \end{align*}

• The LHS is zero by assumption and thus so is the RHS.

• $$m \ll \mu$$ by construction.

• Note that $$gf$$ is positive.

• Define $$A_k = \left\{{x\in X {~\mathrel{\Big\vert}~}gf \cdot \chi_E > {1 \over k} }\right\}$$, for $$k\in {\mathbf{Z}}^{\geq 0}$$

• Then by Chebyshev with $$p=1$$, for every $$k$$ we have

\begin{align*} \mu(A_k) \leq k \int_E gf ~d\mu = 0 \end{align*}

• Then noting that $$A_k \searrow A \coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}gf\cdot \chi_E(x) > 0}\right\}$$, we have $$\mu(A) = 0$$.

• Since $$gf$$ is positive, we have \begin{align*} x\in E \iff gf\chi_E(x) > 0 \iff x\in A \end{align*} so $$E = A$$ and $$\mu(E) = \mu(A)$$.

• But $$m \ll \mu$$ and $$\mu(E) = 0$$, so we can conclude that $$m(E) = 0$$.

## Fall 2016.4 #real_analysis/qual/completed

Let $$(X, \mathcal M, \mu)$$ be a measure space and suppose $$\left\{{E_n}\right\} \subset \mathcal M$$ satisfies \begin{align*} \lim _{n \rightarrow \infty} \mu\left(X \backslash E_{n}\right)=0. \end{align*}

Define \begin{align*} G \coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}x\in E_n \text{ for only finitely many } n}\right\}. \end{align*}

Show that $$G \in \mathcal M$$ and $$\mu(G) = 0$$.

\todo[inline]{Add concepts.}


• Claim: $$G\in {\mathcal{M}}$$.

• Claim: \begin{align*} G = \qty{ \bigcap_{N=1}^\infty \bigcup_{n=N}^\infty E_n}^c = \bigcup_{N=1}^\infty \bigcap_{n=N}^\infty E_n^c .\end{align*}

• This follows because $$x$$ is in the RHS $$\iff$$ $$x\in E_n^c$$ for all but finitely many $$n$$ $$\iff$$ $$x\in E_n$$ for at most finitely many $$n$$.
• But $${\mathcal{M}}$$ is a $$\sigma{\hbox{-}}$$algebra, and this shows $$G$$ is obtained by countable unions/intersections/complements of measurable sets, so $$G\in {\mathcal{M}}$$.

• Claim: $$\mu(G) = 0$$.

• We have \begin{align*} \mu(G) &= \mu\qty{\bigcup_{N=1}^\infty \bigcap_{n=N}^\infty E_n^c} \\ &\leq \sum_{N=1}^\infty \mu \qty{\bigcap_{n=N}^\infty E_n^c} \\ &\leq \sum_{N=1}^\infty \mu(E_M^c) \\ &\coloneqq\sum_{N=1}^\infty \mu(X\setminus E_N) \\ &\overset{N\to\infty}\to 0 .\end{align*}
\todo[inline]{Last step seems wrong!}


## Spring 2016.3 #real_analysis/qual/work

Let $$f$$ be Lebesgue measurable on $${\mathbf{R}}$$ and $$E \subset {\mathbf{R}}$$ be measurable such that \begin{align*} 0

Show that for every $$0 < t < 1$$, there exists a measurable set $$E_t \subset E$$ such that \begin{align*} \int_{E_{t}} f(x) d x=t A. \end{align*}

## Spring 2016.5 #real_analysis/qual/work

Let $$(X, \mathcal M, \mu)$$ be a measure space. For $$f\in L^1(\mu)$$ and $$\lambda > 0$$, define \begin{align*} \phi(\lambda)=\mu(\{x \in X | f(x)>\lambda\}) \quad \text { and } \quad \psi(\lambda)=\mu(\{x \in X | f(x)<-\lambda\}) \end{align*}

Show that $$\phi, \psi$$ are Borel measurable and \begin{align*} \int_{X}|f| ~d \mu=\int_{0}^{\infty}[\phi(\lambda)+\psi(\lambda)] ~d \lambda \end{align*}

## Spring 2016.2 #real_analysis/qual/work

Let $$0 < \lambda < 1$$ and construct a Cantor set $$C_\lambda$$ by successively removing middle intervals of length $$\lambda$$.

Prove that $$m(C_\lambda) = 0$$.

## Fall 2015.2 #real_analysis/qual/work

Let $$f: {\mathbf{R}}\to {\mathbf{R}}$$ be Lebesgue measurable.

• Show that there is a sequence of simple functions $$s_n(x)$$ such that $$s_n(x) \to f(x)$$ for all $$x\in {\mathbf{R}}$$.
• Show that there is a Borel measurable function $$g$$ such that $$g = f$$ almost everywhere.

## Spring 2015.3 #real_analysis/qual/work

Let $$\mu$$ be a finite Borel measure on $${\mathbf{R}}$$ and $$E \subset {\mathbf{R}}$$ Borel. Prove that the following statements are equivalent:

• $$\forall \varepsilon > 0$$ there exists $$G$$ open and $$F$$ closed such that \begin{align*} F \subseteq E \subseteq G \quad \text{and} \quad \mu(G\setminus F) < \varepsilon. \end{align*}
• There exists a $$V \in G_\delta$$ and $$H \in F_\sigma$$ such that \begin{align*} H \subseteq E \subseteq V \quad \text{and}\quad \mu(V\setminus H) = 0 \end{align*}

## Spring 2014.3 #real_analysis/qual/work

Let $$f: {\mathbf{R}}\to {\mathbf{R}}$$ and suppose \begin{align*} \forall x\in {\mathbf{R}},\quad f(x) \geq \limsup _{y \rightarrow x} f(y) \end{align*} Prove that $$f$$ is Borel measurable.

## Spring 2014.4 #real_analysis/qual/work

Let $$(X, \mathcal M, \mu)$$ be a measure space and suppose $$f$$ is a measurable function on $$X$$. Show that \begin{align*} \lim _{n \rightarrow \infty} \int_{X} f^{n} ~d \mu = \begin{cases} \infty & \text{or} \\ \mu(f^{-1}(1)), \end{cases} \end{align*} and characterize the collection of functions of each type.

# Measure Theory: Functions

## Spring 2021.1 #real_analysis/qual/completed

Let $$(X, \mathcal{M},\mu)$$ be a measure space and let $$E_n \in \mathcal{M}$$ be a measurable set for $$n\geq 1$$. Let $$f_n \coloneqq\chi_{E_n}$$ be the indicator function of the set $$E$$ and show that

• $$f_n \overset{n\to\infty}\to 1$$ uniformly $$\iff$$ there exists $$N\in {\mathbb{N}}$$ such that $$E_n = X$$ for all $$n\geq N$$.

• $$f_n(x) \overset{n\to\infty}\to 1$$ for almost every $$x$$ $$\iff$$ \begin{align*} \mu \qty{ \bigcap_{n \geq 0} \bigcup_{k \geq n} (X \setminus E_k) } = 0 .\end{align*}

Part a:

$$\implies$$:

• Suppose $$\chi_{E_n}\to 1$$ uniformly, we want to produce an $$N$$ such that $$n\geq N \implies x\in E_n$$ for all $$x\in X$$.
• Take $${\varepsilon}\coloneqq 1/2$$. By uniform convergence, for $$N$$ large enough, \begin{align*} & \forall n\geq N \quad {\left\lvert {\chi_{E_n}(x) - 1} \right\rvert} < 1/2 && \forall x\in X\\ &\iff \forall n\geq N \quad \chi_{E_n}(x) = 1 && \forall x\in X \\ &\iff \forall n\geq N \quad x\in E_n && \forall x\in X &\iff \forall n\geq N \quad E_n = X ,\end{align*} where we’ve used that $$E_n \subseteq X$$ by definition and this shows $$X \subseteq E_n$$. So this $$N$$ suffices.

$$\impliedby$$:

• Let $${\varepsilon}> 0$$ be arbitrary.
• Choose $$N$$ such that $$n\geq N \implies X = E_n$$. Then \begin{align*} &\forall n\geq N \quad x\in E_n && \forall x\in X \\ &\forall n\geq N \quad \chi_{E_n}(x) = 1 && \forall x\in X \\ &\forall n\geq N \quad {\left\lvert {\chi_{E_n}(x) - 1} \right\rvert} = 0 < {\varepsilon}&& \forall x\in X ,\end{align*} so $$\chi_{E_n} \to 1$$ uniformly.

Part b:

• Define \begin{align*} S &\coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}\chi_{E_k}(x) \to 1}\right\}\\ &\coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}\forall {\varepsilon},\, \exists N\, \text{ s.t. } {\left\lvert {\chi_{E_k}(x) - 1 } \right\rvert} < {\varepsilon},\forall k\geq N}\right\}\\ L &\coloneqq\bigcap_{n\geq 0} \bigcup_{k\geq n} \qty{X\setminus E_k} ,\end{align*} so $$S$$ is the set where $$f_n\to f$$ and $$X\setminus S$$ is the exceptional set where $$f_n\not\to f$$ doesn’t converge pointwise.

• Claim: $$L = X\setminus S$$, so if $$x\in S \iff x\in X\setminus L$$.

• Proof of claim: Suppose there exists an $$N$$ such that the first line below is true. Then for a fixed $$x$$, there are equivalent statements: \begin{align*} &\qquad x \in S \\ &\iff \exists N \text{ s.t. } \forall {\varepsilon}>0,\quad {\left\lvert {\chi_{E_k}(x) - 1 } \right\rvert} < {\varepsilon}&& \forall k\geq N \\ &\iff \exists N \text{ s.t. } {\left\lvert {\chi_{E_k}(x) - 1 } \right\rvert} = 0 && \forall k\geq N \\ &\iff \exists N \text{ s.t. } \chi_{E_k}(x) = 1 && \forall k\geq N \\ &\iff \exists N \text{ s.t. } x\in E_k && \forall k\geq N \\ &\iff \exists N \text{ s.t. } x\not\in X\setminus E_k &&\forall k\geq N \\ &\iff \exists N \text{ s.t. } x\not\in \bigcup_{k\geq N} X\setminus E_k \\ &{\color{blue} \iff} x\not\in \bigcap_{n\geq 0}\bigcup_{k\geq n} X\setminus E_k \\ &\iff x\not\in L \\ &\iff x\in X\setminus L .\end{align*}

• Proving the iff: $$f_n\to f$$ almost everywhere $$\iff \mu(X\setminus S) = 0 \iff \mu(L) = 0$$.

## Spring 2021.3 #real_analysis/qual/work

Let $$(X, \mathcal{M}, \mu)$$ be a finite measure space and let $$\left\{{ f_n}\right\}_{n=1}^{\infty } \subseteq L^1(X, \mu)$$. Suppose $$f\in L^1(X, \mu)$$ such that $$f_n(x) \overset{n\to \infty }\to f(x)$$ for almost every $$x \in X$$. Prove that for every $${\varepsilon}> 0$$ there exists $$M>0$$ and a set $$E\subseteq X$$ such that $$\mu(E) \leq {\varepsilon}$$ and $${\left\lvert {f_n(x)} \right\rvert}\leq M$$ for all $$x\in X\setminus E$$ and all $$n\in {\mathbb{N}}$$.

## Fall 2020.2 #real_analysis/qual/work

• Let $$f: {\mathbf{R}}\to {\mathbf{R}}$$. Prove that \begin{align*} f(x) \leq \liminf_{y\to x} f(y)~ \text{for each}~ x\in {{\mathbf{R}}} \iff \{ x\in {{\mathbf{R}}} \mathrel{\Big|}f(x) > a \}~\text{is open for all}~ a\in {{\mathbf{R}}} \end{align*}

• Recall that a function $$f: {{\mathbf{R}}} \to {{\mathbf{R}}}$$ is called lower semi-continuous iff it satisfies either condition in part (a) above.

Prove that if $$\mathcal{F}$$ is any family of lower semi-continuous functions, then \begin{align*} g(x) = \sup\{ f(x) \mathrel{\Big|}f\in \mathcal{F}\} \end{align*} is Borel measurable.

Note that $$\mathcal{F}$$ need not be a countable family.

## Fall 2016.2 #real_analysis/qual/completed

Let $$f, g: [a, b] \to {\mathbf{R}}$$ be measurable with \begin{align*} \int_{a}^{b} f(x) ~d x=\int_{a}^{b} g(x) ~d x. \end{align*} Show that either

• $$f(x) = g(x)$$ almost everywhere, or
• There exists a measurable set $$E \subset [a, b]$$ such that \begin{align*} \int _{E} f(x) \, dx > \int _{E} g(x) \, dx \end{align*}

• Monotonicity of the Lebesgue integral: $$f\leq g$$ on $$A$$ $$\implies \int_A f \leq \int_A g$$

Take the assumption and the negation of (1) and show (2). The obvious move: define the set $$A$$ where they differ. The non-obvious move: split $$A$$ itself up to get a strict inequality.

• Write $$X\coloneqq[a, b]$$,
• Suppose it is not the case that $$f=g$$ almost everywhere; then letting $$A\coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}f(x) \neq g(x)}\right\}$$, we have $$m(A) > 0$$.
• Write \begin{align*} A = A_1 {\textstyle\coprod}A_2 \coloneqq\left\{{f > g}\right\} {\textstyle\coprod}\left\{{f < g}\right\} .\end{align*}
• Both $$A_i$$ are measurable:
• Since $$f,g$$ are measurable functions, so is $$h\coloneqq f-g$$.
• We can write \begin{align*} A_1 &\coloneqq\left\{{ x\in X {~\mathrel{\Big\vert}~}h > 0 }\right\} = h^{-1}((0, \infty)) \\ A_2 &\coloneqq\left\{{ x\in X {~\mathrel{\Big\vert}~}h < 0 }\right\} = h^{-1}((-\infty, 0)) ,\end{align*} and pullbacks of Borel sets by measurable functions are measurable.
• Then on $$E$$, we have $$f(x)>g(x)$$ pointwise. This is preserved by monotonicity of the integral, thus \begin{align*} f(x) > g(x) \text{ on } E \implies \int_{E} f(x)\,dx > \int_{E} g(x)\, dx .\end{align*}

## Spring 2016.4 #real_analysis/qual/work

Let $$E \subset {\mathbf{R}}$$ be measurable with $$m(E) < \infty$$. Define \begin{align*} f(x)=m(E \cap(E+x)). \end{align*}

Show that

• $$f\in L^1({\mathbf{R}})$$.
• $$f$$ is uniformly continuous.
• $$\lim _{|x| \to \infty} f(x) = 0$$.

Hint: \begin{align*} \chi_{E \cap(E+x)}(y)=\chi_{E}(y) \chi_{E}(y-x) \end{align*}

#real_analysis/qual/work #real_analysis/qual/completed