Solutions to Exercises from Real Analysis: Modern Techniques and Their Applications

Some selected problems from Gerald Folland’s Real Analysis: Modern Techniques and Their Applications that I’ve done.

Chapter 1

Problem 14

Problem: If \(\mu\) is a semifinite measure and \(\mu(E) = \infty\), prove that for any \(C > 0\) there exists \(F\subseteq E\) such that \(C < \mu(F) < \infty\).

Solution: Let \(\alpha := \sup\{\mu(F):F\subseteq E,\, \mu(F)<\infty\}\). It suffices to show that \(\alpha = \infty\). Suppose \(\alpha < \infty\). Choose \(G_n\) such that

\begin{equation*} \alpha -\frac{1}{n} < \mu(G_n) < \infty. \end{equation*}

Note that \(\mu(G_1\cup\dots\cup G_n) < \infty\), so

\begin{equation*} \alpha - \frac{1}{n} < \mu(G_1\cup\dots\cup G_n) \leq \alpha. \end{equation*}

Let \(G:=\bigcup_{n=1}^\infty G_n\) so that

\begin{equation*} \mu(G) = \lim_{n\to\infty}\mu(G_1\cup\dots\cup G_n) = \lim_{n\to\infty}(\alpha-\tfrac{1}{n}) = \alpha. \end{equation*}

Thus, \(\mu(G) = \alpha\).

Let \(E’ = E \,\backslash\, G\) so \(\mu(E’) = \infty\). Because \(\mu^*\) is semifinite, there exists \(F\subseteq E’\) with \(0<\mu(F) <\infty\). Thus,

\begin{equation*} \alpha = \mu(G) < \mu(G) + \mu(F) = \mu(G\cup F) < \infty, \end{equation*}

but this contradictions the definition of \(\alpha\). Thus, \(\alpha = \infty\).

Problem 17

Problem: If \(\mu^*\) is an outer measure on \(X\), and \(\{A_j\}_{j=1}^\infty\) is a collection of disjoint \(\mu^* \)-measurable sets, prove that for any \(E\subseteq X\), \(\mu^*\left(E\cap\bigcup_{j=1}^\infty A_j\right) = \sum_{j=1}^\infty \mu^*(E\cap A_j)\).

Solution: Firstly, note that if \(A_1\cap A_2=\varnothing\) and both are \(\mu^*\)-measurable,

\begin{align*} \mu^*(E\cap (A_1\cup A_2)) &= \mu^*((E\cap (A_1\cup A_2))\cap A_1) +\mu^*((E\cap (A_1\cup A_2))\cap A_1^\complement) \\ &= \mu^*(E\cap A_1) + \mu^*(E\cap A_2). \end{align*}

By induction, we have the result for finite collections of disjoint \(\mu^*\)-measurable sets.

Then, note that for all \(k\in\mathbb{N}\),

\begin{align*} \sum_{j=1}^k\mu^*(E\cap A_j) = \mu^*\left(E\cap\bigcup_{j=1}^k A_j\right) \leq \mu^*\left(E\cap\bigcup_{j=1}^\infty A_j\right) \leq \sum_{j=1}^\infty\mu^*(E\cap A_j). \end{align*}

Letting \(k\to\infty\) we obtain the desired result.

Problem 18

Problem: Let $\mathcal{A}\subset P(X)$ be an algebra, $\mathcal{A}_\sigma$ the collection of countable unions of sets in $\mathcal{A}$, and $\mathcal{A}_{\sigma\delta}$ the collection of countable intersections of sets in $\mathcal{A}_\sigma$. Let $\mu_{0}$ be a premeasure on $\mathcal{A}$ and $\mu^*$ the induced outer measure.

1. For any $E\subset X$ and $\varepsilon > 0$ there exists $A\in \mathcal{A}_\sigma$ with $E\subset A$ and $\mu^*(A) \leq \mu^*(E) + \varepsilon$.
2. If $\mu^{*}(E) < \infty$, then $E$ is $\mu^{*}$-measurable if and only if there exists $B\in A_{\sigma\delta}$ with $E\subset B$ and $\mu^{*}(B\setminus E) = 0$.
3. If $\mu_0$ is $\sigma$-finite, the restriction $\mu^{*}(E) < \infty$ in (b) is superfluous.

Solution:

1. From the definition, \begin{align*} \mu^*(E) = \inf \left\{\sum_{n=1}^\infty \mu_0(A_n) : A_n\in\mathcal{A}, \, E \subseteq \bigcup_{n=1}^\infty A_n\right\}. \end{align*} Simply applying the definition of infimum yields a collection $\{A_n\}_{n=1}^\infty\subseteq\mathcal{A}$ with $E \subseteq \bigcup_{n=1}^\infty A_n \in\mathcal{A}_\sigma$ and $\sum_{n=1}^\infty \mu_0(A_n) \leq \mu^*(E) + \varepsilon$. Then, by Proposition 1.13, we have \begin{align*} \mu^*\left(\bigcup_{n=1}^\infty A_n\right) \leq \sum_{n=1}^\infty \mu^*(A_n) =\sum_{n=1}^\infty \mu_0(A_n) \leq \mu^*(E) + \varepsilon. \end{align*}

2. Suppose $E$ is $\mu^*$-measurable.

   From part (a), choose $B_n\in A_{\sigma}$ such that $E\subseteq B_n$ and $\mu^*(B_n)\leq\mu^*(E)+\frac{1}{n}$. Then, \begin{align*} \mu^*(E)+\frac{1}{n}\geq \mu^*(B_n) = \mu^*(B_n\cap E) + \mu^*(B_n\cap E^\complement) = \mu^*(E) + \mu^*(B_n \setminus E). \end{align*} Because $\mu^*(E)<\infty$, we have $\mu^*(B_n \setminus E)\leq \frac{1}{n}$. Let $B = \bigcap_{n=1}^\infty B_n$, so that $\mu^*(B\setminus E) \leq \mu^*(B_n\setminus E) \leq \frac{1}{n}$ for all $n$. Thus, $\mu^*(B\setminus E) =0$, and since $E\subseteq B\in A_{\sigma\delta}$, we have completed the proof of the forwards direction.

   Suppose there exists $B\in A_{\sigma\delta}$ such that $E\subseteq B$ and $\mu^*(B\setminus E) = 0$. Let $A\subseteq X$ be any set. Note that $\mu^*(A\cap E) \leq \mu^*(A\cap B)$. Also, \begin{align*} \mu^*(A\cap B) &= \mu^*(A\cap E \cup A\cap (B\setminus E)) \leq \mu^*(A\cap E) + \mu^*(A\cap (B\setminus E)) = \mu^*(A\cap E). \end{align*} Thus, $\mu^*(A\cap E) = \mu^*(A\cap B)$.

   Similarly, \begin{align*} \mu^*(A\cap E^\complement) &= \mu^*(A\cap B^\complement \cup A\cap (E^\complement\setminus B^\complement)) \leq \mu^*(A\cap B^\complement) + \mu^*(A\cap (B\setminus E)) = \mu^*(A\cap B^\complement). \end{align*} Since $\mu^*(A\cap B^\complement)\leq \mu^*(A\cap E^\complement)$, we have $\mu^*(A\cap B^\complement) = \mu^*(A\cap E^\complement)$. We also used the fact that $E^\complement\setminus B^\complement = B\setminus E$.

   Lastly, by Proposition 1.13 every set in $\mathcal{A}$ is $\mu^*$-measurable. By Theorem 1.11 (Carathéodory's Theorem), the set of $\mu^*$-measurable sets forms a $\sigma$-algebra. Thus, $B$ is $\mu^*$-measurable. Combining all of the results, \begin{align*} \mu^*(A) = \mu^*(A\cap B) + \mu^*(A\cap B^\complement) = \mu^*(A\cap E) + \mu^*(A\cap E^\complement). \end{align*} Since $A$ was arbitrary, we have that $E$ is $\mu^*$-measurable. This completes the proof of the reverse direction.

3. Let $\{B_n\}_{n=1}^\infty$ be a collection of sets in $\mathcal{A}$ such that $X = \bigcup_{n=1}^\infty B_n$ and $\mu_0(B_n)<\infty$ for all $n$, where $X$ is the ambient space.

   Now, suppose $E$ is $\mu^*$-measurable. Define $E_n:=E\cap \bigcup_{i=1}^n B_i$ so that \begin{align*} \mu^*(E_n) = \mu^*\left(E\cap \bigcup_{i=1}^n B_i\right) \leq \mu^*\left( \bigcup_{i=1}^n B_i\right) \leq \sum_{i=1}^n\mu^*(B_i) = \sum_{i=1}^n\mu_0(B_i) < \infty.\end{align*} Note that we use Proposition 1.13 in the last equality.

   By part (a), choose $A_n^{(k)}\in\mathcal{A}_\sigma$ so that $E_n\subseteq A_n^{(k)}$ and \begin{align*} \mu^*(A_n^{(k)}) \leq \mu^*(E_k)+ \frac{1}{2^{k+n}}.\end{align*} By Theorem 1.11 (Carathéodory's Theorem) and Proposition 1.13, $E_n$ is $\mu^*$-measurable, Repeating a computation from part (b), we find that $\mu^*(A_n^{(k)} \setminus E_k) \leq \frac{1}{2^{k+n}}$. Then, $E \subseteq \bigcup_{k=1}^\infty A_n^{(k)}$, and \begin{align*} \mu^*\left(\bigcup_{k=1}^\infty A_n^{(k)}\setminus E\right) &= \mu^*\left(\bigcup_{k=1}^\infty (A_n^{(k)}\setminus E)\right) \\ &\leq \sum_{k=1}^\infty \mu^*\left(A_n^{(k)}\setminus E\right) \\ &\leq \sum_{k=1}^\infty \mu^*\left(A_n^{(k)}\setminus E_k\right) \\ &\leq \frac{1}{2^n}\sum_{k=1}^\infty \frac{1}{2^{k}} \\ &= \frac{1}{2^n}. \end{align*} Define $A = \bigcap_{n=1}^\infty \bigcup_{k=1}^\infty A_n^{(k)}$, so that \begin{align*} \mu^*(A\setminus E) \leq \mu^*\left(\bigcup_{k=1}^\infty A_n^{(k)}\setminus E\right) \leq \frac{1}{2^n}\end{align*} for all $n$. Thus, $\mu^*(A\setminus E) = 0$. This proves the forwards direction.

   Suppose there exists $G\in \mathcal{A}_{\sigma\delta}$ such that $E\subseteq G$ and $\mu^*(G\setminus E) = 0$. Let $A\subseteq X$ be any set. Define $E_n = E\cap\bigcup_{i=1}^n B_i$, and $G_n = G\cap\bigcup_{i=1}^n B_i$. Note that $G_n\in\mathcal{A}_{\sigma\delta}$ and that for any sets $A,B$, and $C$, \begin{align*} (A\cap C)\setminus (B\cap C) = (A\cap C)\cap (B\cap C)^\complement = (A\cap B^\complement\cap C)\cup (A\cap C^\complement\cap C)=(A\setminus B)\cap C.\end{align*} Applying this fact, we find \begin{align*} \mu^*(G_n\setminus E_n) &= \mu^*\left((G\setminus E)\cap\bigcup_{i=1}^n B_i \right) \leq \mu^*(G\setminus E) = 0. \end{align*} By part (b), $E_n$ is $\mu^*$-measurable. Since $E = \bigcup_{n=1}^\infty E_n$ and from Theorem 1.11 (Carathéodory's Theorem), we have that $E$ is $\mu^*$-measurable. This completes the proof of the reverse direction.