Exercise 22 – Let $(X,M,\mu)$ be a measure space, $\mu^*$ the outer measure induced by $\mu$ according to (1.12), $M^*$ the $\sigma$-algebra of $\mu^*$-measurable sets, and $\overline{\mu} = \mu^*|M^*$.
a.) If $\mu$ is $\sigma$-finite, then $\overline{\mu}$ is the completion of $\mu$ (Use exercise 18 found here)
b.) In general, $\overline{\mu}$ is the saturation of the completion of $\mu$
This exercises is from Follad's Real Analysis (Section 1.4. Outer Measures).
Attempted proof a.) – Let $F\subset N$, where $N$ is a measurable null set, i.e., $N\in M$ and $\mu(N) = 0$. I will prove that $F\in M^*$. Using 18b and 18c, I believe it suffices to show that there is a $B\in \mathcal{A}_{\sigma \delta}$ such that $F\subset B$ and $\mu^*(B\setminus F) = 0$.
Since $M\subset M^*$, $N\in M^*$ and hence there exists a $C\in\mathcal{A}_{\sigma \delta}$ such that $N\subset C$ and $\mu^*(C\setminus N) = 0$. Thus $$\mu^*(C\setminus F) \leq \mu^*(C\setminus N) + \mu^*(N\setminus F) \leq \mu^*(C\setminus N) + \mu^*(N) = 0$$
I am not sure if this is completely right. Also any suggestions on part b would be greatly appreciated. I just don't understand the statement "$\overline{\mu}$ is the saturation of the completion of $\mu$".
Best Answer
Initial Remark: This is a long exercise and it depends on some previous exercises in Folland. I made this answer as self-contained as reasonable (I explictly used only the results Exercise 18). The definition of saturation can be found Exercise 1.3.16.
Notation: Let $(X,M,\mu)$ be a measure space, $\mu^*$ the outer measure induced by $\mu$ according to (1.12), $M^*$ the $\sigma$-algebra of $\mu^*$-measurable sets, and $\overline{\mu} = \mu^*|M^*$. Let $\widehat{M}$ be the $\mu$-completion of $M$ and $\widehat{\mu} $ the completion of $\mu$.
We begin with three lemmas.
Proof
Let $E\subset X$ be any $\mu^*$-measurable set such that $\mu^*(E)< \infty$.
From Exercise 18-b, there exists $B\in M_{\sigma\delta} \subset M$ with $ E\subset B$ and $\mu^{*}(B\setminus E)) = 0$. So
$$\mu^*(B)\leq \mu^*(E)+ \mu^{*}(B\setminus E))= \mu^*(E)$$ Since $ E\subset B$, we have $\mu^*(B)= \mu^*(E)$.
Now, we have that $B\setminus E$ is $\mu^*$-measurable set and $\mu^*(B\setminus E)\leq \mu^*(B)= \mu^*(E) < \infty$. From Exercise 18-b, again, there exists $C\in M$ with $ B\setminus E\subset C$ and $\mu^{*}(C\setminus (B\setminus E)) = 0$
Define $D=C\cap B$. Then, since $B\setminus E\subset C$, we have $B\setminus E\subset D$ and so we have $$B\setminus D \subset E $$ and $$ E\setminus (B \setminus D)= E\cap D= D\setminus (B\setminus E) = (C\setminus (B\setminus E))\cap B \subset C\setminus (B\setminus E)$$ So, since $\mu^{*}(C\setminus (B\setminus E)) = 0$, we have $\mu^{*}(E\setminus (B\setminus D)) = 0$.
So we have that $D\in M$, $B\setminus D\subset E$ and $\mu^{*}(E \setminus (B\setminus D)) = 0$.
So $E =(B \setminus D) \cup (E\setminus (B\setminus D))$ where $B-D \in M$ and $\mu^{*}(E\setminus (B\setminus D)) = 0$. So $E\in \widehat{M}$ (where $\widehat{M}$ is the $\mu$-completion of $M$).
Proof From Exercise 18-a we know that
For any $E\subset X$ and $\epsilon > 0$ there exists $A\in M$ with $E\subset A$ and $\mu^*(E)\leq \mu^*(A) \leq \mu^*(E) + \epsilon$.
So, for each $n\in\mathbb{N}$, $n>0$, let $A_n\in M$ with $E\subset A_n$ and $\mu^*(E)\leq \mu^*(A_n) \leq \mu^*(E) + \frac{1}{n}$.
Then let $B=\bigcap_{n=1}^\infty A_n$. Then we have $B\in M$ and $E\subset B$. Moreover, for all $n\in\mathbb{N}$, $n>0$, $B\subset A_n$ and $$\mu^*(E)\leq \mu^*(B) \leq \mu^*(A_n) \leq \mu^*(E) + \frac{1}{n}$$ So, $\mu^*(B)=\mu^*(E)$.
Remark: $E$ don't need to be measurable. And as a consequence of item Exercise 18 item b.) $\mu^*(B\setminus E)$ may not be zero.
Proof:
Let $A\in \widehat{M}$. So $A =B\cup C$ , where $B\in M$ and $C\subset N$ such that $N \in M$ and $\mu(N)=0$. In particular, we have $0 \leq \mu^*(C)\leq \mu^*(N)=\mu(N)=0$. So $\mu^*(C)=0$ and it follows that $C$ is $\mu^*$-measurable. So we have $A =G\cup H$, $B\in M \subset M^*$ and $C\in M^*$. So $A\in M^*$. Moreover $$\mu*(B) \leq \mu^*(A) \leq \mu*(B) + \mu*(C) = \mu*(B)$$ So $$\mu^*(A)= \mu*(B)= \mu(B)=\widehat{\mu}(A)$$
Proof:
a.) Suppose $\mu$ is $\sigma$-finite. Then there is $\{X_i\}_{i\in \mathbb{N}}$ family of disjoint sets, such that, for all $i$, $\mu(X_i)<+\infty$ and $X=\bigcup_{i=1}^\infty X_i$.
Let $E\subset X$ be any $\mu^*$-measurable set. Then, we define, for all $i$, $E_i=E\cap X_i$, we have that $\{E_i\}_{i\in \mathbb{N}}$ is a family of disjoint sets, such that, for all $i$, $\mu(X_i)<+\infty$ and $E=\bigcup_{i=1}^\infty E_i$.
From the lemma 1, for all $i$, $E_i\in \widehat{M}$ (where $\widehat{M}$ is the $\mu$-completion of $M$).
Since $\widehat{M}$ is a $\sigma$-algebra, $E = E=\bigcup_{i=1}^\infty E_i \in \widehat{M}$.
b.) Let $\widetilde{\widehat{\mu}}$ be the saturation of the completion of $\mu$. We know that $\widetilde{\widehat{\mu}}$ is defined on $$\widetilde{\widehat{M}}= \{E : E \textrm{ is ${\widehat{\mu}}$-locally measurable} \} $$ On the other hand, $\overline{\mu}$ is defined on $M^*$. We want to prove that $\widetilde{\widehat{\mu}}=\overline{\mu}$.
We begin by proving that $\widetilde{\widehat{M}}=M^*$, it means, we begin by proving that $E$ is ${\widehat{\mu}}$-locally measurable if and only if $E$ is $\mu^*$-measurable.
($\Rightarrow$) Suppose $E$ is ${\widehat{\mu}}$-locally measurable. For any $B \in M$ such that $\mu(B)<+\infty$, we have $B\in \widehat{M}$ and $$\widehat{\mu}(B)=\mu(B)<\infty$$
So $B\in \widehat{M}$ and $\widehat{\mu}(B)<\infty$. Since $E$ is ${\widehat{\mu}}$-locally measurable, we have that $E\cap B \in \widehat{M}$. So, by lemma 3, $E\cap B \in M^*$. And we also have $E^c\cap B= B \setminus (E\cap B) \in M^*$.
So we have proved that $B \in M$ and $\mu(B)<+\infty$, then $E\cap B \in M^*$ and $E^c\cap B \in M^*$
Now let us check the $\mu*$-measurability of $E$. Let $A\subset X$. If $\mu^*(A)=\infty$, we trivially have $$\mu^*(A) \geq \mu*(E\cap A)+\mu*(E^c\cap A)$$ If $\mu^*(A)<\infty$ then using lemma 2, we have $B\in M$ such that $A\subset B$ and $\mu(B)=\mu^*(B)=\mu^*(A)<\infty$. So we have that $E\cap B \in M^*$ and $E^c\cap B \in M^*$ and $$\mu^*(A)= \mu^*(B) = \mu^*(E\cap B ) + \mu^*(E^c\cap B) \geq \mu^*(E\cap A ) + \mu^*(E^c\cap A)$$
So E is $\mu^*$-measurable.
($\Leftarrow$) Suppose E is $\mu^*$-measurable. For any $A\in \widehat{M}$ such that $\widehat{\mu}(A)<+\infty$. By lemma 3, $A\in M^*$ and $\mu*(A)= \widehat{\mu}(A)<+\infty$. So $E\cap A\in M^*$ and $\mu*(E\cap A)\leq \mu^*(A)<+\infty$. So by lemma 1, $E \cap A \in \widehat{M}$. So, $E$ is ${\widehat{\mu}}$-locally measurable.
So we have proved $\widetilde{\widehat{M}}=M^*$.
Now we must prove that $\widetilde{\widehat{\mu}}$ coincides to $\overline{\mu}$.
Since the extension of $\mu$ to $\widehat{M}$ (the completion of $\mu$) is unique, we have for all $E \in \widehat{M}$, $$ \widetilde{\widehat{\mu}}(E)= \overline{\mu}(E)= \widehat{\mu}(E) $$
If $E\in \widetilde{\widehat{M}} \setminus \widehat{M}$, then by definition of $\widetilde{\widehat{\mu}}$, we have $\widetilde{\widehat{\mu}}(E)=+\infty$. Note $E\in \widetilde{\widehat{M}} \setminus \widehat{M}= M^*\setminus \widehat{M}$ and, by lemma 1, we have $\mu^*(E)=+\infty$, that is $\overline{\mu}(E)=\mu^*(E)=+\infty$. So, $\widetilde{\widehat{\mu}}(E)=\overline{\mu}(E)$.
So we have proved that $\widetilde{\widehat{\mu}}=\overline{\mu}$.