Measure - exercise 22 from Folland

I'm doing some exercises from Folland's real analysis book. Exercise 18 is done and should help to do exercise 22, but I'm stuck.

The definition of completion is given below.

This is not homework, I'm doing this by myself to learn. If possible, I would appreciate a complete answer. Thank you!

PS: my idea was to try to show that $\mathcal{M}^\ast = \overline{\mathcal{M}}$, then the uniqueness would follow from theorem 1.9 and Carathéodory's theorem (which guarantees that $\mu^\ast$ is complete). I don't see how I can do that and don't know if this is the right idea. That's why I'm asking for help here.

Solution 1:

Have $\mu$ be a $\sigma$-finite measure on the $\sigma$-algebra $\mathscr{M}$. Have $\mu^*$ be the outer measure induced by $\mu$, $\mathscr{M}^*$ the $\mu^*$-measurable sets and $\overline{\mu}$ the measure $\mu^*$ restricted to $\mathscr{M}^*$. We want to prove $\overline{\mu}$ on $\mathscr{M}^*$ is the completion of the measure $\mu$ defined on $\mathscr{M}$. We first prove a lemma.

Lemma. Let $(X,\mathscr{F},\nu)$ be a measure space, and let $\overline{\mathscr{F}}$ be the completion of $\mathscr{F}$ according to the definition$$\overline{\mathscr{F}} := \{E \cup F\text{ }|\text{ }E \in \mathscr{F},\,F \subset N \text{ for some }N \in \mathscr{F} \text{ such that }\nu(N) = 0\}.$$Then,$$\overline{\mathscr{F}} := \{G - D\text{ }|\text{ }G \in \mathscr{F},\text{ }D \subset N \text{ for some }N \in \mathscr{F}\text{ such that }\nu(N) = 0\}.$$

Proof. Take $E \cup F \in \overline{\mathscr{F}}$, where $E \in \mathscr{F}$ and $F \subset N$ for some $N \in \mathscr{F}$ such that $\nu(N) = 0$. Then $$E \cup F = E \cup N - (N - (F \cup E)).$$ Note that $E \cup N \in \mathscr{F}$ and $N - (F \cup E) \subset N$.

Conversely, consider $G - D$, where $G - D$, where $G \in \mathscr{F}$ and there is $N \in \mathscr{F}$ with $\nu(N) = 0$ such that $D \subset N$. Then $$G - D = (G - N) \cup [(N \cap G) - D],$$ and $G - N \in \mathscr{F}$ and $(N \cap G) - D \subset N$. So we are done. $\square$

Remember that $$\overline{M} := \{G \cup F\text{ }|\text{ }B \in \mathscr{M},\,F \subset N\text{ for some }N \in \mathscr{M}\text{ such that }\mu(N) = 0\}$$and for such a set $G \cup F$ as in this definition, $\overline{\mu}(G \cup F) := \mu(G)$.

We want to prove that $\mathscr{M}^* = \overline{\mathscr{M}}$ and $\mu^*(E) = \overline{\mu}(E)$ for $E \in \mathscr{M}^*$. We know from Exercise 18 that if $E$ is $\mu^*$-measurable, there exists $B$ in $\mathscr{M}_{\sigma \delta}$ such that $E \subset B$ and $\mu^*(B - E) = 0$. However, because $\mathscr{M}$ is a $\sigma$-algebra, $\mathscr{M}_{\sigma\delta} = \mathscr{M}$ and $B \in \mathscr{M}$. By similar reasoning, since $B - E$ is $\mu^*$-measurable, there exists $C \in \mathscr{M}$ such that $B - E \subset C$ and $\mu^*(C) = \mu(C) = 0$. Because $E = B - (B - E)$, where $B \in \mathscr{M}$, our lemma tells us that $E \in \overline{\mathscr{M}}$. This estalblishes that $\mathscr{M}^* \subset \overline{\mathscr{M}}$.

The converse, $\overline{\mathscr{M}} \subset \mathscr{M}^*$, is true since any set of outer measure $0$ is in $\mathscr{M}^*$. So if $G \in \mathscr{M} \subset \mathscr{M}^*$ and $F \subset N$, where $N \in \mathscr{M}$ and $\mu(N) = 0$, $F \in \mathscr{M}^*$ as well, and thus $G \cup F \in \mathscr{M}^*$.

Finally, if $G \in \mathscr{M}$ and $F \subset N$, where $N \in \mathscr{M}$ and $\mu(N) = 0$, then we have that $$\mu(G) = \mu^*(G) \le \mu^*(G \cup F) \le \mu^*(G) + \mu^*(F) = \mu^*(G) = \mu(G).$$ Thus $$\mu^*(G \cup F) = \mu(G) = \overline{\mu}(G).$$

Solution 2:

I will show that $\mathscr M^*=\overline{\mathscr M}$. From Theorem 1.9 and Carathéodory's theorem it follows, indeed, that $\mu^*|\mathscr M^*$ is the completion of $\mu$, since $\mu^*|\mathscr M^*$ is a complete measure on the $\sigma$-algebra $\mathscr M^*=\overline{\mathscr M}$ and it extends $\mu$ by Proposition 1.13.

Consider a measure space $(X,\mathscr M,\mu)$ as in Exercise 22 and suppose that $\mu$ is $\sigma$-finite. In the language of Exercise 18, $\mathscr M=\mathscr M_{\sigma}=\mathscr M_{\sigma\delta}$, because $\mathscr M$ is not only an algebra but also a $\sigma$-algebra. Then, by Exercise 18(b) and (c), $E$ is $\mu^*$-measurable (that is, $E\in\mathscr M^*$) if and only if

$$\text{$\exists B\in\mathscr M$ such that $E\subseteq B$ and $\mu^*(B\setminus E)=0$.}\tag{1}$$

Now, $(1)$ is equivalent to the following (note that $B\setminus E=E^c\setminus B^c$): $$\text{$\exists B\in\mathscr M$ such that $B^c\subseteq E^c$ and $\mu^*(E^c\setminus B^c)=0$.}\tag{2}$$

Next, $\mu^*(E^c\setminus B^c)=0$ is equivalent to requiring that $E^c\setminus B^c$ be a subset of a $\mu$-null set (using the definition of $\mu^*$ and its completeness—see below for details). Therefore, $(2)$ is, in turn, equivalent to: $$\text{$\exists B\in\mathscr M$ such that $B^c\subseteq E^c$ and $E^c\setminus B^c\subseteq N$ for some $N\in\mathscr N$,}\tag{3}$$ where $\mathscr N$ is as defined in Theorem 1.9. In other words, $(3)$ requires that $E^c$ be able to be expressed as the union of a $\mathscr M$-set $B^c$ and a subset $E^c\setminus B^c$ of a $\mu$-null set $N$. That is, $(3)$ is equivalent to $E^c\in\overline {\mathscr M}$ in the language of Theorem 1.9, or, since $\overline{\mathscr M}$ is a $\sigma$-algebra, $E\in\overline {\mathscr M}$.

Conclusion: $E\in\mathscr M^*$ iff $(1)$ iff $(2)$ iff $(3)$ iff $E^c\in\overline{\mathscr M}$ iff $E\in\overline{\mathscr M}$. Therefore, $\mathscr M^*=\overline{\mathscr M}$.

In what follows, let me prove that if $B\subseteq X$, then $\mu^*(B)=0$ if and only if $B\subseteq N$ for some $N\in\mathscr M$ such that $\mu(N)=0$. Necessity follows from Carathéodory's theorem: since $\mu^*|\mathscr M^*$ is complete and $\mu^*|\mathscr M=\mu$ (see Proposition 1.13), it follows that if $B\subseteq N$ for some $N\in\mathscr M(\subseteq\mathscr M^*)$ and $\mu(N)=0$, then $B\in\mathscr M^*$ and $\mu^*(B)=0$ given that $\mu^*(N)=\mu(N)=0$.

As for sufficiency, suppose that $\mu^*(B)=0$. Then, using the infimum definition of $\mu^*$, for any $n\in\mathbb N$, there exist sets $\{A_{nk}\}_{k=1}^{\infty}\subseteq\mathscr M$ such that $$\mu\left(\bigcup_{k=1}^{\infty}A_{nk}\right)\leq\sum_{k=1}^{\infty}\mu(A_{nk})<\frac{1}{n}$$ and $B\subseteq\bigcup_{k=1}^{\infty}A_{nk}$. Now, let $N\equiv\bigcap_{n=1}^{\infty}\bigcup_{k=1}^{\infty}A_{nk}\in\mathscr M$. Clearly, $B\subseteq N$ and $\mu(N)=0$ since $\mu(N)<1/n$ for any $n\in\mathbb N$.