# Completion and outer-measure extension of sigma finite measure

I need some help with the proof of the following.

Let $$(X,\mathcal{A},\mu)$$ be a measure space with $$\mu$$ being $$\sigma$$-finite, $$\mu^*$$ be the outer measure given by the formula $$\mu^*(E)=\inf\{\sum_{n}\mu(A_n): E\subset\bigcup A_n, (A_n)\subset\mathcal{A}\}$$ and $$\mathcal{M}$$ the $$\sigma$$-algebra of the $$\mu^*$$-measurable sets. Also let $$\mathcal{A}_{\mu}=\{A\subset X: \exists E,F\in\mathcal{A},$$ with $$E\subset A\subset F$$ and $$\mu(F\setminus E)=0\}$$ and $$\overline{\mu}:\mathcal{A}_\mu\to [0,+\infty]$$ given by $$\overline{\mu}(E)=\sup\{\mu(B): B\in\mathcal{A}, B\subset E\}.$$

Then $$\mathcal{A}_\mu=\mathcal{M}$$ and $$\overline{\mu}=\mu^*\vert_{\mathcal{M}}$$.

Okay, so it is known that the measure space $$(X,\mathcal{A}_\mu, \overline\mu)$$ is the completion of $$(X,\mathcal{A},\mu)$$ and by the facts that $$\mu^*\vert_\mathcal{A}=\mu$$ and $$\mathcal{A}\subset\mathcal{M}$$, we have that $$\mathcal{A}_\mu\subset\mathcal{M}$$ and that $$\mu^*\vert_{\mathcal{A}_\mu}=\overline\mu$$. So in order to prove the statement above, it would be enough to show that $$\mathcal{M}\subset\mathcal{A}_\mu$$. I'm trying to prove that if $$A\in\mathcal{M}$$ and $$\mu^*(A)<\infty$$ then $$A\in\mathcal{A}_{\mu}$$ but I'm stuck. My progress is the following:

For each $$n\in\mathbb{N}$$ there exists a sequence $$(A_k^{(n)})\subset\mathcal{A}$$ such that $$\sum_{k}\mu(A_k^{(n)})<\mu^*(A)+\frac{1}{n}$$. Let $$A_n=\bigcup_{k}A_k^{(n)}\in\mathcal{A}$$. We have that $$\mu(A_n)\leq\sum_{k}\mu(A_k^{(n)})<\mu^*(A)+\frac{1}{n}$$. Finally let $$F=\bigcap_{n} A_n$$. Then $$\mu(F)\leq\mu(A_n)<\mu^*(A)+\frac{1}{n}$$ for all $$n$$. Taking limits we have that $$\mu(F)\leq\mu^*(A)$$, but $$A\subset F$$, so $$\mu^*(A)\leq\mu^*(F)=\mu(F)$$ so we found the first desirable set. But what about the other? I'm stuck here and I can't seem to be able to use the sigma-finiteness. Any help?

EDIT: Maybe considering the fact that the sigma-algebra $$\mathcal{A}_1=\{A\cup E: A\in\mathcal{A}, E\subset F,$$ where $$F\in\mathcal{A}, \mu(F)=0\}$$ is equal to $$\mathcal{A}_\mu$$ helps. It is easy to show this equality.

Denote by $(X_n)_{n \in \mathbb{N}} \subseteq \mathcal{A}$ a sequence of increasing sets, $X_n \uparrow X$, such that $\mu(X_n)< \infty$ for all $n \in \mathbb{N}$. You have already shown that

$$B \in \mathcal{M} \implies \exists F \in \mathcal{A}, F \supseteq B: \mu(F) = \mu^*(B). \tag{1}$$

Roughly speaking, the idea is to use $(1)$ for $B:=A^c \in \mathcal{M}$ and then define $E:=F^c$; if $\mu$ is a finite measure, you can easily verify that the so-defined set has all desired properties. For $\sigma$-finite measures the reasoning is somewhat more complicated.

Fix $A \in \mathcal{M}$. Applying $(1)$ for $B:=A^c \cap X_n \in \mathcal{M}$ we find that there exists a set $H_n \in \mathcal{A}$, $H_n \supseteq A^c \cap X_n$, such that

$$\mu(H_n) = \mu^*(A^c \cap X_n)<\infty.$$

Note that $E_n := X_n \backslash H_n \in \mathcal{A}$ satisfies $E_n \subseteq A \cap X_n$ and

$$\mu(E_n) = \mu(X_n)-\mu(H_n) = \mu^*(X_n)-\mu^*(A^c \cap X_n) = \mu^*(X_n \backslash A^c) = \mu^*(X_n \cap A)$$

where we have used in the penultimate step that $A^c \in \mathcal{M}$. Combining this with the fact that $X_n \cap A^c \in \mathcal{M}$, we find that

$$\mu^*((X_n \cap A) \backslash E_n)=0. \tag{2}$$

Now set $E:= \bigcup_{n \geq 1} E_n \in \mathcal{A}$, then $E \subseteq A$ and, by the monotonicity and subadditivity of $\mu^*$,

\begin{align*} \mu^*(A \backslash E) &= \mu^* \left( \left( \bigcup_{n \geq 1} A \cap X_n \right) \backslash \left( \bigcup_{k \geq 1} E_k \right) \right) \\ &\leq \mu^* \left( \bigcup_{n \geq 1} (A \cap X_n) \backslash E_n \right) \\ &\leq \sum_{n \geq 1} \mu^*((A \cap X_n) \backslash E_n) \\ &\stackrel{(2)}{=} 0.\end{align*}

Thus,

$$\mu^*(A) = \mu^*(E) = \mu(E).$$