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.