Question about the measure induced from another measure, problem 1.22 from Folland I'm self-learning Measure Theory using Real Analysis book of Folland. Unfortunately, I got stuck in this problem and couldn't find any clue to solve this. Can someone help me, or give me some hint so I can solve it. Thanks so much:


Let ($X$, $M$, $\mu$) be a measure space, $\mu^*$ the outer space induced by $\mu$ according to the formula: $$\mu^*(E) = \inf\{\sum_1^\infty \mu_0(A_j): A_j \in M, E \subset \bigcup_1^\infty A_j\}$$
    $M^*$ the $\sigma$-algebra of $\mu^*$-measurable sets, and $\overline{\mu} = \mu^*|M^*$. Prove that $\overline{\mu}$ is the saturation of the completion of $\mu$


Although I can solve the first problem(which states that if $\mu$ is $\sigma$-finite, then $\overline{\mu}$ is the completion of $\mu$), but I still got no clue to solve this general problem. Please help me, I really appreciate.
 A: To avoid confusion, let's clarify some notation. Let $\mathcal{M}^{*}$ denote the $\sigma$-algebra of $\mu^{*}$-measurable subsets of $X$. Let $\bar{\mathcal{M}}$ denote the completion of $\mathcal{M}$ with respect to $\mu$. Let $\widetilde{\bar{\mathcal{M}}}$ denote the $\sigma$-algebra of locally measurable subsets of $(X,\bar{\mathcal{M}},\bar{\mu})$. Since $\mu^{*}$ restricts to a complete measure on $\mathcal{M}^{*}$, we know that $\mu^{*}|{\mathcal{M}}^{*}$ coincides with the completion of $\mu$ on the completion $\bar{\mathcal{M}}$ of $\mathcal{M}$ with respect to $\mu$, so there is no ambiguity in using the notation $\bar{\mu}$.


Lemma. Let $(X,\mathcal{M},\mu)$ be a measure space, and let $\mu^{*}$ be the outer measure induced by $\mu$. For $E\in\mathcal{M}^{*}$, with $\mu^{*}(E)<\infty$, there exists $A\in\mathcal{M}$ such that 
  \begin{align*} E\subset A, \quad \mu^{*}(E)=\mu(A) \end{align*} In particular, if $\mu^{*}(E)=0$, then $E$ is contained in a $\mu$-null set $N\in\mathcal{M}$.

Proof. By definition of infimum, there exists a countable collection of sets $\left\{A_{j}\right\}\subset\mathcal{M}$ such that
\begin{align*}
E\subset A:=\bigcup_{j=1}^{\infty}A_{j}, \quad \sum_{j=1}^{\infty}\mu(A_{j})<\mu^{*}(E)+\epsilon
\end{align*}
Since $\mathcal{M}$ is a $\sigma$-algebra, $A\in\mathcal{M}$, and by subadditivity, $\mu(A)\leq\mu^{*}(E)+\epsilon$.
We can apply the preceding result for each $\epsilon_{n}=1/n$, to obtain a collection of sets $\left\{A^{n}\right\}\subset\mathcal{M}$ such that $E\subset A^{n}$ and $\mu(A^{n})\leq\mu^{*}(E)+1/n$. If we set $A:=\bigcap_{n}A^{n}$, then $E\subset A$ and by monotonicity,
\begin{align*}
\mu(A)\leq\mu(A^{n})\leq\mu^{*}(E)+\dfrac{1}{n},\qquad\forall n
\end{align*}
Letting $n\rightarrow\infty$, we obtain $\mu(A)\leq\mu^{*}(E)$. The reverse inequality holds by monotonicity. $\Box$

Let $E$ be a locally measurable subset of $(X,\bar{\mathcal{M}},\bar{\mu})$. I claim that $E\in\mathcal{M}^{*}$. It suffices to show that for any $F\subset X$ with $\mu^{*}(F)<\infty$, we have
\begin{align*}
\mu^{*}(F)\geq\mu^{*}(E\cap F)+\mu^{*}(E^{c}\cap F)
\end{align*}
By the lemma, there exists a set $A\in\mathcal{M}$ such that
\begin{align*}
F\subset A, \quad \mu^{*}(F)=\mu(A)
\end{align*}
So $E\cap A$ is measurable, whence $E^{c}\cup A^{c}\cap A=E^{c}\cap A$ is measurable. By monotonicity and additivity, we see that
\begin{align*}
\mu^{*}(E\cap F)+\mu^{*}(E^{c}\cap F)\leq\mu^{*}(E\cap A)+\mu^{*}(E^{c}\cap A)=\mu(A)=\mu^{*}(F)
\end{align*}
The reverse inclusion also holds: $E\in\mathcal{M}^{*}$ implies $E\in\widetilde{\bar{\mathcal{M}}}$. For any set $A\in\bar{\mathcal{M}}$ with $\bar{\mu}(A)<\infty$, $\mu^{*}(E\cap A)<\infty$, whence there exists a set $B\in\mathcal{M}$ such that $E\cap A\subset B$ and $\mu^{*}(E\cap A)=\mu(B)$. Since $E\cap A, B\in\mathcal{M}^{*}$, $\mu^{*}(B\setminus (E\cap A))=0$. But then there exists $N\in\mathcal{M}$, such that
\begin{align*}
B\setminus (E\cap A)\subset N, \quad \mu^{*}(B\setminus (E\cap A))=\mu(N)=0
\end{align*}
We conclude that $B\setminus (E\cap A)\in\bar{\mathcal{M}}$, whence
\begin{align*}
E\cap A=B\setminus (B\setminus (E\cap A))\in\bar{\mathcal{M}}
\end{align*}
With $E$ as above, suppose $\mu^{*}(E)<\infty$. Then, as asserted before, there exists a set $A\in\mathcal{M}$ such that $E\subset A$ and $\bar{\mu}(E)=\mu(A)$. But then $E=E\cap A\in\bar{\mathcal{M}}$. We conclude that
\begin{align*}
\bar{\mu}(E)=\begin{cases}\bar{\mu}(E) & {E\in\bar{\mathcal{M}}}\\ \infty & {E\in\widetilde{\bar{\mathcal{M}}}\setminus\bar{\mathcal{M}}} \end{cases},
\end{align*}
which is the definition of the $\widetilde{\bar{\mu}}$, the saturation of $\bar{\mu}$.
