I have some doubts about the proof of the last two points of the following theorem. Before the theorem we give a definition.
Definition. Let $(X,\mathcal{A},\mu)$ a measure space. We will say that: $N\in\mathcal{A}$ is a set of zero measure if $\mu(N)=0$, we denote with $\mathcal{N_\mu}$ the family of zero measure sets.
A set $E\subset X$ is said negligible if exists $N\in\mathcal{A}$ such that $E\subset N$ and $\mu(N)=0$, we denote with $\mathcal{T}_\mu$ the family of negligible sets.
A measure space $(X,\mathcal{A},\mu)$ is said complete if $\mathcal{T}_\mu\subseteq \mathcal{A}$.
$$$$ We define $$\overline{\mathcal{A}}=\bigg\{E\subseteq 2^{X}\;|\;\text{exists}\;F, G\in\mathcal{A}, \text{with}\;F\subseteq E\subseteq G\;\text{and such that}\;\mu(G\setminus F)=0\bigg\},$$ and $$\overline{\mu}\colon\overline{\mathcal{A}}\to[0,+\infty]$$ such that taken $E\in\overline{\mathcal{A}},\quad$$\overline{\mu}(E):=\mu(F)$.
Theorem. Let $(X,\mathcal{A},\mu)$ a measure space, then exists a triad $(X,\overline{\mathcal{A}},\overline{\mu})$ such that:
$(i)\quad (X,\overline{\mathcal{A}},\overline{\mu})$ is a measure space;
$(ii)\quad (X,\overline{\mathcal{A}},\overline{\mu})$ is a complete space;
$(iii)\quad (X,\overline{\mathcal{A}},\overline{\mu})$ is an extension of $(X,\mathcal{A},\mu)$. That is $\mathcal{A}\subseteq\overline{\mathcal{A}}$ and $\overline{\mu}|_\mathcal{A}=\mu$.
$(iv)$ The measure space $(X,\overline{\mathcal{A}},\overline{\mu})$ is the smallest measure space which contains $(X,\mathcal{A},\mu)$, that is if $(X,\mathcal{A}_1,\mu_1)$ is a complete measure space such that $\mathcal{A}\subseteq \mathcal{A}_1$, $\mu_1|_{\mathcal{A}}=\mu$, then $\overline{\mathcal{A}}\subseteq \mathcal{A}_1$, and $\mu_1| _{\overline{\mathcal{A}}}=\overline{\mu}$.
Proof. The points $(i)$ and $(ii)$ are ok. We prove $(iii)$: Let $E\in\mathcal{A}$, then $E\subseteq E\subseteq E$ and $\mu(E\setminus E)=\mu(\emptyset)=0$, that is $E\in\overline{\mathcal{A}}$. Moreover, let $E\in\mathcal{A}\subseteq\overline{\mathcal{A}}$, then $E\subseteq E\subseteq E$ and $\overline{\mu}(E)=\mu(E)$. It's correct?
We prove, now, the point $(iv)$. First we show that $\overline{\mathcal{A}}\subseteq \mathcal{A}_1$. Let $E\in\overline{\mathcal{A}}$, then exists $F, G\in\mathcal{A}$ with $F\subseteq E\subseteq G$ and $\mu(G\setminus F)=0$. We observe that $G\setminus F\in\mathcal{A}$ and, for hypotesis, $\mu_1|_{\mathcal{A}}=\mu$; therefore $\mu_1(G\setminus F)=\mu(G\setminus F)=0$. We note that $G\setminus E\subseteq G\setminus F$ and $\mu_1(G\setminus F)=0$, then $G\setminus E\in\mathcal{T}_{\mu_1}\subseteq \mathcal{A}_1$. Then $E=G\setminus(G\setminus E)\in\mathcal{A}_1$. It's correct?
Could someone help me show that $\mu_1| _{\overline{\mathcal{A}}}=\overline{\mu}$?
My attempt
Let $E\in\overline{\mathcal{A}}\subseteq\mathcal{A}_1$, exist $F,G\in\mathcal{A}$ with $F\subseteq G\subseteq E$ such that $\mu(G\setminus F)=0$. Then \begin{equation} \begin{split} \overline{\mu}(F)&\le\overline{\mu}(E)\le\overline{\mu}(G)\\ \mu_1(F)&\le\mu_1(E)\le\mu_1(G). \end{split} \end{equation} Then
\begin{equation} \begin{split} \mu(F)&\le\overline{\mu}(E)\le\mu(G)\\ \mu(F)&\le\mu_1(E)\le\mu(G). \end{split} \end{equation} Therefore, $\mu_1(E)=\mu_1(E\setminus F)+\mu_1(F)\le\mu_1(F)=\mu(F)\le\overline{\mu}(E)$, moreover $\overline{\mu}(E)=\overline{\mu}(E\setminus F)+\overline{\mu}(F)\le\mu(F)\le\mu_1(E)$.
Then $\overline{\mu}(E)=\mu_1(E)$.
Thanks!
Since $F\subseteq E\subseteq G$, $$\bar\mu(F) \le \bar\mu(E) \le \bar\mu(G)$$and $$\mu_1(F) \le \mu_1(E) \le \mu_1(G)$$ But $\mu_1(F) = \mu(F) = \bar\mu(F)$ and similarly for $G$. And of course, $$\mu(G) = \mu(G\setminus F) + \mu(F) = \mu(F)$$