Suppose $<X,M,\mu>$ is a measure space and $\overline{M}=\{A:\exists B,C \in M, B\subset A \subset C\ \text{and}\ \mu(C \smallsetminus B)=0\}$ is the closure of the $\sigma$-algebra $M$. I'm trying to understand the proof of this theorem: $M=\overline{M}$ if and only if $<X,M,\mu>$ is complete.
Proof
$(\rightarrow)$ Suppose $M=\overline{M}$, then, choose $E=A\cup N$ where $A \in M$ and $N$ is a null subset, then $E \in M$.
Suppose $E \subset A \in M$ and $\mu(A)=0$, take $N=A\smallsetminus E$, then $E \in M$.
Is the first line used for something?, Why $E \in M$ in the second line?
Let me try to reword the argument for $(\implies)$. We are supposing $\mathcal{M} = \overline{\mathcal{M}}$, and we have to show $(X,\mathcal{M}, \mu)$ is complete.
Ok, so let $N \in \mathcal{M}$ be any set with $\mu(N) = 0$, and take any subset $A \subset N$. We have to show that $A \in \mathcal{M}$ (showing completeness means we have to show every subset of a measurable-null set is measurable). Now, notice that \begin{align} \emptyset \subset A \subset N \qquad \text{and} \qquad \mu(N \setminus \emptyset) = \mu(N) = 0. \end{align} This says exactly that $A \in \overline{\mathcal{M}}$ (I took $B = \emptyset$ and $C=N$ in the definition of $\overline{\mathcal{M}}$). So, $A \in \overline{\mathcal{M}} = \mathcal{M}$, where the last equal sign is by hypothesis. This is exactly what we had to show to prove $(X,\mathcal{M}, \mu)$ is complete.