The Lebesgue $\sigma$-algebra $L(\mathbb{R}^n)$ is the completion of the Borel $\sigma$-algebra $B(\mathbb{R}^n)$ I came across the following proof of the completion of Borel $\sigma$-algebra to $\sigma$-algebra comprised of Lebesgue measurable sets which I cannot understand quite clearly. Can anyone elaborate this proof in a detailed manner and what is the whole point of the proof?
Theorem 2.28. The Lebesgue $\sigma$-algebra $L(\mathbb{R}^n)$ is the completion of the Borel $\sigma$-algebra $B(\mathbb{R}^n)$.
Proof. Lebesgue measure is complete. If $A \subset \mathbb{R}^n$ is Lebesgue measurable, then there is an $F_\sigma$ set $F \subset A$ such that $M = A \setminus F$ has Lebesgue measure zero. It follows by the approximation theorem that there is a Borel set $N \in G_\delta$ with $\mu (N) = 0$ and $M \subset N$. Thus, $A = F\cup M$ where $F \in B$ and $M \subset N \in B$ with $\mu (N ) = 0$, which proves that $L(\mathbb{R}^n)$ is the completion of $B(\mathbb{R}^n)$.
 A: (1) $\mu$ is complete if $\mu$ is induced by an outer measure $\mu^*$.
proof. $\mu$ is a measure on $\bar{\alpha}$, where
$$
\bar{\alpha}=\{E\subset X| 
\mu^*(A)=\mu^*(A\setminus E)+\mu^*(A\cap E),\;\forall\;A\subset X\}
$$
Check: $(E\in \bar{\alpha},\mu^*(E)=0,F\subset E)
\Rightarrow(F\in \bar{\alpha})$.
$\le$: $\forall\;A\subset X,\;\mu^*(A)\le \mu^*(A\setminus F)+\mu^*(A\cap F)$
$\ge$: $\forall\;A\subset X,$
\begin{align*}
    \mu^*(A)&=\mu^*(A)+\mu*(E)\\
&\ge \mu^*(A\setminus F)+\mu^*(A\cap E)\\
&\ge \mu^*(A\setminus F)+\mu^*(A\cap F)
\end{align*}
Thus,Lebesgue measure is complete
(2)A is Lebesgue measurable($A\in L$). Then A=F$\cup$M, where $F$ is Borel, $\mu(M)=0.$
proof. $\forall\; n\ge 1,\;\exists$
closed $F_n\subset A,\mu(A\setminus F_n)<\frac{1}{n}$.
Let $F=\bigcup_n F_n$. $F\subset A$ is Borel.
Let $M=A\setminus F\in L$.
$$\mu(M)=\mu(A\setminus \bigcup_n F_n)\le \mu(A\setminus F_n)<\frac{1}{n},\forall\;n\ge 1$$
That is, $\mu(M)=0$.
(3)$M\in L,\mu(M)=0$. Then $\exists$ Borel set $N$, where $F\subset N$, $\mu(N)=0$.
proof.$\forall\; n\ge 1,\;\exists$
open $O_n\supset M,\mu(O_n\setminus M)<\frac{1}{n}$.
Let $N=\bigcap_n O_n$. $N\supset M$ is Borel.
$$\mu(N)=\mu(N\setminus M)+ \mu(N\cap M)\le
\mu(O_n\setminus M)+\mu(M)<\frac{1}{n},\forall\;n\ge 1$$
That is, $\mu(N)=0$.
