# 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 $$B(\mathbb{R}^n)$$.

(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, $$\mu(N)=0$$.