Show that the completion of the Borel $\sigma$-algebra gives the collection of Lebesgue measurable subsets of $\Bbb R$

Show that the completion of the Borel $$\sigma$$-algebra gives the collection of Lebesgue measurable subsets of $$\Bbb R$$ (with respect to Lebesgue measure)

I know that $$\mathcal B \subset \mathcal L$$ and $$\mathcal L$$ is complete implies $$\mathcal{\overline {B}} \subset \mathcal L$$. For the reverse inclusion, I chose some element of $$\mathcal L$$ and wanted to show it could be approximated on the outside by open sets and on the inside by closed sets and thus show it was an element of $$\mathcal{\overline {B}}$$ but got confused along the way.

• what's the definition of completion?
– user798113
Commented Dec 18, 2020 at 13:59
• Ah ok I see now why $\mathcal{\overline {B}} \subset \mathcal L$ holds but am still confused on the reverse inclusion Commented Dec 18, 2020 at 14:06

$$\tag{1}\overline{\mathcal{B}}=\{A\subseteq \mathbb{R}:\, \text{there exists}\ F,G\in \mathcal{B}\, \text{ with }\ F\subseteq A\subseteq G\ \text{and}\ \mu(G\setminus F)=0\}$$ Now, consider some $$A\in \mathcal{L}$$ with $$\mu(A)<\infty$$. We know, that the Lebesgue measure satisfies some normality conditions, i.e. as you said $$A$$ can be approximated above by open sets and below from closed sets. More formally, for $$\epsilon>0$$ there is an open set $$G_\epsilon$$ and a closed set $$F_\epsilon$$ such that $$F_\epsilon\subseteq A\subseteq G_\epsilon$$ and $$\mu(G_\epsilon)<\mu(A)+\epsilon\\ \mu(F_\epsilon)>\mu(A)-\epsilon$$ Applying this for $$\epsilon=1/n$$ we may find a sequence of open sets $$G_n$$ and closed sets $$F_n$$ such that $$F_n\subseteq A\subseteq G_n$$ and $$\tag{*}\mu(G_n)<\mu(A)+1/n,\\ \mu(F_n)>\mu(A)-1/n$$ for all $$n$$. Let $$F=\bigcup_{n=1}^{\infty}F_n$$ and $$G=\bigcap_{n=1}^{\infty}G_n$$. Observe that both $$F,G$$ belong to $$\mathcal{B}$$ ( $$F$$ is an $$F_\sigma$$ set and $$G$$ a $$G_\delta )$$. By $$(*)$$ we obtain that $$\mu(G)\leq \mu(A)$$ and $$\mu(F)\geq \mu(A)$$. Since, $$F_n\subseteq A\subseteq G_n$$ it follows that $$F\subseteq A\subseteq G$$. Since $$\mu(A)<\infty$$ we have that \begin{align} \mu(G\setminus F)&=\mu(A\cap G\setminus F)+\mu(A^c\cap G\setminus F)\\ &\leq \mu(A\setminus F)+\mu(G\setminus A)\\ &=\mu(A)-\mu(F)+\mu(G)-\mu(A)=0 \end{align} From $$(1)$$ we get that $$A\in \overline{\mathcal{B}}$$. Now if $$\mu(A)=\infty$$ consider the sets $$A_n=[-n,n]\cap A$$. The $$A_n's$$ have finite measure, hence from the previous step they belong to $$\overline{\mathcal{B}}$$. Since, $$A=\bigcup_{n=1}^{\infty}A_n$$ it follows that $$A\in \overline{\mathcal{B}}$$ and the inclusion $$\mathcal{L}\subseteq \overline{\mathcal{B}}$$ is justified.