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.
 A: I suspect as i read your approach that your definition of completion is the following:
$$\tag{1}\overline{\mathcal{B}}=\{A\subseteq \mathbb{R}:\, \text{there exists}\ F,G\in \mathcal{B}\, \text{ with }\ F\subseteq A\subseteq G\ \text{and}\ \mu(G\setminus F)=0\} $$
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.
