Approximation property with F sigma and G delta Sets to show a set is measurable

[ryo0071]

Prove that a set $A\subset\mathbb{R}^n$ is (Lebesgue) measurable $\iff$ there exist a set $B$ which is an $F_{\sigma}$ and a set $C$ which is a $G_{\delta}$ such that $B\subset A\subset C$ and $C$~$B$ (C without B) is a null set.

$F_{\sigma}$ is a countable union of closed sets, and $G_{\delta}$ is a countable intersection of open sets.

I have proven the forward implication already. For the second one I know that I can pick one of the closed sets that make up $B$, say the closed set $F$ with $F \subset B$, and one of the open sets that make up $C$, say the open set $G$ with $C \subset G$. It then follows that $F \subset A \subset G$. I am having trouble showing that the measure of $G$~$F$ is less than an arbitrary $\epsilon > 0$. I know I need to use the fact that $C$~$B$ is a null set but I am not sure how.

Any help is appreciated.

 

[Opalg]

Prove that a set $A\subset\mathbb{R}^n$ is (Lebesgue) measurable $\iff$ there exist a set $B$ which is an $F_{\sigma}$ and a set $C$ which is a $G_{\delta}$ such that $B\subset A\subset C$ and $C$~$B$ (C without B) is a null set.

$F_{\sigma}$ is a countable union of closed sets, and $G_{\delta}$ is a countable intersection of open sets.

I have proven the forward implication already. For the second one I know that I can pick one of the closed sets that make up $B$, say the closed set $F$ with $F \subset B$, and one of the open sets that make up $C$, say the open set $G$ with $C \subset G$. It then follows that $F \subset A \subset G$. I am having trouble showing that the measure of $G$~$F$ is less than an arbitrary $\epsilon > 0$. I know I need to use the fact that $C$~$B$ is a null set but I am not sure how.

Any help is appreciated.

First, \(\displaystyle A = \bigcup_{k\geqslant1}\bigl(A\cap[-k,k]^n\bigr).\) A countable union of measurable sets is measurable, so it will be sufficient to show that $A\cap[-k,k]^n$ is measurable. So we might as well assume that $A$ is bounded. (That will eliminate any problems that might arise from sets with infinite measure.)

Next, $B$ is a countable union of closed sets, say $B = \bigcup F_r$, and is therefore measurable. Replacing each $F_r$ by $\bigcup_{1\leqslant j\leqslant r}F_j$, we may assume that the sets $F_r$ form an increasing sequence. Then the measure $m(B)$ of $B$ is the sup of the measures $m(F_r).$ So we can find a closed set $F\subset B$ with $m(F) > m(B) - \varepsilon/2.$ By the same argument we can find an open set $G\supset C$ with $m(G)<m(C) + \varepsilon/2$. But $C$ is the union of $B$ and a null set, so $m(C) = m(B).$ It follows that $m(G) - m(F) < \varepsilon.$