Approximation of Lebesgue Measurable Sets by compact and G-delta?

Question

Just to check if this is true:

Let $E$ be a Lebesgue measurable set on $\mathbb{R}^n$. Does there exists a sequence of open sets $G_m$, compact sets $K_m$, with $K_m\subset E\subset G_m$ and $\mu((\cap G_m)\setminus (\cup K_m))=0$?

I am aware of the usual approximation by $G_\delta$ and $F_\sigma$ sets, however here we need compactness which may not be true if the sets are not bounded?

Thanks for any help.

Answer 1

Yes. For $n\geq 0$ let $B_n=\{x\in E: \|x\|<n\}.$ Then $B_0=\emptyset$ and $$\mu (E)=\sum_{n=0}^{\infty}\mu ( B_{n+1}\backslash B_n)=$$ $$=\sup_{m\in N}\sum_{n=0}^m \mu (B_{n+1}\backslash B_n)=$$ $$=\sup_{m\in N}\mu (\cup_{n=0}^{m+1}( B_n)=\sup_{m\in N}\mu ( B_{m+1}).$$ Let $C_m$ be a compact subset of $B_m$ with $\mu (C_m)>\mu ( B_m)-2^{-m}.$ Then $\mu (E)=\sup_{m\in N}\mu (C_m).$

Let $D=\cup_{m\in N}C_m.$ For each $n\in N$ we have $\mu (B_n\backslash D)=0.$ So we have $$\mu(E\backslash D)=\sum_{n\in N}\mu (B_n\backslash D)=\sum_{n\in N}0=0.$$