From Bass, Real Analysis for Graduate Students:
Let $m$ be the Lebesgue measure and $A$ a Lebesgue measurable subset of $\mathbb{R}$ with $m(A)< \infty$. Let $\epsilon>0$. Show there exist $G$ open and $F$ closed such that $F\subset A\subset G$ and $m(G\setminus F)<\epsilon$.
Using the definition of Lebesgue measure as an outer measure I can easily build an open $G$ such that $A\subset G$ and $m(G\setminus A)<\epsilon/2$.
I'm stuck with finding a closed set $F$ such that $F\subset A$ and $m(A\setminus F)<\epsilon/2$.
I know how to build one whenever $\textbf A$ is bounded. Nevertheless, $m(A)<\infty$ need not imply that $A$ is bounded ($A=\mathbb Q$ is an example).
Is there a way to circumvent this looser setting ?
Best Answer
Since $A$ is measurable so $A^c$ is measurable
Hence there exists an open set $A^c\subseteq O$ such that $m^*(O-A^c)<\epsilon$ for all $\epsilon>0$.
Now consider $O^c$ which is closed.Also $ O^c\subseteq A$ and $m^*(A-O^c)=m^*(O-A^c)<\epsilon$