Let $m$ be a Lebesgue measure and $A$ a Lebesgue measurable subset of $\mathbb{R}$ with $m(A)<\infty$. Let $\varepsilon>0$. Show there exist $G$ open and $F$ closed such that $m(G-F)<\varepsilon$.
My attempt at a proof, I have only done one part so far: Since $A$ is lebesgue-measurable then $m(E)=m(E\cap A)+m(E\cap A^\complement)$ for all $E\subset\mathbb{R}$. Choose $E$ to be an open set containing $A$. Then we can write
$m(E\setminus A)=m(E\cap A^\complement)=m(E)-m(E\cap A)=m(E)-m(A).$
Here I am stuck. How can I further choose $E$ so that I know the difference is going go be less than the given $\varepsilon$? If I can get this, and I set $G=E$, I have the first part done. Furthermore, is there a similar way to show the $F\subset A$ part?
Best Answer
Using the definition of $m$ as an outer measure, there exist $A_i=(a_i,b_i]$ such that $A\subset \cup_i A_i$ and $\sum_i (b_i-a_i) \leq m(A) + \epsilon/2$. Let $b_i' = b_i + \epsilon 2^{-i-1}$. $G:=\cup_i (a_i,b_i')$ is an open set that contains $A$ and $m(G)\leq \sum_i (b_i'-a_i) =\epsilon/2 + \sum_i (b_i-a_i) \leq m(A) + \epsilon $.
Since $m(A)<\infty$, $m(G \setminus A)=m(G) - m(A)\leq \epsilon$.
Since $m(A)<\infty$, $A$ is approcheable by a bounded set. Indeed, $m(A) = m(\cup_n (A\cap [-n,n])) = \lim_n m(A \cap [-n,n])$. There is therefore some $N$ such that $$m(A \cap [-N,N])\geq m(A)-\epsilon/2 \quad (\star)$$
Let $A' = A \cap [-N,N]$. Since $[-N,N]\setminus A'$ has finite measure, there exists some open $G'$ such that $[-N,N]\setminus A' \subset G'$ and $m(G'\setminus ( [-N,N]\setminus A'))\leq \epsilon/2$. Let us prove that the closed set $[-N,N]\setminus G'$ fits the bill.
It's easy to prove $[-N,N]\setminus G'\subset A'$. Furthermore, $$\begin{aligned} m(A'\setminus ([-N,N]\setminus G')) &= m(A'\cap ([-N,N]^c \cup G'))\\ &= m(A'\cap G') \end{aligned}$$ and $$\begin{aligned} \epsilon/2 \geq m(G'\setminus ( [-N,N]\setminus A')) &= m((G'\cap [-N,N]^c) \cup (G'\cap A'))\\ &\geq m(G'\cap A')\end{aligned}$$ Hence $m(A'\setminus ([-N,N]\setminus G')) \leq \epsilon/2$. Let $F = [-N,N]\setminus G'$, then $$m(A') - m(F) \leq \epsilon/2 \quad (\star \star)$$
$(\star)$ and $(\star \star)$ yield $$m(A\setminus F) = m(A) - m(F) \leq (m(A') - m(F)) + \epsilon /2 \leq \epsilon $$
Finally, $F\subset A \subset G$ and $m(G\setminus F) = m(G\setminus A) + m(A\setminus F) \leq 2\epsilon$