Approximation of finite Lebesgue measurable set by closed subset

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 ?

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 $m^*(A-O^c)=m^*(O-A^c)<\epsilon$

• This is true, but it doesn't quite fit my needs. – Gabriel Romon Nov 9 '16 at 9:24
• Why does not it fit your needs@LeGrandDODOM – Learnmore Nov 9 '16 at 13:07
• At the time of writing, I didn't know how to prove existence of open $O$ such that $A\subset O$ and $m(O\setminus A)\leq \epsilon$ when A is not assumed to have finite measure. Now I know how to prove it and your answer makes perfect sense. – Gabriel Romon Nov 9 '16 at 13:29

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 $m(G\setminus F) = m(G\setminus A) + m(A\setminus F) \leq 2\epsilon$