TITLE: Inner regularity of Lebesgue measurable sets
QUESTION [9 upvotes]: This is an exercise in real analysis:

Let $E\subset{\Bbb R}^d$ be Lebesgue measurable. Show that 
  $$
m(E)=\sup\{m(K):K\subset E, K \text{compact}\}.
$$

When $E$ is bounded, this can be done by the following proposition:  

$E\subset{\Bbb R}^d$ is Lebesgue measurable if and only if for every $\varepsilon>0$, one can find a closed set $F$ contained in $E$ with $m^*(E\setminus F)\leq\varepsilon$. 

How can I deal with the case that $E$ is unbounded?

REPLY [14 votes]: You've done the case when $E$ is bounded. When $E$ is unbounded, there are two cases:

$m(E)=+\infty$
$m(E)<+\infty$

Now we handle with the first case. Consider the closed ball in ${\Bbb R}^d$ $A_m=\{x\in{\Bbb R}^d: |x|\leq m\}$. Then $$
E=\cup_{m=1}^{\infty}E_m
$$
with $E_m=E\cap A_m$, and $E_1\subset E_2\subset\cdots\subset{\Bbb R}^d$. By monotone convergence theorem for measurable sets, 
$$
\lim_{n\to\infty}m(E_n)=m(E)=+\infty.
$$
Note that since $E_m$ is bounded, by the case you've done, for every $m$, we have a compact set $K_m\subset E_m\subset E$ such that $m(K_m)+1\geq m(E_m)\to\infty$. Hence 
$$\sup\{m(K):K\subset E,K\text{compact}\}=+\infty=m(E).$$
Now assume that $m(E)<+\infty$. For any $\varepsilon>0$, we can choose $N$ such that
$$
m(E)\leq m(E_N)+\varepsilon/2.
$$
We also have a compact set $K\subset E_N\subset E$ with 
$$
m(E_N)\leq m(K)+\varepsilon/2
$$
since $E_N$ is bounded. It follows that
$$
m(E)\leq m(K)+\varepsilon.
$$
We are done.