Inner regularity of Lebesgue measurable sets 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?
 A: The standard method to deal with the unbounded case is to do something clever involving epsilon and powers of $1/2$. In your case I would suggest letting $A_i=B(0,i)\setminus B(0,i-1)$. Then set $E_n=A_n \cap E$ and find an appropriate $F_n$ such that $m^\ast(E_n \setminus  F_n)\leq \varepsilon/2^n$. Then taking unions should give you what you want. 
Edit: There's a little more subtlety here than I originally noticed. There's two cases $m(E)$ is finite and $m(E)$ is infinite. If $m(E)$ is finite then notice that
$$m(E)=\sum_{n=1}^\infty m(E_n)$$
so we can find some $N$ such that
$$\sum_{n=N}^\infty m(E_n) < \varepsilon/2.$$
Then we can handle $\bigcup_{i=1}^N E_n$ using the bounded case. In the case that $m(E)$ is finite we pick our $F_n$ appropriately and let $F$ be their union. Now $F$ is not compact, but we do know that
$$m(E \setminus F) < \varepsilon.$$
Furthermore each finite union of the $F_n$ is compact and 
$$\lim_{n\rightarrow \infty} \bigcup_{i=1}^n F_i=F$$
in particular $m(\bigcup_{i=1}^n F_i)$ is unbounded so our desired sup is infinte. 
A: 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.
