Looking for a bounded set in a set with finite measure lebesgue. Let $A\subseteq\mathbb{R}^{n}$ with $\mu^{*}\left(A\right)<\infty$. Show that for each $\varepsilon >0$  there is $A_{\varepsilon}\subseteq\mathbb{R}^{n}$  bounded such that $A_{\varepsilon}\subseteq A$  and $\mu^{*}\left(A-A_{\varepsilon}\right)<\varepsilon$.
Remark: The meausure $\mu^{*}$ is defined by
$\mu^{*}\left(E\right)=\inf\left\{ \sum_{k=1}^{\infty}\mu\left(I_{k}\right)\::\; I_{k}\:\mbox{is a } n\mbox{-celd and }\bigcup_{k=1}^{\infty}I_{k}\supseteq E\right\} $ 
where
$\mu \left(I_{k}\right)=\prod_{1\leq i\leq n}\left(b_{i}^{(k)}-a_{i}^{(k)}\right).$
The numbers $b_{i}^{(k)}$ and $a_{i}^{(k)}$ are the ends of the $n$-celd $I_{k}$.
 A: I thought for several days on this problem and the following is the proof of this fact:
Let $p\in A$ fixed, we define the sets $A_n$ by:
$A_{n}=\left\{ x\in A\::\: n\leq d\left(p,x\right)\leq n+1\right\} $
Note that $\bigcup_{n\in\mathbb{N}}A_{n}=A$.

Let $\delta>0$ fixed,  there is $\left\{I_{k}\right\}_{k \in \mathbb{N}}$ a sequence of $n$-celds such that $\bigcup_{k=1}^{\infty}I_{k}\supseteq A$  and 
$\sum_{k=1}^{\infty}\mu\left(I_{k}\right)<\delta/2+\mu^{*}\left(A\right)$  
Furthermore, we can consider $\left\{I_{k}\right\}_{k \in \mathbb{N}}$ a sequence of open $n$-celds such that if $I_{k}\cap A_{n}\neq\phi$ y $I_{k}\cap A_{n+1}\neq\phi$ then $I_{k}\cap A_{j}=\phi$ for all $j \in \mathbb{N}\setminus\left\{ n,n+1\right\} $.

We consider the set:
$M_{n}=\left\{k \in \mathbb{N} \::\: I_{k}\cap A_{n} \neq \phi \right\}$
Then $A_n  \subseteq \bigcup_{k\in M_{n}}I_k$, hence  $\mu^{*}\left(A_n\right)  \leq \mu^{*}\left(\bigcup_{k\in M_{n}}I_k\right )\leq \sum_{k\in M_{n}}\mu\left(I_{k}\right)$. So, we have that 
$\sum_{n=0}^{\infty}\mu^{*}\left(A_{n}\right)\leq2\sum_{k=1}^{\infty}\mu\left(I_{k}\right)<\delta+2\mu^{*}\left(A\right)$
Then $\sum_{n=0}^{\infty}\mu^{*}\left(A_{n}\right)$ converge. So, for each $\varepsilon>0$ there is $N>0$ such that if $m>N$ then 
$\sum_{n=m+1}^{\infty}\mu^{*}\left(A_{n}\right)<\varepsilon$
Considering $A_{\varepsilon}=\bigcup_{n=0}^{m}A_{n}$,  of the above it follows that:
$\mu^{*}\left(A-A_{\varepsilon}\right)=\mu^{*}\left(\bigcup_{n=0}^{\infty}A_{n}-\bigcup_{n=0}^{m}A_{n}\right)\leq\mu^{*}\left(\bigcup_{n=m+1}^{\infty}A_{n}\right)\leq\sum_{n=m+1}^{\infty}\mu^{*}\left(A_{n}\right)<\varepsilon.$
