$E\subset\mathbb{R}^n$ is measurable $\Leftrightarrow$ $\exists$ a sequence of measurable subsets $\{E_k\}_{k\geq1}$ s.t. $m(E_k)\to m^*(E)$. Let $E\subset\mathbb{R}^n$ and $m^*(E)<\infty$, prove that $E$ is measurable if and only if there exists a sequence of measurable subsets $\{E_k\}_{k\geq1}$ s.t. $m(E_k)\to m^*(E)$. (Here we are talking about the Lebesgue measure.)
For one direction($\Rightarrow$): $\forall k\geq1$, there exists a closed set $F_k\subset E$ such that $m(E-F_k)\le\frac{1}{k}$. Taking $\{F_k\}_{k\geq1}$, the proof is completed.
For the other direction($\Leftarrow$), I can't prove it. Can you give me some tips? And is my proof for the first direction correct? 

As stated, the converse cannot be true. Given the number $m^*(E)$, you can always find a measurable set $E'$ with $m(E')=m^*(E)$; and this can be done for any non-measurable bounded set. 
If you add the condition that $E_k\subset E$ for all $k$, let $F_k=\bigcup_{j=1}^k E_j$. Then we have an increasing sequence of measurable sets with $F_k\subset E$, so $m(F_k)\leq m^*(E)$. Being increasing and bounded, the sequence $\{m(F_k)\}$ is convergent, say $m(F_k)\to c$. By continuity of the measure you have $m(\bigcup_k F_k)=c$. Then
$$
m^*(E)=\lim_km(E_k)\leq\lim_km(F_k)=c\leq m^*(E). 
$$
So $F=\bigcup_kF_k$ is a measurable set with $F\subset E$ and $m(F)=m^*(E)$. Now, since $F$ is measurable, 
$$
m^*(E)=m^*(E\cap F)+m^*(E\cap F^c)=m(F)+m^*(E\setminus F). 
$$
So $m^*(E\setminus F)=0$, which makes $E\setminus F$ measurable. Then $E=F\cup(E\setminus F)$ is a union of measurable sets, so measurable. 
