A measurable set will have a sequence converges to its characteristic function Let $E$ be Lebesgue measurable with finite Lebesgue measure. Then there exsits a sequence of continuous functions $\{f_n\}$ such that $f_n : \mathbb{R}^n \rightarrow [0,1]$ converges to $1_E$ almost everywhere. (That is, there exists a set $N$ such that $m(N) = 0$ and $\lim f_n(x) = 1_E(x)$ for $x \in E\setminus N.$)
For me, I feel like the statement is, more or less, like the converse of the Egorov's thm. Also, the sequence $f_n$ is continuous and has range in $[0,1]$, pretty much remind me of the Urysohn funtion :
If $F$ is a closed set and $G$ is an open set such that $F\subseteq G$, then the function $f: \mathbb{R}^n \rightarrow [0,1]$ defined by $$f(x) = \frac{d(x, G^c)}{d(x, G^c)+d(x,F)}$$ is continuous and $f(x) = 1$ on $F$ and $f(x) = 0$ on $G^c$. The range of the urysohn function is similar to $1_A$ which take values $1$ on $A$ and $0$ outside $A$. 
So I try to construct a sequence of urysohn functions, but first need to come up with a set $F$ and $G$. I also know that if $E$ is measurable then there exsit a null set $N$ and a Borel set $B$ such that $B$ and $N$ are disjoint and $A = B \cup N$. I think I might choose this set $N$ as a null set in the theorem and work with $B$. However, even the collection of Borel sets is generated by all open sets, I dont think its element does need to open or closed. So I think I cant take $B$ to construct Urysohn function directly.
Not sure how to move on from this point. Any suggestion or helps please ?

Use the regularity of the Lebesgue measure: there are opens $U_n$ and compact (hence closed) $K_n$ such that $K_n\subseteq E\subseteq U_n$ with $\lambda (U_n-K_n)=1/n.$ Now, apply Urysohn, or just use your suggestion, $f_n(x) = \frac{d(x, U_n^c)}{d(x, U_n^c)+d(x,K_n)}.$ Then, $0\le f_n(x)\le 1$ and $f_n(x)=1$ if $x\in K_n$ and $0$ if $x\in U^c_n.$ Then it is easy to show that $f_n\to \chi_E$ in the $L^1$ norm, and so by a familiar result, there is a subsequence $f_{n_k}$ that converges pointwise a.e to $\chi_E$. 
