It is a well known fact that if $E$ is Lebesgue measurable, then $E$ can be written as a union of an $F_{\sigma}$ set $F$ and a measure zero set $N$: $E=F\cup N$, and the converse is also true.
I encountered a real analysis exercise that, if it were still true that $F$ is changed into merely a closed set:
If $E$ is measurable, then $E=F\cup N$ for some closed set $F$ and $|N|=0$.
I believe this is false, but cannot figure out any counterexample.
The answer is essentially due to John Doe in the comment session.
The center issue is to look for a measurable set $E$ such that it cannot be written as a union of a closed set $F$ and a measure zero set $N$.
Take $E=(0,1)$. If $F$ is a closed subset (closed relatively to $\mathbb{R}$) of $(0,1)$, then it is compact. Consider the covering $\{(1/n,1-1/n)\}_{n\geq 3}$ of $(0,1)$. Then $F\subseteq(1/n_{0},1-1/n_{0})$. But then
\begin{align*} 1=|(0,1)|=|F\cup N|=|F|\leq|(1/n_{0},1-1/n_{0})|=1-2/n_{0}<1, \end{align*} a contradiction.