If $E \subset \mathbb R$ is Lebesgue measurable, does there exist a closed set $F\subset E :\ m(E)=m(F)$?

Question

My question is the following.

If $E \subset \mathbb R$ is a Lebesgue measurable set, does there exist a closed set $F\subset E :\ m(E)=m(F)$?

$m$ stands for the Lebesgue measure of the real line.

I tried to prove this with the known approximation:

$$m(E)=\sup\{m(K):K\subset E,\ K\ \text{compact}\}$$

and I could only prove that there exists an $F \in F_{\sigma}:\ m(E)=m(F)$. The problem is that an $F_{\sigma}$ set is not necessarily a closed set.

Is the answer to my question affirmative? If it is, can you give me a hint for a proof?

Thanks in advance.

EDIT: By a comment below we can check that when we do not have any restriction for $F$, the answer is trivial. What if we want $F\subset E$?

Answer 1

A closed non-empty subset of the open unit interval $(0,1)$ would have to contain its own infimum, which would have to be $\epsilon > 0$. But then the measure of this closed subset is at most $1 - \epsilon < 1$.