Existence of closed $F\subset E$ such that $m(F)=\frac{1}{3}m(E)$ and $m(E)\in(0,\infty),E\subset \mathbb{R}$

Question

I am stuck in this quite difficult for me exercise: If $E\subset \mathbb{R}$, $E$ measurable and $0<m(E)<\infty$, then there exists a closed set $F\subset E$ such that $m(F)=\frac{1}{3}m(E)$. I thought to take the function $f(x)=m(E\cap (-\infty,x])$. With this function I can prove that there exists a measurable set $F\subset E$ such that $m(F)=\frac{1}{3}m(E)$, but I can't find a closed set with the same property. I will be glad for any help.

Answer 1

Here is an outline: (I take it that your are talking about the Lebesgue measure). You need to know that the Lebesgue measure is inner regular, i.e. $$\mu(E) = \sup\{\mu(F): F\subset E \text{ is closed }\}$$ In case of finite measure you can take the $F$ to be compact.

Now choose such a set such that $\mu(F) > \mu(E)/3$ and repeat your reasoning to find a subset of $F$ which has the desired value for its measure. Note that $(-\infty, x]\cap F$ is closed and that $x\mapsto \mu((-\infty, x]\cap F)$ is continuous (why is this true and how does that help?).