# Show that there is a measurable subset A such that $\mu(A) = \frac{1}{2}\mu(E)$

Problem Statement: If E is a Lebesgue measurable subset of $[0,1]$ and $\mu$ Lebesgue measure. Then there is a measurable subset $A\subset E$ such that $\mu(A) = \frac{1}{2} \mu(E)$.

I have a proof that there is a measurable subset of A of $\mathbb{R}$ such that for each open interval $(a,b)$ we have that $\mu(A \cap (a,b)) = \frac{1}{2}a-b$. I'm not sure how to use this to help me though. Any tip on how to begin would be appreciated.

Hint: define $f(x)=\mu(E\cap[0,x])$. This function is continuous (why?) and has $f(0)=0$ and $f(1)=\mu(E)$. Then apply the intermediate value theorem.
• @BenRay I would argue like this. Let $\varepsilon>0$ and consider $x_0\in [0,1]$. Let $\delta=\varepsilon$. Then if $|x-x_0|<\delta$ (WLOG let $x<x_0$) then $|f(x)-f(x_0)|=|\mu(E\cap[0,x])-\mu(E\cap[0,x_0])|=\mu(E\cap[x,x_0])\leq \mu([x,x_0])<\delta=\varepsilon$.