Find an interval such that the intersection has measure $\epsilon$ Suppose $E$ is measurable with finite lebesgue measure. Show that for each $0<\epsilon<m(E)$, there exists $x>0$ such that $m(E\cap (-x,x))=\epsilon$
I tried to use the $m(E)=\inf\{\sum_{k=1}^{\infty}l(I_k): \{I_k\}$ is a cover of open intervals of $E\}$, but this would only give me the estimation. I can rewrite $E$ as a disjoint union of a finite number of measurable sets each of which has measure at most $\epsilon$.
Now define $g(x)=m(E_x)$, $E_x=E\cap (-x,x)$, I know that $g$ is non-decreasing, so $g$ is differentiable almost everywhere. But when I calculate the derivative of $g$, I start from the definition, $$g'(x)=\lim_{h\to0} \frac{g(x+h)-g(x)}{h}$$
I have no idea how to deal with the parts $E\cap(-x-h,-x)$ and $E\cap(x,x+h)$ 

Hint If $0<x <y$ then $g(x) \leq g(y)$ and
$$g(y)-g(x) \leq 2(y-x)$$
Use this to show that $g$ is continuous.
