If E in R is measurable and 0

Question

If E in R is measurable and it holds that $0<u<m(E)$ i need to prove that exists a subset of E that is measurable and has measure u.

Answer 1

Consider $E_x=[-x,x]\cap E$. For some large $x=x_0$ the measure of this set is greater than $u$. Claim, $f(x)=m(E_x)$ is a continuous function with $f(0)=0$ and $f(x_0)>u$. Then, by intermediate value theorem, there exists $x$ such that $f(x)=u$ as needed. $f$ is continuous as it is Lipschitz with constant 2.

If $m(E)>0$ then for some interval $[n,n+1)$, $n$ an integer, $m(E\cap[n,n+1))>0$. Then, $E+\mathbb{Q}$ contains countably many disjoint translated copies of that intersection and so has infinite measure.