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.
2026-04-13 04:26:13.1776054373
If E in R is measurable and 0<u<m(E) exists a subset of E that has measure u
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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.