# Prove for any set $E\subset R$ with lebesgue measure 1 there exists a subset with lebesgue measure 1/2.

Prove for any set $E\subset R$ with Lebesgue measure 1 there exists a subset with Lebesgue measure 1/2. It looks easy but I have tried for an hour and could not find a way to prove it. Can anyone provide a hint? Like what theorem or technique to use? Thank you!

Hint: Show that the function $f:\mathbb{R}\to\mathbb{R}^+\cup\{0\}$ defined as $f(x) = m\left(E\cap[-x,x]\right)$ if $0\leq x$ and $f(x)=0$ if $x<0$ is continuous.
• Thank you! I got it. One more question, does this hold for not just $R$ but also $R^n$? It looks so. – Tony Jul 6 '15 at 13:55
As hj implied, this is true for any value in $[0,1]$. I remember this from one of the Rudin books. A similar approach to his/her post, for any value $\alpha$ in between (for Lebesgue measure): consider a compact subset $K$ with $m(K)> \alpha$. Then $K$ is contained in a interval $I=[c,d]$ , then use the function $M(x):m([c,x] \cap I )$, and the intermediate value theorem, that we can use, as hj said, by continuity.