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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!

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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.

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  • $\begingroup$ Thank you! I got it. One more question, does this hold for not just $R$ but also $R^n$? It looks so. $\endgroup$ – Tony Jul 6 '15 at 13:55
  • $\begingroup$ I don't see why it shouldn't. $\endgroup$ – hjhjhj57 Jul 6 '15 at 15:48
  • $\begingroup$ How do you get the intuition for such a function? $\endgroup$ – Bijayan Ray May 29 '19 at 5:31
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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.

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