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Let $E$ be a set of positive Lebesgue measure on the real line. Let $1>\epsilon>0$ be given. Show that there exists an interval $I$ such that $m(E\cap I)>\epsilon m(I)$ where $m$ is the Lebesgue measure on the real line.

I tried answering by contradiction method without success. Then tried writing the inequality as $m(I\backslash E)<(1-\epsilon)m(I)$ which also didn't help much either.

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Are you sure that what you're trying to prove is true? What about $E$ being the union of two disjoint intervals and $\epsilon$ close to one? –  Eckhard Dec 28 '12 at 17:15
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See this post. –  David Mitra Dec 28 '12 at 17:27
    
This is better than I solution. –  Bombyx mori Dec 28 '12 at 17:39
    
I've removed qualifying-exam tag from the questions where it was used; 1, 2, 3. Meta-tags and dependent tags are generally discouraged. –  Martin Sleziak Dec 30 '12 at 7:14
    
If you think the tag would be useful, feel free to ask about opinion of other users at meta or to join the discussion in tagging chatroom. –  Martin Sleziak Dec 30 '12 at 7:17

1 Answer 1

If $E$ is a set of positive Lebesgue measure, then there exists a Borel set $D$ of type $F_{\delta}$ such that $\mu(E\Delta D)=0$. So it suffice to prove this for $F_{\sigma}$ type sets. And it is clear that we can reduce this to a counterable intersection of open sets such that one contains the other.

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apply the lebesgue theorem to the $X_E$( characteristic of E ) –  user52188 Dec 28 '12 at 17:21
    
@user32240: Should that $G_\sigma$ not be $F_\sigma$? –  Host-website-on-iPage Dec 28 '12 at 17:33
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You can find at Real and Complex Analysis, walter rudim, pg 141 –  user52188 Dec 28 '12 at 17:40

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