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Does there exist a set $E\subset [0,1]$ with $m(E)<1$ such that $m(E\cap I)\geq m(I)/2$ for all measurable sets $I\subset [0,1]$? I am not able to construct one, but it seems possible. Any help would be appreciated! Thanks!

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Is $I$ an interval or an arbitrary measurable subset of $[0,1]$? – JSchlather Dec 10 '12 at 3:48
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Duplicate of A Lebesgue measure question - the question is different, but the answers address this question (after taking complements). – Nate Eldredge Dec 10 '12 at 3:57
@Jacob. $I$ is an arbitrary measurable subset of $[0,1]$. – stephen Dec 10 '12 at 5:40
@Nate. I am not sure I see how the answers address this question. What do you mean by taking complements? – stephen Dec 10 '12 at 5:43
@Nate: I too fail to see how the linked question contains an answer for the present one. – Martin Argerami Dec 10 '12 at 6:04
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Such set cannot exist.

Since $m(E)<1$, $m(E^c)>0$. Then $m(E\cap E^c)=0<m(E^c)/2$.

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The idea is to take a fat cantor set, and fill in the blank area by other smaller fat cantor sets. Iterating this process, you should be able to produce $E$. You then approximate the measurable set by a Borel set, and the intersection of the Borel set with $E$ should have measure greater than the prescribed measure. The post others cited give a different proof.

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I think you might be answering a different question from the one being asked? – Nate Eldredge Dec 10 '12 at 13:58
I don't think so. The complement of $E$ is nowhere dense, and $E\cap I$'s measure relative to $I$'s measure can be bounded somehow by the fat cantor sets. – user32240 Dec 11 '12 at 3:04
What exactly do you mean by "bounded somehow"? And how do you reconcile this with the disproofs given in the other answers and comments? – Nate Eldredge Dec 11 '12 at 4:27
I mean the $E$ should have a complement which is nowhere dense. I see strictly speaking this "example" is wrong by above disproof. But I think this example is what the problem wants. – user32240 Dec 11 '12 at 23:44

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