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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1Is $I$ an interval or an arbitrary measurable subset of $[0,1]$? – JSchlather Dec 10 '12 at 03:48
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1Duplicate of A Lebesgue measure question - the question is different, but the answers address this question (after taking complements). – Nate Eldredge Dec 10 '12 at 03:57
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@Jacob. $I$ is an arbitrary measurable subset of $[0,1]$. – stephen Dec 10 '12 at 05:40
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@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 05:43
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@Nate: I too fail to see how the linked question contains an answer for the present one. – Martin Argerami Dec 10 '12 at 06:04
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@MartinArgerami: The "general fact" in xen's answer, applied to $E^c$ instead of $E$, and with, say, $c=3/4$, shows that there is an interval $I$ with $m(E^c \cap I) \ge \frac{3}{4} m(I)$, which means $m(E \cap I) \le \frac{1}{4} m(I)$. Of course if $I$ is allowed to be any measurable set, this becomes trivial as in Martin Argerami's answer below. – Nate Eldredge Dec 10 '12 at 13:57
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Such set cannot exist.
Since $m(E)<1$, $m(E^c)>0$. Then $m(E\cap E^c)=0<m(E^c)/2$.
Martin Argerami
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