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!
On one property of the Lebesgue Measure
1
$\begingroup$
real-analysis
measure-theory
-
1Is $I$ an interval or an arbitrary measurable subset of $[0,1]$? – 2012-12-10
-
1Duplicate of [A Lebesgue measure question](http://math.stackexchange.com/questions/89424/a-lebesgue-measure-question) - the question is different, but the answers address this question (after taking complements). – 2012-12-10
-
0@Jacob. $I$ is an arbitrary measurable subset of $[0,1]$. – 2012-12-10
-
0@Nate. I am not sure I see how the answers address this question. What do you mean by taking complements? – 2012-12-10
-
0@Nate: I too fail to see how the linked question contains an answer for the present one. – 2012-12-10
-
0@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. – 2012-12-10
1 Answers
1
Such set cannot exist.
Since $m(E)<1$, $m(E^c)>0$. Then $m(E\cap E^c)=0