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Consider Lebesgue measure $m$ on $[0, 1]$. Fix a countable sequence $a_i, 0 < a_i < 1$ such that $\sum_i a_i = 1$. Is there a sequence of disjoint measurable subsets of $[0, 1]$, $E_i$ whose measure in every open interval $I$ respectively is $a_i m(I)$?

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There is no measurable subset $E$ of $[0,1]$ such that $m(E\cap I)=m(I)/2$ for every open interval $I\subseteq [0,1]$.

Indeed, assume there is such $E$. Then $m(E)=1/2$, so there is an open set $U$, $E\subseteq U \subseteq [0,1]$ such that $m(U)=3/4$. But $U$ is a union of a sequence of pairwise disjoint open intervals $I_j$, $j=0,1,2,\dots$. Since $m(E\cap I_j)=m(I_j)/2$ for every $j$, it follows $m(E)=m(E\cap U)=m(U)/2=3/8\neq 1/2$, a contradiction.

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    $\begingroup$ And the statement in the answer is obvious if you know that almost every point of a measurable set is a density point. $\endgroup$ – user95282 Aug 8 '19 at 13:36

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