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 $a\_i m(I)$?



Existence of sequence of measurable sets with prescribed densities
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.


