Does there exist for arbitrary $\alpha$, $0<\alpha<1$, a measurable subset $A$ of the closed unit interval $[0,1]$ such that Lebesgue measure $m(A)=\alpha$ and the following "homogeneity" condition is satisfied: for any subinterval $[a,b]\subseteq[0,1]$ one has: $m(A\cap[a,b])=\alpha\cdot (ba)$?
closed as offtopic by Nate Eldredge, Bill Johnson, Andres Caicedo, Anthony Quas, Andrey Rekalo Jul 2 '13 at 7:00This question appears to be offtopic. The users who voted to close gave this specific reason:



It turned out that such sets do not exist, due to the Lebesgue density theorem, as prompted by Nate Eldridge. Applied to the unit interval, the Lebesgue density theorem states that If $A$ is a measurable subset of $[0,1]$, then $$ \lim_{\epsilon\to 0} \frac{m(A\cap (x\epsilon,x+\epsilon))}{2\epsilon} = 1 $$ for almost all $x\in A$. So if we take $[a,b]=[x\epsilon,x+\epsilon]$, the condition from the original question says that $m(A\cap[a,b])/m([a,b])=\alpha$ for all $a,b$, so $\alpha$ should be equal to 1. Intuitively, this means that measurable subsets of nonfull measure cannot be spread homogeneously onto the whole set, but must be 'lumpy' like generalized Cantor sets  have dense regions in some places and holes in the other. 

