Q:

homogeneous subset of [0,1] of arbitrarily small Lebesgue measure

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: $m(A\cap[a,b])=\alpha\cdot (b-a)$? 

A:

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 non-full 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.

