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

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

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## closed as off-topic by Nate Eldredge, Bill Johnson, Andrés E. Caicedo, Anthony Quas, Andrey RekaloJul 2 '13 at 7:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Nate Eldredge, Bill Johnson, Andrés E. Caicedo, Anthony Quas, Andrey Rekalo
If this question can be reworded to fit the rules in the help center, please edit the question.

This is a standard homework question. But since you have 310 rep: the Lebesgue density theorem. – Nate Eldredge Jul 2 '13 at 4:48
@NateEldredge I guess no one gave me proper homework, then... – Yemon Choi Jul 2 '13 at 6:20

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.