# 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, Andres 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, Andres 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.