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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 Rekalo Jul 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.

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This is a standard homework question. But since you have 310 rep: the Lebesgue density theorem. –  Nate Eldredge Jul 2 '13 at 4:48
5  
@NateEldredge I guess no one gave me proper homework, then... –  Yemon Choi Jul 2 '13 at 6:20

1 Answer 1

up vote 2 down vote accepted

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.

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