Take the tour ×  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:
$m(A\cap[a,b])=\alpha\cdot (b-a)$?  share|improve this question
This is a standard homework question. But since you have 310
rep: the Lebesgue density theorem. –  Nate Eldredge Jul 2 at
4:48 @NateEldredge I guess no one gave me proper homework,
then... –  Captain Oates Jul 2 at 6:20 add comment  closed as
off-topic by Nate Eldredge, Bill Johnson, Andres Caicedo,
Anthony Quas, Andrey Rekalo Jul 2 at 7:00     • "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  1 Answer
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