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If $A$ is a subset of $\mathbb{R}$ with Lebesgue measure strictly greater than $0$, does it follow then that are there $a$ and $b$ such that the measure of $[a,b]\cap A$ is $b-a$?

Thank you.

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If this fails, you need a measurable set that contains no intervals. – Ross Millikan Dec 5 '10 at 19:32
No. Hint: fat Cantor sets. – Asaf Karagila Dec 5 '10 at 19:38
@Ben : I rettaged your question since this is not really a problem in set theory but rather in measure theory (though one may learn these topics as part of a classical analysis course). – Andrés E. Caicedo Dec 5 '10 at 20:36
@Ben: Please don't use the title as an integral part of the message; make the body of the message self-contained as a service to readers. – Arturo Magidin Dec 5 '10 at 22:04
@Asaf Karagila: Post as an answer? – Nate Eldredge Dec 5 '10 at 22:35
up vote 4 down vote accepted

For the sake of completeness, the answer is no; a counterexample is given by the Smith-Volterra-Cantor set, or fat Cantor set.

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What is actually true is this: for every set $C$ of positive measure and every $\epsilon < 1$ there is some open interval $(a,b)$ such that $\mu(C \cap (a,b)) \geq \epsilon |b-a|$.

I have always viewed this as an instance of one of Littlewood's three principles for analysis: a measurable set is almost an open set.

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Any hint as to how one could see this? – user9352 Sep 27 '11 at 12:14
It's called the Lebesgue density theorem and is often proved in textbooks.'s_density_theorem is a little cryptic because it covers the theorem for $\mathbb{R}^n$. – Carl Mummert Sep 27 '11 at 12:27

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