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Is it possible for there to be a measurable set $E\subseteq\mathbb{R}$ such that for every open interval $I$, $0<m(I\cap E)<m(I)$ (where $m$ is the Lebesgue measure of a set)?

The first thing I notice is that this set must be dense, since if it were not dense we would be able to find an interval with no members of $E$. It of course also must be uncountable in size, since it cannot have zero measure itself. It is also pretty clear that $\mathbb{R}\setminus E$ must also have this property if $E$ has this property.

Aside from that I'm stumped. I've tried a couple of toy sets but they all turned out to not have this property. Then I tried to show no set could have this property but I hit a brick wall pretty quickly.

Sriotchilism O'Zaic
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  • https://math.stackexchange.com/questions/57317/construction-of-a-borel-set-with-positive-but-not-full-measure-in-each-interval/57326#57326 – ibnAbu Sep 27 '18 at 21:38

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EDITED:

Hint: it suffices to prove for countably many open intervals with rational endpoints. Construct your set as the union of a sequence of "fat Cantor sets".

Robert Israel
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