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I made up this question before, but hardly gave it any thought until now.

Is there a measurable set $A\subset[0,1]$ such that $\displaystyle\int_0^r 1_A(x)\;\text{d}x=\dfrac{1}{2}r$?

Loosely, this means that half of the points of any open neighbourhood of $[0,1]$ belong to $A$.

Cheers!

Chris Sanders
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    Looks like a duplicate of https://math.stackexchange.com/q/1163417/42969 to me. – Martin R Jul 14 '22 at 14:19
  • thanks @MartinR – Chris Sanders Jul 14 '22 at 14:19
  • I've just hastily added a comment on the linked question because the answer there isn't actually quite right. It's not a direct duplicate of your question but you may find that Lebesgue density does answer your question (because you ask for a measurable set) – SBK Jul 14 '22 at 14:22
  • I think @T_M that "$A\cap I$ has Lebesgue measure" means that $A$ is supposed to be measurable in that question. If you have insight on the equivalent question for outer measure I'm happy to edit my question and ask for re-opening, unless you want a different arrangement? – Chris Sanders Jul 14 '22 at 14:25
  • Well for the purposes of your question above, nothing needs to be changed because you ask for a measurable set. One can show that no such set exists. But if you ask for any set and just allow outer measure, then such sets can be constructed (using the axiom of choice e.g. a non-principal ultrafilter iirc). And as I added over on the question, sometimes in analysis an author will just call every outer measure a 'measure' and then separately make a point of stating whether or not sets are measurable. So leaving it kind of 'implicit' in the phrasing of "$A \cap I$ is measurable" is a bit unclear – SBK Jul 15 '22 at 09:58

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