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Is there a subset $A\subset\mathbb R$ such that for any interval $I$ of length $a$ the set $A\cap I$ has lebesgue measure $a/2$?

Can it be constructed explicitly?

Klaas
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  • Relevant: https://terrytao.wordpress.com/2008/10/14/non-measurable-sets-via-non-standard-analysis/ – SBK Jul 14 '22 at 14:25

1 Answers1

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You can use Lebesgue's density theorem. If such a set exists then for every point of $A$ the Lebesgue density would be $\frac{1}{2}$. Since, by Lebesgue's theorem the Lebesgue density must be $0$ or $1$ almost everywhere this is a contradiction.

  • Would it be possible to have a type of measure for which this would hold? – Sambo Jul 20 '17 at 17:36
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    This is misleading because the question doesn't say measurable. Such sets do exist afaik so long as you say "outer measure" at the end, so clarification from either you or the questioner would be needed. – SBK Jul 14 '22 at 14:21
  • @T_M: Pardon my ignorance: Does it make sense to speak of the Lebesgue measure of $A \cap I$ if the set is not measurable? – Martin R Jul 14 '22 at 14:25
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    It's a little ambiguous in my opinion and sort of depends on the flavour of measure theory you are used to and how you learned it etc. In some treatments of measure theory, an author will basically just call outer measures 'measures' and then apply them to any set they like. (Then when a set is measurable, you can use the Caratheodory criterion and when it's not, you can't). And since it actually affects the answer to the question, it probably needs to be made clear. – SBK Jul 15 '22 at 09:54
  • A related paper: https://web.archive.org/web/20170811020516id_/http://www.ams.org/journals/proc/1960-011-03/S0002-9939-1960-0141755-8/S0002-9939-1960-0141755-8.pdf Other references are in the comments of this question: https://math.stackexchange.com/questions/1367873/construction-of-a-set-with-density-of-half-at-0 – Pavel Kocourek Jan 06 '23 at 06:43