I was wondering if there is a dense set in $\mathbb{R}$ measurable such that $m(A∩I)=1/2|I|$ for any interval this property also tell us that its complement is a set of the same kind.
A dense scission of $\mathbb{R}$
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measure-theory
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0@HenningMakholm$I$was thinking about exterior measure but this case is not interesting. From that we know that Lebesgue Theorem avoid some quite reasonable things. – 2012-03-03
1 Answers
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Here's another way to see no such set can exist. If it did, let A' = A \cap (0,1), so that m(A') = 1/2. Since A' is measurable and Lebesgue measure is outer regular, there is an open set $U$ with A' \subset U \subset (0,1) and $m(U) < 1$. (Alternatively, note that the measure of A' is equal to its outer measure.) But $U$ can be written as a countable disjoint union of open intervals $I_n$, so $\sum_n m(I_n) = m(U) < 1$. On the other hand, m(A') = \sum m(A' \cap I_n) = \sum m(A \cap I_n) = \frac{1}{2} \sum m(I_n) < \frac{1}{2}, which is absurd.
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0Yeah, that also proof the problem for a non-mensurable set (replacing $m$ by the exterior measure). – 2012-03-04