# A dense scission of $\mathbb{R}$

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.

-
I am sure you were wondering. – user21436 Mar 3 '12 at 20:33
@Kann: One can wonder while wandering, and vice versa. – Did Mar 3 '12 at 20:35
Well, yes I agree! +1. – user21436 Mar 3 '12 at 20:37
If you mean "is there any interval $I$ such that $m(A\cap I) = 1/2|I|$", that's a different question from "Is there a dense set such that $m(A\cap I)=1/2|I|$ for every interval $I$?". Merely changing "any" to "every" would resolve the ambiguity. – Michael Hardy Mar 3 '12 at 21:20
Well guys Lebesgue differentiation theorem will not let it be measurable, but you can still think about a non-measurable set. – checkmath Mar 3 '12 at 21:47

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.
Yeah, that also proof the problem for a non-mensurable set (replacing $m$ by the exterior measure). – checkmath Mar 4 '12 at 2:34