4
$\begingroup$

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.

  • 0
    I am sure you were wondering.2012-03-03
  • 2
    @Kann: One can wonder while wandering, and vice versa.2012-03-03
  • 0
    Well, yes I agree! +1.2012-03-03
  • 0
    Of course, density of $A$ will follow automatically if $\mu(A\cap [a,b])=\frac{b-a}2$ for all $a (as will the same property for $\mathbb R\setminus A$).2012-03-03
  • 1
    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.2012-03-03
  • 0
    Well, $I$ is the interval, but fixed the interval the question is quite easy to be solved, the harder part is that I am thinking about any interval.2012-03-03
  • 3
    Well guys Lebesgue differentiation theorem will not let it be measurable, but you can still think about a non-measurable set.2012-03-03
  • 0
    @chessmath: for a non-measurable set your defining equation isn't even meaningful.2012-03-03
  • 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 1

4

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.

  • 0
    Yeah, that also proof the problem for a non-mensurable set (replacing $m$ by the exterior measure).2012-03-04