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Consider $[0,1]^2$ with Lebesgue measure $\mu$. Let $D\subseteq [0,1]^2$ be measurable with $0<\mu(D)<1$. Can you find $A,B\subseteq[0,1]$ measurable with $\mu(A)\mu(B)>0$ and yet $\mu(D\cap (A\times B))=0$? (That is, I want to find a non-null rectangle which is essentially inside the complement of $D$).

Some thoughts:

  • By regularity, we may suppose that $D$ is open, and that $A$ and $B$ are closed
  • If $D$ is not dense, then the answer is easily "yes". So we may suppose that $D$ is open and dense.
  • You can then write $D$ as a countable union of open rectangles with rational coordinates. Treating each one at a time, we can delete null sets from A or B. Repeating countably many times shows that if $A$ and $B$ exist, then actually, we can choose them with $D \cap (A\times B) = \emptyset$.

I somewhat suspect that this is false, but I cannot find a counter-example (I can't get the usual trick of, say, covering the rationals by small open balls to work).

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2 Answers 2

up vote 5 down vote accepted

The existence of a counterexample is implied by Theorem 2.1 of Kendall and Montana's paper "Small sets and Markov transition densities". Below is an excerpt from pages 180-182, including the beginning of the section for context.

This question is answered with a different approach in Halmos's Problems for mathematicians, young and old. Problem, Hint, and Solution 14.E are given below the Kendall and Montana excerpt.

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And here's Halmos:

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Brilliant! And if I could +1 twice for including scans of the article, I would. That is indeed more complicated than I expected... –  Matthew Daws Jun 2 '11 at 19:05
    
Btw, did you just happen to know this paper? Or did you manage to search for it? If the latter, any hints on how you found it?? –  Matthew Daws Jun 2 '11 at 19:11
    
@Matthew: I did not know about it in advance. I don't remember the exact Google search, but it included "measurable rectangle", and probably null. This also led me to Halmos. –  Jonas Meyer Jun 2 '11 at 19:15

I don't see the need to work so hard. Let $S$ be the unit square $[0, 1]^2$, $D = \{(x, y) \in S : x + y \notin \mathbb{Q}\}$. Then both $D$ and $S \backslash D$ are positive rectangle free (for $D$, use Steinhaus theorem). Now divide $S$ into four equal squares and choose $E \subset S$ which is the union of $D$ relativized to 1st and 3rd square and $S \backslash D$ relativized to 2nd and 4th square. Then, $E$ has area $1/2$ and both $E$ and $S \backslash E$ are positive rectangle free.

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