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 $D$).
- 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).