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Let $P\subset \mathbb{R}^2$ be a positive Lebesgue measure set. Then $P$ may not contain a subset of the form $A\times B$ where $A,B\subset \mathbb{R}$ are of positive Lebesgue measure.

For example consider $P=\{(x,y)\in [0,1]\times[0,1]:x-y\notin \mathbb{Q}\}.$

This example leads me to ask:

Given any $P\subset \mathbb{R}^2,$ a positive Lebesgue measure set, does there exists a measure zero set $U\subset \mathbb{R}^2$ such that $P\cup U$ contains a subset of the form $A\times B$ where $A,B\subset \mathbb{R}$ are of positive Lebesgue measure?

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No, in general this is not true. Denote by $\lambda^d$ Lebesgue measure in $\mathbb{R}^d$.

It is possible to construct a Borel set $A \in \mathcal{B}(\mathbb{R}^2)$ with strictly positive Lebesgue measure such that $\lambda^2(A^c \cap R)>0$ for any non-degenerate rectangle $R$, i.e. for any $R = S \times T$ where $S,T \in \mathcal{B}(\mathbb{R})$ have positive Lebesgue measure; the idea is to define

$$A:= \{(x,y) \in \mathbb{R}^2\:;\: x-y \in B\}$$

for $B \in \mathcal{B}(\mathbb{R})$ such that $\lambda^1(B \cap I)>0$ and $\lambda^1(B^c \cap I)>0$ for any interval $I \neq \emptyset$. See [1] for a proof that this set does the job. In particular, if $N \subseteq \mathbb{R}^2$ is any Lebesgue null set, then $A \cup N$ still does not contain any rectangle $R$ since

$$\lambda^2((A \cup N)^c \cap R) = \lambda^2(A^c \cap R)>0.$$

[1] Darst, R. and Goffman, C.: A Borel Set which Contains no Rectangles. The American Mathematical Monthly 77 (1970) 728–729.

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  • $\begingroup$ Is $\lambda^m$ Lebesgue measure in $\mathbb{R}^m?$ $\endgroup$ Commented Sep 30, 2020 at 13:47
  • $\begingroup$ @Mathemajician Yes, exactly. $\endgroup$
    – saz
    Commented Sep 30, 2020 at 14:26
  • $\begingroup$ Thanks for the reference. It was immensely helpful. $\endgroup$ Commented Oct 1, 2020 at 2:00

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