Does every positive Lebesgue measure set in $\mathbb{R}^2$ contain a product of positive Lebesgue measure sets up to a null set?

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

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?

• Commented Sep 30, 2020 at 7:27
• How about the product of fat Cantor sets rotated by $45$ degrees?
– user140541
Commented Sep 30, 2020 at 8:15

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.

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