Let $P\subset \mathbb{R}^2$ be a positive Lebesgue measure set. Then $P$ does not necessarily 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?



Regarding a positive Lebesgue measure set in $\mathbb{R}^2$
The question was answered by Robert Israel 1995 on Usenet, essentially by the set mentioned in fedja's comment. The proof that this set has the required property is carried out in detail in Example 4.3.1 of M. Väth, Ideal Spaces, [Springer 1997](https://www.springer.com/de/book/9783540631606).


Here is a sketch of the proof: Let $F\subseteq[-1,1]$ be closed with empty interior and positive measure and $P=\{(t,s)\in[0,1]\times[-1,2]:t-s\in F\}=\{(t,t+s):\text{$t\in[0,1]$, $s\in F$}\}$. Then $P$ has positive measure by the Cavalieri principle. Assume by contradiction that there are sets $A,B$ of positive measure such that $N=(A\times B)\setminus P$ is a null set. Consider $$x(s)=\int\_A\chi\_B(s+t)dt.$$ Then $x$ is continuous (translation is $L\_1$-continuous) and vanishes a.e. on the complement $C$ of $P$, because by Fubini-Tonelli $$\int\_Cx(s)ds=\text{mes}N=0.$$ Since $C$ is dense, it follows that $x$ is the null function. But again by Fubini-Tonelli $$\int\_{-\infty}^{\infty}x(s)ds=\text{mes}A\,\text{mes}B,$$ a contradiction.


Robert Israel originally formulated the proof in terms of a convolution (IIRC $\chi\_A\*\chi\_B$); the argument above is a variant of that.


