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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?

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  • $\begingroup$ I think the good example might be work is the A=B= Empty set $\endgroup$ Sep 30 '20 at 11:54
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    $\begingroup$ Isn't it essentially the same example? Just take the set $\{x,y:x-y\in F\}$ where $F$ is a closed set of positive measure without inner points. $\endgroup$
    – fedja
    Sep 30 '20 at 16:35
  • $\begingroup$ Did you mean "interior set empty" by "without inner points"? If that is so then you are saying to take $F$ to be fat Cantor set. Let me check that and I will come here again. $\endgroup$
    – Duplicate
    Oct 1 '20 at 3:11
  • $\begingroup$ Thanks. It is working because of the fact that adding a measure zero set to $F$ wouldn't make it interval. Am I right? $\endgroup$
    – Duplicate
    Oct 1 '20 at 3:22
  • $\begingroup$ Can you give some simple reason or hints why your above defined set is positive measure @fedja? $\endgroup$
    – Duplicate
    Oct 1 '20 at 4:51
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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.

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.

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    $\begingroup$ By "has the required property", you mean that this set is a counter-example to the property stated in the question ? $\endgroup$ Sep 30 '20 at 19:22
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    $\begingroup$ Yes. I added a sketch of the proof. $\endgroup$ Oct 1 '20 at 5:25

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