# Regarding a positive Lebesgue measure set in $\mathbb{R}^2$

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

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?

• I think the good example might be work is the A=B= Empty set Sep 30 '20 at 11:54
• 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. Sep 30 '20 at 16:35
• 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. Oct 1 '20 at 3:11
• Thanks. It is working because of the fact that adding a measure zero set to $F$ wouldn't make it interval. Am I right? Oct 1 '20 at 3:22
• Can you give some simple reason or hints why your above defined set is positive measure @fedja? Oct 1 '20 at 4:51

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.