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Let $Q\subset\mathbb{R}^n$ be a compact set (if you want say $Q=[0,1]^n$). My question is: If $S\subset Q\times Q\subset\mathbb{R}^{2d}$ is a measurable set with $|(Q\times Q)\setminus S|<\epsilon$, does there exist a set $T\subset Q$ such that $T\times T\subset S$ and $|Q\setminus T|<\delta(\epsilon)$, where $\delta(\epsilon)$ is a function satisfying $\delta(\epsilon)\to 0$ as $\epsilon\to 0$. Edit: Also assume $\{(x,x):x\in Q\}\subset S$, otherwise there is an easy counterexample as was pointed out in the comments.

My attempt so far: For simplicity let $Q=[0,1]$, so $Q^2=[0,1]^2$. Consider the function $f(x,y)=(y,x)$. Set $Z:=S\cap f(S)$. By assumption $|Q^2\setminus Z|=|Q^2\setminus(S\cap f(S))|\leq |Q^2\setminus S|+|Q^2\setminus f(S)|=2|Q^2\setminus S|<2\epsilon$. Then $Z$ is symmetric in the sense that $(x,y)\in Z$ implies $(y,x)\in Z$. However I don't know how to construct a suitable set $T$ from $Z$ (or if this is even possible). Maybe some kind of projection works, but I am unsure.

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    $\begingroup$ What about $S=(Q\times Q)\setminus \Delta(Q)$ where $\Delta(Q)=\{(x,x)\mid x\in Q\}$?This set has full measure and if $T$is non-empty, then $T\times T$ is never contained in $S\times S$. $\endgroup$
    – dialegou
    May 15 at 16:00
  • $\begingroup$ Thanks for your comment. Actually in my situation we have $\{(x,x):x\in Q\}\subset S$, I forgot to write that. Does that change anything? Are there other counterexamples in this case? $\endgroup$ May 22 at 14:51
  • $\begingroup$ Take any measure-preserving Borel isomorphism $\phi: Q\rightarrow Q$ with no fixed points, let $graph(\phi)=\{(x,\phi(x)\mid x\in Q\}$. Then take $S=(Q\times Q)\setminus graph(\phi)$. If the measure of $T$ is large enough, then $T\cap \phi(T)\neq\emptyset$ and thus $graph(\phi)\cap T\times T\neq \emptyset$. Thus $T\times T$ is not contained in $S$. $\endgroup$
    – dialegou
    May 22 at 16:56

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