# Relation between $2$-dimensional and $1$-dimensional Lebesgue measure

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.

• 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 $S\times S$. May 15 at 16:00
• 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? May 22 at 14:51
• 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 $S$. May 22 at 16:56