I was asked (by myself) to give a proof of the following seemingly simple geometric statement, but after thinking a little I now suspect it could be less elementary than I thought (or am I being silly?). Does anybody know it, and can give an answer or a reference to it? Of course, I'm quite sure it should fit within a larger theory in combinatorics or in probability, but an elementary answer would be appreciated.



> 
> Let $S$ be a (say open) subset of a square
>  $[0,1]^2$ with Lebesgue measure
>  $|S|>1/2$. Then, there exists a rectangle with a vertex on the diagonal, and the other three vertices in $S$ (in other words, there are three points of $S$ of the form
>  $(x,y)$, $(x,z)$ and $(y,z)\\ $).
> 
> 
> 


The constant $1/2$ cannot be lowered, as the example of the subset $S^\* :=(0,1/2)\times(1/2,1)\cup(1/2,1)\times(0,1/2)$ shows (for any three points $x,y,z$ in $[0,1]$, at least 2 of them are both either smaller or larger than $1/2$, so the corresponding pair is not in $S^\*$.



A puzzle about finding three points $(x,y)$, $(x,z)$ and $(y,z)$ in a subset of a square.
Let $S(x)=\{y\mid(x,y)\in S\}$ and $S^{-1}(y)=\{x\mid(x,y)\in S\}$, and let $\lambda\_n$ denote the Lebesgue measure on $[0,1]^n$. We have


$$\begin{align\*}
\int\_S(\lambda\_1S(x)+\lambda\_1S^{-1}(y))\,dx\,dy
&=\int\_S\lambda\_1S(x)\,dx\,dy+\int\_S\lambda\_1S^{-1}(y)\,dx\,dy\\
&=\int(\lambda\_1S(x))^2\,dx+\int(\lambda\_1S^{-1}(y))^2\,dy\\
&\ge\left(\int\lambda\_1S(x)\,dx\right)^2+\left(\int\lambda\_1S^{-1}(y)\,dy\right)^2\\
&=2(\lambda\_2S)^2>\lambda\_2S=\int\_S1\,dx\,dy,
\end{align\*}$$


hence there exists $(x,z)\in S$ such that $\lambda\_1S(x)+\lambda\_1S^{-1}(z)>1$. This implies $S(x)\cap S^{-1}(z)\ne\varnothing$, i.e., there exists $y$ such that $(x,y),(y,z)\in S$.


