3 clarify the notation

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 S(x)+\lambda S^{-1}(y))\,dx\,dy &=\int_S\lambda S(x)\,dx\,dy+\int_S\lambda S^{-1}(y)\,dx\,dy\\ &=\int(\lambda S(x))^2\,dx+\int(\lambda S^{-1}(y))^2\,dy\\ &\ge\left(\int\lambda S(x)\,dx\right)^2+\left(\int\lambda S^{-1}(y)\,dy\right)^2\\ &=2(\lambda S)^2>\lambda S=\int_S1\,dx\,dy, 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 $\lambda S(x)+\lambda S^{-1}(z)>1$. \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$. 2 simplify the argument Let $S(x)=\{y\mid(x,y)\in S\}$ and $S^{-1}(y)=\{x\mid(x,y)\in S\}. We have \begin{align*} \int_S(\lambda S(x)&+\lambda S(y))\,dx\,dy+\int_S(\lambda S^{-1}(x)+\lambda S^{-1}(y))\,dx\,dy\\ &=\int_S(\lambda S(x)+\lambda S^{-1}(x))\,dx\,dy+\int_S(\lambda S(y)+\lambda S^{-1}(y))\,dx\,dy\\ &=\int\lambda S(x)(\lambda S(x)+\lambda S^{-1}(x))\,dx+\int\lambda S^{-1}(y)(\lambda S(y)+\lambda S^{-1}(y))\,dy\S^{-1}(y))\,dx\,dy &=\int_S\lambda S(x)\,dx\,dy+\int_S\lambda S^{-1}(y)\,dx\,dy\\ &=\int(\lambda S(x)+\lambda S^{-1}(x))^2\,dx\S(x))^2\,dx+\int(\lambda S^{-1}(y))^2\,dy\\ &\ge\left(\int(\lambda S(x)+\lambda S^{-1}(x))\,dx\right)^2=(2\lambda &\ge\left(\int\lambda S(x)\,dx\right)^2+\left(\int\lambda S^{-1}(y)\,dy\right)^2\\ &=2(\lambda S)^2>\lambda S=\int_S1\,dx\,dy, \end{align*} hence $$\int_S(\lambda S(x)+\lambda S(y))\,dx\,dy\ge2(\lambda S)^2=\int_S2\lambda S\,dx\,dy$$ or $$\int_S(\lambda S^{-1}(x)+\lambda S^{-1}(y))\,dx\,dy\ge2(\lambda S)^2=\int_S2\lambda S\,dx\,dy.$$ In the former case, there exists(x,y)\in (x,z)\in S$such that$\lambda S(x)+\lambda S(y)\ge2\lambda S>1$, hence S^{-1}(z)>1$. This implies $S(x)\cap S(y)\ne\varnothing$S^{-1}(z)\ne\varnothing$, i.e., there exists$z$y$ such that $(x,z),(y,z)\in (x,y),(y,z)\in S$. The other case is symmetric.

1

Let $S(x)=\{y\mid(x,y)\in S\}$ and $S^{-1}(y)=\{x\mid(x,y)\in S\}$. We have

\begin{align*} \int_S(\lambda S(x)&+\lambda S(y))\,dx\,dy+\int_S(\lambda S^{-1}(x)+\lambda S^{-1}(y))\,dx\,dy\\ &=\int_S(\lambda S(x)+\lambda S^{-1}(x))\,dx\,dy+\int_S(\lambda S(y)+\lambda S^{-1}(y))\,dx\,dy\\ &=\int\lambda S(x)(\lambda S(x)+\lambda S^{-1}(x))\,dx+\int\lambda S^{-1}(y)(\lambda S(y)+\lambda S^{-1}(y))\,dy\\ &=\int(\lambda S(x)+\lambda S^{-1}(x))^2\,dx\\ &\ge\left(\int(\lambda S(x)+\lambda S^{-1}(x))\,dx\right)^2=(2\lambda S)^2, \end{align*}

hence

$$\int_S(\lambda S(x)+\lambda S(y))\,dx\,dy\ge2(\lambda S)^2=\int_S2\lambda S\,dx\,dy$$

or

$$\int_S(\lambda S^{-1}(x)+\lambda S^{-1}(y))\,dx\,dy\ge2(\lambda S)^2=\int_S2\lambda S\,dx\,dy.$$

In the former case, there exists $(x,y)\in S$ such that $\lambda S(x)+\lambda S(y)\ge2\lambda S>1$, hence $S(x)\cap S(y)\ne\varnothing$, i.e., there exists $z$ such that $(x,z),(y,z)\in S$. The other case is symmetric.