A real $x\in2^\omega$ is random (or Solovay-random) over a model $M$ iff it's obtained via forcing by closed sets of reals, of positive measure, coded in $M$. Similarly, a pair $(x,y)$ of reals is random over a model $M$ iff it's obtained via forcing by closed sets in $2^\omega\times2^\omega$, of positive product measure, still coded in $M$. Thus such a pair is **not** a product-forcing pair. (To establish this, one has to define a closed set $X\subseteq2^\omega\times2^\omega$, of positive product measure, which does not include any positive-measure rectangle, not an easy exercise!) Rather $(x,y)$ is a random pair over a model $M$ iff $x$ is random over $M$ and $y$ is random over $M[x]$ iff *vice versa*.

**Question**. In spite of not being product-forcing-generic, prove that if $(x,y)$ is a random pair over a model $M$ then $M[x]\cap M[y]=M$.

A proof of $M[x]\cap M[y]\cap2^\omega\subseteq M$ can be manufactured by reduction, via continuous reading of names, to the claim that if $X\subseteq2^\omega\times2^\omega$, is a closed set of positive product measure and $f,g:2^\omega\to2^\omega$ are continuous maps such that $f(x)=g(y)$ for almost all $(x,y)\in X$, then $X=X'\cup\bigcup_nX_n$, where $X'$ is null and each $X_n$ is Borel non-null on which both $f$ and $g$ are constants. What about the general case?