Given two measure preserving systems $(X,\mathcal{B},\mu,T)$ and $(Y,\mathcal{C},\nu,S)$, a joining $\rho$ of those is defined as a $ T \times S$-invariant measure on the product $\sigma$-algebra, such that $\rho(A \times Y)=\mu(A)$ and $\rho(X \times B)=\nu(B)$, for any $A \in \mathcal{B}$ and $B \in \mathcal{C}$ (according to Einsiedler-Ward, Ergodic Theory; Definition 6.7).
If there are sets $A,B$ for which $\rho(A \times Y)>0$ and $\rho(X \times B)>0$, can we conclude that $\rho(A \times B)>0$ as well? This is obviously true for the product joining, but seems too strong to hold in general, however, I can't think of a counterexample.
This is not an exercise in the book, nor is it hinted by the theory presented there, but it is a natural question that came to my mind.
Thinking about it, if true, perhaps this is a known property of measures on product measure spaces, the projections of which on each coordinate are the measures you start with. I.e. given probability spaces $(X,\mathcal{B},\mu)$ and $(Y,\mathcal{C},\nu)$, and a measure $\rho$ on the product space $(X \times Y, \mathcal{B} \bigotimes \mathcal{C})$, such that $\rho(A \times Y)=\mu(A)$ and $\rho(X \times B)=\nu(B)$, for any $A \in \mathcal{B}$ and $B \in \mathcal{C}$, if there are sets $A,B$ for which $\rho(A \times Y)>0$ and $\rho(X \times B)>0$, can we conclude that $\rho(A \times B)>0$?
From the joining condition we know that $\rho(A\times B) + \rho(A\times B^c)>0$ So in order to find a counter example we need to find a joining that assigns all mass in the product to the complement of $B$.
Take the probability spaces $(\{0,1\},\mathcal{P}(\{0,1\}),P)$ given by a Bernoulli random variable $X$. And now take the probability spaces in the product given by the random vector $(X,1-X)$. The resulting probability $\rho$ fulfills the joining condition and $\rho(\{0\}\times \{0,1\})=\rho(\{0,1\}\times \{0\})=\frac{1}{2}$ but $\rho(\{0\}\times\{0\})=0$.