Let $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\nu)$ be measure spaces. Let $\lambda$ and $\rho$ be signed measures on $\mathcal{M}$ and $\mathcal{N}$ respectively. Suppose $\lambda \perp \mu$ and $\rho \perp \nu$. Then is it true that $\lambda \times \rho \perp \mu \times \nu$?
I want to say that this is false. I tried thinking of an example, something like $\nu$ and $\lambda$ are the Lebesgue measures on $\mathbb{R}$ and $\mu$ and $\rho$ are Dirac measures on $\mathbb{R}$ at $0$, but I can't get it to work.
In this example, $\lambda \perp \mu$ and $\rho \perp \nu$. If we think of the first component space $\mathbb{R}$ as the $x$-axis, and the second component space $\mathbb{R}$ as the $y$-axis, then $\lambda$ gives a mass of $1$ to the $x$-axis and $\nu$ gives a mass of $1$ to the $y$-axis. So in the $xy$-plane, that is, $\mathbb{R} \times \mathbb{R}$, let $E$ be a set that contains all of the $y$-axis except $(0,0)$ and none of the $x$-axis. Then $X = E \sqcup E^C$ and $(\lambda \times \rho)(E) = 0$ since $E$ does not contain the $x$-axis, and $(\mu \times \nu)(E^C) = 0$ since $E^C$ does not contain the $y$-axis apart from the single point $\{(0,0)\}$, which has Lebesgue measure $0$. Thus, $\lambda \times \rho \perp \mu \times \nu$ and this is not a counterexample. (But I'm not sure if this reasoning is totally correct).
It seems that the statement is true. Because $\lambda \perp \mu $ then there exists some $A\in \mathcal{M}$ such that $\lambda (A)=0$ and $\mu (A^\complement )=0$. Similarly there is some $B\in \mathcal{N}$ such that $\rho (B)=0$ and $\nu (B^\complement )=0$.
Now let some $C\in \mathcal{M}\otimes \mathcal{N}$, then $C$ can be written as the following disjoint union $$ C=(C\cap (A\times B))\cup (C\cap( A\times B^\complement ))\cup (C\cap (A^\complement \times B))\cup (C\times (A^\complement \times B^\complement )) $$ Therefore $$ \lambda \times \rho (C)=\lambda \times \rho (C\cap (A^\complement \times B^\complement ))\\ \mu \times \nu (C)=\mu \times \nu (C\cap (A\times B)) $$ where we used the fact that $A\times H$ and $G\times B$ are null sets for $\lambda \times \rho $ for any chosen $G\in\mathcal{M}$ and any chosen $H\in \mathcal{N}$, and a similar thing happens for $\mu \times \nu $. Hence $\lambda \times \rho \perp \mu \times \nu $.