Null sets w.r.t. product measure

Question

Let $(X,A,m)$ and $(Y,B,n)$ be two measure spaces, and for every $x\in X$, let $B_x\in B$ have zero measure: $n(B_x)=0$. Is it true that $\cup_{x\in X}(\{x\}\times B_x)$ has zero outer measure w.r.t to the product measure $m\otimes n$ (and is therefore measurable in the Caratheodory sense)? We may assume that the measure spaces are sigma-finite, if necessary.

Answer 1

No. There's a counterexample in Rudin Real and Complex Analysis (example (c) in section 8.9).

The example depends on the Continuum Hypothesis, so one might object that we haven't really shown a counterexample exists, since we don't know that CH is true. But note this:

Suppose that CH implies some proposition P is false. Then it's impossible to prove P in ZFC (unless of course ZFC is inconsistent).

Proof: It's known that (if ZFC is consistent then) ZFC does not prove not-CH.

Corollary A counterexample depending on CH is worth more than you might think, even if you don't believe CH.