(For all of this post, at least Countable Choice is assumed to hold.)

For all Tychonoff spaces $\langle X,\mathcal{T}\hspace{.06 in}\rangle$ :

Define $\mathbf{Z}(\langle X,\mathcal{T}\hspace{.06 in}\rangle)$ to be the set of subsets $S$ of $X$ such that there exists a

continuous $\: f : X\to (-\infty,\scriptsize+\normalsize\infty) \:$ such that $\;\;\; \{x\in X : f(x)=0\} \; = \; S \;\;\;$.

Define $\; \operatorname{Ba}(\langle X,\mathcal{T}\hspace{.06 in}\rangle) \;$ to be the sigma-algebra on $X$ generated by $\mathbf{Z}(\langle X,\mathcal{T}\hspace{.06 in}\rangle)$.

Define $\operatorname{UM}(\langle X,\mathcal{T}\hspace{.06 in}\rangle)$ to be the set of subsets $S_0$ of $X$ such that

for all probability measures $\: \mu : \operatorname{Ba}(\langle X,\mathcal{T}\hspace{.06 in}\rangle) \to (-\infty,\scriptsize+\normalsize\infty) \:$,

there exists a member $S_1$ of $\operatorname{Ba}(\langle X,\mathcal{T}\hspace{.06 in}\rangle)$ such that the symmetric difference between $S_0$ and $S_1$ is $\mu$-null.

It follows that $\operatorname{UM}(\langle X,\mathcal{T}\hspace{.06 in}\rangle)$ is a sigma-algebra on $X$.

Let $\langle X,\mathcal{T}_0\rangle$ and $\langle Y,\mathcal{T}_1\rangle$ be Tychonoff spaces. $\;\;$ Let $\: \mu_0 : \operatorname{UM}(\langle X,\mathcal{T}_0\rangle) \to (-\infty,\scriptsize+\normalsize\infty) \:$

and $\: \mu_1 : \operatorname{UM}(\langle Y,\mathcal{T}_1\rangle) \to (-\infty,\scriptsize+\normalsize\infty) \:$ be probability measures. $\;\;$ Let $\mathcal{T}_2$ be the product topology

on $\: X\times Y \:$. $\;\;$ It follows that $\langle X\times Y,\mathcal{T}_2\rangle$ is a Tychonoff space and for all members $S_0$ of

$\operatorname{UM}(\langle X,\mathcal{T}_0\rangle)$, for all members $S_1$ of $\operatorname{UM}(\langle Y,\mathcal{T}_1\rangle)$, $\;\; S_0 \times S_1 \: \in \: \operatorname{UM}(\langle X\times Y,\mathcal{T}_2\rangle) \;\;$.

If there is a measurable Aleph, then there is an example where (*), defined as

there exists a unique probability measure $\: \mu_2 : \operatorname{UM}(\langle X\times Y,\mathcal{T}_2\rangle) \to (-\infty,\scriptsize+\normalsize\infty) \:$ such that

for all members $S_0$ of $\operatorname{UM}(\langle X,\mathcal{T}_0\rangle)$, for all members $S_1$ of $\operatorname{UM}(\langle Y,\mathcal{T}_1\rangle)$,

$\mu_2(S_0 \times S_1) \; = \; \mu_0(S_0) \cdot \mu_1(S_1)$

does not hold.

${}$1. $\:$ Does there provably-in-ZFC exist an example where (

) does not hold?) to hold?

1a. $\:$ If yes, how much of AC (beyond Countable Choice) does such a proof need to use?

${}$2.

What amount of Choice (beyond Countable Choice) and/or

conditions on $\langle X,\mathcal{T}_0\rangle$ and $\langle Y,\mathcal{T}_1\rangle$ are known to force (

I don't know any way full AC makes this easier, and the only condition I already know for 2 is:

$\langle X,\mathcal{T}_0\rangle$ and $\langle Y,\mathcal{T}_1\rangle$ both have countable networks $\;\; \implies \;\;$ (*)