For a Polish space $X$, let $C_b(X)$ denote the real Banach space of bounded continuous real-valued functions on $X$. Let $M(X)$ denote the space of all finite signed Borel measures on $X$, equipped with the topology of weak convergence, i.e. the weakest topology such that the map $\mu \mapsto \int f \,d\mu$ is continuous for every $f \in C_b(X)$. Note that this topology is not sequential.

Let $Y$ be another Polish space, and consider the product map $$M(X) \times M(Y) \ni (\mu, \nu) \mapsto \mu \times \nu \in M(X \times Y)$$ where $\mu \times \nu$ is the product measure.

Is this map continuous with respect to the weak topologies?

If not, what about the special case where $X,Y$ are compact metric spaces?

A positive answer would also give a positive answer to Is a specific sequentially closed subset of $M([0,1])$ closed?

In the compact case, we can use the Stone-Weierstrass theorem to show that the map is *sequentially* continuous. Let $\mu_n \to \mu_0$, $\nu_n \to \nu_0$ be weakly convergent sequences. It is sufficient to show that for arbitrary $f \in C(X \times Y)$, we have $\int f \,d(\mu_n \times \nu_n) \to \int f\,d(\mu_0 \times \nu_0)$. Thanks to the uniform boundedness principle, we can find a constant $C$ such that $\|\mu_n\|, \|\nu_n\| \le C$. By the Stone-Weierstrass theorem, for any $\epsilon > 0$ we can find functions $g_1, \dots, g_k \in C(X)$ and $h_1,\dots, h_k \in C(Y)$ such that if we set $\tilde{f}(x,y) = g_1(x) h_1(y) + \dots + g_k(x) h_k(y)$, then we have $\|f - \tilde{f}\|_\infty < \epsilon$.

Now for any $n$ we have $\left|\int (f- \tilde{f})\,d(\mu_n \times \nu_n)\right| < C^2 \epsilon$, and we also have $$\begin{align*} \int \tilde{f} \,d(\mu_n \times \nu_n) &= \int (g_1(x) h_1(y) + \dots + g_k(x) h_k(y)) \,d(\mu_n \times \nu_n)(x,y) \\ &= \int g_1\,d\mu_n \int h_1\,d\nu_n + \dots + \int g_k\,d\mu_n \int h_k\,d\nu_n \\ &\to \int g_1\,d\mu_0 \int h_1\,d\nu_0 + \dots + \int g_k\,d\mu_0 \int h_k\,d\nu_0 \\ &= \int \tilde{f}\,d(\mu_0 \times \nu_0) \end{align*}$$ so by a standard triangle inequality argument we conclude $\int f\,d(\mu_n \times \nu_n) \to \int f\,d(\mu \times \nu)$. This shows sequential continuity, but of course full continuity does not follow. (We cannot use the same argument with nets, because the uniform boundedness argument breaks down: a weak-* convergent net of linear functionals is not necessarily pointwise bounded.)