Is taking the product of signed measures weakly continuous?

Question

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.)

Answer 1

It seems that it is not continuous, even in the compact case. I think this is a proof, loosely following Anthony Quas's outline.

Take $X = Y = [0,1]$ and set $f(x,y) = e^{xy}$. Let $$A = \left\{(\mu, \nu) \in M([0,1]) \times M([0,1]) : \left| \int f\,d (\mu \times \nu) \right| < 1\right\}.$$ If the product map is to be jointly continuous then $A$ must be open in $M([0,1]) \times M([0,1])$ (with respect to the product of the weak topologies). In particular, it must contain a set of the form $U \times U$ where $U$ is a basic open neighborhood of the 0 measure. Such a $U$ could be written in the form $U = \left\{ \mu : \left|\int g_i\,d\mu \right| < 1, i = 1, \dots, n\right\}$ for some $g_1, \dots, g_n \in C([0,1])$. We will contradict this by producing measures $\mu, \nu \in U$ with $(\mu, \nu) \notin A$. Specifically, we will show:

Proposition. For any $n$ and any $g_1, \dots, g_n \in C([0,1])$, there exist signed measures $\mu,\nu$ such that $\int g_i\,d\mu = \int g_i\,d\nu = 0$ for all $i$, and $\int f\,d(\mu \times \nu) = 1$.

Let $E = \operatorname{span}\{g_1, \dots, g_n\} \subset C([0,1])$. We can assume without loss of generality that $g_1, \dots, g_n$ are linearly independent; if not, replace them with a basis for $E$ (reducing $n$ as needed).

As suggested in Daniel Fischer's answer to this Math.SE question of mine, by the Vandermonde determinant, the functions $\{f(\cdot, y) : y \in [0,1]\} \subset C([0,1])$ are all linearly independent, so they span an infinite dimensional linear subspace of $C([0,1])$. In particular, we may choose $y_1, \dots, y_{n+1} \in [0,1]$ such that the functions $$\{g_1, \dots, g_n, f(\cdot, y_1), \dots, f(\cdot, y_{n+1})\}$$ are linearly independent.

Now since $E$ has dimension $n$, the linear map taking $g \in E$ to $(g(y_1), \dots, g(y_{n+1})) \in \mathbb{R}^{n+1}$ is not surjective. Hence there exist numbers $c_1, \dots, c_{n+1} \in \mathbb{R}$ such that if $h(y_j) = c_j$ for $j = 1,\dots, n+1$, then $h \notin E$.

Using the Hahn-Banach and Riesz representation theorems, we can find a signed measure $\mu$ such that $$\begin{align*}\int g_i(x)\, \mu(dx) &= 0, && i = 1, \dots, n \\ \int f(x,y_j) \,\mu(dx) &= c_j, && j = 1, \dots, n+1.\end{align*}$$ Set $h(y) = \int f(x,y) \,\mu(dx)$. By the dominated convergence theorem (valid for signed measures), $h$ is continuous. Moreover, $h(y_j) = c_j$ for $j = 1, \dots, n+1$, hence $h \notin E$. Thus by Hahn-Banach and Riesz again, we can find a signed measure $\nu$ such that $$\begin{align*}\int g_i(y)\, \nu(dy) &= 0, && i = 1, \dots, n \\ \int h(y) \,\nu(dy) &= 1.\end{align*}$$ By Fubini's theorem, we have $$\int f\,d(\mu \times \nu) = \int \int f(x,y) \,\mu(dx)\,\nu(dy) = \int h(y) \nu(dy) = 1$$ completing the proof.

Note that this method, as it stands, cannot be used to resolve the question Is a specific sequentially closed subset of $M([0,1])$ closed?; to do so, we would need to be able to take $\nu = \mu$. It's not clear how to do that, since we use one to construct the other.

Answer 2

Too long for a comment, but not fully worked out yet.

So I think both answers are negative. Consider the case when $X=Y=[0,1]$ and take the function $f(x,y)=e^{-|x-y|}$ for example. Let $\mu_0$ and $\nu_0$ be a pair of measures on the interval and take neighbourhoods of $\mu_0$ and $\nu_0$ of the form $\{\mu\colon |\int g_i\,d\mu-\int g_i\,d\mu_0|<1,\ i=1\ldots,n\}$ and $\{\nu\colon |\int g_i\,d\nu-\int g_i\,d\nu_0|<1\}$ (you can assume the same functions are used). I think you can then find a measure $\Delta$ such that $\int g_i\,d\Delta=0$ (even supported on finitely many points), but so that $\int f(x,y)\,d\Delta\times\Delta\ne 0$. Now you can build measures in the neighbourhoods of $\mu_0$ and $\nu_0$ by taking $\mu=\mu_0+M\Delta$ and $\nu=\nu_0+M\Delta$. For large enough $M$, the product does not belong to the neighbourhood of $\mu_0\times\nu_0$.