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](https://mathoverflow.net/a/209038/4832).


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.



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> Is this map continuous with respect to the weak topologies?
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> If not, what about the special case where $X,Y$ are compact metric spaces?
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A positive answer would also give a positive answer to [Is a specific sequentially closed subset of $M([0,1])$ closed?](https://mathoverflow.net/questions/210742/is-a-specific-sequentially-closed-subset-of-m0-1-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.)



Is taking the product of signed measures weakly continuous?
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 $\int f\,d(\mu \times \nu) = 1$.
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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](https://math.stackexchange.com/a/1355351/822), 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?](https://mathoverflow.net/questions/210742/is-a-specific-sequentially-closed-subset-of-m0-1-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.


