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.)

  • $\begingroup$ I'm skimming the question at 2am, so apologies if you specify this somewhere, but do you want the product map to be jointly continuous wrt the weak topologies (i.e. continuous for the product of the weak topologies) or merely separately continuous? $\endgroup$
    – Yemon Choi
    Jul 9, 2015 at 1:20
  • 1
    $\begingroup$ @YemonChoi: I'm looking for joint continuity. I think separate continuity is almost immediate. Fix $f \in C_b(X \times Y)$ and $\mu \in M(X)$ and set $G(y) = \int f(x,y) \mu(dx)$; then $G \in C_b(Y)$ (dominated convergence). Now by Fubini, $\int f\,d(\mu \times \nu) = \int G\,d\nu$ which is a weakly continuous function of $\nu$. $\endgroup$ Jul 9, 2015 at 1:40

2 Answers 2


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.


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 $\mu_0\times\nu_0$.

  • $\begingroup$ The hard part, of course, is finding the measure $\Delta$. But I think I can maybe at least see how to find two measure $\mu_1, \nu_1$ such that $\int g_i \,d\mu_1 = \int g_i \,d\nu_1 = 0$ for all $i$, yet $\int f\,d(\mu_1 \times \nu_1) \ne 0$. I'll try and write this up. One piece that seems to be needed: to show that there exists an $f$ (either your specific choice or another one) which cannot be written in the form $f(x,y) = a_1(x) b_1(y) + \dots + a_n(x) b_n(y)$ for bounded continuous $a_i, b_i$. This must be true but I can't quite see how to prove it. $\endgroup$ Jul 9, 2015 at 15:58
  • $\begingroup$ I asked the latter question on Math.SE: math.stackexchange.com/questions/1355311/… $\endgroup$ Jul 9, 2015 at 16:52
  • $\begingroup$ I think you should expect to have $\Delta$ being a measure supported on $n$ points. $\endgroup$ Jul 9, 2015 at 21:32

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