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Let $X$, $Y$ and $Y'$ be (standard) Borel spaces. We let $\mathcal B(X)$ be the Borel $\sigma$-algebra of $X$ and $\mathcal P(X)$ to be the space of all Borel probability distributions on $X$ endowed with the topology of weak convergence, so that it is a Borel space as well. Consider any Borel set $A\in \mathcal B(X\times \mathcal P(Y))$ and a pushforward operator $\pi_*:\mathcal P(Y\times Y')\to \mathcal P(Y)$ defined as $$ (\pi_*p)(B) = p(B\times Y') \qquad \forall B\in \mathcal B(Y). $$ What can we say about measurability of $A':=\{(x,p):\pi_*p\in A_x\}\subseteq X\times \mathcal P(Y\times Y')$, where $$ A_x:=\{q\in \mathcal P(Y):(x,q)\in A\}. $$

If I am not mistaken, we obtain $A' = (\mathrm{id}_X\times \pi_*)^{-1}(A)$, hence the question can be reduced to the measurability of the map $\pi_*$.

I think, a more general result holds true. Let $(\Omega,\mathcal F)$ and $(\Omega',\mathcal F')$ be arbitrary measurable spaces, and let $\mathcal P$ be a set of all probability measures on $(\Omega,\mathcal F)$ endowed with a $\sigma$-algebra $\mathcal A$ generated by evaluation maps $\theta_F:\mathcal P \to \Bbb R$ given by $\theta_F(p) = p(F)$ for any $F\in \mathcal F$ and $p\in \mathcal P$. Let $(\mathcal Pi',\mathcal A')$ be a corresponding measurable space of probability measures for $(\Omega',\mathcal F')$ and denote by $\theta'_{F'}$ the corresponding evaluation maps. For any measurable $\varphi:\Omega\to\Omega'$ it holds that $\varphi_*:\mathcal P\to\mathcal P'$ is measurable.

Proof: note that for any $p\in \mathcal P$ and $F'\in \mathcal F'$ it holds that $$ \theta'_{F'}(\varphi_*p) = p(\varphi^{-1}(F')) = \theta_{\varphi^{-1}(F')}(p) $$ hence $\theta'_{F'}\circ \varphi_* = \theta_{\varphi^{-1}(F')}$. Since $\mathcal A'$ is generated by evaluation maps, for the measurability of $\varphi_*$ it is necessary and sufficient that $\varphi_*^{-1}((\theta'_{F'})^{-1}(B))\in \mathcal A$ for any Borel $B\subseteq \Bbb R$. The latter fact is true since $$ \varphi_*^{-1}((\theta'_{F'})^{-1}(B)) = (\theta'_{F'}\circ \varphi_*)^{-1}(B) = (\theta_{\varphi^{-1}(F')})^{-1}(B)\in \mathcal A $$ since $\varphi^{-1}(F')\in \mathcal F$ and $\mathcal A$ is generated by maps $\theta_{F}$ with $F\in \mathcal F$.

hardmath
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SBF
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  • Isn't $\varphi_*$ a contraction for the total variation distance? Edit: false, since you use the weak topology, and not the strong topology. Sorry. – D. Thomine Nov 20 '13 at 12:16
  • I think everything you wrote here is correct. The proof is a bit simple if you note that the $\sigma$-algebra generated by the evaluations is generated by sets of the form ${\mu:\mu(F)\geq r}$ for measurable sets $F$ and numbers $r$. Also, in your example $\pi_*=\textrm{marg}_Y$. – Michael Greinecker Nov 20 '13 at 12:43
  • Also, mapping $X$ to $\mathcal{P}(X)$ and $f$ to $f_*$ gives you an endofunctor of the category of measurable spaces and maps. – Michael Greinecker Nov 20 '13 at 12:46
  • @MichaelGreinecker: actually, the latter argument indicates the type of sources where I shall look for the references. – SBF Nov 20 '13 at 13:00

1 Answers1

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In this case, you can say even more: under compatible Polish topologies, $\pi_*$ is not only measurable but continuous.

Indeed, suppose $U,V$ are Polish spaces and $F : U \to V$ is continuous. Then $F_* : \mathcal{P}(U) \to \mathcal{P}(V)$ is continuous in the weak topologies. The proof is immediate: suppose $\mu_n, \mu \in \mathcal{P}(U)$ with $\mu_n \to \mu$ weakly. Let $g : V \to \mathbb{R}$ be bounded and continuous. Then $g \circ F$ is a bounded continuous function on $U$, so we have $$\int_V g\,dF_* \mu_n = \int_U g \circ F \,d\mu_n \to \int_U g \circ F\,d\mu = \int_V g\,dF_*\mu.$$

For your example, fix compatible Polish topologies on $Y,Y'$, and set $U = Y \times Y'$, $V = Y$, and $F = \pi$, the projection map.

Nate Eldredge
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  • Nice, thanks Nate – SBF Nov 20 '13 at 15:15
  • @NateEldredge: Why is it enough to check the continuity of $F_*$ with sequences, as opposed to nets (weak topologies usually not being first-countable)? Also, what does it mean for two topologies to be "compatible"? – Alex M. Jan 20 '18 at 13:22
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    The space $\mathcal{P}(X)$ of probability measures on a Polish space $X$, endowed with the weak topology (induced by bounded continuous functions), is first countable - indeed Polish. This is a standard result that you can find in, say, Billingsley's Convergence of Probability Measures. Note here that the "weak topology" is really a weak-* topology. You can think about the fact that, given a separable Banach space $W$, the weak-* topology on $W^$ is not first countable, but the topology it induces on the ball of $W^$ is first countable (indeed, Polish!) – Nate Eldredge Jan 20 '18 at 15:11
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    @AlexM. I'm not saying the topologies are "compatible" with each other - that would make no sense. Recall that a standard Borel space is a measurable space - a pair $(X,\mathcal{B})$ of a set $X$ with a $\sigma$-algebra $\mathcal{B}$ such that there exists a Polish topology $\tau$ on $X$ for which $\sigma(\tau) = \mathcal{B}$. Any Polish $\tau$ with this property is said to be a "compatible topology" for $(X,\mathcal{B})$ (it is typically not unique). So I am saying we should fix a compatible topology for $Y$, and a compatible topology for $Y'$. – Nate Eldredge Jan 20 '18 at 15:16
  • It seems to me that the hypotheses that $U$ and $V$ be Polish spaces, and that the measures be probabilities, are not necessary. It is sufficient that $U$ and $V$ be Borel spaces, $\mu_i$ and $\mu$ be Radon measures and $F$ continuous: it follows that if $\mu_i \to \mu$ in the weak* topology (with $(\mu_i){i \in I}$ being a net), then $F* \mu_i \to F_* \mu$, so $F_$ is weak to weak* continuous. Am I overlooking something? – Alex M. Jan 27 '18 at 10:43
  • @AlexM.: The word "continuous" has no meaning for a map between Borel spaces, since a Borel space does not have any topology. Likewise there is no canonical weak-* topology on a space of measures on a Borel space. You need to fix topologies to make sense of the question. – Nate Eldredge Jan 27 '18 at 15:17
  • @NateEldredge: My bad, I meant to say "topological spaces endowed with their respective Borel $\sigma$-agebras". The rest of my comment remains valid. – Alex M. Jan 27 '18 at 15:19
  • @AlexM.: Yes, then I agree. – Nate Eldredge Jan 27 '18 at 15:26
  • Does this still hold true if we restrict to Wasserstein spaces and consider the Wasserstein p topology thereon? – ABIM Jun 18 '20 at 20:15