# Measurability of Markov kernel wrt the Borel $\sigma$-algebra generated by the weak topology

Consider two Polish metric probability spaces $$(\mathcal{A}, \Sigma_\mathcal{A})$$ and $$(\mathcal{B}, \Sigma_\mathcal{B})$$, endowed with their Borel $$\sigma$$-algebras. Denote as $$\mathcal{P}_\mathcal{B}$$ the space of probability measures on $$(\mathcal{B}, \Sigma_\mathcal{B})$$. We can endow it with the Borel $$\sigma$$ algebra $$\Sigma_{\mathcal{P}_\mathcal{B}}$$ generated by the weak topology (wrt the weak convergence of measures).

Consider a Markov kernel $$\kappa:\mathcal{A}\times\Sigma_\mathcal{B}\to [0,1]$$. In particular we can see $$a\mapsto \mu_a=\kappa(a, \cdot)$$ as a mapping $$\mathcal{A}\to\mathcal{P}_\mathcal{B}$$. Is this mapping measurable (wrt $$\Sigma_{\mathcal{P}_\mathcal{B}}$$)?

The main motivation behind this question is that it is related to the following problem (cf. https://math.stackexchange.com/questions/4360086). Denoted as $$\mathcal W$$ the 1-Wasserstein distance between probabilities in $$\mathcal P_\mathcal B$$, defined via the metric on $$\mathcal B$$, fixed $$\nu\in\mathcal P_\mathcal B$$, is the mapping $$\mathcal A\to\mathbb R$$ $$a\mapsto \mathcal W(\nu, \mu_a)$$ always measurable?

I think that one way to prove the measurability of this last function would be to exploit Corollary 5.22 in [1], which essentially tells you that if $$a\mapsto\mu_a$$ is measurable wrt $$\Sigma_{\mathcal{P}_\mathcal{B}}$$, then $$a\mapsto\pi_a$$ is measurable, where $$\pi_a$$ is the optimal coupling between $$\mu_a$$ and $$\nu$$. It would then follow that $$a\mapsto \mathcal{W}(\mu_a, \nu) = \mathbb E_{(A,A')\sim\pi_a}[d(A, A')]$$ is measurable. So we are back to the first question: is it actually true that $$a\mapsto\mu_a$$ is measurable?

I am interested in this since I have often encountered expressions like $$\int_\mathcal A \mathcal{W}(\mathbb P_B, \mathbb P_{B|A=a})\,\mathrm d\mathbb P_A(a)$$ (where $$A, B$$ are coupled random variables, with marginals $$\mathbb P_A$$ and $$\mathbb P_B$$, and $$\mathbb P_{B|A=a}$$ is a regular conditional probability) without any formal justification, see for instance [2] and the papers it builds on. But does this expression actually make sense? Is the integrand always measurable?

[1] Villani, Optimal transport, old and new, 2008.
[2] Rodríguez-Gálvez, Tighter expected generalization error bounds via Wasserstein Distance, 2021.

• The answer is yes. I don't have them at hand, but you can find a proof in the discrete time optimal control book by Bertsekas and Shreve, or (an appendix of) the Bayesian nonparametrics book by Ghosal and van der Vaart. Jan 19, 2022 at 18:16
• Thanks! I'll have a look!
– ECL
Jan 19, 2022 at 18:20
• To be precise: if $B$ is any separable metric space, then the Borel $\sigma$-algebra of the topology of weak convergence on $\mathcal{P}_B$ is precisely the $\sigma$-algebra generated by the set of evaluation mappings $\{\mu \mapsto \mu(A) : A \in \Sigma_B\}$. [Any proof where the result is stated for "Polish spaces" should work for more general separable metric spaces - i.e. being completely metrisable shouldn't play a role.] Apr 4, 2022 at 23:11