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.

  • 2
    $\begingroup$ 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. $\endgroup$ Jan 19, 2022 at 18:16
  • $\begingroup$ Thanks! I'll have a look! $\endgroup$
    – ECL
    Jan 19, 2022 at 18:20
  • $\begingroup$ 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.] $\endgroup$ Apr 4, 2022 at 23:11

1 Answer 1


As already mentioned in the comments, this is really a standard result, see, e.g., Lemma 3.1. in Kallenberg's book.


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