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Let $(X, d, \mu)$ be a metric measure space. Let $P^1(X)$ denote the space of probability measures on $(X,d)$, which have finite first moments, that is: \begin{equation} \nu \in P^1(X) \implies \int d(x, x_0) \ d \nu < \infty \end{equation} for some $x_0 \in X$. Let us endow $P^1(X)$ with the Wasserstein distance $W_1$, that is: \begin{gathered} \forall \nu, \nu' \in P^1(X) \quad W_1( \nu, \nu') = \inf_{\eta} \int_{X \times X} d(x, y) \ d \eta(x,y), \end{gathered} where the infimum is taken over all joint probability distributions $\eta$ which have $\nu$ and $\nu'$ as their marginals.

Now, let $\gamma \colon [0,1] \to P^1(X)$ be an absolutely continuous curve. I would like to know what are the minimal requirements that a function $g \colon X \to [0, \infty]$ has to satisfy so that a function \begin{equation} [0,1] \ni t \mapsto <g, \gamma(t)> = \int_X g \ d \gamma(t) \end{equation} is measurable.

Since convergence in the Wasserstein distance implies the weak convergence of measures along with the convergence of the first moments, if $g$ is continuous then $t \mapsto <g, \gamma(t)>$ is also continuous. However, I wonder whether this condition could be relaxed to, for example, just the measurability of integrability (with respect to $\mu$) of $g$.

RobPratt
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Kakuro
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  • Just a quick remark: I do not see how the measure $\mu$ relates to your problem. So I don't think the $\mu$-measurability of $g$ could be of any help – vizietto Dec 18 '22 at 19:12
  • Admittedly, the presence of $\mu$ is mostly due to the fact that the question appeared when I wanted $<g, \gamma(t)>$ to be well-defined for $g\in L^p(\mu)$ for some $p \ge 1$. The only real connection is that $\mu$ and $\gamma(t)$ should be defined on the same $\sigma$-filed of borel subsets of $X$. – Kakuro Dec 19 '22 at 00:47

1 Answers1

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The following proof uses a non trivial fact from descriptive set theory. Maybe there is a more straightforward proof.

In Kechris' Classical Descriptive Set Theory, the following is shown (Exercise 11.7): let $X$ be a metric space and $\mathscr{C}$ be a collection of functions $X \to \mathbb{R}$ such that

  • $\mathscr{C}$ contains the bounded continuous functions
  • if $(f_n)$ is a sequence in $\mathscr{C}$ that is uniformly bounded ($\sup_n \|f_n\|_\infty < \infty$) and converges pointwise to a function $f$, then $f \in \mathscr{C}$

then $\mathscr{C}$ contains all bounded Borel functions.

Now we turn back to your problem and denote by $\mathscr{C}$ the collection of bounded Borel functions $g \colon X \to \mathbb{R}$ such that $\langle g, \gamma(\cdot)\rangle$ is Borel measurable. This collection satisfies the properties stated above.

Now you can remove the boundedness assumption on $g$, provided it has values in $[0, \infty]$. Indeed, by the monotone convergence theorem

$$t \mapsto \langle g, \gamma(t) \rangle = \lim_{n \to \infty} \int_X \min(g, n) d\gamma(t)$$

is a Borel function, as a pointwise limit of Borel functions.

vizietto
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