Absolutely continuous curves in Wasserstein distance and measurability.

absolute-continuityfunctional-analysismeasurable-functionsprobability theorywasserstein

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 $g$.

Best Answer

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.

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