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Let $X,Y$ be Polish spaces and $\mathcal P(X)$ the space of all Borel probability measures on $X$.

  • Fix $\mu\in \mathcal P(X), \nu \in \mathcal P(Y)$. Let $\gamma \in \Pi(\mu, \nu)$, i.e., $\gamma \in \mathcal P(X \times Y)$ whose marginal on $X$ is $\mu$ and that on $Y$ is $\nu$.
  • Let $c:X \times Y \to \mathbb R$ and $\varphi:X \to \mathbb R \cup\{-\infty\}$ be measurable. Let $\psi:Y \to \mathbb R \cup\{-\infty\}$ which is not necessarily measurable.
  • We assume $\varphi (x) + \psi (y) = c(x, y)$ for $\gamma$-a.e. $(x, y) \in X \times Y$. This means there is a Borel subset $S$ of $X \times Y$ such that $\gamma (S) = 1$, $\varphi (x), \psi(y) \neq \pm\infty$, and $\varphi (x) + \psi (y) = c(x, y)$ for all $(x, y) \in S$.

At page 84 of Villani's Optimal Transport: Old and New, the author constructs a measurable $\psi': Y \to \mathbb R \cup \{\pm\infty\}$ such that $\psi (y) = \psi' (y)$ for $\nu$-a.e. $y\in Y$ as follows.

The measurability of $\psi$ is subtle also, and at the present level of generality it is not clear that this function is really Borel measurable. However, it can be modified on a $\nu$-negligible set so as to become measurable. Indeed, $\varphi (x) + \psi (y) = c(x, y)$ for $\gamma$-almost surely, so if one disintegrates $\gamma (\mathrm dx, \mathrm dy)$ as $\gamma(\mathrm d x | y) \nu(\mathrm d y)$, then $\psi(y)$ will coincide, $\nu(\mathrm d y)$-almost surely, with the Borel function $$ \psi'(y):=\int_{X}[c(x, y) - \varphi (x)] \gamma(\mathrm d x | y). $$

The disintegration theorem applies when the integrand is either non-negative measurable or bounded measurable. I could not see how the map $$ (x, y) \mapsto c(x, y) - \varphi (x) $$ qualifies here. Could you elaborate on my confusion?

Akira
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  • I think you might mistake the disintegration theorem for a different result. The simplest version, sufficient for this case, says that for $X$, $Y$ Polish and $\gamma\in\Pi(\mu,\nu)$, there is a measurable function $\rho:X\to \Delta(Y)$ such that for $F\subseteq X\times Y$ Borel one has $\gamma(F)=\int\int 1_F(x,y)~\mathrm d\rho_x(y)~\mathrm d\mu(x)$. – Michael Greinecker Nov 19 '22 at 11:18
  • @MichaelGreinecker I understand that disintegration theorem leads to the collection ${\gamma(\mathrm d x | y) | y \in Y }$. But i don't get how the map $\psi'$ is well-defined and measurable. Could you please explain it? – Akira Nov 19 '22 at 13:48
  • It seems what you want to know is why the integrals are well-defined. This does not follow from the part you quote; it follows from the assumption made that $c\geq a+b$ with $a$ $\mu$-integrable and $b$ $\nu$-integrable. – Michael Greinecker Nov 19 '22 at 15:25
  • @MichaelGreinecker I assume $c:X \times Y \to [0, \infty)$, so we can pick $a \equiv 0$ and $b \equiv 0$ and that requirement is trivially satisfied. Could you explain how the integral is well-defined and measurable? [...] – Akira Nov 19 '22 at 22:09
  • [...] Let $f(x, y) := c(x, y) - \varphi (x)$. Then $f$ is measurable. I think the author deduces that $y \mapsto \int f(x,y) \gamma(\mathrm d x | y)$ is measurable from claim 1. of this version of disintegration theorem. However, we don't know if $f$ is bounded or integrable... This confuses me. – Akira Nov 19 '22 at 22:10
  • That is the standard machinery of going from indicator functions to simple functions to increasing limits of simple functions, which works with all nonnegative functions. And this still works if the integral is infinite. – Michael Greinecker Nov 19 '22 at 22:18
  • @MichaelGreinecker So you meant $\psi'$ is well-defined and measurable even if $f$ is just measurable (in this case)? – Akira Nov 19 '22 at 22:20
  • If all your functions are nonnegative, yes. – Michael Greinecker Nov 19 '22 at 22:26
  • @MichaelGreinecker If $f$ is non-negative, then disintegration theorem still holds, but we don't know if $f$ is non-negative... – Akira Nov 19 '22 at 22:27
  • In Villani's book, the value of the function at $y$ is defined to be the integral of$\psi(x)+c(x,y)$ with respect to the value of the disintegration at $y$. Villani assumes both $c$ to be (and shows this is wlog under his assumptions) nonnegative and $\psi$ to be integrable. Since you you use different notation, that is not easy to square with your question. – Michael Greinecker Nov 19 '22 at 23:08
  • @MichaelGreinecker I'm sorry for the different notations. I'm reading the proof of Theorem 5.10 (ii), the part $(b) \implies (c)$. I could not find where Villani assumes $\psi$ (in his notation) is integrable. Could you elaborate on where you find this essential assumption? – Akira Nov 19 '22 at 23:22
  • Sorry, I looked again and it is less clear to me how Villani does it. But the problem should be avoidable as long as either the positive or negative part of $\varphi$ is integrable, only $\infty-\infty$-problems are troublesome. – Michael Greinecker Nov 19 '22 at 23:45
  • @MichaelGreinecker I have a closely related question here about the measurability of $\varphi^c$. Please have a look at it. – Akira Nov 19 '22 at 23:48

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