# Gluing lemma in optimal transport

This thread is meant to record a question that I feel interesting during my self-study. I'm very happy to receive your suggestion and comments.

Gluing Lemma: Let $$X,Y,Z$$ be Polish spaces and $$\mathcal P(X), \mathcal P(Y), \mathcal P(Z)$$ the spaces of Borel probability measures on $$X, Y, Z$$ respectively. Let $$\mu \in \mathcal P(X), \nu \in \mathcal P(Y), \omega \in \mathcal P(Z)$$. Let $$\pi_1 \in \Pi(\mu, \nu)$$ and $$\pi_2 \in \Pi(\nu, \omega)$$. Let $$P^{X \times Y}$$ and $$P^{Y \times Z}$$ be the projection maps from $$X \times Y \times Z$$ to $$X \times Y$$ and $$Y \times Z$$ respectively. Then there is $$\gamma \in \mathcal P(X \times Y \times Z)$$ such that $$P^{X \times Y}_\sharp \gamma = \pi_1 \quad \text{and} \quad P^{Y \times Z}_\sharp \gamma = \pi_2.$$

• What is the question? Commented Jul 10, 2022 at 3:22
• @YuvalPeres I have no question but just want to share a proof of this lemma. Please see here and here. Commented Jul 10, 2022 at 3:29

Disintegration of Measures: Let $$X, Z$$ be Polish spaces, $$f:X \to Z$$ measurable, and $$\pi \in \mathcal P (X)$$. Let $$\omega = f_\sharp \pi \in \mathcal P(Z)$$ and $$X_z := f^{-1} (z) \subset X$$ for $$z \in Z$$. Then there is a family $$\{\pi(\cdot |z)\}_{z\in Z}$$ of Borel probability measures such that $$\pi(\cdot |z) \in \mathcal P(X_z)$$ and $$\int_X g \mathrm d \pi = \int_Z \int_{X_z} g(x) \mathrm d \pi (x|z) \mathrm d \omega (z)$$ for all measurable function $$g:X \to [0, +\infty]$$. Moreover, $$\{\pi(\cdot |z)\}_{z\in Z}$$ is unique up to $$\omega$$-a.e.
By disintegration of measures, $$\pi_1 (A \times B) =\int_B \pi_1(A |y) \mathrm d \nu (y)$$ for some family $$\{\pi_1(\cdot |y)\}_{y \in Y} \subset \mathcal P(Y)$$. Similarly, $$\pi_2 (B \times C) =\int_B \pi_2(C |y) \mathrm d \nu (y)$$ for some family $$\{\pi_2(\cdot |y)\}_{y \in Y} \subset \mathcal P(Y)$$. Define a non-negative finite Borel measure $$\gamma$$ such that $$\gamma (A \times B \times C) := \int_B \pi_1(A |y) \pi_2(C |y) \mathrm d \nu (y).$$
It's clear that $$\gamma$$ is the required measure.