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 $$ P^{X \times Y}_\sharp \gamma = \pi_1 \quad \text{and} \quad P^{Y \times Z}_\sharp \gamma = \pi_2. $$

  • 2
    $\begingroup$ What is the question? $\endgroup$ Commented Jul 10, 2022 at 3:22
  • $\begingroup$ @YuvalPeres I have no question but just want to share a proof of this lemma. Please see here and here. $\endgroup$
    – Akira
    Commented Jul 10, 2022 at 3:29

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .