Disintegration theorem: how do the authors prove that $\mu_y$ is supported on $\pi^{-1} (y)$ for $\nu$-a.e. $y \in Y$?

Question

Recently, I came across Tao's blog about disintegration theorem.

Disintegration theorem Let

• $X$ be a Polish space, $\mathcal X$ its Borel $\sigma$-algebra, and $\mu$ a Borel probability measure on $X$.
• $(Y, \mathcal Y)$ be a measurable space and $\pi:X\to Y$ a measurable map.
• $\nu := f_\sharp \mu$ be the push-forward of $\mu$ by through $f$.

Then there is a collection $(\mu_y)_{y\in Y}$ of Borel probability measures on $X$ with the following properties.

1. For all bounded measurable $f:X\to \mathbb C$, the map $$ y \mapsto \int_X f\mathrm d\mu_y $$ is measurable.
2. For all bounded measurable $f:X\to \mathbb C$ and $\nu$-integrable $g:Y\to \mathbb C$, $$ \int_X f (g\circ \pi) \mathrm d\mu = \int_Y \left(\int_X f\mathrm d\mu_y\right)g(y)\mathrm d\nu(y). \quad (\star) $$
3. For all bounded measurable $g:Y\to \mathbb C$, for $\nu$-a.e. $y \in Y$, $$ g\circ \pi = g(y) \quad \mu_y\text{-a.e.} \quad (\star\star) $$

At page 80 of Dellacherie/Meyer's Probabilities and Potential, the authors prove further that

if $Y$ is a separable metric space and $\mathcal Y$ its Borel $\sigma$-algebra, then $\mu_y$ is supported on $\pi^{-1} (y)$ for $\nu$-a.e. $y \in Y$.

Their reasoning is as follows, i.e.,

Let $(\mu_y)_{y\in Y}$ be the family obtained by above theorem. Let $G := X \times Y$ and $\mathcal G$ its product Borel $\sigma$-algebra. We define a map $$ \phi : X \to G, x \mapsto (x, \pi (x)). $$ Let $\lambda := \phi_\sharp \mu$ be the push-forward of $\mu$ through $\phi$. Then $$ \lambda (B) = \int_Y (\mu_y \otimes \delta_y) (B) \mathrm d \nu (y) \quad \forall B \in \mathcal G. $$ Let $K$ be a countable union of compact subsets of $X$ that supports $\mu$; $K$ is obviously Lusin, so $\phi (K)$ is Souslin and hence universally measurable in $G$, and finally it supports $\lambda$. We deduce that for $\nu$-a.e. $y \in Y$, $\mu_y$ is supported by the section $K_y$ of $\phi(K)$ by $y$ and this is contained in $\pi^{-1} (y)$.

It seems to me

• "...universally measurable in $G$..." means that $\phi(K)$ is Borel in $G$, i.e., $\phi(K) \in \mathcal G$, and
• "...finally it supports $\lambda$..." means that $\lambda (\phi (G)) = 1$.

However, I could not understand the sentence

We deduce that for $\nu$-a.e. $y \in Y$, $\mu_y$ is supported by the section $K_y$ of $\phi(K)$ by $y$ and this is contained in $\pi^{-1} (y)$.

Could you explain the reasoning behind it? Thank you so much!

Answer 1

Let $(\mu_y)_{y\in Y}$ be the family obtained by above theorem. Let $G := X \times Y$ and $\mathcal G$ the Borel $\sigma$-algebra generated by the product topology. We define a map $$ \phi : X \to G, x \mapsto (x, \pi (x)). $$

Let $\lambda := \phi_\sharp \mu$ be the push-forward of $\mu$ through $\phi$. Then $$ \lambda (C) = \int_Y (\mu_y \otimes \delta_y) (C) \mathrm d \nu (y) \quad \forall C \in \mathcal G. $$

Notice that $\mathcal G$ coincides with $\mathcal X \otimes \mathcal Y$, so the product measure $\mu_y \otimes \delta_y$ is compatible with $\mathcal G$. Let $K$ be a countable union of compact subsets of $X$ that supports $\mu$; $K$ is Lusin and thus Suslin.

Suslin-Lusin theorem (page 49) Let $\left(E, \mathcal E \right)$ and $\left(F, \mathcal{F}\right)$ be separable metric spaces together with their Borel $\sigma$-algebras. Let $h:E \to F$ be injective measurable. If $E$ is Suslin, then $h$ is a Borel isomorphism of $E$ onto $h(E)$.

By Suslin-Lusin theorem, $H := \phi (K)$ is Suslin.

Lemma (page 48) Let $\left(F, \mathcal F \right)$ be a separable metric space together with its Borel $\sigma$-algebra. If $B \subset E$ is Suslin, then $B$ is analytic.

Notice that $G$ is separable. By Lemma, $H$ is analytic in $G$. Let $\hat{\mathcal G}$ be the universal completion of $\mathcal G$ and $\hat \lambda$ the unique extension of $\lambda$ from $\mathcal G$ to $\hat{\mathcal G}$. It is mentioned here that measurable sets are analytic, and all analytic sets are universally measurable. So $H\in \hat{\mathcal G}$. Clearly, $\hat \lambda (H) = 1$. Then there is $M \in \mathcal G$ such that $M \subset H$ and $\lambda(M)= 1$. We have $$ 1=\int_Y (\mu_y \otimes \delta_y) (M) \mathrm d \nu (y). $$

It follows that $$ (\mu_y \otimes \delta_y) (M) = 1 \quad \nu\text{-a.e.} $$

Let $M_y := \{x \in X \mid (x, y) \in M\}$. By Fubini's theorem, $M_y \in \mathcal X$ and $$ \begin{align} (\mu_y \otimes \delta_y) (M) &= \int_X \int_Y 1_{M} (x, z) \mathrm d \delta_y (z) \mathrm d \mu_y (x) \\ &= \int_X 1_{M_y} (x) \mathrm d \mu_y (x) \\ &= \mu_y (M_y). \end{align} $$

It follows that $$ \mu_y (M_y) = 1 \quad \nu\text{-a.e.} $$

Notice that $M_y \subset \pi^{-1} (y)$. This completes the proof.

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Update: Below I prove the equality involving $\lambda$. For $C \in \mathcal G$, we define a map $$ f_C: y \to \mathbb R, y \mapsto (\mu_y \otimes \delta_y) (C) . $$

Let $\mathcal C := \{A\times B \mid A \in \mathcal X, B \in \mathcal Y\}$ and $\mathcal D := \{ C \in \mathcal G \mid f_C \text{ is measurable}\}$. If $C = A\times B \in \mathcal C$, then $$ f_C (y) = \left ( \int_X 1_A \mathrm d \mu_y \right ) 1_B(y). $$

By (1.), $y \mapsto \int_X 1_A \mathrm d \mu_y$ is measurable. As such, $f_C$ is measurable for all $C \in \mathcal C$. This implies $\mathcal C \subset \mathcal D$. Clearly, $\mathcal C$ is a $\pi$-system. Let's prove that $\mathcal D$ is a $\lambda$-system. Clearly, $X \times Y \in \mathcal D$. If $C \in \mathcal D$ then $f_{C^c} = 1- f_C$ is measurable and thus $C^c \in \mathcal D$. If $(C_n) \subset \mathcal D$ is pairwise disjoint, then $f_{C} = \sum_n f_{C_n}$ with $C := \bigcup_n C_n$ is measurable and thus $C \in \mathcal D$. By Dynkin's $\pi$−$\lambda$ theorem, we get $\sigma(\mathcal C) \subset \mathcal D$. Hence $f_C$ is measurable for every $C \in \mathcal G$.

Let $\mathcal E := \{C \in \mathcal G \mid \lambda(C) = \int_Y f_C \mathrm d \nu\}$. For $C = A \times B \in \mathcal C$. By (2.), $$ \int_Y f_C \mathrm d \nu = \int_Y \left ( \int_X 1_A \mathrm d \mu_y \right ) 1_B(y) \mathrm d \nu (y) = \int_X 1_A (1_B \circ \pi) \mathrm d \mu = \mu(A \cap \pi^{-1} (B)). $$

On the other hand, $$ \lambda (C) = \mu (\phi^{-1} (C)) = \mu (A \cap \pi^{-1} (B)). $$

As such, $\mathcal C \subset \mathcal E$. Just as above, we can prove that $\mathcal E$ is a $\lambda$-system. By Dynkin's $\pi$−$\lambda$ theorem, we get $\sigma(\mathcal C) \subset \mathcal E$. Hence $\lambda(C) = \int_Y f_C \mathrm d \nu$ for every $C \in \mathcal G$. This completes the proof.