Disintegration theorem: how to construct the $\nu$-null set $E$ and pick the collection $\{\tilde{\pi}_{\sharp} f \mid f \in C(X)\}$?

Question

Theorem $4$ of this blog entry of Terrence Tao states the following:

Let

• $X$ be a compact metric space, $\mathcal X$ its Borel $\sigma$-algebra, and $\mu$ a Borel probability measure on $X$.
• $(Y, \mathcal Y)$ a measurable space, $\pi:X\to Y$ a measurable map, and $\nu := f_\sharp \mu$.

Then there is a collection $(\mu_y)_{y\in Y}$ of Borel probability measures on $X$, such that $$ \int_X f\cdot (g\circ \pi) \mathrm d\mu = \int_Y \left(\int_X f\mathrm d\mu_y\right)g(y)\mathrm d\nu(y) $$ for all bounded measurable maps $f:X\to \mathbb C$ and $g:Y\to \mathbb C$. For $\nu$-a.e. $y \in Y$, we have $$ g\circ \pi=g(y) \quad \mu_y\text{-a.e.} $$

Proof: We have the pullback map $$ \pi^\sharp:L^2(Y, \mathcal Y, \nu)\to L^2(X, \mathcal X, \mu), g \mapsto g \circ \pi. $$

We take its adjoint $\pi_\sharp:L^2(X, \mathcal X, \mu)\to L^2(Y, \mathcal Y, \nu)$, and have the duality $$ \int_X f(\pi^\sharp g) \mathrm d \mu = \int_Y\left(\pi_{\sharp} f\right) g \mathrm d \nu \quad \forall f \in L^2(X, \mathcal X, \mu), \forall g \in L^2(Y, \mathcal Y, \nu). $$

From duality that, we have $\|\pi_\sharp f\|_{L^\infty(Y)}\le \|f\|_\infty$ for all $f\in C(X)$. Since $C(X)$ is separable, we find a measurable representative $\tilde{\pi}_{\sharp} f$ of $\pi_{\sharp} f$ to every $f \in C(X)$ which varies linearly with $f$, and is such that $|\tilde{\pi}_{\sharp} f(y)| \le \|f\|_{\infty}$ for all $y$ outside of a set $E$ of $\nu$-measure zero and for all $f \in C(X)$.

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So basically, from the separability of $C(X)$, the author constructed a $\nu$-null set $E \in \mathcal Y$ and a collection $\{\tilde{\pi}_{\sharp} f \mid f \in C(X)\}$ such that

• $\tilde{\pi}_{\sharp}$ is a representative of $\pi_{\sharp} f$.
• For each $y \in Y \setminus E$, the map $$ L_y:C(X) \to \mathbb C, f \mapsto \tilde{\pi}_{\sharp} f(y) $$ is linear.
• For each $y \in Y \setminus E$, $$ |L_y(f)| \le \|f\|_{\infty} \quad \forall f \in C(X). $$

Can somebody explain how to construct $E$ and pick $\{\tilde{\pi}_{\sharp} f \mid f \in C(X)\}$?

Thank you so much for your help!

Answer 1

Because $X$ is compact, $C(X)$ is separable. Let $F$ be a countable dense subset of $C(X)$. Let $\mathcal F := \operatorname{span}_{\mathbb Q} (F)$. For each $f \in F$, we have $\pi_\sharp f$ is an equivalence class of $L_\infty (Y)$, so we fix a representative $\tilde \pi_\sharp f \in \mathcal L_\infty (Y)$. For each $y \in Y$, we define a map $L_y:\mathcal F \to \mathbb C$ by $$ L_y \bigg (\sum_{i=1}^n \lambda_i f_i \bigg) := \sum_{i=1}^n \lambda_i (\tilde \pi_\sharp f_i) (y) \quad \forall (\lambda_i)_{i=1}^n \subset \mathbb Q, \forall (f_i)_{i=1}^n \subset F. $$

1. Let $\pmb \lambda = (\lambda_i)_{i=1}^n \subset \mathbb Q$ and $\pmb f=(f_i)_{i=1}^n \subset F$ such that $f := \sum_{i=1}^n \lambda_i f_i \in F$. Because $\pi_\sharp$ is linear, $$ \tilde\pi_\sharp f = \tilde\pi_\sharp \bigg (\sum_{i=1}^n \lambda_i f_i \bigg) = \sum_{i=1}^n \lambda_i (\tilde\pi_\sharp f_i) \quad \nu \text{-a.e.} $$ So there is a $\nu$-null set $N_{(\pmb \lambda, \pmb f)} \in \mathcal Y$ such that $$ \tilde\pi_\sharp f (y) = \sum_{i=1}^n \lambda_i (\tilde\pi_\sharp f_i) (y) \quad \forall y \in Y \setminus N_{(\pmb \lambda, \pmb f)}. $$ Because the set of such pairs $(\pmb \lambda, \pmb f)$ is countable, there is a $\nu$-null set $N_1 \in \mathcal Y$ such that for each $y\in Y \setminus N_1$, $L_y$ is well-defined and $\mathbb Q$-linear on $\mathcal F$.

2. Let $\pmb \lambda = (\lambda_i)_{i=1}^n \subset \mathbb Q$, $\pmb f=(f_i)_{i=1}^n \subset F$, and $f:= \sum_{i=1}^n \lambda_i f_i$. Because $\pi_\sharp$ is $\mathbb Q$-linear and $\|\pi_\sharp f\|_{L_\infty(Y)} \le \|f\|_\infty$, we get $$ \sum_{i=1}^n \lambda_i (\tilde\pi_\sharp f_i) = \tilde\pi_\sharp (f) \le \|f\|_\infty \quad \nu \text{-a.e.} $$ So there is a $\nu$-null set $N_{(\pmb \lambda, \pmb f)} \in \mathcal Y$ such that $$ \sum_{i=1}^n \lambda_i (\tilde\pi_\sharp f_i) (y) \le \|f\|_\infty \quad \forall y \in Y \setminus N_{(\pmb \lambda, \pmb f)}. $$ Because the set of such pairs $(\pmb \lambda, \pmb f)$ is countable, there is a $\nu$-null set $N_2 \in \mathcal Y$ such that for each $y\in Y \setminus N_2$, $L_y (f) \le \|f\|_\infty$ for all $f \in \mathcal F$.

Let $N := N_1 \cup N_2$. Then $N \in \mathcal Y$ is a $\nu$-null set such that for each $y \in N^c := Y \setminus N$, we have

• $L_y$ is well-defined and $\mathbb Q$-linear.
• $L_y$ is bounded, i.e., $L_y (f) \le \|f\|_\infty$ for all $f \in \mathcal F$.

Let $\mathcal G := \operatorname{span}_{\mathbb R} (F)$. Then $\mathcal F$ is dense in $\mathcal G$. So for each $y \in N^c$, we extend $L_y$ to a $\mathbb R$-linear continuous function on $\mathcal G$. In particular, if $(f_n) \subset \mathcal F$ and $f \in \mathcal G$ such that $f_n \to f$, then $$ L_y (f) := \lim_{n \to \infty} L_y (f_n). $$

Answer 2

I have found a cleaner approach and posted it below.

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Because $X$ is compact, $C(X)$ is separable. Let $F$ be a countable dense subset of $C(X)$. Let $\mathcal F := \operatorname{span}_{\mathbb Q} (F)$. For each $f \in C(X)$, we have $\pi_\sharp f$ is an equivalence class of $L_\infty (Y)$, so we fix a representative $\tilde \pi_\sharp f \in \mathcal L_\infty (Y)$. For each $y \in Y$, we define a map $L:Y \times \mathcal F \to \mathbb C$ by $$ L(\cdot, f) := \tilde\pi_\sharp f. $$

Because $\mathcal F$ is countable and $\|\pi_\sharp f\|_{L_\infty(Y)} \le \|f\|_\infty$, there is a $\nu$-null set $N \in \mathcal Y$ such that for each $y\in N^c := Y \setminus N$ the map $L(y, \cdot)$ is $\mathbb Q$-linear continuous on $\mathcal F$.

Because $\mathcal F$ is dense in $C(X)$, there is a map $L':N^c \times C(X) \to \mathbb C$ such that for each $y \in N^c$, we have $L'(y, \cdot)$ is a $\mathbb R$-linear continuous extension of $L(y, \cdot)$. In particular, if $(f_n) \subset F$ and $f \in C(X)$ such that $f_n \to f$ in $\|\cdot\|_\infty$, then $$ L'(\cdot, f) := \lim_{n \to \infty} L(\cdot, f_n). $$

Notice that $L(\cdot, f_n)$ is Borel measurable, so is $L'(\cdot, f)$. Let's prove that for each $f \in C(X)$ we have $$ L'(\cdot, f) = \tilde\pi_\sharp f \quad \nu \text{-a.e.} $$

Because $\|f_n-f\|_\infty \to 0$, we get $\|f_n-f\|_{L_\infty (X)} \to 0$. Because $\pi_\sharp$ is continuous, we get $$ \| \pi_\sharp f_n - \pi_\sharp f \|_{L_\infty (Y)} \to 0. $$

It follows that $$ \tilde \pi_\sharp f_n \to \tilde \pi_\sharp f \quad \nu \text{-a.e.} $$

On the other hand, $$ L(\cdot, f_n) = \tilde\pi_\sharp f_n. $$

This completes the proof.