Theorem 4 of this blog entry of Terrence Tao states that:
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 $f \in C(X)$. We can then apply the Riesz representation theorem to obtain a Radon probability measure $\mu_y$ such that
$$
\tilde{\pi}_{\sharp} f(y)=\int_X f \mathrm d \mu_y
$$
for all such $y$. We set $\mu_y$ equal to some arbitrarily fixed Radon probability measure for $y \in E$. We then observe that the required properties (including the measurability of $y \mapsto \int_X f \mathrm d \mu_y$ are already obeyed for $f \in C(X)$.
My understanding: Riesz theorem gives us a regular Borel measure $\mu_y$. Could you please explain why $\mu_y$ is a probability measure for $\nu$-a.e. $y\in Y$?
Best Answer
Let $\pmb 1 \in C(X)$ be the constant function. It suffices to prove that $$ \int_X \pmb 1 \mathrm d \mu_y = 1 \quad \text{or equivalently} \quad \tilde{\pi}_{\sharp} \pmb 1 = 1 \quad \nu \text{-a.e.} $$
By the duality, we have for all $g \in L_1(Y)$, $$ \int_Y (\pi_\sharp \pmb 1) g \mathrm d \nu = \int_X \pmb 1(\pi^\sharp g) \mathrm d \mu = \int_X g \circ \pi \mathrm d \mu = \int_Y g \mathrm d \nu. $$
It follows that for all $g \in L_1(Y)$ $$ \int_Y (\pi_\sharp \pmb 1 - 1) g \mathrm d \nu = 0. $$
Because $\nu$ is finite, we get $L_\infty(Y) \subset L_1(Y)$. So $g := \pi_\sharp \pmb 1 - 1 \in L_1(Y)$. As such, $$ \int_Y (\pi_\sharp \pmb 1 - 1)^2 \mathrm d \nu = 0. $$
It follows that $\pi_\sharp \pmb 1 - 1 =0$ $\nu$-a.e. This completes the proof.