By change-of-variables formula, the bounded linear operator
$$
\pi^\sharp: L_1(Y) \to L_1(X), g \mapsto g \circ \pi
$$
is well-defined and $\|\pi^\sharp\| = 1$. We denote its adjoint by
$$
\pi_\sharp : L_\infty(X) \to L_\infty(Y).
$$
We have the duality
$$
\int_Y (\pi_\sharp f) g \mathrm d \nu = \int_X f (\pi^\sharp g) \mathrm d \mu \quad \forall f \in L_\infty(X), \forall g \in L_1(Y).
$$
Also,
$$
\|\pi_\sharp f\|_{L_\infty(Y)} \le \|\pi^\sharp\| \cdot \|f\|_{L_\infty(X)} \le \|f\|_\infty \quad \forall f \in \mathcal C(X).
$$
For each $f \in \mathcal C(X)$, we have $\pi_\sharp f$ is an equivalence class of $L_\infty (Y)$, so we fix a representative $\tilde \pi_\sharp f \in \mathcal L_\infty (Y)$. Let $\pmb 1 \in \mathcal C(X)$ be the constant function.
- Claim 1. $\tilde{\pi}_{\sharp} \pmb 1 = 1$ $\nu$-a.e.
Proof: By the duality,
$$
\int_Y (\pi_\sharp \pmb 1) g \mathrm d \nu = \int_X \pmb 1(\pi^\sharp g) \mathrm d \mu = \int_Y g \mathrm d \nu \quad \forall g \in L_1(Y).
$$
It follows that $\int_Y (\pi_\sharp \pmb 1 - 1) g \mathrm d \nu = 0$ for all $g \in L_1(Y)$. Because $\nu$ is finite, we get $L_\infty(Y) \subset L_1(Y)$. So we pick $g := \pi_\sharp \pmb 1 - 1 \in L_1(Y)$, and obtain $\int_Y (\pi_\sharp \pmb 1 - 1)^2 \mathrm d \nu = 0$. The claim then follows.
Because $X$ is compact, $\mathcal C(X)$ is separable. Let $F$ be a countable dense subset of $\mathcal C(X)$ and $\mathcal F := \operatorname{span}_{\mathbb Q} (F)$. For each $y \in Y$, we define a map
$$
L: \mathcal F \times Y \to \mathbb C
$$
by
$$
L(f, \cdot) := \tilde\pi_\sharp f \quad \forall f \in \mathcal F.
$$
Because $\mathcal F$ is countable, $\pi_\sharp$ is linear, and $\|\pi_\sharp f\|_{L_\infty(Y)} \le \|f\|_\infty$, there is a $\nu$-null set $N \in \mathcal Y$ such that for each $y\in N^c := Y \setminus N$,
- $L(\pmb 1, y) = 1$, and
- $L(\cdot, y)$ is $\mathbb Q$-linear continuous on $\mathcal F$.
We fix some $\overline y \in N^c$ and re-define $L$ by
$$
L(\cdot, y) := L(\cdot, \overline y) \quad \forall y \in N.
$$
Then for each $y\in Y$,
- $L(\pmb 1, y) = 1$, and
- $L(\cdot, y)$ is $\mathbb Q$-linear continuous on $\mathcal F$.
Because $\mathcal F$ is dense in $\mathcal C(X)$, there is an extension
$$
L: \mathcal C(X) \times Y \to \mathbb C
$$
such that for each $y \in Y$,
- $L(\pmb 1, y) = 1$, and
- $L(\cdot, y)$ is $\mathbb R$-linear continuous on $\mathcal C(X)$.
In particular, if $(f_n) \subset F$ and $f \in \mathcal C(X)$ such that $f_n \to f$ in $\|\cdot\|_\infty$, then
$$
L(f, \cdot) = \lim_{n \to \infty} L(f_n, \cdot).
$$
In other words, the sequence $\{L(f_n, \cdot) \mid n \in \mathbb N\}$ converges pointwise and everywhere to $L(f, \cdot)$. It follows that $L(f, \cdot)$ is measurable.
- Claim 2. $L(f, \cdot) = \tilde\pi_\sharp f$ $\nu$-a.e. for each $f \in \mathcal C(X)$.
Proof: 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 $\tilde \pi_\sharp f_n \to \tilde \pi_\sharp f$ $\nu$-a.e. On the other hand, $L(f_n, \cdot) = \tilde\pi_\sharp f_n$. The claim then follows.
By Riesz–Markov–Kakutani theorem, for each $y \in Y$, there is a unique regular Borel measure $\mu_y$ on $X$ such that
$$
L(f, y) = \int_X f \mathrm d \mu_y \quad \forall f \in \mathcal C(X).
$$
It follows from $L(\pmb 1, y) = 1$ that $\mu_y$ is a probability measure. Also, for all $f \in \mathcal C(X)$ and for all $g \in L_1(Y)$,
$$
\int_X f (\pi^\sharp g) \mathrm d \mu = \int_Y (\pi_\sharp f) g \mathrm d \nu = \int_Y (\tilde \pi_\sharp f) g \mathrm d \nu = \int_Y L(f, \cdot) g \mathrm d \nu = \int_Y \left ( \int_X f \mathrm d \mu_y \right ) g (y) \mathrm d \nu (y).
$$
Let $\tau$ be the metric topology of $X$, the map $g:Y\to \mathbb C$ be $\nu$-integrable map. and
$$
\mathcal H_g := \left \{f:X \to \mathbb C \text{ bounded measurable} \,\middle\vert\, y \mapsto\int_X f \mathrm d \mu_y \text{ is measurable, and } (\star) \text{ holds for } f \right\}.
$$
- Claim 3.a. If $O \in \tau$ then $1_A \in \mathcal H_g$.
Proof: There is a sequence $(f_n) \subset \mathcal C(X)$ of non-negative functions such that $f_n \nearrow 1_O$ pointwise and everywhere. By DCT,
$$
\int_X f_n (g \circ \pi) \mathrm d \mu \to \int_X 1_O (g \circ \pi) \mathrm d \mu \quad \text{and} \quad \int_X f_n \mathrm d \mu_y \to \int_X 1_O \mathrm d \mu_y.
$$
We have $y \mapsto \int_X f_n \mathrm d \mu_y$ is equal to $L(f_n, \cdot)$ which is measurable, so $y \mapsto \int_X 1_O \mathrm d \mu_y$ is also measurable. By DCT again,
$$
\int_Y \left ( \int_X f_n \mathrm d \mu_y \right ) g (y) \mathrm d \nu (y) \to \int_Y \left ( \int_X 1_O \mathrm d \mu_y \right ) g (y) \mathrm d \nu (y).
$$
The claim then follows from
$$
\int_X f_n (g \circ \pi) \mathrm d \mu = \int_Y \left ( \int_X f_n \mathrm d \mu_y \right ) g (y) \mathrm d \nu (y) \quad \forall n.
$$
- Claim 3.b. Let $(f_n) \subset \mathcal H_g$ be an increasing sequence of non-negative functions. If $f_n$ converges pointwise and everywhere to a bounded function $f$, then $f \in \mathcal H_g$.
Proof: Notice that $f$ is measurable. We then obtain the claim by applying DCT (as in the proof of Claim 3.a.).
- Claim 3.c. $(\star)$ holds for all bounded measurable map $f:X\to \mathbb C$.
Proof: By linearity of integral, if $f, g \in \mathcal H$ and $c \in \mathbb R$, then $f +c g \in \mathcal H_g$. The claim then follows from monotone class theorem for functions.
- Claim 4. For $\nu$-a.e. $y \in Y$, we have $g\circ \pi=g(y)$ $\mu_y$-a.e.
Proof: Now let $f:X \to \mathbb C$ and $g,h:Y \to \mathbb C$ be bounded measurable. We apply $(\star)$ with $(f(g\circ\pi), h)$ in place of $(f, g)$ and get
$$
\int_Y\left(\int_X f(g \circ \pi)\mathrm d\mu_y\right) h(y)\mathrm d \nu(y) = \int_X f((gh)\circ \pi)\mathrm d\mu.
$$
We apply $(\star)$ with $(f, gh)$ in place of $(f, g)$ and get
$$
\int_X f((gh)\circ \pi)\mathrm d\mu = \int_Y\left(\int_X f \mathrm d\mu_y\right) (gh)(y)\mathrm d \nu(y) = \int_Y\left(\int_X f g(y)\mathrm d\mu_y\right) h(y)\mathrm d \nu(y).
$$
Thus
$$
\int_Y\left(\int_X f(g \circ \pi)\mathrm d\mu_y\right) h(y)\mathrm d \nu(y) = \int_Y\left(\int_X f g(y)\mathrm d\mu_y\right) h(y)\mathrm d \nu(y).
$$
So
$$
\int_Y\left(\int_X f(g \circ \pi)\mathrm d\mu_y - \int_X f g(y)\mathrm d\mu_y\right) h(y)\mathrm d \nu(y) = 0.
$$
We pick a particular $\bar h$ by
$$
\bar h (y ) := \int_X f(g \circ \pi)\mathrm d\mu_y - \int_X f g(y)\mathrm d\mu_y.
$$
Clearly, $\bar h:Y \to \mathbb C$ is bounded measurable. So
$$
\int_X f(g \circ \pi)\mathrm d\mu_y = \int_X f g(y)\mathrm d\mu_y \quad \nu\text{-a.e.}
$$
As such, for $\nu$-a.e. $y \in Y$, we get
$$
\int_X f [g \circ \pi - g(y)] \mathrm d\mu_y=0.
$$
Again, we pick a particular $\bar f: X \to \mathbb C$ by $\bar f := g \circ \pi - g(y)$. Clearly, $\bar f$ is bounded measurable. Then $g \circ \pi - g(y) =0$ $\mu_y$-a.e.
This completes the proof.
