Theorem $4$ of this blog entry of Terrence Tao states the following:
Let $X$ be a compact metric space, $\mathcal X$ be the Borel $\sigma$-algebra of $X$, and $\mu$ be a probability measure on $X$. Let $(Y, \mathcal Y, \nu)$ be a measure space and $\pi:X\to Y$ be a measure preserving (measurable) map. Then there is a map $y\mapsto \mu_y:Y\to \mathscr P(X)$, where $\mathscr P(X)$ is the set of all the probability measures on $X$, such that $$\int_X f\cdot (g\circ \pi)\ d\mu = \int_Y \left(\int_X f\ d\mu_y\right)g(y)\ d\nu(y)$$ for all bounded measurable maps $f:X\to \mathbb C$ and $g:Y\to \mathbb C$. Further, we have $$g\circ \pi=g(y),\ \mu_y\text{-a.e. in } X$$ for y $\nu$-a.e. in $Y$.
The proof given is as follows: We have the pullback map $\pi^\sharp:L^2(Y, \mathcal Y, \nu)\to L^2(X, \mathcal X, \mu)$. We take it's adjoint to get a map $\pi_\sharp:L^2(X, \mathcal X, \mu)\to L^2(Y, \mathcal Y, \nu)$. It is clear that $\|\pi_\sharp f\|_{L^\infty(Y)}\leq \|f\|_{C(X)}$ for all $f\in C(X)$.
Now here is what I do not follow:
Since $C(X)$ is separable, we can find representatives $\tilde \pi_\sharp f$ of $\pi_\sharp f$ for every $f\in C(X)$ such that these representatives vary linearly with $f$.
Can somebody explain to me how the last line holds? Thanks.