How to write conditional expectation as integral with respect to regular conditional distribution 
Can I write a conditional expectation as an integral with respect to a regular conditional distribution?

Set Up

*

*$(\Omega, \mathcal{F}, \mathbb{P})$ probability space

*$(\mathsf{E}, \mathcal{E})$ measurable space with reference measure $\lambda_{\mathcal{E}}$

*$X:\Omega\to\mathsf{E}$ random variable with $\mathbb{E}[|X|] < \infty$ and distribution $P_X = X_*\mathbb{P}$ with $P_X \ll \lambda_{\mathcal{E}}$ and density $p_x = dP_X / d\lambda_{\mathcal{E}}$

*$\mathcal{G}\subseteq \mathcal{F}$ sub-sigma algebra

*$\nu:\mathsf{E}\times\mathcal{F}\to[0, 1]$ is a regular conditional probability. In other words a kernel satisfying
$$
\mathbb{P}(\mathsf{A}\cap X^{-1}(\mathsf{B})) = \int_{\mathsf{B}} \nu(x, A) P_x(dx)
$$

*$h:\mathsf{E}\to\mathbb{R}$ be a sufficiently well-behaved function (probably measurable?)

Claim
According to here and here it is possible to write $\mathbb{E}[h(X) \mid \mathcal{G}]$ as an integral of $h(X)$ with respect to the regular conditional probability.
$$
\mathbb{E}[h(X) \mid \mathcal{G}] = \int_? h \circ X\,\, \nu(?, ?)
$$
 A: Yes it is possible. Suppose $\nu : \Omega \times \mathcal{E} \to [0, 1]$ is a regular conditional distribution of $X$ given $\mathcal{G}$. This means that

*

*For every $\omega \in \Omega$, $\nu(\omega, \cdot)$ is a probability measure on $\mathcal{E}$.

*For every $A \in \mathcal{E}$, $\nu(\cdot, A)$ is $\mathcal{G}-\mathcal{B}(\mathbb{R})$ measurable.

*For every $A \in \mathcal{E}$, $P(X \in A \mid \mathcal{G})(\omega) = \nu(\omega, A)$ for a.e. $\omega \in \Omega$.

Condition 3 says that for any $A \in \mathcal{E}$,
$$E(1_A(X) \mid \mathcal{G})(\omega) = \int_{E}1_{A}(x)\nu(\omega, dx) \hspace{20pt} \text{ for a.e. } \omega \in \Omega.$$
By nonnegative linearity of conditional expectations and integrals we get that for any simple function $f : (E, \mathcal{E}) \to [0, \infty)$,
$$E(f(X) \mid \mathcal{G})(\omega) = \int_{E}f(x)\nu(\omega, dx) \hspace{20pt} \text{ for a.e. } \omega \in \Omega.$$
Then for a measurable $f : E \to [0, \infty)$ with $E(f(X)) < \infty$, take a sequence of simple functions $f_n \to f$ and apply the previous result and the monotone convergence theorem for conditional expectations and integration to deduce the result for this $f$. Then apply linearity to deduce the result for real valued $f$ with $E(|f(X)|) < \infty$.
