This is cross-posted at [MSE](https://math.stackexchange.com/q/3347019). 


I'm looking for a reference for the following result. It seems like it must be known, or follow quickly from something known, but I have not been able to find it in any of the textbooks I have.


Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. Let $\mathcal G$ be a sub-$\sigma$-field. Let $\mu$ be a regular conditional probability for $\mathbb P$ given $\mathcal G$. By this I mean that $\mu: \mathcal F \times \Omega \to [0,1]$ is a probability measure (denoted $\mu\_\omega$) in its first argument, $\mathcal G$ measurable in its second argument and satsifies
$$\int\_G \mu\_\omega(A)\mathbb P(d\omega) = \mathbb P(A \cap G)$$
for all $A \in \mathcal F$ and $G \in \mathcal G$. 
Let $X: \Omega \times \Omega \to \mathbb R$ be bounded and $\mathcal G \otimes \mathcal F$ measurable. Let 
$$EX(\omega) = \int X(\omega, v)\mu\_\omega(dv).$$


*Theorem.* $EX$ is $\mathcal G$ measurable and 
$$\int EX(\omega)\mathbb P(d\omega) = \int X(\omega,\omega)\mathbb P(d\omega).$$



What is this disintegration-like theorem?
The measure-theoretic formulation of conditional probability/expectation is cumbersome, so it is often easy to lose connection between intuitive facts and the formal demonstrations of them. I personally often have difficulty with arguing for seemingly obvious facts.


In this case, I think more than your theorem is true: your function $EX(\omega)$ is a regular version of the conditional expectation of $f(\omega):=X(\omega,\omega)$ given $\mathcal{G}$, i.e., $\mathbb{E}[f\,|\,\mathcal{G}]$. By definition, this means that $EX(\omega)$ is $\mathcal{G}$-measurable and
\begin{align}
 \underbrace{\int\_G EX(\omega)\,\mathrm{d}\mathbb{P}(\omega)}\_{\int\_G \mathbb{E}[f\,|\,\mathcal{G}]\,\mathrm{d}\mathbb{P}(\omega)} &=
 \underbrace{\int\_G X(\omega,\omega)\, \mathrm{d}\mathbb{P}(\omega)}\_{\int\_G f(\omega) \mathrm{d}\mathbb{P}(\omega)} \;,
\end{align}
for every $G\in\mathcal{G}$.


This is almost immediate from your definition. Note that


$\quad$($\star$) for each $\omega$, we have $X(\omega,v)=X(v,v)$ for $\mu\_\omega$-almost every $v$ (see @1 below).


Thus, $EX(\omega)=\int f(v)\,\mathrm{d}\mu\_\omega(v)$. That $\int f(v)\,\mathrm{d}\mu\_\omega(v)$ is $\mathcal{G}$-measurable for every $f$ is standard (see @2 below).
Furthermore,
\begin{align}
 \int\_G EX(\omega)\,\mathrm{d}\mathbb{P}(\omega) &=
 \int\_G\int f(v)\,\mathrm{d}\mu\_\omega(v)\mathrm{d}\mathbb{P}(\omega) \\
 &= \int\_G f(v)\,\mathrm{d}\mathbb{P}(v) \;.
\end{align} 


**Further details:**


@1: I believe this can be shown using the usual measure theory trick. Namely, first show that ($\star$) holds if $X$ is the indicator of a rectangle $G\times F$ with $G\in\mathcal{G}$ and $F\in\mathcal{F}$. Then, show that the family of sets $A\in\mathcal{G}\otimes\mathcal{F}$ for which $X(\cdot,\cdot):=1\_A$ satisfies ($\star$) is a $\sigma$-algebra. Then, show that $(\star)$ holds if $X$ is a simple function. Finally, show that if $X$ is the uniform limit of an increasing sequence of functions satisfying ($\star$), then $X$ itself satisfies ($\star$).


@2: The argument for this again uses the usual trick. We already know (by definition) that $\int f(v)\,\mathrm{d}\mu\_\omega(v)$ is $\mathcal{G}$-measurable if $f$ is an indicator function. First, show that the same holds for simple function. Then, show that if $f$ is the uniform limit of an increasing sequence $f\_1\leq f\_2\leq\cdots$ such that $\int f\_1(v)\,\mathrm{d}\mu\_\omega(v)$ is $\mathcal{G}$-measurable, then $\int f(v)\,\mathrm{d}\mu\_\omega(v)$ is also $\mathcal{G}$-measurable.


