Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $\mathcal F\subseteq\mathcal A$ be a $\sigma$-algebra on $(\Omega,\mathcal A)$
- $(E,\mathcal E)$ be a measurable space
- $f:\Omega\times E\to\mathbb R$ be $\mathcal A\otimes\mathcal E$-measurable
I'll write $f(x)$ instead of $f(\;\cdot\;,x)$ for $x\in E$ and $f(X)$ instead of $f(\;\cdot\;,X(\;\cdot\;))$ for $X:\Omega\to E$. Assume $$\operatorname E\left[\left|f(x)\right|\right]<\infty\;\;\;\text{for all }x\in E.\tag1$$ By a monotone class argument, it's easy to show that there is a $\mathcal F\otimes\mathcal E$-measurable $g:\Omega\times E\to\mathbb R$ with $$\operatorname E\left[f(x)\mid\mathcal F\right]=g(x)\;\;\;\text{almost surely for all }x\in E.\tag2$$
I've found the following exercise in the book Stochastic Flows and Stochastic Differential Equations (Exercise 1.4.11): Let $X:\Omega\to E$ be $\mathcal F$-measurable with $$\operatorname E\left[\left|f(X)\right|\right]<\infty.\tag3$$ Show that $$\operatorname E\left[f(X)\mid\mathcal F\right]=g(X)\;\;\;\text{almost surely}.\tag4$$ How can we prove that?
Unfortunately, I don't have any idea for an argument. I don't think that the author has forgotten to add a crucial assumption, but nevertheless I've tried to consider the case where $E$ is a separable metric space, $\mathcal E$ is the Borel $\sigma$-algebra on $E$ and $f(\omega,\;\cdot\;)$ is continuous for allmost all $\omega\in\Omega$.
With that assumption it's easy to see that $(4)$ is at least satisfied if $X$ has finite range. Now, since $E$ is separable, we can find a sequence $(X_n)_{n\in\mathbb N}$ of $\mathcal F$-measurable functions $X_n:\Omega\to E$ with finite range such that $$d(X_n,X)\le d(X_{n+1},X)\;\;\;\text{for all }n\in\mathbb N\tag5$$ and $$d(X_n,X)\xrightarrow{n\to\infty}0.\tag6$$ However, I don't know how we need to proceed from here.
So, the question is: How can we prove the claim in the general case and, if that's not possible, how do we need to proceed in the described special case?
Firstly, we notice that $g$ is not unique, so we denote any valid candidate by $g_{f}$.
Consider the following ``formula'' with free variable $f$:
$\varphi(f):$ If $f:\Omega\times E\rightarrow\mathbb{R}$ is $\mathcal{A\otimes\mathcal{E}}$-measurable and $E\left[|f(x)|\right]<\infty$ for all $x\in E$, then for any $\mathcal{F}/\mathcal{E}$-measurable map $X:\Omega\rightarrow E$ with $E\left[|f(X)|\right]<\infty$, we have that $E\left[f(X)\mid\mathcal{F}\right]=g_{f}(X)$ $P$-a.e. for some valid candidate $g_{f}$.
We go to prove that $\varphi(f)$ is true for all $\mathcal{A}\otimes\mathcal{E}$-measurable map $f:\Omega\times E\rightarrow\mathbb{R}$ that satisfies $\forall x\in E$, $E\left[|f(x)|\right]<\infty$ by considering the following cases.
Case 1: $f=1_{A\times B}$, where $A\in\mathcal{A}$ and $B\in\mathcal{E}$. By direct verification, $g_{f}(x)=1_{B}(x)E\left[1_{A}\mid\mathcal{F}\right]$ is a valid candidate. (Here, we fix a choice for the conditional expectation $E\left[1_{A}\mid\mathcal{F}\right]$. Let $X$ be such a map, then \begin{eqnarray*} E\left[f(X)\mid\mathcal{F}\right] & = & E\left[1_{A}1_{B}\circ X\mid\mathcal{F}\right]\\ & = & 1_{B}\circ X\cdot E\left[1_{A}\mid\mathcal{F}\right]\\ & = & g_{f}(X) (a.e.) \end{eqnarray*} by observing that $1_{B}\circ X:\Omega\rightarrow\mathbb{R}$ is $\mathcal{F}$-measurable. This shows that $\varphi(f)$ is true whenever $f$ is of the form $f=1_{A\times B},$ where $A\in\mathcal{A}$ and $B\in\mathcal{E}$.
Case 2: $f=1_{C}$, where $C\in\mathcal{A}\otimes\mathcal{E}$. Define $\mathcal{P=}\{A\times B\mid A\in\mathcal{A}\mbox{ and }B\in\mathcal{E}$} and $\mathcal{D}=\{C\in\mathcal{A}\otimes\mathcal{E}\mid\varphi(1_{C})\mbox{ is true.}\}$. Clearly $\mathcal{P}$ is a $\pi$-class (i.e., $\mathcal{P}$ is closed under finite intersection.) It is routine to check that $\mathcal{D}$ is a $\lambda$-class (in the sense that: $\emptyset\in\mathcal{D}$, $C^{c}\in\mathcal{D}$ whenever $C\in\mathcal{D}$, and $\cup_{i=1}^{\infty}C_{i}\in\mathcal{D}$ whenever $C_{i}\in\mathcal{D}$ with $C_{i}\cap C_{j}=\emptyset$ whenever $i\neq j$.). By Case 1, $\mathcal{P}\subseteq\mathcal{D}$. By Dynkin Theorem, we have $\sigma(\mathcal{P})\subseteq\mathcal{D}$. Note that $\mathcal{D}\subseteq\mathcal{A}\otimes\mathcal{E}=\sigma(\mathcal{P})$. Therefore $\mathcal{D}=\mathcal{A}\otimes\mathcal{E}$.
Case 3: $f$ is a simple $\mathcal{A}\otimes\mathcal{E}$-measurable function. Argue by linearity and by the result of Case 2.
Case 4: $f:\mathcal{A}\otimes\mathcal{E}\rightarrow[0,\infty)$ is a non-negative $\mathcal{A}\otimes\mathcal{E}$-measurable function. Argue by monotone convergence theorem (Here, we also need monotone convergence theorem, conditional expectation version).
Case 5: $f:\mathcal{A}\otimes\mathcal{E}\rightarrow\mathbb{R}$. Write $f=f^{+}-f^{-}$.