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Let us discuss the following assertion, where $(\ast)$ stands for some extra assumption to be specified.

--- (A) --- Let $(\Omega,\mathfrak A,\mathbb P)$ be a probability space. Let $\mathfrak B\subset \mathfrak A$ be a sub-$\sigma$-field, $n\in\mathbb N$. Let $(X,Y) \in L^0((\Omega,\mathfrak B),\mathbb R^n) \times L^0((\Omega,\mathfrak A),\mathbb R^n)$ be, respectively, $\mathfrak B$- and $\mathfrak A$-measurable random vectors of our space, and $f$ a bounded Borel function on $\mathbb R^{2n}$. Then, if $(\ast)$ holds, we have, for almost any $\omega\in\Omega$,

\begin{equation} \mathbb E[f(X,Y) \mid \mathfrak B] (\omega) = \mathbb E[f(X(\omega),Y) \mid \mathfrak B] (\omega). \end{equation}

Let us continue the discussion.

--- (1) --- If the probability space if finite (= $(\ast)$), the assertion is easy to prove since every random vector takes only a finite number of values, so it suffices to prove the assertions for functions having the form $f = \mathbf 1_{A\times B}$ where $A,B\in \mathfrak B(\mathbb R^n)$ are Borel sets.

--- (2) --- This can be extended to the case where $\Omega$ is countable (= $(\ast)$), by approximation of $f$ with simple functions being linear combinations of the above-mentioned type. But the argument is based upon the assumption that the union of $|\Omega|$ many null sets is still a null set.

--- (3) --- It is rather easy to see that the assertion holds for general probability spaces if $Y$ and $\mathfrak B$ are independent (= $(\ast)$).

--- (4) --- Whence the question: Is (A) true in general, i.e. $(\ast)$ being void? I suppose no, but I did not find a convenient counterexample. Do you know one? It would be great to obtain a useful answer to this.

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Here is the proof (of the general case) given in Gikhman & Skorohod: The Theory of Stochastic Processes I:

Define $g(x,\omega):=\mathsf{E}[f(x,Y)\mid \mathfrak{B}](\omega)$. Then your assertion is equivalent to $$ \mathsf{E}[f(X,Y)\mid \mathfrak{B}](\omega)=g(X(\omega),\omega) \quad\text{a.s.}\tag{1}\label{1} $$ First, it is clear that $\eqref{1}$ holds for functions of the form $f_n(x, y)=\sum_{k=1}^n \varphi_k(x)h_k(y)$. For an arbitrary function $f(x,y)$ with $\mathsf{E}|f(X,Y)|<\infty$ the assertion follows from the existence of a sequence of functions $f_n(x,y)$ of the previous type s.t. $f_n(X,Y)\to f(X,Y)$ a.s. and in $L^1$.

The joint measurability of $g(x,\omega)$ w.r.t. $\mathcal{B}(\mathcal{X})\otimes \mathfrak{B}$ follows from the result of this paper. Specifically, if for each $x\in \mathcal{X}$, $\mathsf{E}|f(x,Y)|<\infty$, then there exists a (jointly) measurable process $g$ s.t. $g(x,\cdot)$ is a version of $\mathsf{E}[f(x,Y)\mid \mathfrak{B}]$ for each $x\in\mathcal{X}$.

Update: A more useful result, which holds for any random elements $X$ and $Y$ s.t. $Y$ admits a regular version of the conditional probability $\mathsf{P}(Y\in \cdot\mid \mathfrak{B})$, is given in Kallenberg: Foundations of Modern Probability (Theorem 5.4 on page 85):

Theorem. Suppose that the random elements $X$ and $Y$ take values in $S$ and $T$, and $\mathsf{P}(Y\in \cdot\mid \mathfrak{B})$ has a regular version $\nu$. Then for any measurable function $f$ on $S\times T$, $$ \mathsf{E}[f(X,Y)\mid \mathfrak{B}](\omega)=\int f(X(\omega),y)\nu(dy,\omega) \quad\text{a.s.} $$

  • Last step uses Dominated Convergence right ? Then you need $|f_n|\leq |f|$ or something similar right ? I still agree such a sequence exists in $L^1$. – P. Quinton Nov 26 '19 at 11:03
  • You use that $Z_n \longrightarrow Z$ a.s., where $Z(\omega) = g(X(\omega),\omega)$, $\omega\in\Omega$, $Z_n$ defined similarly (define $g_n$ via $f_n$ and proceed as above). Why does this hold true? That was exactly the crucial point of the question. --- What is your argument? It is no more possible in the general setting to make use of the following argument: For any $\omega$ there is a null set $N_\omega$ such that on its complement, $f_n(X(\omega),Y) \longrightarrow f(X(\omega),Y)$, so that $Z_n\longrightarrow Z$ on the complement of $\bigcup_{\omega\in\Omega} N_\omega$. – Kehrwert Nov 26 '19 at 11:06
  • By the way: Is there an argument for proving the assertion that $g$ is measurable on $(\mathbb R^n×\Omega,\mathfrak B(\mathbb R^n)⊗\mathfrak B)$? If $\mathfrak B$ contains only two elements, it follows from the results on so-called "parameter integrals", I think. But in general? – Kehrwert Nov 26 '19 at 11:40