I've asked this question also on mathematics stackexchange, but despite nearly two dozen views, there isn't a single comment, nevermind an answer. Any help would be appreciated.
Update: See update 1 at bottom.
Let $(X,\mathcal A)$ and $(Y,\mathcal F)$ be measurable spaces.
Consider a probability kernel $\kappa : X \times \mathcal F \to [0,1]$.
I need to formalize the notion of a nonrandomized probability kernel. Two natural definitions are:
- for all $x \in X$, exists $y \in Y$, $\kappa(x,\{y\}) = 1$.
- for all $x \in X$, exists $y \in Y$, for all $A \in \mathcal F$, $(\kappa(x,A) = 1 \iff y \in A)$.
I believe the two definitions are equivalent if the singletons are measurable in $\mathcal F$. (Agree?)
In either case, consider the double integral: $$ \Phi = \int_X \Bigl \{ \int_Y f(x,y) \kappa(x,dy) \Bigr \} \mu(dx), $$ where $f$ is product measurable and $\mu$ is a probability measure on $(X,\mathcal A)$.
If $\kappa$ is nonrandomized (as in Defn 1 or 2 above), when can I assume that there exists a(n ostensibly measurable?) function $g : X \to Y$ such that $$ \Phi = \int_X f(x,g(x)) \mu(dx) $$ holds? (We can assume $f$ is integrable with respect to $\mu \otimes \kappa$, or alternative that $f$ is nonnegative (or nonpositive).)
Update 1
Commenters rightly pointed out that the question seems trivial. Indeed, the case where we assume definition 1 is straightforward and I followed the outline provided by Nate Eldredge to give sketched proofs.
The case where we assume merely definition 2 is still not clear to me. There may not be a unique $y$ for each $x$, and then we would need some sort of measurable selection, and I'm not versed in the requisite theorems. It would seem that I would need some structure on $Y$ beyond a $\sigma$-algebra. E.g., the Kuratowski and Ryll-Nardzewski measurable selection theorem would seem to require $(Y,\mathcal F)$ to be a Polish space with its Borel $\sigma$-algebra, but I believe that would imply that singletons are measurable, and so then the definitions collapse.