Nonrandomized probability kernels 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. 
 A: This is a partial answer.  See updated question for seemingly trickier question.
Claim. Definitions 1 and 2 are equivalent if singletons are measurable.  
Proof. To see this, consider the second definition, pick $x \in X$, and let $y \in Y$ be as in the definition.  Since $y \in \{y\}$ and $\{y\}$ is measurable, $\kappa(x,\{y\}) = 1$. In the other direction, $\kappa(x,\{y\}) = 1$, which then implies, by the monotonicity of probability measures, that $\kappa(x,A) = 1$ if $y \in A$. If $y \not\in A$, then, again because $\{y\}$ is measurable and basic facts about probability distributions, $\kappa(x,A) = 1 - \kappa(x,Y \setminus A) \le 1 - \kappa(x,\{y\}) = 0$.
Note: I believe that the measurability of singletons is not necessary for the equivalence because it suffices that for all $x \in X$, there exists $y \in Y$, such that the singleton $\{y\}$ is measurable.
If we adopt definition 1, the comments by Nate Eldredge lead to the following proof.
Claim. Assume definition 1. Then such a $g$ exists and is measurable.
Proof. Define $g(x)$ to be the unique $y$ satisfying $\kappa(x,\{y\})=1$. To see that $g$ is measurable, note that, for all $A \in \mathcal F$, $\kappa(\cdot,A)$ is measurable and $$\begin{align}g^{-1}(A) &= \{ x \in X : (\exists y \in A)\, \kappa(x,\{y\}) = 1 \}\\ &= \{ x \in X : \kappa(x,A) = 1 \}\\ &= \kappa(\cdot,A)^{-1}(\{1\}).\end{align}$$ 
Then, for all $x \in X$,
$$
\int_Y f(x,y) \kappa(x,dy) = f(x,g(x)),
$$
and this quantity is $\mathcal A$-measurable because the l.h.s. was assumed to be.
