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:

  1. for all $x \in X$, exists $y \in Y$, $\kappa(x,\{y\}) = 1$.
  2. 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.

  • $\begingroup$ Maybe I am missing something, but this seems pretty trivial? You have assumed that for each $x$ there exists $y$ such that $\kappa(x,\{y\})=1$, and by definition of a probability kernel, such $y$ is unique. So just define $g(x)$ to be that unique $y$, and it's clear that $\int f(x,y) \kappa(x,dy) = f(x, g(x))$. Since probability kernels are measurable in the first variable, you will get that $g^{-1}(\{y\})$ is measurable for each singleton $\{y\}$, but $g$ need not be measurable. $\endgroup$ Commented Dec 28, 2016 at 17:02
  • $\begingroup$ Correction: $g$ is measurable, since for any measurable $A \subset Y$, we have $g^{-1}(A) = \kappa(\cdot, A)^{-1}(\{1\})$. $\endgroup$ Commented Dec 28, 2016 at 17:30
  • 3
    $\begingroup$ I suggest you work out the details and post your own answer. $\endgroup$ Commented Dec 28, 2016 at 17:30
  • $\begingroup$ I will do that. $\endgroup$ Commented Dec 28, 2016 at 18:42

1 Answer 1


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.


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