# Atoms for Markov kernels

Let $$X$$ and $$Y$$ be standard Borel measurable spaces. A Markov kernel $$f : X \rightsquigarrow Y$$ is a map $$f(-|-) : \Sigma_Y \times X \to [0,1]$$ such that:

• $$f(-|x)$$ is a probability measure on $$Y$$ for every $$x \in X$$,
• $$f(S|-)$$ is a measurable function $$X \to [0,1]$$ for every $$S \in \Sigma_Y$$.

The set of atoms of a Markov kernel $$f$$ is $$A := \{y \in Y \mid \exists x \in X : \: f(\{y\}|x) > 0 \}.$$

Question: Is $$A$$ necessarily a Borel subset of $$Y$$? If not, is it at least universally measurable?

Clearly if $$X$$ or $$Y$$ is countable, then $$A$$ is indeed Borel. The non-obvious case is $$X \cong Y \cong \mathbb{R}$$.

For a concrete example, let $$f : X \rightsquigarrow X$$ be the identity kernel defined by $$f(S|x) = \delta_x(S)$$. Then the set of atoms is clearly $$Y$$ itself, hence trivially Borel.

• What is $\Sigma_Y$? Commented Jun 9, 2023 at 6:22
• @Joel The Borel $\sigma$-algebra on $Y$. Commented Jun 9, 2023 at 6:24
• Thanks, that's what I thought, but I wasn't sure. Commented Jun 9, 2023 at 6:27
• Yes, $\Sigma_Y$ is the Borel $\sigma$-algebra on $Y$. Commented Jun 9, 2023 at 6:53

I claim that the set $$A$$ is an analytic subset of $$Y$$ and thus universally measurable.

If we assume that $$X$$ and $$Y$$ are uncountable standard Borel spaces, then wlog we can assume that $$X=Y=[0,1]$$.

There we can then consider the cumulative distribution functions: \begin{align} F(y|x) &:= f([0,y]|x), & G(y|x) &:= f([0,y)|x). \end{align} One can show by making use of the their one-sided continuity in the first argument that both $$F$$ and $$G$$ are jointly measurable maps w.r.t. the product $$\sigma$$-algebra: \begin{align} F,G: Y \times X &\to [0,1]. \end{align} This shows that also the map: \begin{align} Y \times X &\to [0,1], & (y,x) & \mapsto f(\{y\}|x)=F(y|x) - G(y|x), \end{align} is measurable w.r.t. the product $$\sigma$$-algebra. This shows that the following set: \begin{align} C&:= \{ (y,x) \in Y \times X \,|\, f(\{y\}|x) >0 \} \end{align} is a Borel set of $$Y \times X$$. Since both $$X$$ and $$Y$$ are standard Borel spaces we see that the set: \begin{align} A &= \mathrm{pr}_Y(C), \end{align} is an analytic subset of $$Y$$ and thus universally measurable.

• Universal measurability is perfectly enough for our purposes, so thank you! Let me know in case that you'd like to be acknowledged with your full name. We'll include a link to this answer in our paper in any case. Commented Jun 9, 2023 at 12:22
• @NateEldredge: For Lusin-Novikov to apply, don't we need the sections $C_y$ to be countable for every $y \in Y$, in contrast to the countability $C_x$ for $x \in X$, which we would get by the countably-many-atoms-only argument? Commented Jun 9, 2023 at 17:14
• Oh, you're right, I got X and Y swapped in my head. Sorry. Commented Jun 9, 2023 at 17:47

No, the set of atoms is not necessarily a Borel set. Therefore Packo's answer is the best that one can hope for.

For a simple counterexample, let $$g : X \to Y$$ be any measurable function with non-measurable image, and consider the associated Markov kernel given by $$f(S|x) = \delta_{g(x)}(S)$$. Then the set of atoms of $$f$$ is exactly the image of $$g$$.