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.

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

2 Answers 2


Here probably a partial answer.

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.

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    $\begingroup$ 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. $\endgroup$ Jun 9 at 12:22
  • $\begingroup$ @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? $\endgroup$
    – Packo
    Jun 9 at 17:14
  • $\begingroup$ Oh, you're right, I got X and Y swapped in my head. Sorry. $\endgroup$ Jun 9 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 $g$.


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