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.