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.



Atoms for Markov kernels
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.


