Borel kernel over an analytic set implies existence of a Borel map Let $X$ and $Y$ be standard Borel spaces, and let $A\subseteq X\times Y$ be an analytic set with a full projection on $X$: that is $\pi_X(A) = X$. Suppose that there exists a Borel-measurable kernel $\mu:X\to\mathcal P(Y)$ such that $\mu(x,A_x) = 1$ for all $x\in X$ where 
$$
  A_x = \{y\in Y:(x,y)\in A\}
$$ 
is an $x$-section of $A$. Does it follow that there exist a Borel map $f:X\to Y$ such that whose graph is a subset of $A$, i.e. $\mathrm{Gr}[f]\subseteq A$?
Some comments:


*

*There always exists a universally measurable $g:X\to Y$ whose graph is a subset of $A$.

*Clearly, the converse fact always hold true. Given the existence of such $f$ one can come up with a Borel kernel $\mu(C|x) = \delta_{f(x)}(C)$.

*In case $A$ is Borel, the existence of Borel $f$ follows from the existence of Borel $\mu$ as proven e.g. in Controlled Markov processes or can be also found here.
 A: The answer is no since, according to excercise 36.25 in A. Kechris: Classical Descriptive Set Theory, there should be an analytic set $A\subseteq \omega^\omega\times\omega^\omega$ whose vertical sections exclude at most a single point (i.e. $|\omega^\omega\setminus A_x|\leq 1$ for each $x$) which nevertheless does not admit a Borel uniformization. Right now, however, I don't see how to find the set :-)
A: The following is a construction of an analytic set $A\subseteq\omega^\omega\times\omega^\omega$ whose vertical sections exclude at most a single point, and which does not admit a Borel uniformization. This is as stated in jonathanverner's answer, although that did not include a construction - only mentioning exercise 36.25 in Kechris: Classical Descriptive Set Theory. The construction below is after viewing the hint in Kechris's book. I'll use the notation $\mathcal{N}$ for Baire space $\omega^\omega$.

Construction. Let $S$ be a universal subset of $\mathcal{N}\times\mathcal{N}^3$ (i.e., $S$ is closed and every closed subset of $\mathcal{N}^3$ is a vertical section $S_x$ of $S$ for some $x\in\mathcal{N}$). Then, let $A\subseteq\mathcal{N}^2$ consist of the points $(x,y)$ such that whenever, for any such $x$, there is a unique $(u,v)\in\mathcal{N}^2$ with $(x,x,u,v)\in S$ we have $y\not=u$. The set $A$ satisfies the required properties.

It is clear the sections $A_x$ exclude at most a single point. More precisely, it excludes the point $y$ if and only if there is a unique $(u,v)\in\mathcal{N}^2$ with $(x,x,u,v)\in S$ and $y=u$.
$A$ does not have a Borel section: Suppose that $\Gamma\subseteq\mathcal{N}^2$ is Borel. I'll now use the fact that every Borel subset of a Polish space is a continuous bijective image of a closed subset of $\mathcal{N}$ (Kechris, Theorem 13.7). By taking the graph of such a continuous bijection, there exists a closed set $C\subseteq\mathcal{N}^2\times\mathcal{N}$ whose projection onto $\mathcal{N}^2$ is one-to-one and has image $\Gamma$. Write $C=S_x$ for some $x\in\mathcal{N}$. Then, $(x,y)\in\Gamma$ if and only if $(x,x,y,v)\in S$ for some $v$, which is then unique. If $\Gamma_x$ consists of the single point $y$ then $(y,v)$ is unique such that $(x,x,y,v)\in S$ and, by construction, $(x,y)\not\in A$. So, $\Gamma$ is not a section of $A$.
$A$ is analytic: The set $T=\lbrace(x,y,z)\colon(x,x,y,z)\in S\rbrace$ is a closed subset of $\mathcal{N}^3$ and $(x,y)\in A$ iff whenever there is unique $(u,v)\in\mathcal{N}^2$ with $(x,u,v)\in T$ we have $y\not=u$. We can write $A=B\cup(\mathcal{N}^2\setminus C)$ where
$$
\begin{align}
&B = \left\lbrace(x,y)\in\mathcal{N}^2\colon(x,y^\prime,z)\in T, {\rm some\ }y^\prime\not=y\right\rbrace\\
&C=\left\lbrace(x,y)\in\mathcal{N}^2\colon\exists ! z\in\mathcal{N}{\rm\ with\ }(x,y,z)\in T\right\rbrace.
\end{align}
$$
As $B$ is the projection onto $\mathcal{N}^2$ of the (Borel) set of $(x,y,y^\prime,z)\in\mathcal{N}^2\times\mathcal{N}^2$ with $(x,y^\prime,z)\in T$ and $y\not=y^\prime$, it is analytic. Finally, $\mathcal{N}^2\setminus C$ is analytic by the following (surprising) result from Kechris (originally by Lusin).
Theorem (Kechris, Thm 18.11). Let $X,Y$ be standard Borel spaces and $B\subseteq X\times Y$ be Borel. Then,
$$
\left\lbrace x\in X\colon\exists ! y\in Y{\rm\ s.t.\ }(x,y)\in B\right\rbrace
$$
is coanalytic.
