# Existence of an universally measurable pullback

Let $X,Y$ and $Z$ be standard Borel spaces: topological spaces homeomorphic to Borel subsets of complete separable metric spaces. Let $K\subseteq X\times Y$ be analytic. Assume that $K_x$ is not empty for any $x\in X$ where $$K_x := \{y\in Y:(x,y)\in K\}$$ is the $x$-section of $K$ (that is, $K$ has a full projection onto $X$). Consider the following diagram:

Suppose that $g$ is a Borel map, and $h$ is an analytically measurable map with a property that $$h(x) \in g(x,K_x)\qquad \forall x\in X \tag{\star}$$ I wonder whether there does exist a map $f:X\to Y$ satisfying the following properties:

1. the diagram commutes: $g(x,f(x)) = h(x)$ for all $x\in X$,

2. it holds that $(\mathrm{id}_X,f)(X) \subseteq K$ or equivalently $\mathsf{Gr}[g]\subseteq K$ where the graph of $f$ is $$\mathsf{Gr}[f]:=\{(x,f(x)):x\in X\}\subseteq X\times Y.$$

3. $f$ is analytically measurable.

Note that if we consider a set $A\subseteq X\times Y$ defined by $$A:=\{(x,y):g(x,y) = h(x)\},$$ then conditions 1,2 together are equivalent to the statement that $\mathsf{Gr}[f]\subseteq A\times K$. Note that $(\star)$ implies that $A_x\neq\emptyset$ and thus conditions 1 and 2 can be always satisfied. The question is thus whether it is possible to satisfy an additional condition 3.

Some thoughts. Note also that $$A = \{(x,y): d_Z(g(x,y),h(x)) = 0\} \tag{\star\star}$$ where $d_Z$ is any metric compatible with the topology of $Z$. If we are able to show that $A$ is analytical, then $A\cap K$ is analytical as well and hence contains a graph of an analytical map. However, from $(\star\star)$ it only follows that $A$ is analytically measurable.

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I would be happy if anybody could at least help me simplifying the setting. I guess, that for the solution of the problem it does not matter which are the spaces $X^n$ and $\mathscr P(X)$ and one can go just for some continuous surjective map in place of $\pi_X$. However, I don't know how to simplify/make neater constraints related to $K$. – Ilya Jul 15 '13 at 16:04
I don't think that there is any need to mention $P(X)$ at all. You can just put an arbitrary Borel space $Z$ there instead. Might make it a bit easier to read (I'm having a bit of difficulty parsing the question). – George Lowther Jul 17 '13 at 11:56
@GeorgeLowther: I understand your point - readability is something I was worried about. I think that if we replace $\mathscr P(X)$ by $Z$, then the case $n=1$ is equivalent to the general case (at least it is necessary). Due to this reason, I simplified the notation and also fixed a couple of mistakes (e.g. the formula for the set $A$). Is it a bit easier to read now? – Ilya Jul 17 '13 at 13:23
I don't think it is any less general with $Z$ in place of $P(X)$, since all uncountable standard Borel spaces are Borel isomorphic. Also, I think that the answer to the question is no. I had to google the definition of analytical functions which, apparently means measurable with respect to the sigma algebra generated by the analytic sets. – George Lowther Jul 17 '13 at 19:24
@GeorgeLowther: I think, you are right - also, indeed analytical functions are used here to mean analytically measurable (if that's unusual, I can fix it). Anyways, using a metric on $Z$ we can show that $A$ is analytically measurable so that even in special case $K = X\times Y$, the question would be whether $A$ contains a graph of a universally measurable map. I guess, tuning $g$ and $h$ one can make $A$ be pretty arbitrary, so unlikely it does contain a graph of a universally measurable map. – Ilya Jul 18 '13 at 12:58