A question on Borel measurability Let $X$ be a compact metric space. Given a map $x\mapsto E_x$ where $E_x\subset X$ is a Borel set in X. What can be said about the Borel measurability of the set $F_E:=\{(x,y)\in X\times X:y\in E_x\}$? What are some (possibly useful) sufficient conditions for $F_E$ to be Borel?
 A: Since this is quite open-ended and have not received any answer yet, I'll give it a go. Please correct me if I get something wrong!

It seems there are quite a number of non-examples. Say each $E_x$ is a singleton, then the map $x \mapsto E_x$ becomes a function, call it $f$. Observe that $F_E$ is the graph of $f$. Also, each $E_x$ is measurable since singleton is closed in $T_1$ space (in particular metric space).
From this question, any Borel sigma algebra has cardinality at most continuum. In particular, $|B(X \times X)| \le \mathfrak{c}$
Suppose $X$ has cardinality $|X| \ge \mathfrak{c}$ (e.g. $X = [0, 1]$), then there are $|X^X| = |X|^{|X|} \ge 2^{|X|} \ge 2^\mathfrak{c}$ many $f: X \to X$ functions.
We thus conclude $|X^X| \ge 2^\mathfrak{c} > \mathfrak{c} \ge |B(X \times X)|$ which means that there are more $f: X \to X$ functions than Borel sets in $X \times X$. Since different functions corresponds to different graphs, there must be a $f: X \to X$ function with non-Borel graph (i.e. $F_E$ is not Borel).
In short, $F_E$ can be non-Borel even when each $E_x$ is a singleton! (assuming $|X| \ge \mathfrak{c}$)

On the other hand, if $f$ is Borel measurable, then $F_E$ being the graph of $f$ is Borel (I use the hint by saz). But this is probably too restrictive to be useful. (since we have to assume each $E_x$ is a singleton)
