1

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?

deb
  • 414
  • I think $F_E = {(x, y) \in X \times X \mid y \in E} = X \times E$ is always Borel in $X \times X$ for any topological space $X$ and $E$ Borel in $X$. The reason is that (finite) product of Borel sets is Borel in the product space. A hint of the proof can be found in Cartesian Product of Borel Sets is Borel Again. Feel free to ask for more details. – Alex Vong Mar 26 '19 at 16:42
  • I meant $x\mapsto E_x$ describes a map $X\to\mathcal{B}(X)$, from $X$ to the collection of Borel subsets of X (the sets $E_x$ vary with $x\in X$). – deb Mar 26 '19 at 16:49
  • I don't understand, we are talking about the Borel measurability of each $E_x$ in $X$ instead of the Borel measurability of the map $x \mapsto E_x$, right? If so, then each $E_x$ being Borel in $X$ implies that $F_E = X \times E_x$ being Borel in $X \times X$, right? – Alex Vong Mar 26 '19 at 16:59
  • Yes, the set $X\times E_x$ is a Borel set. But the set that I denoted $F_E$ (defined above) is a different set. Are you saying that $X\times E_x$ being Borel, implies $F_E$ is Borel? How, if so? – deb Mar 26 '19 at 17:02
  • The Borel measurability of $E_x$ is given, yes. It is not clear what you mean by the Borel measurability of the map $x\mapsto E_x$ and $F_E$ is not equal to $X\times E_x$. – deb Mar 26 '19 at 17:07
  • I understand what's my mistake now. We have $F_E = {x \in X, y \in E_x}$ which is not the same as $X \times E_x$ because each $y$ comes from different $E_x$. Thanks for your explanation! – Alex Vong Mar 26 '19 at 17:11

1 Answers1

0

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)

Alex Vong
  • 1,654