Let $p:X \to Y$ be a measurable surjection and assume that for each $y \in Y$ the set $p^{-1}(y)$ is at most countable. Define $Y_n$ (for $n \in \mathbb{N} \cup \{ \infty \}$) to be the set of those $y \in Y$ for which there are exactly $n$ distinct $x \in X$ such that $p(x)=y$. Is it clear that each $Y_n$ is measurable?

EDIT: I forgot to add, $X,Y$ are assumed to be standard Borel spaces: i.e. sigma algebras are the sigma algebras of Borel set and $X$ and $Y$ are assumed to be complete, separable metric spaces

  • $\begingroup$ Sure, You are right: I forgot to add that $X,Y$ are standard Borel spaces. I'm sorry $\endgroup$ – truebaran Oct 20 '18 at 13:33
  • $\begingroup$ What about taking Y to be the reals R, and X to be a disjoint union of R with R, and p the identity on one copy of R, and on the other copy some example of a measurable map from R to R that doesn't have a measurable image. Then take n = 2. $\endgroup$ – Lorenzo Najt Oct 20 '18 at 14:25
  • $\begingroup$ @Lorenzo is X a complete metric space? $\endgroup$ – Tim kinsella Oct 20 '18 at 14:30
  • $\begingroup$ I guess it doesn't matter. You can just do what you said on two halves of $\mathbb{R}$. Provided there exists an injective measurable map whose image isn't measurable, as you say. $\endgroup$ – Tim kinsella Oct 20 '18 at 14:48

As in the answer I provided in MO, this follows from Exercise 18.15 in Kechris' Classical Descriptive Set Theory, which states (adapted to your notation) what you need that but for a subset $P\subseteq Y\times X$ with countable sections $P_y$ (you can take $P:=(\mathrm{graph}(p))^{-1}$).

For $n=0,\dots,\infty$, 18.15 says that $Y_n$ is Borel and there are Borel functions $f_i^{(n)}:Y_n\to X$ with disjoint graphs such that for $x\in Y_n$, $p^{-1}(x) = \{f_i^{(n)}(x) : i <n\}$. This exercise follows from the Lusin-Novikov Theorem (18.10 in the same book) that states the projection of such a $P$ must be Borel in $X$.

In particular, there are no injective, Borel measurable maps between complete, separable metric spaces, with non Borel image.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.