The set of points reached exactly $n$ times is measurable

Question

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

Answer 1

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.