Let $f:X \rightarrow Y$ be a finite morphism of schemes. For $y \in Y$ let $n(y)$ be the cardinality of the set-theoretic fiber $f^-1(y)$. This is finite since $f$ is finite.
I think that the function $n$ is not upper semicontinuous, is this correct? (I think in the comments of Cardinality of the Fiber of a Finite Morphism Vs. Degree (via Vakil) there is a counter-example but I'm not 100% sure).
My question is: are there any nice conditions implying that $n$ is semicontinuous?
Assuming $Y$ is integral with generic point $\xi$ I guess in general the set $\lbrace y \in Y \mid n(y) = n(\xi) \rbrace$ is not open in $X$, is it? Is it constructible in general? Perhaps normality of $Y$ is an important ingredient (see Morphisms between schemes such that every point in the codomain has at most $n$ preimages.)?