In Kechris' textbook "Classical Descriptive Set Theory", the following exercise is stated (pp. $77$, Exercise $(12.14)$):
"Let X be a measurable space and Y a Polish space. Show that a function $f\colon X \to F(Y)$ is measurable iff $f^{-1}(\{\emptyset\})$ is measurable and there is a sequence of measurable functions $f_n\colon X\to Y$ such that $\{f_n(x)\}$ is a dense subset of $f(x)$ when $f(x)\ne \emptyset$."
I'm able to prove $(\Rightarrow)$ as an immediate consequence of the Selection Theorem for $F(X)$, but I don't get the right idea to prove the converse $(\Leftarrow)$.
Thanks in advance for your help.
It is enough to show that the preimages of generators of the measurable space $F(Y)$ are Borel in $X$. Namely, for every open set $U\subseteq Y$, we need to check that $$ Q:=f^{-1}(\{F\in F(Y) : F\cap U \neq \emptyset\}) = \{x\in X : f(x)\cap U \neq \emptyset\} $$ is Borel in $X$. Then we have $$ f(x)\cap U \neq \emptyset \iff f(x) \neq \emptyset \land \exists n . f_n(x)\in U, $$ hence $Q = (X\setminus f^{-1}(\{\emptyset\})) \cap \bigcup_n f_n^{-1}(U)$, which is measurable since the $f_n$ are.