Let $(X,\mu)$ be a standard probability space, and $(Y,\tau)$ an uncountable Polish space. Then the set $L^0(X,\mu,Y)$ of measurable maps from $X$ to $Y$ identified up to measure 0 is Polish w.r.t. the topology of convergence in measure.

It is then not hard to see that the subset of maps $X\to Y$ with countable range is analytic, but is it Borel?

*To see that this set is analytic, note that if $A$ belongs to the measure algebra of $(X,\mu)$, then $f:X\to Y$ is (a.e.) constant on $A$ iff for all $B\subseteq A$, $\frac{\int_A f}{\mu(A)}=\frac{\int_B f}{\mu(B)}$, which is a closed condition on $(f,A)$. Then $f$ has countable range iff there exists a sequence $(A_n)$ of elements of the measure algebra such that $\bigcup_n A_n=X$ and such that the restrictions of $f$ to the $A_n$'s are constant functions.*