This is not a research level question, but I have not received any feedback on the other side, so I thought I'd try here. My apologies if the answer (counterexample) is a triviality: Suppose $(X,\mathscr A)$ is a measurable space, $Y$ is Polish and $f\_n:X\to Y$ is a sequence of measurable functions. Then the set $L\subseteq X$ for which $\lim f\_n$ exists [is measurable.](https://math.stackexchange.com/questions/2344579/measurability-of-limit-function-into-a-polish-space/%22measurable.%22) The proof relies on the fact that $L$ is just the set of $x\in X$ for which $(f\_n(x))\_n$ is Cauchy in $Y$. Of course, this only works because $Y$ is complete. Moreover, it is clear that separability is necessary, to ensure that $\mathscr B(Y\times Y)=\mathscr B(Y)\times \mathscr B(Y).$ Now I am wondering what happens if we remove the condition of completeness of $Y?$ Obviously the forementioned proof won't work. In fact, I think that completeness is necessary, and am looking for a counterexample. i.e.; a sequence of measurable $f\_n$ such that $L=\left \{ x:\lim f\_n(x)\ \text {exists} \right \}$ is not measurable. 



When is the set of points for which limit of $f_n $ exists measurable?
This may be the trivial counterexample you are looking for: 


Let $A\subseteq \mathbb R$ be a nonmeasurable set. 
Let $X=\mathbb R$, and let $Y = (\mathbb R \times \{1,1/2,1/3,\ldots\} )\cup (A \times \{0\})$ with the Euclidean metric. 


Let $f\_n(x) = (x,1/n)$ for all $n\ge 1$. These functions are continuous. 


Then $\{ x\in \mathbb R: \exists \lim\limits\_{n\to\infty} f\_n(x) \} = A$. 


