# Is the set of points for which $\lim f_n$ exists still measurable?

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. 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.

Let $X$ be a measurable space, $S\subset X$ a non-measurable subset. Let $Y=X\times\mathbb{R}-S\times\{0\}$, and define $f_n(x)=(x,n^{-1})$. Then $$\{x:\lim_n\, f_n(x)\text{ exists}\}=X- S$$ is not measurable.