measurability of metric space valued functions Let's say that we have a measure space $(X, \Sigma)$ and a metric space $(Y, d)$ with its Borel sigma algebra.  If $f_n: X\rightarrow Y$ is an arbitrary sequence of measurable functions, then I already know that if $f$ is a pointwise everywhere limit, then $f$ is measurable.  But if I don't assume that it converges everywhere, and instead ask where does it converge, is the set of points at which the $f_n$ converge measurable?

Not in general. Consider the functions $f_n:X\to X\times [0,1]$ defined by $f_n(x)=(x,1/n)$. This sequence converges everywhere to $x\mapsto (x,0)$. Now I pull a trick: define $Y=(X\times [0,1])\setminus (E\times \{0\})$ where $E\subset X $ is a nonmeasurable set. The functions $f_n:X\to Y$, defined as above, converge only on the set $X\setminus E$, which is not measurable. 
But it is true (as Davide Giraudo said) that the set of all $x\in X$ such that $(f_n(x))$ is a Cauchy sequence is measurable. Specifically, it is $\bigcap_k \bigcup_N \bigcap_{m\ge N} \bigcap_{n\ge N} A(m,n,k)$ where 
$$A(m,n,k)=\{x\in X : d(f_m(x),f_n(x))<1/k\}$$
Consequently, the set of convergence is measurable when $Y$ is complete.
