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Let $(X,\Sigma)$ be a measurable space, and let $Y$ be a topological space. Let $f_n:X \to Y$ be a measurable sequence, when we take on $Y$ the Borel sigma algebra. In other words the $f_n$ are measurable as mappings between the measurable spaces $$ (X,\Sigma) \to (Y,B(Y)).$$

Is $E= \{ x\in X : f_n(x) \text{ is convergent in } Y \}$ measurable?

This is well-known (and easy) in the case of real-valued functions. Indeed,

$$ E=\bigcap_{k\in\mathbb{N}}\bigcup_{n\in\mathbb{N}}\bigcap_{r,s >n}\left\{ x\in X:\left|f_{r}\left(x\right)-f_{s}\left(x\right)\right|<\frac{1}{k}\right\}. $$

I think this proof could be adapted to the case where $Y$ is a complete metric space; just replace $$ \left|f_{r}\left(x\right)-f_{s}\left(x\right)\right|$$ with $$ d\left(f_{r}(x),f_{s}(x)\right) .$$

Does this hold for non-complete metric spaces as targets? What about when $Y$ is not a metric space?

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This is a really instructive question: apparently, completeness is the key!

First, let $X=Y=[0,1]$ with the usual topology. We define an example for a sequence of functions $f_n: X\rightarrow Y$, which converges everywhere point-wise. Then we ruin $Y$.

Let $f_1$ be constant $0$ on $[0,1/2[$ and constant $1/2$ on $[1/2,1]$. Let $f_2$ be constant $0$ on $[0,1/4[$, constant $1/4$ on $[1/4,1/2[$, constant $1/2$ on $[1/2,3/4[$, and constant $3/4$ on $[3/4,1]$. In general, divide $[0,1]$ into $2^n$ intervals of equal length (closed from the left and open from the right, except for the last one which is closed), and let $f_n$ be the step function whose value on each interval is the left endpoint. It is clear that this sequence of functions converges point-wise to the identity function on $[0,1]$.

We keep these functions, but modify $Y$ now. Pick any non-measurable set $S\subseteq [0,1]$, and let $D$ consist of those rational numbers in $[0,1]$ whose denominator is a power of $2$. Let $Y$ be the topological subspace of $[0,1]$ induced by $S\cup D$. This is of course a metric space (as it is a subspace of a metric space), but it is no longer complete!

Define the same sequence of functions as above; they are still measurable (and also functions $X\rightarrow Y$, because we were careful enough to put all elements of $D$ in $Y$). However, the domain of convergence is $Y= S\cup D$: a sequence in $Y$ is convergent if and only if it is convergent as a sequence in $[0,1]$ with limit in $Y$. As $S$ is not measurable and $D$ is countable, we have that $Y$ is not measurable (as a subset of $X$).

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