Pointwise Limit of a Sequence of Measurable Functions

Question

Let $X$ be a measurable space and $Y$ a topological space. I am trying to show that if $f_n : X \to Y$ is measurable for each $n$, and the pointwise limit of $\{f_n\}$ exists, then $f(x) = \lim_{n \to \infty} f_n(x)$ is a measurable function. Let $V$ be some open set in $Y$. I was able to show that $\bigcup_{N=1}^\infty \bigcap_{n \ge N} f^{-1}_n(V)$ is contained in $f^{-1}(V)$ but not the other set inclusion. I could use some help.

Answer 1

Let $x \in f^{-1}(V)$ where $V$ is an open subsete of $Y$

Then $f(x) \in V$ and because of the convergence of $f_n(x)$ exists $m \in \Bbb{N}$ such that $f_n(x) \in V,\forall m \geq n$ thus $x \in \bigcup_{m=1}^{\infty} \bigcap_{n=m}^{\infty} f_n^{-1}(V)$

So you have the other desired set inclusion.