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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.

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  • $\begingroup$ If $f_n (x) \in V$ for $n \geq N$ it does not follow that $f(x) \in V$. I think some assumption on the space Y is required for this to work, but I don't remember a counter-example at this moment. $\endgroup$ – Kabo Murphy Dec 8 '17 at 7:54
  • $\begingroup$ @KaviRamaMurthy That's precisely the problem I was encountering. What is the minimal assumption on $Y$ that will help prove this theorem? $\endgroup$ – user193319 Dec 8 '17 at 15:24
  • $\begingroup$ Whoops...I mispoke. I was not able to show that $\bigcup_{N=1}^\infty \bigcap_{n \ge N} f^{-1}_n(V)$ is contained in $f^{-1}(V)$. I was able to show that $f^{-1}(V) \subseteq \bigcup_{N=1}^\infty \bigcap_{n \ge N} f^{-1}_n(V)$. $\endgroup$ – user193319 Dec 8 '17 at 15:35
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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.

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