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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.