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Fellows. I'm trying to prove some measurability result and I figured out a solution using the following and now I wonder if this is actually true.

Let $X$ be a topological space and $Y$ be a Polish space and let $f_n:(X,\mathcal{B}_X)\to(Y,\mathcal{B}_Y)$ be a sequence of measurable functions. Assume the pointwise limit $f(x):=\lim_{n\to\infty}f_n(x)$ exists for all $x\in X$. Then the function $f$ is again measurable.

Is this true? I know this result holds for $Y=\mathbb{R}$, but is it also satisfied for more general spaces? My intuition says yes, since Polish spaces have a countable base and are completely metrizable, but I have no clue how to prove it. Just to be clear, I'm not asking for a proof of this result (unless super-easy), but a reference would be great. Thanks a lot!

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  • $\begingroup$ The title of your question doesn't fit - the "continuous functions" mentioned there do not appear in your question, as far as I can see. $\endgroup$ – saz Sep 6 '14 at 17:21
  • $\begingroup$ Thank's, I edited it. No idea, where that came from... $\endgroup$ – Mr. Barrrington Sep 7 '14 at 8:41
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Since $Y$ is a Polish space, the Borel-$\sigma$-algebra $\mathcal{B}(Y)$ is generated by the open balls

$$\mathcal{G}_Y := \{B_Y(y,r); y \in Y, r>0\}.$$

Consequently, it suffices to show that $f^{-1}(G) \in \mathcal{B}(X)$ for any $G \in \mathcal{G}_Y$. For $G:=B_Y(y,r)$, we have

$$f(x) = \lim_{n \to \infty} f_n(x) \in B_Y(y,r)$$

if, and only if, $$\exists k=k(x) , N=N(x) \in \mathbb{N} \, \, \forall n \geq N: f_n(x) \in B_Y\left(y,r- \frac{1}{k} \right).$$

Hence,

$$\{x; f(x) \in B_Y(y,r)\}= \bigcup_{\substack{k \in \mathbb{N} \\ \frac{1}{k} < r}} \bigcup_{N \in \mathbb{N}} \bigcap_{n \geq N} \underbrace{\{f_n \in B_Y(y,r-1/k)\}}_{\in \mathcal{B}(X)}.$$

Since the right-hand side is a countable union of Borel sets, we conclude $$\{f \in G\} = \{f \in B_Y(y,r)\} \in \mathcal{B}(X).$$

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  • $\begingroup$ Hello Saz : I had a similar question to that of the OP and I found your answer helpful. The question is : under what circumstances is the pointwise limit $f$ of a sequence of measurable functions $f_n:(E,\mathcal E)\to (X,\mathcal{B}(X))$ still measurable. Am I correct in thinking that a sufficient condition is that $X$ be second countable regular ? Are there other known sufficient conditions on $X$ ? I have just posted a question on that topic if you are interested. $\endgroup$ – Olivier Bégassat Jan 1 '18 at 0:54
  • $\begingroup$ Is it essential to this argument that the space $Y$ be Polish? What if were merely a complete metric space and $\mathcal{B}(Y)$ was induced by the open balls w.r.t. the $Y$ metric? $\endgroup$ – user2379888 Nov 13 '18 at 3:29
  • $\begingroup$ @user2379888 Under these assumptions the proof should work as well. $\endgroup$ – saz Nov 13 '18 at 6:21

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