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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
    Commented Sep 6, 2014 at 17:21
  • $\begingroup$ Thank's, I edited it. No idea, where that came from... $\endgroup$ Commented Sep 7, 2014 at 8:41

2 Answers 2

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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. $\endgroup$ Commented Jan 1, 2018 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? $\endgroup$ Commented Nov 13, 2018 at 3:29
  • $\begingroup$ @user2379888 Under these assumptions the proof should work as well. $\endgroup$
    – saz
    Commented Nov 13, 2018 at 6:21
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Perhaps a little late to the party, but I just want to point out that the only thing needed for this result to hold is for the topology of $Y$ to have the following property: that for each open set $U$ there exists a sequence $(A_k)_{k\in\mathbb{N}}$ of open sets such that $U=\bigcup_{n\in\mathbb{N}}A_k$ and $\overline{A_k}\subseteq U$ for all $k\in\mathbb{N}$.

Indeed, let $\mathcal{T}$ be the set of open sets in $Y$, and let $U\in\mathcal{T}$. By our assumption, there exists a sequence $A\in\mathcal{T}^\mathbb{N}$ such that $\overline{A_k}\subseteq U$ for each $k\in\mathbb{N}$ and $U=\bigcup_{k\in\mathbb{N}}A_k$. We can now prove that:

$$f^{-1}(U) = \bigcup_{k\in\mathbb{N}}\;\bigcup_{n\in\mathbb{N}}\;\bigcap_{p\in\mathbb{N}}f^{-1}_{n+p}(A_k) $$

In order to see this, first let $x\in f^{-1}(U)$; then $f(x)\in U$, whence there exists $k\in\mathbb{N}$ with $f(x)\in A_k$. Now, since $f_n(x)\rightarrow f(x)$, there exists $n\in\mathbb{N}$ such that for all $p\in\mathbb{N}$ we have $f_{n+p}(x)\in A_k$. Thus, $x$ is an element of the right-hand side. Now, let $x$ be an element of the right-hand side ; this implies that there exist $k,n\in\mathbb{N}$ such that, for all $p\in\mathbb{N}$, $f_{n+p}(x)\in A_k$. But this implies that a tail of the sequence $(f_n(x))_{n\in\mathbb{N}}$ is contained in $A_k$, whence $f(x)\in\overline{A_k}\subseteq U$. Thus, $x\in f^{-1}(U)$.

With this equality in hand, the $\mathcal{B}(X)/\mathcal{B}(Y)$-measurability of $f$ follows from the fact that $f^{-1}(U)\in\mathcal{B}(X)$ for all open sets $U$ (since $f^{-1}(U)$ is constructed as countable unions and intersections of elements of $\mathcal{B}(X)$, via the $\mathcal{B}(X)/\mathcal{B}(Y)$-measurability of the $f_n$ and our previous equality), and from the fact that $\mathcal{B}(Y)=\sigma(\mathcal{T})$.

Edited for typos.

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    $\begingroup$ It's worth noting that this condition is equivalent to $Y$ being perfectly normal. It is open/closed dual version of "regular $G_\delta$" defined here, which is noted to imply perfectly normal. On the other hand, for an open set $U$ in a perfectly normal space, we may write it as a countable union of closed sets $C_i$, and use normality to get open sets $A_i$ satisfying $C_i \subset A_i \subset \overline{A_i} \subset U$, which satisfies your condition. $\endgroup$
    – elihs
    Commented Oct 12, 2021 at 4:35
  • $\begingroup$ Great comment, thanks! I wasn't aware of perfectly normal spaces. At the risk of being nitpicky though: does the topology really need to be T1? I'm now reading that a space is perfectly normal iff it is T4 and every open set is an $F_\sigma$, but maybe we could replace "T4" with "normal" here... $\endgroup$ Commented Oct 21, 2022 at 14:06
  • $\begingroup$ There must be different conventions -- I was working under the definition that "perfectly normal" does not require the space be $T_1$, rather it means what you said, that it is "normal and every closed set is $F_\sigma$", which is equivalent to "every pair of closed sets $E$ and $F$ in the space $Y$ can be precisely separated by a continuous function $f: Y \rightarrow [0,1]$ (i.e. $f^{-1}(\{0\})=E$ and $f^{-1}(\{1\})=F$)." Then I would call a $T_6$ space a perfectly normal space which is also $T_1$. $\endgroup$
    – elihs
    Commented Oct 23, 2022 at 17:05

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