# Measurability of a pointwise limit of measurable functions

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!

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

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 $$\{f \in G\} = \{f \in B_Y(y,r)\} \in \mathcal{B}(X).$$

• 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. Commented Jan 1, 2018 at 0:54
• 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? Commented Nov 13, 2018 at 3:29
• @user2379888 Under these assumptions the proof should work as well.
– saz
Commented Nov 13, 2018 at 6:21

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.

• 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. Commented Oct 12, 2021 at 4:35
• 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... Commented Oct 21, 2022 at 14:06
• 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 $T_1$. Commented Oct 23, 2022 at 17:05