Suppose $(\Omega, \cal F)$ is a measurable space and $(X, \mathcal B_X)$ is a topological space with its Borel sigma algebra.

If $f_n: \Omega \to X$ is a sequence of $(\cal F , B$$_X)$-measurable functions and if $f_n \to f$ pointwise, then is it true that $f$ is $(\cal F , B$$_X)$-measurable?

Of course, we know it is true if $X = \Bbb R$ with the usual topology. This is just a standard result in real analysis which can be proved easily using the order structure of $\Bbb R$.

I am more interested in what happens when $X$ is not some Euclidean Space.

I claim it is still true for metrizable $X$. Indeed, supposes $d$ induces the topology of $X$, and $C \subset X$ is closed. For $\varepsilon >0$, let $C_{\varepsilon} = \{x \in X: d(x,C) < \varepsilon \}$, which is open. Then $$f^{-1}(C) = \bigcap_{n \in \Bbb N} \bigcup_{N \in \Bbb N} \bigcap_{ k \geq N} f_k^{-1}\big(C_{2^{-n}}\big)$$ which is in $\cal F$. Since preimages of closed sets are in $\cal F$, it easily follows $f$ is $(\cal F, B$$_X)$-measurable. I guess the crucial thing here was that any closed set in a metrizable space is $G_{\delta}$.

Does the result still hold for any first countable Hausdorff space? What about uniformizable spaces? I guess the answer would probably be no if $X$ is not Hausdorff since the limit function wouldn't necessarily be unique.

I doubt this would be a useful thing to know, but I'm curious nonetheless.

  • $\begingroup$ I don't understand why the identity on $f^{-1}(C)$ holds. Would you please elaborate? $\endgroup$
    – user254385
    Jul 15, 2015 at 5:22
  • $\begingroup$ @Billford: Let $B_n := \bigcup_{N \in \Bbb N} \bigcap_{k \geq N} f_k^{-1}(C_{2^{-n}}) = \liminf_{k \to \infty} f_k^{-1}(C_{2^{n}})$. Notice that $B_n$ is precisely the set of all $x \in X$ such that $f_k(x) \in C_{2^{-n}}$ for all but finitely many $k$. (See this question if that is unclear:math.stackexchange.com/questions/107931/…). Since $f_k \to f$ and $C$ is closed, we see that $f(x) \in C \iff \big[ f_k(x) \in C_{2^{-n}}$ for all but finitely many $k$, for all $n \big] \iff x \in B_n$ for all $n \iff x \in \bigcap_{n \in \Bbb N} B_n$. $\endgroup$
    – shalop
    Jul 15, 2015 at 9:23
  • $\begingroup$ I still don't understand that$[f_k(x) \in C_{2^{-n}}$ for all but finitely many $k$, for all $n]$ implies that $f(x) \in C$. Again, would you please elaborate on it? $\endgroup$
    – user254385
    Jul 15, 2015 at 9:42
  • $\begingroup$ @billford: suppose that $f_k(x) \in C_{2^{-n}}$ for all but finitely many $k$, for all $n$. Then it follows that $d(f(x),C)=\lim_k d(f_k(x),C) \leq 2^{-n}$, for all $n$. Hence $d(f(x),C)=0$. But $C$ is assumed to be a closed set, and so $d(f(x),C)=0$ implies that $f(x) \in C$. $\endgroup$
    – shalop
    Jul 15, 2015 at 9:47
  • $\begingroup$ Got it. Thanks. $\endgroup$
    – user254385
    Jul 16, 2015 at 11:33

2 Answers 2


To me it seems that there are two crucial factors for your proof. First is the space being regular, by which I mean just that any closed set and an outside point can be separated by disjoint neighbourhoods, without requiring $X$ to be $T_0$. Second, that any closed set has a countable neighbourhood base.

Given the above assumptions, we can use the same argument. If $C\subseteq X$ is closed, let $\{V_j\}_{j\in\mathbb{N}} \subseteq \mathcal{T}$ be its neighbourhood base. Now we prove that: $$ f^{-1}(C)=\bigcap_{j\in\mathbb{N}}\bigcup_{k\in\mathbb{N}}\bigcap_{n\geq k}f_n^{-1}(V_j) $$

If $\omega\in f^{-1}(C)$, then $\{f_n(\omega)\}$ is eventually in any neighbourhood of $C$. Hence for all $j\in \mathbb{N}$, there exists $k_j \in \mathbb{N}$, such that $$\omega\in \bigcap_{n\geq k_j}f_n^{-1}(V_j)$$ implying that $\displaystyle{\omega\in \bigcup_{k\in\mathbb{N}}\bigcap_{n\geq k}f_n^{-1}(V_j)}$. $\;$ Hence we obtain:$\;$ $\displaystyle{\omega\in \bigcap_{j\in\mathbb{N}}\bigcup_{k\in\mathbb{N}}\bigcap_{n\geq k}f_n^{-1}(V_j)}$

Conversely, if $\displaystyle{\omega\in \bigcap_{j\in\mathbb{N}}\bigcup_{k\in\mathbb{N}}\bigcap_{n\geq k}f_n^{-1}(V_j)}$, then for each $j\in\mathbb{N}$: $\;\displaystyle{\omega\in \bigcap_{n\geq k_j}f_n^{-1}(V_j)}$ for some $k_j\in\mathbb{N}$, meaning that $\{f_n(\omega)\}$ is eventually in $V_j$. Suppose now that $f(\omega)\in C^c$ and let $W_1$ and $W_2$ be some neighbourhoods of $\omega$ and $C$ respectively. Since there exists some $s\in\mathbb{N}$ such that $C\subseteq V_s \subseteq W_2$ and since $f(\omega)$ is a limit of $\{f_n(\omega)\}$, it follows that eventually $\{f_n(\omega)\}$ is in both $W_1$ and $W_2$, meaning that $W_1 \cap W_2 \neq \varnothing$. Since the neighbourhoods are arbitrary, it means that $f(\omega)$ cannot be separated from $C$, contradicting regularity. Therefore $f(\omega)$ must be in $C$.

Interestingly, the above assumptions do not imply that $X$ is Hausdorff, unless it also happens to be $T_0$, in which case the countability condition will also be stronger than first countability.

EDIT (Weaker assumption)==========================================

Let $\mathcal{B}_X$ be the Borel sigma-algebra of a topological space $(X,\mathcal{T})$. In what follows $\varphi(\mathscr{C})$ denotes the filter generated by a subbase $\mathscr{C} \subset \mathcal{P}(X)$ and $\mathscr{N}(A)$ - the neighbourhood filter of a subset $A$


  • $\mathcal{T}$ is regular (not assuming $T_0$)
  • For any nonempty closed $C \subseteq X$ there exists $\{V_j\}_{j\in\mathbb{N}} \subseteq \mathscr{N}(C) \cap \mathcal{B}_X$, such that any convergent filter containing $\{V_j : j\in\mathbb{N}\}$ contains $\mathscr{N}(C)$

Note that such $\{V_j : j\in\mathbb{N}\}$ is necessarily a filter subbase, since it has the finite intersection property, so there do exist filters that contain it. However it is not necessarily a base for $\mathscr{N}(C)$.

As before, since each $V_j$ is a neighbourhood of $C$, we have $$f^{-1}(C) \subseteq \bigcap_{j\in\mathbb{N}} \bigcup_{k\in\mathbb{N}} \bigcap_{n\geq k} f^{-1}_n(V_j)$$

On the other hand $$\omega \in \bigcap_{j\in\mathbb{N}} \bigcup_{k\in\mathbb{N}} \bigcap_{n\geq k} f^{-1}_n(V_j) \implies \{f_n(\omega)\} \text{ is eventually in each }V_j \implies$$ $$\implies \mathscr{F}_\omega =: \varphi \Big( \Big\{ \{f_n(\omega) : n\geq k\} : k\in\mathbb{N}\Big\} \Big) \supseteq \{V_j : j\in\mathbb{N}\} \implies$$ $$\implies \mathscr{F}_\omega \supseteq \mathscr{N}(C) \quad \text{ since } \mathscr{F}_\omega \text{ is convergent}$$ $$\implies \forall \quad U\in\mathscr{N}(C), W\in\mathscr{N}(f(\omega)): \Big( \{f_n\} \text{ is eventually in } U\cap W \implies U \cap W \neq \varnothing \Big)$$ $$\implies f(\omega) \in C \quad \text{by regularity assumption}$$


The sets $\{C_{2^{-n}}\}$ in Shalop's proof satisfy the second assumption (which can be verified using continuity of $d(\cdot, C)$), while not necessarily being a neighbourhood base at $C$.

  • $\begingroup$ Great answer! It seems that another set of conditions is that $X$ be second countable regular. I posted an eerily similar question to that of OP yesterday. $\endgroup$ Jan 1, 2018 at 14:23
  • $\begingroup$ Thanks) I read your post and think it's great too, so thank you for sharing $\endgroup$
    – Artem
    Jan 1, 2018 at 16:47
  • $\begingroup$ @OlivierBégassat Any second-countable regular space is metrizable (see Munkres, Chapter 4). So it falls within the category of spaces mentioned by Artem. $\endgroup$
    – shalop
    Feb 24, 2018 at 3:01
  • $\begingroup$ It has just occurred to me that in the metric space the sets $\{C_{2^{-n}}\}$ do not necessarily form a neighbourhood base at $C$. It's not hard to come up with counterexamples in $\mathbb{R}^2$. This leads me to think that the above countability condition is not plausible even for metric spaces. I will try to formulate a weaker one in an edit, so please let me know what you think) $\endgroup$
    – Artem
    Jul 22, 2018 at 3:38

I think I can confirm your suspicion that this doesn't necessarily hold if the target space is non-Hausdorff, assuming I haven't made a mistake somewhere...

Let $\mathbb{R}$ be the real line in its standard topology. Let $\mathbb{R}_0$ be the real line with the topology whose non-empty open sets $U$ are precisely the standard open sets $U \subseteq \mathbb{R}$ such that $0 \in U$. This topology is non-Hausdorff. It is not too difficult to check that the Borel $\sigma$-algebra of $\mathbb{R}$ and $\mathbb{R}_0$ are actually the same. I used this observation previously here.

Fix a non-Borel measurable set $S \subseteq \mathbb{R}$. Let $f_n : \mathbb{R} \to \mathbb{R}_0$ be identically $0$ for all $n \in \mathbb{N}$. Let $f:\mathbb{R} \to \mathbb{R}_0$ be the characteristic function of $S$. Clearly each $f_n$ is measurable, while $f$ is nonmeasureable. Furthermore, $f_n \to f$ pointwise (because the sequence $0,0,0,\ldots$ converges simultaneously to every point of $\mathbb{R}_0$).

  • $\begingroup$ Yes, I believe this example is correct. +1. But I'm still interested in counterexamples or proofs in the other cases, for example first-countable and uniformizable spaces. Also interested in the compact Hausdorff case. Like I said before, for a proof it would suffice to show that closed sets are $G_{\delta}$, but I'm not sure if this condition is equivalent to metrizability. $\endgroup$
    – shalop
    Jun 30, 2015 at 5:12
  • $\begingroup$ @Shalop: I agree that these other questions are more interesting, not intended as a complete answer :) $\endgroup$
    – Mike F
    Jun 30, 2015 at 5:32
  • $\begingroup$ Gotcha. Thanks for this part at least :) $\endgroup$
    – shalop
    Jun 30, 2015 at 5:34
  • $\begingroup$ @MikeF The problem with your example is that $f_k$ does not converge to $f$ since $(1/2,3/2)$ is a neighborhood of $1$ not containing $0$. Recall that $f_k$ converge to $f$ pointwise if for all $x\in X$ and all neighborhood of $f(x)$ $U$, there is a $n_{x,U}$ such that for all $k\geq n_{x,U}$, $f_k(x)\in U$. One should be careful in non-Hausdorff contexts, $f_k(x)$ is in a neighborhood of $f(x)$ is not the same as $f(x)$ is in a neighborhood of $f_k(x)$. $\endgroup$ Jun 30, 2015 at 5:37
  • $\begingroup$ @JosuéTonelli-Cueto: You're right, of course. Let me think a bit to see if there's a similar (working) example nearby, or if this is too wrong to be rightened. $\endgroup$
    – Mike F
    Jun 30, 2015 at 5:51

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.