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
  • 1
    $\begingroup$ 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

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