Limit of measurable functions is measurable?

Question

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.

Answer 1

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 $A$

Assumptions:

• $\mathcal{T}$ 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 $C$.

Answer 2

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 $\mathbb{R}_0$).