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 $C$.

## Best Answer

Yes, the theorem is correct. Let us prove it: $\def\AAA {\mathcal{A}} \def\BBB {\mathcal{B}} \def\NN {\mathbb{N}}$

Proof:(1 $\Rightarrow$ 2) Just take $f_n=f$ for all $n$.

(2 $\Rightarrow$ 1) Since $f_n$ are measurable functions, $f_n$ converges a.e. to $f$ and $(X,\AAA,\mu)$ is a complete measure space, there is a set $E \in \AAA$ such that $\mu(E)=0$ and $f_n$ converges to $f$ in $X \setminus E$.

Let $g_n = \chi_{X\setminus E} f_n$. It is clear that $\chi_{X\setminus E}$ is a measurable function and so $g_n$ are measurable functions.

It is easy to see that $g_n$ converges everywhere to $\chi_{X\setminus E} f$.

Since $(Y,\BBB)$ is a measurable space where $\BBB$ is the Borel $\sigma$-algebra of some metrizable topology of $Y$, we know that the pointwise (everywhere) limit of a sequence of measurable functions is a measurable function, so we have that $\chi_{X\setminus E} f$ is measurable.

Since $f = \chi_{X\setminus E} f$ a.e., $\mu(E)=0$ and $(X,\AAA,\mu)$ is a complete measure space, we have that $f$ is measurable.