Let $(X, \mathcal F, \mu)$ be a $\sigma$-finite measure space and $(E, |\cdot|)$ a Banach space. Here we use the Bochner integral. Then we have Theorem 1.40 in Rudin's Real and Complex Analysis, i.e.,
Theorem 1.40 Let $f \in L_1(X, \mu, E)$ and $F$ be a non-empty closed subset of $E$. If $$ \varphi(A) :=\frac{1}{\mu(A)} \int_A f \mathrm d \mu \in F \quad \forall A \in \mathcal F \text{ s.t. } \mu(A) \in (0, \infty), $$ then $f(x) \in F$ for $\mu$-a.e. $x \in X$.
I would like to strengthen above theorem when $X$ is given more structure, i.e.,
Proposition Let $X$ be a metric space, $\tau$ its metric topology, $\mathcal F$ its Borel $\sigma$-algebra, and $\mathcal C$ the collection of all closed subsets of $X$. Theorem 1.40 still holds true if we restrict the test sets $A \in \mathcal F$ to those $A \in \tau$ (or to those $A \in \mathcal C$).
Could you confirm if my below attempt is fine?
Proof Assume that $\varphi (A) \in F$ for every $O \in \tau$ with $\mu(O) \in (0, \infty)$. Fix $A \in \mathcal F$ with $\mu(A) \in (0, \infty)$. Because $\mu$ is outer regular, there is a sequence $(O_n) \subset \tau$ such that $A \subset O_{n+1}\subset O_n$ and that $\lim_n \mu(O_n) = \mu(A)$. Clearly, $\frac{1}{\mu(O_n)} \to \frac{1}{\mu(A)}$. Also, $1_{O_n} f \to 1_A f$ $\mu$-a.e. By dominated convergence theorem, $\int_{O_n} f \mathrm d \mu \to \int_A f \mathrm d \mu$. As such, $$ \varphi (O_n) \to \varphi (A). $$
Because $\varphi (O_n) \in F$ for all $n$ and $F$ is closed, we get $\varphi (A) \in F$. The claim then follows from Theorem 1.40. $\tag*{$\blacksquare$}$
