We know the more general definition of a measurable funnction: let $(X,\mathcal{X})$ and $(Y,\tau)$ a measurable and topological space, respectively. The function $f:X\to Y$ is a $\mathcal{X}$-measurable function if $f^{-1}(V) \in \mathcal{X}$, $\forall V \in \tau$.
I would like to know if it is true that we can define the measurability of the function $f$ using a base of the topology $\tau$. That is, suposse that $\beta$ is a base of $\tau$. Then, I wold like to prove that
$$f \hbox{ is } \mathcal{X}\hbox{-measurable function} \Leftrightarrow f^{-1}(B) \in \mathcal{X},\quad \forall B \in \beta $$
We know that any $V \in \tau$ can be written as
$$V = \bigcup_{A \in \mathcal{g}} A,\quad \mathcal{g} \subset \beta$$
with $g$ not necessarily enumerable. One implication of my goal is trivial. Any element of $\beta$ is open, that is, it is a element of $\tau$. Then, if $f$ is $\mathcal{X}$-measurable function, then $f^{-1}(B) \in \mathcal{X}, \forall B \in \beta$. But, because $g \in \beta$ is not necessarily enumerable, I can not conclude the other implication. That is, I can not conclude that
$$f^{-1}(V) = \bigcup_{A \in \mathcal{g}} f^{-1}(A) \in \mathcal{X}$$
I would like to know if it is possible to circumvent this or in what cases this result is valid.
This is too long for the comment section.
You are probably after the following result. On any set $\mathrm{Z},$ denote by $\sigma(\mathscr{A})$ the minimal sigma algebra containing the subset $\mathscr{A}$ of the power set of $\mathrm{Z}$. If $\varphi$ is a function with values in $\mathrm{Z},$ we also denote by $\varphi(\mathscr{A})$ the set of $\varphi^{-1}(\mathrm{A})$ as $\mathrm{A}$ runs through $\mathscr{A}.$
Proposition. Let $f$ be any function $\mathrm{X} \to \mathrm{Y}$ and let $\mathscr{Y}$ be any subset of the power set of $\mathrm{Y}.$ Then, $f^{-1}(\sigma(\mathscr{Y})) = \sigma(f^{-1}(\mathscr{Y})).$
Proof. You probably already know that the set of preimages of a sigma algebra conforms a sigma algebra, this proves at once the $\supset.$ For the other side, consider the set $\mathscr{Y}'$ of sets $\mathrm{E}$ in $\sigma(\mathscr{Y})$ such that $f^{-1}(\mathrm{E}) \in \sigma(f^{-1}(\mathscr{Y})).$ Straightforwardly, $\mathscr{Y}'$ is a sigma field containing $\mathscr{Y},$ hence, it also contains $\mathscr{Y},$ this proves $\subset.$ Q.E.D.