Proof that $\operatorname{sup}_{n}f_{n}$ is measurable

Question

Let $\{f_{n}\}_{n\in \mathbb{N}}$ be a sequence of measurable functions $f_{n}: (X, \mathcal{F}) \to \overline{\mathbb{R}}$. In my notes, my professor proved $g(x) = \sup_{n}f_{n}(x)$ is measurable just by noticing that: $$g^{-1}((a,\infty]) = \bigcup_{n=1}^{\infty}f_{n}^{-1}((a,\infty]) \tag{1}\label{1}$$ But why is this enough? I mean, the Borel $\sigma$-algebra $\mathbb{B}_{\overline{\mathbb{R}}}$ on $\overline{\mathbb{R}}$ is the $\sigma$-algebra generated by open sets of $\overline{\mathbb{R}}$, where the topology in this set is defined by the order topology. Equivalently, $$\mathbb{B}_{\overline{\mathbb{R}}} = \{E \subseteq \overline{\mathbb{R}}: A \cap \mathbb{R} \in \mathbb{B}_{\mathbb{R}}\}$$ And it seems to me that (\ref{1}) is implying that $\mathbb{B}_{\overline{\mathbb{R}}}$ is generated by the sets $\{(a,\infty]: a \in \mathbb{R}\}$, but I don't see why is this the case.