prove in details that if $f \in \mathcal{S}$ (a Schwartz function) where$$ \mathcal{S}\left(\mathbb{R}^{n}\right):=\left\{f \in C^{\infty}\left(\mathbb{R}^{n}\right): \forall \alpha \in \mathbb{N}_{0}^{n}, \forall k \in \mathbb{N}_{0}:|x|^{k}\left|\partial^{\alpha} f(x)\right| \lesssim 1 \quad \forall x \in \mathbb{R}^{n}\right\} $$ then $f$ is measurable (i.e. $\left\{x \in \mathbb{R}^{n}: f(x)<a\right\}$ is measurable)
Does it follows directly since $f$ is continuous (and thus measurable)? or maybe it stems from the fact that $f$ decays faster than any inverse polynom?
Yes, measurability of $f$ follows directly from its continuity.