Let $\mathcal{G}=\left\{G \subseteq \mathbb{R}^{+}_0 \mid G\right.$ denumerable or $\bar{G}$ denumerable $\}$ and $f: \mathbb{R}^{+} \longrightarrow(0,1): x \mapsto \begin{cases}\exp (-x) & \text {} x \text { rational } \\ 0 & \text { else }\end{cases}$
I am wondering if the function $f$ is $\mathcal{G}-\mathcal{B}((0,1))$ measurable, because as the $exp(-x)$ is continuous and $\mathbb Q$ is denumerable, normally all $f^{-1}\left(\mathcal{\mathcal{B}((0,1)}\right) \subset \mathcal{G}$ for the exponential function, but I am lacking the final idea to finish the proof. My idea was to use the following: As $\mathcal{B}((0,1))$ is created by $\sigma(\mathcal{E})$ with $\mathcal{E}= (0,1)$ then we can use $\mathcal{E}$ to proof that $$ f^{-1}(\mathcal{E}) \subset \mathcal{G} $$ Any help is much appreciated.
For any set $E$ the inverse image $g^{-1}(E)$ is the union of the set of all irrational numbers and a countable set or a subset of the set of rational numbers (depending on whether $0 \in E$ or not). Hence, $g$ is measurable.