Wronsonia8g

2022-06-30

Let $\mathcal{G}=\left\{G\subseteq {\mathbb{R}}_{0}^{+}\mid G$ denumerable or $\overline{G}$ denumerable $\right\}$ and
I am wondering if the function $f$ is $\mathcal{G}-\mathcal{B}\left(\left(0,1\right)\right)$ measurable, because as the $exp\left(-x\right)$ is continuous and $\mathbb{Q}$ is denumerable, normally all ${f}^{-1}\left(\mathcal{B}\mathcal{\left(}\mathcal{\left(}\mathcal{0}\mathcal{,}\mathcal{1}\mathcal{\right)}\right)\subset \mathcal{G}$ for the exponential function, but I am lacking the final idea to finish the proof. My idea was to use the following: As $\mathcal{B}\left(\left(0,1\right)\right)$ is created by $\sigma \left(\mathcal{E}\right)$ with $\mathcal{E}$=(0,1) then we can use $\mathcal{E}$ to proof that
${f}^{-1}\left(\mathcal{E}\right)\subset \mathcal{G}$
Any help is much appreciated.

Isla Klein

For any set $E$ the inverse image ${g}^{-1}\left(E\right)$ 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.