Wronsonia8g

2022-06-30

Let $\mathcal{G}=\{G\subseteq {\mathbb{R}}_{0}^{+}\mid G$ denumerable or $\overline{G}$ denumerable $\}$ and $f:{\mathbb{R}}^{+}\u27f6(0,1):x\mapsto \{\begin{array}{ll}\mathrm{exp}(-x)& x\text{rational}\\ 0& \text{else}\end{array}$

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 ${f}^{-1}\left(\mathcal{B}\mathcal{(}\mathcal{(}\mathcal{0}\mathcal{,}\mathcal{1}\mathcal{)}\right)\subset \mathcal{G}$ 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.

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 ${f}^{-1}\left(\mathcal{B}\mathcal{(}\mathcal{(}\mathcal{0}\mathcal{,}\mathcal{1}\mathcal{)}\right)\subset \mathcal{G}$ 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.

Isla Klein

Beginner2022-07-01Added 12 answers

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.

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