Show that $$ f\colon\mathbb{R}\to\mathbb{R}, x\mapsto\chi_{[1,\infty)}(x)\frac{1}{x}\mbox{ and }g\colon [1,\infty)\to\mathbb{R}, x\mapsto\frac{1}{x} $$ are measurable.

Concerning $f$, I guess $\mathcal{B}(\mathbb{R})\setminus\mathcal{B}(\mathbb{R})$ - measurability is meant?

If yes: For all $r\in\mathbb{R}$ it is $$ f^{-1}((-\infty,r])=\begin{cases}\mathbb{R}, & r\geq 1\\ \emptyset, & r<0\\ [1/r,\infty), & 0<r<1\\ [1,\infty)^C, & r=0\end{cases} $$ and this are all Borelsets, so $f$ is measurable.

Concerning $g$ it probably is meant $\mathcal{D}\setminus\mathcal{B}(\mathbb{R})$ - measurability with $\mathcal{D}=\left\{B\subseteq [1,\infty) : B\in\mathcal{B}(\mathbb{R})\right\}$?

If yes, it is $\forall r\in\mathbb{R}$: $$ g^{-1}((-\infty,r])=\begin{cases}[1,\infty), & r\geq 1\\ \emptyset, & r\leq 0\\ [1/r,\infty), & 0<r<1\end{cases} $$ and this are all sets that are in $\mathcal{D}$, so $g$ is measurable in the sense how I understood it.

Wish you a nice weekend and would be great to get a kind of feedback!

With kind regards



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.