veirarer

2022-06-19

A measure $\mu :{B}_{\mathbb{R}}\to \left[0,\infty \right]$ is locally finite if $\mu \left(K\right)<\infty$ for all K compact. Show that any locally finite measure is $\sigma$-finite. Give an example of a $\sigma$-finite measure on $\mathbb{R}$, which is not locally finite.
Attempt: Since $\mathbb{R}=\bigcup _{n\in \mathbb{Z}}\left[n,n+1\right]$ and each $\left[n,n+1\right]$ is compact, we have $\mu \left(\left[n,n+1\right]\right)<\mathrm{\infty }$ for each $n$ and $\mu$ is $\sigma$-finite. Is this part of the problem correct?
For the second part, I have no idea. I wanted to take counting measure on a collection of subsets of $\mathbb{R}$, but I know this measure is locally finite on the integers and not so with the usual topology. But I know this measure is not sigma finite on $\mathbb{R}$. Any suggestions?

Layla Love

Your solution to the first part is correct. For the second part, e.g. $\mu \left(X\right)=m\left(\mathrm{tan}\left(X\cap \left(-\pi /2,\pi /2\right)\right)\right)$ where $m$ is the Lebesgue measure.
The intuition is clear: We ignore any part of $X$ outside $\left(-\pi /2,\pi /2\right)$, and since tan is a bijection from $\left(-\pi /2,\pi /2\right)$ to $\mathbb{R}$, the Lebesgue measure on $\mathbb{R}$ induces a natural measure on $\left(-\pi /2,\pi /2\right)$. To formally prove this won't be hard. In particular, $\mu \left(\left[\pi /2,\pi /2\right]\right)=\mathrm{\infty }$, so $\mu$ is not locally finite.
If this is not intuitive enough. We can do the following modified version of the counting measure, let $A=\left\{1,1/2,1/3,1/4,\cdots \right\}$, and define $\mu \left(X\right)=|A\cap X|$, then $\mu$ is $\sigma$-finite but not locally finite as $\mu \left(\left[0,1\right]\right)=\mathrm{\infty }$.

Davon Irwin

No, your answer to 1. is not completely correct. For $\sigma$-additivity, you are in need of disjoint sets, i. e. consider $\left(n,n+1\right]$ (with parenthesis). These sets are not compact themselves, but are subsets of compact sets $\left[n,n+1\right]$. Now, since $\mu \left(\left(n,n+1\right]\right)\le \mu \left(\left[n,n+1\right]\right)$, you get the result by an inequality. EDIT: I am wrong.