veirarer

2022-06-19

A measure $\mu :{B}_{\mathbb{R}}\to [0,\infty ]$ is locally finite if $\mu (K)<\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 $\mathbb{R}=\bigcup _{n\in \mathbb{Z}}[n,n+1]$ and each $[n,n+1]$ is compact, we have $\mu ([n,n+1])<\mathrm{\infty}$ 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?

Attempt: Since $\mathbb{R}=\bigcup _{n\in \mathbb{Z}}[n,n+1]$ and each $[n,n+1]$ is compact, we have $\mu ([n,n+1])<\mathrm{\infty}$ 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

Beginner2022-06-20Added 29 answers

Your solution to the first part is correct. For the second part, e.g. $\mu (X)=m(\mathrm{tan}(X\cap (-\pi /2,\pi /2)))$ where $m$ is the Lebesgue measure.

The intuition is clear: We ignore any part of $X$ outside $(-\pi /2,\pi /2)$, and since tan is a bijection from $(-\pi /2,\pi /2)$ to $\mathbb{R}$, the Lebesgue measure on $\mathbb{R}$ induces a natural measure on $(-\pi /2,\pi /2)$. To formally prove this won't be hard. In particular, $\mu ([\pi /2,\pi /2])=\mathrm{\infty}$, so $\mu $ is not locally finite.

If this is not intuitive enough. We can do the following modified version of the counting measure, let $A=\{1,1/2,1/3,1/4,\cdots \}$, and define $\mu (X)=|A\cap X|$, then $\mu $ is $\sigma $-finite but not locally finite as $\mu ([0,1])=\mathrm{\infty}$.

The intuition is clear: We ignore any part of $X$ outside $(-\pi /2,\pi /2)$, and since tan is a bijection from $(-\pi /2,\pi /2)$ to $\mathbb{R}$, the Lebesgue measure on $\mathbb{R}$ induces a natural measure on $(-\pi /2,\pi /2)$. To formally prove this won't be hard. In particular, $\mu ([\pi /2,\pi /2])=\mathrm{\infty}$, so $\mu $ is not locally finite.

If this is not intuitive enough. We can do the following modified version of the counting measure, let $A=\{1,1/2,1/3,1/4,\cdots \}$, and define $\mu (X)=|A\cap X|$, then $\mu $ is $\sigma $-finite but not locally finite as $\mu ([0,1])=\mathrm{\infty}$.

Davon Irwin

Beginner2022-06-21Added 5 answers

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