Let $\zeta\colon\mathcal{P}(\Omega)\to [0,\infty]$ be the counting measure. Show that for $\Omega:=\mathbb{R}$ it is not $\sigma$-finite but for $\Omega:=\mathbb{N}$.
Hello and good afternoon! Here are my tries:
(1) Consider $\Omega:=\mathbb{N}$.
Define $A_n:=\left\{1,\ldots,n\right\}, n\geq 2$. Then $A_n\in\mathcal{P}(\mathbb{N}), A_n\nearrow\mathbb{N}$ and $\zeta(A_n)=n<\infty$ for $n\geq 1$. So the measure $\zeta$ is $\sigma$-finite.
(2) Now $\Omega:=\mathbb{R}$. First of all it is $\zeta(\mathbb{R})=\infty$ because the counting
measure of all sets that are not finite, is $\infty$.
Assume there are $A_i\in\mathcal{P}(\mathbb{R}), i\geq 1, A_i\nearrow \mathbb{R}$ and $\zeta(A_i)<\infty, i\geq 1$. Then for $A:=\bigcup_{i\geq 1}A_i$ it is because of the $\sigma$-additivity of $\zeta$ and the subtractivity
$$
\zeta(A)=A_1\uplus\biguplus_{k\geq 2}(A_k\setminus A_{k-1})=\zeta(A_1)+\sum_{k\geq 2}\zeta(A_k\setminus A_{k-1})=\zeta(A_1)+\sum_{k\geq 2}\zeta(A_k)-\zeta(A_{k-1})=\zeta(A_{\infty})<\infty,
$$
so because of the continuity of the measure it is
$$
\zeta(A_i)\nearrow\zeta(A)<\infty.
$$
On the other hand it is $\zeta(A_i)\nearrow\zeta(\mathbb{R})=\infty$. So $\mathbb{R}\neq A$.
Contradiction: It was assumed that $A_i\nearrow\mathbb{R}$, i.e. $A_i\nearrow A=\mathbb{R}$.
So there do not exist $A_i\in\mathcal{P}(\mathbb{R}), i\geq 1$ with $A_i\nearrow\mathbb{R}$ and $\zeta(A_i)<\infty$, i.e. $\zeta$ is not $\sigma$-finite for $\Omega:=\mathbb{R}$.
Maybe you can say me, if I am right.
Miro
Best Answer
If you have an arbitrary set $\Omega$ and counting measure thereupon, the counting measure is $\sigma$-finite if and only if the set $\Omega$ is countable.
The argument for part 1 is fine. For part 2, it is much easier to do the following
Suppose that $A_n$ is of finite counting measure for all $n$. Then $\bigcup_n A_n$ is countable, so $\bigcup_n A_n \not = \mathbb{R}.$ Hence, $\mathbb{R}$ is not $\sigma$-finite under the counting measure.
This argumet will also work in the abstract case.