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.


  • 2
    $\begingroup$ "the countable measure of all sets that are not finite" makes no sense. I have never heard of the "subtractivity" either. To show that counting measure on $\mathbb{R}$ is not $\sigma$-finite, you just have to note that a set has finite counting measure iff it is finite, the countable unions of finite sets is countable, and $\mathbb{R}$ is not countable. $\endgroup$ Commented Feb 5, 2014 at 15:47
  • 1
    $\begingroup$ The first part is fine. $\endgroup$ Commented Feb 5, 2014 at 15:47
  • $\begingroup$ With "countable measure" I mean the counting measure, sorry, it was a mistake in translation and I have corrected it above. And subtractivity means that I can write $\zeta(A_k\setminus A_{k-1})=\zeta(A_k)-\zeta(A_{k-1})$ if $\zeta(A_{k-1})<\infty$. $\endgroup$
    – mathfemi
    Commented Feb 5, 2014 at 15:49
  • $\begingroup$ Is it right then?... I think I have shown what you wrote - but more complicated. $\endgroup$
    – mathfemi
    Commented Feb 5, 2014 at 16:15

1 Answer 1


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.

  • $\begingroup$ Thanks for that information but that does not say me if my proof is allright or not. $\endgroup$
    – mathfemi
    Commented Feb 5, 2014 at 15:58
  • $\begingroup$ Nevertheless I would like to know if my proof is at least right. That it is too complicated is another thing. $\endgroup$
    – mathfemi
    Commented Feb 5, 2014 at 16:22
  • $\begingroup$ It looks correct but you are doing far too much heavy lifting. $\endgroup$ Commented Feb 5, 2014 at 18:58

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