Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let X be any countable! set and and let F be the cofinite set, i.e., $A \in F $ if A or $A^{c}$ is finite (this is an algebra).

Then show that the set function $\mu: F \rightarrow [0,\infty)$ defined as $\mu(A)=0$ if A is finite $\mu(A)=1$ if $A^{c}$ is finite is finitely additive.

I have started the argument by letting $A=\sqcup_i A_i$. If all $A_i$ are finite, then $\mu(A)=0=\sum_i\mu(A_i)$ and finite additivity follows. If there is at least one $A_i$ not finite then $\sqcup A_i$ is not finite. But $\sqcup A_i \in F$, which implies $(\sqcup_i A_i)^{c}$ is finite. But then $\mu(\sqcup_i A_i)=1 \neq \sum_i\mu(A_i)$.

Could anyone let me know where am I going wrong with the second part of the argument and how to finish this off? I can imagine that finite additivity of $\mu$ relies on the fact that X is countable.

share|cite|improve this question
What exactly is the exclamation mark following the word countable supposed to tell us? Also, did you intend to write "i.e., $A\in F$ if ..."? – user20266 Aug 24 '12 at 12:47
Sorry, that was a typo. Also, if X were uncountable then $\mu$ would be countably additive as well. – adamG Aug 24 '12 at 13:40
up vote 4 down vote accepted

If there is at least one infinite $A_i$ then it is the only one: Let $j \neq i$, then since $A_i \cap A_j = \emptyset$ we have $A_j \subseteq A_i^C$, so $A_j$ is finite. Hence in the sum $\sum_i \mu(A_i)$ there is exactly one $1$, giving $$\sum_i \mu(A_i) = \mu\left(\bigcup_i A_i\right) = 1.$$

share|cite|improve this answer
Thanks marlu for the answer! – adamG Aug 24 '12 at 12:56
Can you show how can $\mu$ be countably additive if X is uncountable? – adamG Aug 24 '12 at 13:53
$F$ is not closed under countable unions, so the notion of countable additivity is not well-defined for $\mu: F \to \mathbb [0,\infty)$. – marlu Aug 24 '12 at 14:05

The finite unionion of finite sets is finite. Now if $\bigcup_i A_i$ is a finite union that is not finite, than there must be already an inde $i^*$ such that $A_{i^*}$ is not finite. Since $A_{i^*}$ is not finite but in this algebra, it must have a complement that is finite, so $\mu(A_{i^*})=1$. This shows that $$\sum_i \mu(A_i)\geq \mu(A_{i^*})=1=\mu\Big(\bigcup_i{A_i}\Big).$$

It remains to show for you that it cannot be that there is another index $i^{**}$ such that $A_{i^{**}}$ has finite complement too. That is, you have to show that there cannot be two disjoint subsets of a countably infinite set that both have a finite complement.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.