# Finitely but not countably additive set function

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.

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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

## 2 Answers

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 $$\sum_i \mu(A_i) = \mu\left(\bigcup_i A_i\right) = 1.$$

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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.

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