# Show that $P$ is Countably additive problem.

Let $F$ be the field consisting of the finite and the cofinite ($A$ is cofinite if $A^c$ is finite) sets in an infinite and countable $\Omega$, and define $P$ on $F$ by taking $P(A)$ to be $0$ if $A$ is finite or $1$ if $A$ is infinite. (Note that $P$ is not well defined if $\Omega$ is finite) Show that $P$ is not countably additive.

Second. Suppose $A_1,A_2,\ldots$ in $F$ and $A_i\cap A_j=\emptyset$ $\forall i\neq j$. If $A_i$ is finite $\forall i\in\mathbb N$ then $P(\cup A_i)=\sum P(A_i)=0$. But if $A_i$ is cofinite only for one $i$, say $A_j$ for some $j\in\mathbb{N}$, then $P(\cup A_i)=1=P(A_j)+\sum_{i\neq j} P(A_i)=1+0$.

The problem would arise if we that $A_i$ is cofinite for more than one $i$, say $n$, for $n=2,3,\ldots$. In this case $P(\cup A_i)=1\neq \sum P(A_i)=n$. And then we could say that $P$ is not countably additive. But I think that we can't have more than one $A_i$ cofinite such that $A_i\cap A_j=\phi$. Suppose that infinites $A_i$ is cofinite. Than for each $i\neq j$ we have that $A_i\cap A_j=\emptyset\rightarrow A_i^c\cup A_j^c=\Omega$. But as $A_i$ and $A_j$ is cofinite, their complements are finit, otherwise they wouldn't belongs to $F$. But it is impossible that two finite sets form one infinite set. So I conclude that we would have only one cofinite $A_i$ in the sequence.

What am I missing? There is something wrong here. Help me please! Thank you

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By the definition, the measure of singletons is $0$. But $\Omega$ is a countable union of singletons. So if we had countable additivity, the measure of $\Omega$ would be $0$. But the measure of $\Omega$ has been declared to be $1$.
God how i missed that. You are so right! André $P(\cup A_i)$ is not necessarily finite, it is necessaraly countable. Obviously. Now things makes sense. Thank you! – Rodolfo Mar 29 '12 at 0:22
To show that $P$ is not countably additive you only have to provide one example where countable additivity fails. Your second paragraph does exactly that, provided all the sets are finite.