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 $F$ 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

share|cite|improve this question
up vote 1 down vote accepted

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 $1$.

share|cite|improve this answer
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.

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.