Well defined measure Let $X$ be a nonempty set and let $\mathcal{M}$ be the $\sigma$-algebra of countable subsets and cocountable subsets of $X$.  Let $\mu(A) = 0$ if $A$ is countable and 1 if $A$ is cocountable.  I want to prove that $\mu$ is a measure, but I'm worried that it might not be well defined since $X$ can be countable, and thus any subset of $X$ is both countable and cocountable.  Any ideas?
 A: Suppose that $X$ is uncountable.
The first part of showing that this measure is well-defined is to verify that $\mathcal M$ is indeed a $\sigma$-algebra. It is clearly closed under complements. To show that it is closed under countable intersections, there are two cases to consider: either the intersection involves only cocountable sets (in which case it is also cocountable), or it involves some countable set (in which case it is countable). Thus $\mathcal M$ is indeed a $\sigma$-algebra.
To show the measure is well-defined, we must show that $\mu$ assigns a unique value to every $A\in \mathcal M$. Then either $A$ or $A^c$ is countable, by definition of $\mathcal M$. Moreover $A$ and $A^c$ cannot be simultaneously countable, since otherwise $X$ would be countable. Hence the measure is well-defined.
Lastly, to verify that $\mu$ is a measure, one must check that $\mu(\varnothing)=0$ (clear) and that $$\mu(\bigcup_{n\in \mathbb N} A_n)=\sum_{n\in \mathbb N}\mu(A_n)$$
for any disjoint countable family of sets $\{A_n\}_{n\in \mathbb N}\subset \mathcal M$. First, observe that there is at most one cocountable set among the $\{A_n\}$ (which follows by disjointness and the fact that $X$ is uncountable). Next if there is precisely one cocountable set, observe that both sides are $1$. In the remaining case when all sets are countable, both sides evaluate to $0$. Thus we have a well-defined measure.
Suppose that $X$ is countable.
In this case, the measure is not well-defined, because $0=\mu(X)=1$.
A: One way to make sure that the measure is well-defined is to assign to cocountable sets infinity instead of $1$1. In other words, set $\mu(A) = \infty$ if $A$ is cocountable, and $\mu(A) = 0$ if $A$ is countable. This will only work if $X$ is uncountable to begin with, otherwise cocountable set will be countable by itself.
