# Question about a peculiar sigma-algebra and a measure

Let $X$ be an uncountable set and let $\mathcal A$ be the collection of subsets $A$ of $X$ such that either $A$ or $A^c$ is countable. Define $\mu(A)=0$ if $A$ is countable and $\mu(A)=1$ if $A$ is uncountable. Prove that $A$ is a measure.

I think the statement of the problem is flawed, as it seems $\mu$ isn't even finitely additive. Indeed, if $A$ and $B$ are both uncountable disjoint elements of $\mathcal A$, then $A\cup B$ is not countable, hence $\mu(A \cup B) = 1$, but $\mu(A)+ \mu(B) = 1+1=2$...

I think this issue is resolved if one defines $\mu(A)=\infty$ if $A$ is uncountable.

Did I miss something ? What do you think about my revision ?

Carefully reread the assumptions: either $A$ or $A^c$ must be countable. It follows that you can't have two disjoint uncountable sets $A,B$ in your sigma algebra, since the complement of one would contain the other and therefore $A$ and $A^c$ would be uncountable.