measure that satisfies finite subadditivity but not countable subbaditivity

Question

I try to construct a function $\mu$ from the set of all sets of real numbers into the nonnegative real numbers such that $A ⊂ B$ implies $\mu(A) ⊂ \mu(B)$ and such that $\mu$ satisfies finite subadditivity but not countable subbaditivity.

I know the example when $$ \mu(A)=\begin{cases}1& \text{ if } A \text{ infinite}\\ 0&\text{ if } A \text{ finite}. \end{cases} $$ for $A \subset \Bbb{N}$. Is there anything else?

Thanks

Answer 1

Let $f : \mathbb{R} \rightarrow [0, \infty]$ be any function. Then $$ \mu(A) : = \sum_{a \in A} f(a)$$ is such an example.

EDIT: So I missed the part of the question that was asking for $\mu$ to satisfy finite subadditivity but not satisfy countable subadditivity. As defined, $\mu$ is a measure and thus satisfies countable subadditivity.