Additive but not $\sigma$-additive function 
Give an example of a measure space $(\Omega, \mathit{F})$ and a function $\mu$ on $\mathit{F}$ that is additive but not $\sigma$-additive, i.e. $\mu(\cup A_i)= \sum\mu(A_i)$ for a finite collection of disjoint $A_i$ but not for some infinite collections.

I know a measure function defined on $\sigma$-algebra is $\sigma$-additive, but I struggle finding a function that would not be additive for infinite collections. Can someone give me an example and show me why?
 A: Take $\Omega$ to be the set of all natural numbers, $F$ to be the family of all subsets of $\Omega$ and let $\mu(A) = 0$ if $A$ is a finite set and $\mu(A) = \infty$ if $A$ is infinite, I leave it to you to check that it's additive but not $\sigma$-additive.
A: Hint Consider $\Omega = \mathbb{N}$, the power set $F=\mathcal{P}(\mathbb{N})$  and the mapping $\mu: F\to [0,\infty]$, $$\mu(A) := \begin{cases} 0, & \text{$A$ is a finite set} \\ \infty, & \text{otherwise}. \end{cases}$$
A: We set $\Omega=N$, $F$=$2^{N}$, $\mu (A)=0$, if $|A|<\infty$; $\mu (A)=\infty$, if $|A|=\infty$.
We can see that $\mu$ is finite additive. Given a finite collection of disjoint $A_k$,1$\leq$ k $\leq$ n. If every $|A_k|<\infty$, then $|\bigcup_{k=1}^n A_k|<\infty$, so\ $\mu(\bigcup_{k=1}^n A_k)=0=\sum_{k=1}^n \mu(A_k)$; if there is at least one ${A_k}$ has infinite elements, $\mu (A_k)=\infty$, then $\bigcup_{k=1}^n A_k$ also has infinite elements,  $\mu(\bigcup_{k=1}^n A_k)=\infty=\sum_{k=1}^n \mu(A_k)=\infty$.
On the other hand, we would like to say that it is not $\sigma$-additive. Let $A_n=\{n\}$ for every integer $n\ge1$, we have $\mu(A_n)=0$ and $\mu(\bigcup_{n=1}^{\infty} A_n)=\infty \ne 0=\sum_{n=1}^{\infty}\mu(A_n)$.
