Measure space $(\Omega ,\mathcal F)$ and $\mathbb P:\mathcal F\to [0,1]$ that is finite additive but not $\sigma -$additive I'm looking for a space $(\Omega ,\mathcal F)$ and a function $\mathbb P:\mathcal F\to [0,1]$ s.t. $\mathbb P(\Omega )=1$ and $\mathbb P$ is addtitive but not $\sigma -$additive.
Of course I got inspired but several post on mathstack exchange by $\Omega =\mathbb R$, $$\mathcal F=\{A\subset \mathbb R\mid A\text{ or $A^c$ is countable}\},$$
and $\mathbb P(A)=0$ if $A$ is countable an $\mathbb P(A)=1$ if $A$ is not countable. But unfortunately this is a probability measure. Any example ?

My question is not a duplicate of this post (as several thought) because the answers don't provide probability spaces.
 A: This answer will show only that such a set function must exist. The reason that it is difficult to write one down explicitly is that the existence of an example on $\mathbb{N}$ requires something more than ZF. I suspect that the existence of any example will require more than just ZF but I am not sure of this. I will use the ultrafilter lemma$^*$.
Let $\Omega = \mathbb{N}$ and let $\mathcal{F} = \mathcal{P}(\Omega)$. Let $\mathcal{G}$ be a non-principal ultrafilter on $\mathbb{N}$ (recall that an ultrafilter on $\mathbb{N}$ is called principal if it is of the form $\mathcal{G} = \{A \subseteq \mathbb{N} : a \in A\}$ for some fixed $a$).
Define
$$\mathbb{P}(A) = \begin{cases} 1 \qquad A \in \mathcal{G} \\ 0 \qquad \text{otherwise} \end{cases}$$
Then $\mathbb{P}$ is a finitely additive set function (see this question). However since $\mathcal{G}$ is non-principal, it cannot contain any finite set. In particular, $\mathbb{P}(\{i\}) = 0$ for every $i \in \mathbb{N}$. Hence $\mathbb{P}$ is not countably additive since
$$1 = \mathbb{P}(\mathbb{N}) \neq \sum_{i=0}^\infty \mathbb{P}(\{i\}) = 0$$

$^*$ which is equivalent to the Boolean Prime ideal theorem and hence strictly weaker than choice. I need the ultrafilter lemma to see that a non-principal ultrafilter exists on $\mathbb{N}$. One way to get this is to use the ultrafilter lemma to get an ultrafilter extending the cofinite filter. This ultrafilter then cannot be principal. 
As pointed out in the answer here and the references therein, it is consistent with ZF+DC that there are no "non-principal measures" on $\mathbb{N}$. A non-principal measure on the powerset of a set $X$ is a set function $\mu: \mathcal{P}(X) \to [0,1]$ such that $\mu$ is finitely additive, $\mu(X) = 1$ and $\mu(\{x\}) = 0$ for all $x$. The $\mathbb{P}$ in this answer is exactly an example of a non-principal measure on $\mathbb{N}$. The answer to my follow up question here then shows that it is consistent with ZF+DC that you cannot construct an example on $\mathbb{N}$.
