Each probability measure on a countable space comes from a weight function In my lecture notes on probability theory it says that given a countable set $\Omega$ with a $\sigma$-algebra $\cal{F}$ each probability measure on $(\Omega,\cal{F})$ is induced by a function $p:\Omega \to \mathbb{R}$ such that $\sum_{\omega \in \Omega}p(\omega)=1$.
In case that $\cal{F}$ is the power set of $\Omega$ this is clear to me, but in the general setting it is not. 
 A: On $\Omega$ define the relation:$$x\sim y\iff \forall A\in\mathcal F[\{x,y\}\subseteq A\vee\{x,y\}\subseteq A^{\complement}]$$
This can be shown to be an equivalence relation. Reflexivity and symmetry is evident and if $x\sim y\wedge y\sim z$ then the existence of a set $A\in\mathcal F$ with $x\in A\wedge z\notin A$ leads to a contradiction of $x\sim y$ if $y\notin A$ and to a contradiction of $y\sim z$ if $y\in A$.
Let $[x]$ denote the equivalence class.  For every $y\notin[x]$ there is a set $A_y\in\mathcal F$ that does not contain $x$ so that $[x]^{\complement}=\bigcup_{y\notin[x]}A_y$. So if $\Omega$ is countable then $[x]\in\mathcal F$ because $\bigcup_{y\notin[x]}A_y$ is then a countable union of elements in $\mathcal F$.
So we end up with a partition of elements of $\mathcal F$ that are not empty and are such that a non-empty subset of such an element is not an element of $\mathcal F$. Then $\mathcal F$ will be exactly the collection of the unions of such sets.
As described in the comments you can now define $p:\Omega\to\mathbb R$ as a function prescribed by: $$x\mapsto\frac{P([x])}{|[x]|}$$if $[x]$ is a finite set. 
Next to that it needs to be defined on $\{x\in\Omega\mid [x]\text{ not finite}\}$ as well. If $[x]$ is infinite then just let $p$ be defined on its elements on such a way that: $$\sum_{y\in[x]}p(y)=P([x])$$ Actually that works for finite sets also.
