# Why can a probability measure be defined over power set of countable sample space?

Let $\Omega$ be the sample space, $p: \Omega \rightarrow [0,1]$, be any function satisfying $\sum_{w\in\Omega} p(w) = 1$. Then there is a valid probability triple $(\Omega, \mathcal{F},P)$, where $\mathcal{F}$ is the collection of all subsets of $\Omega$ (the power set), and for $A \in \mathcal{F}, P(A) = \sum_{w\in A} p(w)$.

My question is why/how we know that, when $\Omega$ is finite or countable, a probability measure can be defined over ALL possible subsets (vs, say, the fact that a probability measure -- satisfying the standard axioms, of course -- cannot be defined over all subsets of [0,1]). Is there a proof of this somewhere (if so, please let me know where I can look), or could someone provide one?

• There are probability measure defined over all subsets of $[0,1]$. For instance $P(E)=1$ iff $1 \in E$ and $P(E)=0$ iff $1 \notin E$. – Ramiro Nov 11 '15 at 1:46

You can always define certain measures on the whole power set, for example the dirac measure $\delta_x (A)=1$ if $x\in A$ and $\delta_x (A)=0$ otherwise.
A series converges absolutely iff every subseries converges. For countably infinite $\Omega$, it follows that if $p: \Omega \rightarrow [0,1]$ is such that $\sum_{\omega\in \Omega} p(\omega)$ converges, then $\sum_{\omega\in A} p(\omega)$ converges for every $A\subseteq\Omega$.