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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?

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  • $\begingroup$ 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 $1 \notin E$. $\endgroup$ – Ramiro Nov 11 '15 at 1:46
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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.

Also, countable sums of Diracs work well. On a countable sample space, any measure is of this form (countable linear combination if Diracs).

Only when we consider more interesting measures like the Lebesgue measure, we can't define it on the whole powerset while maintaining desired properties like translation invariance.

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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 $A\subseteq\Omega$.

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