Finite probability spaces are exactly finite full-power-set probability spaces

If $(\mathrm{\Omega },w)$ is a finite probability space, then [L1] makes every subset an event and gives the displayed formula. It is nonnegative, has ${\mathbb{P}}_w(\mathrm{\varnothing })=0$, and has ${\mathbb{P}}_w(\mathrm{\Omega })=1$. For a sequence of pairwise disjoint subsets $(A_j{)}_{j\ge 0}$, at most $\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }$ of them are nonempty. Each $\omega$ in their union occurs in exactly one $A_j$, so rearranging this finite sum gives ${\mathbb{P}}_w({\bigcup }_j{A}_j)={\sum }_j{\mathbb{P}}_w(A_j)$. Thus ${\mathbb{P}}_w$ is a probability measure on $(\mathrm{\Omega },\mathcal{P}(\mathrm{\Omega }))$.

Conversely, let $\mathbb{P}$ be a probability measure on $(\mathrm{\Omega },\mathcal{P}(\mathrm{\Omega }))$ and put $w(\omega )=\mathbb{P}(\{\omega \})$. Each singleton is an atom of the full power-set sigma-algebra, and every $A\subseteq \mathrm{\Omega }$ is the union of the singletons it contains. Thus [L3] gives $\mathbb{P}(A)={\sum }_{\omega \in A}\mathbb{P}(\{\omega \})={\sum }_{\omega \in A}w(\omega )\mathrm{.}$ Taking $A=\mathrm{\Omega }$ yields ${\sum }_{\omega \in \mathrm{\Omega }}w(\omega )=\mathbb{P}(\mathrm{\Omega })=1$, and nonnegativity of $w$ comes from the measure axioms inside [L2]. So $(\mathrm{\Omega },w)$ is a finite probability space.

Step 1.1 constructs a full-power-set probability measure from any finite weight model, and step 1.2 recovers exactly those singleton weights from any full-power-set probability measure. Hence the two descriptions are equivalent, including the boundary case of outcomes with weight $0$.