Suppose $\Omega$ is a finite set and $\mathscr{F}$ is a $\sigma$-algebra on $\Omega$. Define a probability measure $\mathbb{P} : \mathscr{F} \to [0;1]$.

Since $\Omega$ is finite, measurability is not an issue.

Can we extend $\mathbb{P}$ to the whole powerset $\mathcal{P}(\Omega)$ of $\Omega$ so that the extension is a probability measure on $\mathcal{P}(\Omega)$?

i.e. from : $$\mathbb{P} : \mathscr{F} \to [0;1]$$ define : $$\mathbb{P}' : \mathcal{P}(\Omega) \to [0;1]$$ such that : $$\forall F \in \mathscr{F}, \quad \mathbb{P}'(F) = \mathbb{P}(F) $$ ?


Yes. I'm going to call the elements of $\mathcal F$ "measurable".

Say an atom is a measurable set $A$ such that $\emptyset$ and $A$ are the only measurable subsets of $A$. If $A$ and $B$ are atoms then $A=B$ or $A\cap B=\emptyset$. And given $x\in\Omega$, the intersection of all the measurable sets containing $x$ is an atom.

So $\Omega$ is the union of a disjoint collection of atoms; this makes it clear how to extend $\Bbb P$...


Yes. Assume $\Omega=\{1,2,\cdots,n\}$. A probability on $\mathcal P(\Omega)$ is completely and freely determined by an assignment of non-negative values $p_k=p(\{k\})$ such that $p_1+\cdots+p_n=1$.

Consider $\mathcal U$ the family of sets that are minimal with respect to set-inclusion in $\mathcal F\setminus\{\emptyset\}$. Since $\mathcal F$ is an algebra, by minimality the elements of $\mathcal U$ must be pairwise disjoint; in other words (since for all $x\in \Omega$ there is a minimal element of $\mathcal F\setminus\{\emptyset\}$ that contains it), they must form a partition of $\Omega$. For any $U\in\mathcal U$, consider $U=\{k_1,\cdots, k_t\}$ and assign $p_{k_i}$ at leisure, with the only attention that $p_{k_1}+\cdots+p_{k_t}=\Bbb P(U)$.


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