[proofplan]
We define the measure of a subset by summing the assigned point masses over that subset, using the supremum of finite subsums as the definition of a nonnegative sum over a [countable set](/page/Countable%20Set). This immediately gives nonnegativity, total mass one, and the required singleton values. The main point is countable additivity: for pairwise disjoint subsets, finite subsums on the union are controlled by sums over the pieces, and conversely finite approximations inside finitely many pieces give the reverse inequality. Uniqueness follows because every subset of a countable space is a countable disjoint union of its singleton subsets.
[/proofplan]
[step:Define the set function by finite subsums]
For each subset $A\subset\Omega$, define
\begin{align*}
\mathbb P(A)=\sup\left\{\sum_{\omega\in E}p(\omega): E\subset A,\ E\text{ finite}\right\}.
\end{align*}
Here the empty finite sum is $0$. Since $p(\omega)\ge 0$ for every $\omega\in\Omega$, the displayed supremum is well-defined in $[0,\infty]$. Since every finite subset $E\subset A$ is also a finite subset of $\Omega$, the hypothesis gives
\begin{align*}
\sum_{\omega\in E}p(\omega)\le \sum_{\omega\in\Omega}p(\omega)=1.
\end{align*}
Therefore $\mathbb P(A)\in[0,1]$ for every $A\subset\Omega$.
In particular,
\begin{align*}
\mathbb P(\varnothing)=0
\end{align*}
because the only finite subset of $\varnothing$ is $\varnothing$. Also,
\begin{align*}
\mathbb P(\Omega)=\sum_{\omega\in\Omega}p(\omega)=1
\end{align*}
by the assumed countable-sum convention.
[/step]
[step:Verify countable additivity on pairwise disjoint subsets]
Let $(A_n)_{n\in\mathbb N}$ be a sequence of pairwise disjoint subsets of $\Omega$, and define
\begin{align*}
A=\bigcup_{n=1}^{\infty}A_n.
\end{align*}
We prove
\begin{align*}
\mathbb P(A)=\sum_{n=1}^{\infty}\mathbb P(A_n).
\end{align*}
First let $E\subset A$ be finite. For each $n\in\mathbb N$, define the finite set
\begin{align*}
E_n=E\cap A_n.
\end{align*}
Because $E$ is finite and the sets $A_n$ are pairwise disjoint, only finitely many of the sets $E_n$ are nonempty, and $E$ is their disjoint union. Hence finite additivity of finite sums gives
\begin{align*}
\sum_{\omega\in E}p(\omega)=\sum_{n=1}^{\infty}\sum_{\omega\in E_n}p(\omega).
\end{align*}
For each $n$, the inclusion $E_n\subset A_n$ gives
\begin{align*}
\sum_{\omega\in E_n}p(\omega)\le \mathbb P(A_n).
\end{align*}
Therefore
\begin{align*}
\sum_{\omega\in E}p(\omega)\le \sum_{n=1}^{\infty}\mathbb P(A_n).
\end{align*}
Taking the supremum over all finite $E\subset A$ yields
\begin{align*}
\mathbb P(A)\le \sum_{n=1}^{\infty}\mathbb P(A_n).
\end{align*}
For the reverse inequality, fix $m\in\mathbb N$ and $\varepsilon>0$. For each $n\in\{1,\dots,m\}$, by the definition of $\mathbb P(A_n)$ as a supremum, choose a finite set $E_n\subset A_n$ such that
\begin{align*}
\sum_{\omega\in E_n}p(\omega)> \mathbb P(A_n)-\frac{\varepsilon}{m}.
\end{align*}
Define the finite set
\begin{align*}
E=\bigcup_{n=1}^{m}E_n.
\end{align*}
Since the sets $A_n$ are pairwise disjoint, the sets $E_n$ are pairwise disjoint, and $E\subset A$. Hence
\begin{align*}
\sum_{n=1}^{m}\mathbb P(A_n)<\sum_{\omega\in E}p(\omega)+\varepsilon.
\end{align*}
Since $E\subset A$, the definition of $\mathbb P(A)$ gives
\begin{align*}
\sum_{\omega\in E}p(\omega)\le \mathbb P(A).
\end{align*}
Thus
\begin{align*}
\sum_{n=1}^{m}\mathbb P(A_n)\le \mathbb P(A)+\varepsilon.
\end{align*}
Letting $\varepsilon\downarrow 0$ gives
\begin{align*}
\sum_{n=1}^{m}\mathbb P(A_n)\le \mathbb P(A).
\end{align*}
Taking the supremum over $m\in\mathbb N$ gives
\begin{align*}
\sum_{n=1}^{\infty}\mathbb P(A_n)\le \mathbb P(A).
\end{align*}
Combining the two inequalities proves countable additivity.
[guided]
The only non-formal part of the construction is countable additivity. We must prove that if $(A_n)_{n\in\mathbb N}$ is a sequence of pairwise disjoint subsets of $\Omega$ and
\begin{align*}
A=\bigcup_{n=1}^{\infty}A_n,
\end{align*}
then
\begin{align*}
\mathbb P(A)=\sum_{n=1}^{\infty}\mathbb P(A_n).
\end{align*}
We first prove the inequality from left to right. Take an arbitrary finite set $E\subset A$. For each $n\in\mathbb N$, define
\begin{align*}
E_n=E\cap A_n.
\end{align*}
Each $E_n$ is finite, and because $E$ is finite, only finitely many of the sets $E_n$ are nonempty. Since the sets $A_n$ are pairwise disjoint, the sets $E_n$ are pairwise disjoint and their union is exactly $E$. Therefore the finite sum over $E$ decomposes as
\begin{align*}
\sum_{\omega\in E}p(\omega)=\sum_{n=1}^{\infty}\sum_{\omega\in E_n}p(\omega),
\end{align*}
where the right-hand side has only finitely many nonzero terms. Since $E_n\subset A_n$, the definition of $\mathbb P(A_n)$ as the supremum over finite subsums gives
\begin{align*}
\sum_{\omega\in E_n}p(\omega)\le \mathbb P(A_n)
\end{align*}
for every $n\in\mathbb N$. Hence
\begin{align*}
\sum_{\omega\in E}p(\omega)\le \sum_{n=1}^{\infty}\mathbb P(A_n).
\end{align*}
Because this holds for every finite $E\subset A$, taking the supremum over all such $E$ gives
\begin{align*}
\mathbb P(A)\le \sum_{n=1}^{\infty}\mathbb P(A_n).
\end{align*}
For the reverse inequality, we approximate each $\mathbb P(A_n)$ by a finite subsum inside $A_n$. Fix $m\in\mathbb N$ and $\varepsilon>0$. For every $n\in\{1,\dots,m\}$, the definition of supremum gives a finite set $E_n\subset A_n$ satisfying
\begin{align*}
\sum_{\omega\in E_n}p(\omega)> \mathbb P(A_n)-\frac{\varepsilon}{m}.
\end{align*}
Now define
\begin{align*}
E=\bigcup_{n=1}^{m}E_n.
\end{align*}
This set is finite because it is a finite union of finite sets. It is also a subset of $A$, because each $E_n$ is contained in $A_n$ and each $A_n$ is contained in $A$. Since the $A_n$ are pairwise disjoint, the $E_n$ are pairwise disjoint, so finite additivity of finite sums gives
\begin{align*}
\sum_{\omega\in E}p(\omega)=\sum_{n=1}^{m}\sum_{\omega\in E_n}p(\omega).
\end{align*}
Using the chosen lower bounds for the finite subsums, we obtain
\begin{align*}
\sum_{\omega\in E}p(\omega)> \sum_{n=1}^{m}\mathbb P(A_n)-\varepsilon.
\end{align*}
Since $E\subset A$, the definition of $\mathbb P(A)$ gives
\begin{align*}
\sum_{\omega\in E}p(\omega)\le \mathbb P(A).
\end{align*}
Combining the two inequalities yields
\begin{align*}
\sum_{n=1}^{m}\mathbb P(A_n)<\mathbb P(A)+\varepsilon.
\end{align*}
Because $\varepsilon>0$ was arbitrary, we get
\begin{align*}
\sum_{n=1}^{m}\mathbb P(A_n)\le \mathbb P(A).
\end{align*}
Finally, the infinite series $\sum_{n=1}^{\infty}\mathbb P(A_n)$ is the supremum of its finite partial sums, since its terms are nonnegative. Taking the supremum over $m\in\mathbb N$ gives
\begin{align*}
\sum_{n=1}^{\infty}\mathbb P(A_n)\le \mathbb P(A).
\end{align*}
Together with the [first inequality](/theorems/2897), this proves countable additivity.
[/guided]
[/step]
[step:Check the singleton masses]
Let $\omega_0\in\Omega$. The finite subsets of $\{\omega_0\}$ are $\varnothing$ and $\{\omega_0\}$. Therefore
\begin{align*}
\mathbb P(\{\omega_0\})=\sup\{0,p(\omega_0)\}=p(\omega_0),
\end{align*}
because $p(\omega_0)\ge 0$.
[/step]
[step:Conclude that the set function is a probability measure]
We have shown that $\mathbb P:\mathcal P(\Omega)\to[0,1]$ is countably additive, that
\begin{align*}
\mathbb P(\Omega)=1,
\end{align*}
and that $\mathbb P(A)\ge 0$ for every $A\subset\Omega$. Since $\mathcal F=\mathcal P(\Omega)$, this proves that $\mathbb P$ is a probability measure on $(\Omega,\mathcal F)$.
[/step]
[step:Prove uniqueness from the singleton values]
Let $\mathbb Q:\mathcal F\to[0,1]$ be a probability measure such that
\begin{align*}
\mathbb Q(\{\omega\})=p(\omega)
\end{align*}
for every $\omega\in\Omega$. We prove that $\mathbb Q=\mathbb P$.
Fix $A\subset\Omega$. Since $\Omega$ is countable, the subset $A$ is countable. If $A$ is finite, then finite additivity of the measure $\mathbb Q$ gives
\begin{align*}
\mathbb Q(A)=\sum_{\omega\in A}\mathbb Q(\{\omega\})=\sum_{\omega\in A}p(\omega)=\mathbb P(A).
\end{align*}
If $A$ is countably infinite, choose a bijection $a:\mathbb N\to A$. The singleton sets $\{a(n)\}$ are pairwise disjoint and
\begin{align*}
A=\bigcup_{n=1}^{\infty}\{a(n)\}.
\end{align*}
By countable additivity of $\mathbb Q$,
\begin{align*}
\mathbb Q(A)=\sum_{n=1}^{\infty}\mathbb Q(\{a(n)\}).
\end{align*}
Using the prescribed singleton values,
\begin{align*}
\mathbb Q(A)=\sum_{n=1}^{\infty}p(a(n)).
\end{align*}
By the definition of the countable sum over $A$, this equals
\begin{align*}
\sum_{\omega\in A}p(\omega)=\mathbb P(A).
\end{align*}
Thus $\mathbb Q(A)=\mathbb P(A)$ for every $A\subset\Omega$, so $\mathbb Q=\mathbb P$. This proves uniqueness and completes the proof.
[/step]
