Third axiom of Kolmogorov axioms

Question

Let us define for a countably infinite set $S$ of real numbers that can be enumerated as $x_1,x_2,\cdots$,

$$P(S) = \sum_{x \in S}p(x) = \sum_{i=1}^\infty p(x_i) = \lim_{n \to \infty}\sum_{i=1}^n p(x_i),$$

if $\sum_{i=1}^\infty p(x_i)$ converges absolutely. Suppose we are only given $$P(\Omega) = 1$$ for a countably infinite set $\Omega$ and $p(l) \geq 0$ for all $l \in \Omega$. Is it true that for any countable sequence $E_1, E_2, \cdots$ of disjoint subsets of $\Omega$, we must have

$$P\left(\bigcup_{i=1}^\infty E_i\right) = \sum_{i=1}^\infty P(E_i)$$

? This is the third axiom of the Kolmogorov axioms but I'm wondering if we can "replace" it with the given condition above for discrete probability spaces. I managed to only show this for a finite sequence of disjoint subsets by induction.

Answer 1

This fails for the infinite case. Consider the following $P$ $$P(A) = \begin{cases}0 & \text{If $A$ is finite}\\ 1 & \text{If $A$ is infinite}\end{cases} $$

then consider $P$ defined on the subsets of $\Bbb{N}$.

you can see that $\Bbb{N} = \cup_{n} \{n\}$ but $$1 = P(\Bbb{N}) = P(\cup_{n} \{n\}) \neq \sum_n P(\{n\}) = 0 $$

Answer 2

Let $\Omega$ be any set and $p:\Omega\to[0,1]$. For any subset $A\subseteq \Omega$ define $$P(A)=\sup\left\{\sum_{x\in F}p(x): F\subseteq A, F\mbox{ is finite}\right\}.$$

Exercise: If $P(\Omega)=1$, then $P$ defines a probability measure on the $\sigma$-algebra of all subsets of $\Omega.$

Is this your question? If so, let me know where you get stuck and I'll try to help.