Why must probability fields be closed under countable unions?

Assume a probability triplet $$(\Omega, \mathcal{F}, \mathbb{P})$$.

My current understanding of $$\mathcal{F}$$ is that it must define events i.e. the subsets of $$\Omega$$ where probability is defined. I also understand that $$\mathcal{F}$$ must be closed under countable complements and unions. Is it correct to say that the reasoning is as follows:

1. We need closure under complements by Kolmogorov's axioms. If you know $$\mathbb{P}(A)$$ then you know $$\mathbb{P}(A^c) = 1 - \mathbb{P}(A)$$.

What I don't understand is why we need closure under countable unions. Even, if we know $$\mathbb{P}(A), \mathbb{P}(B)$$, we cannot arrive at $$\mathbb{P}(A \cup B)$$ without knowing $$\mathbb{P}(A \cap B)$$. I have read this answer but I am not satisfied. Help?

• You seem to be asking if subsets of $\Omega$ can behave like rational numbers insofar as a countably infinite series of them could represent a number that is not rational/not in the subset of $\Omega$. I guess $\Omega$ behaves more like $\mathbb{R}$ than $\mathbb{Q}$, in that sense. A proof would probably be similar. But I like the question! – Peter Leopold Jul 13 '19 at 13:16
• I believe I got a comprehensive answer on this link! – Vykta Wakandigara Jul 14 '19 at 11:08

Closure under countable unions follows from the closure with respect to disjoint countable unions, which is also a Kolmogorov axiom. For disjoint A and B you don't need $$P(A \cap B)$$ as $$P(A \cup B) = P(A) + P(B)$$.
Your picture is a bit upside-down however. The axioms are a result of the measure-theoretic representation of events. And you need a $$\sigma$$-algebra to have a measure.
I.e. first comes the $$\sigma$$-algebra and on top of it the probability measure, which has to satisfies the axioms by virtue of being a measure that sums to 1.