5

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?

Vykta Wakandigara
  • 423
  • 3
  • 8
  • 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](https://math.stackexchange.com/a/3292510/550037)! – Vykta Wakandigara Jul 14 '19 at 11:08

1 Answers1

3

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.

kjetil b halvorsen
  • 63,378
  • 26
  • 142
  • 467
Grada Gukovic
  • 231
  • 1
  • 7