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?

  • $\begingroup$ 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! $\endgroup$ – Peter Leopold Jul 13 '19 at 13:16
  • $\begingroup$ I believe I got a comprehensive answer on this link! $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.