Suppose $\mathcal{B}$ is a collection of subsets of $\mathbb{R}$ which contains the open sets, and is closed under complements and countable disjoint unions. Then $\mathcal{B}$ contains the Dynkin system generated by open intervals. I want to show that $\mathcal{B}$ contains all the Borel sets.

According to http://www.ams.org/journals/proc/2000-128-02/S0002-9939-99-05507-0/S0002-9939-99-05507-0.pdf the Dynkin system generated by open balls in $\mathbb{R}^d$ is the Borel $\sigma$-algebra, and that the $d = 1$ case is easy and well-known. Can anyone provide a proof or reference for this special case?

| cite | improve this question | | | | |
  • $\begingroup$ Try to prove the following lemma: If $\mathcal{O}$ is some collection of sets that it closed under intersections, then $d(\mathcal{O}) = \sigma( \mathcal{O})$, where $d(\mathcal{O})$ is the Dynkin system generated by $\mathcal{O}$. $\endgroup$ – Nigel Overmars Oct 15 '15 at 20:02

For $d=1$, the open balls (intervals) form a $\pi$-system (closed under intersections).

By Dynkins $\pi-\lambda$-theorem (see https://en.wikipedia.org/wiki/Dynkin_system#Dynkin.27s_.CF.80-.CE.BB_theorem), it follows that the generated $\lambda$ system (Dynkin system) is the sigma algebra generated by the class of open intervals, which is easily seen to be the Borel sigma algebra.

| cite | improve this answer | | | | |

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.