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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?

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  • $\begingroup$ 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
    Commented Oct 15, 2015 at 20:02

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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.

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