I am reading Theorem 26.2 in the book "Probability and Measure" (2nd ed.) by Patrick Billingsley. In the proof of that theorem, the author claims that:
Fact Let $\mu$ and $\nu$ be probability measures on the Borel subsets of $\mathbb{R}$, and set $$A=\left\{\;\left(a,b\right]\;\middle|\;\mu(\{a\})=\mu(\{b\})=\nu(\{a\})=\nu(\{b\})=0\right\}.$$ Then $\sigma(A)$ coincides with the Borel sigma-algebra on $\mathbb{R}$.
I have been thinking about this result, but have absolutely no clues on this. It does not seem straightforward. So my question is:
Question What is a clean way to prove this?
Any hint/help is highly appreciated.
Hint: $\mu \{x\}=\nu \{x\}=0$ for all but countable number of $x$. So there is a countable set $\{t_1,t_2,...\}$ such that the given class contains all intervals $(a,b]$ whose end points are not in this countable set. Can you express any interval in terms of these using countable unions /intersections? (The complement of any countable set is dense). Will be glad to provide more details if needed.