A monotone class in $\mathbb{R}$ which is closed under complement but is not a sigma-algebra

Question

Like the title says I'm looking for an example of a monotone class $ \mathcal{M}\subseteq\mathcal{P}\left(\mathbb{R}\right)$ such that $\mathbb{R}\in\mathcal{M}$ and $ \mathcal{M}$ is closed under complement but is not sigma-algebra.

I'm guessing the idea is to find such a family of sets that isn't closed under finite intersection but I haven't come up with anything thus far.

Thanks in advance!

Try it with intervals. (Not all intervals, just some intervals.) — Harald Hanche-Olsen, Oct 18 '13 at 21:12
I would appreciate a slightly thicker hint about which intervals I should be looking at :) — Serpahimz, Oct 19 '13 at 09:12

score 4 · Answer 1 · answered Aug 08 '14 at 00:34

Following the hint by Harald Hanche-Olsen: take the collection of all unbounded intervals of $\mathbb R$. This includes $\mathbb R$, so we should throw in the empty set too, to keep the class closed under complements. The monotone class is not hard to verify. And finite intersections can create bounded intervals, not in the class.

A monotone class in $\mathbb{R}$ which is closed under complement but is not a sigma-algebra

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