Let $M$ be a monotone class over a set $X$. Suppose that $M$ is:
- Closed under complements.
- Closed under finite disjoint unions.
Is it true that $M$ is a sigma-algebra?
Does anything change if we add the assumption that $M$ is closed under countable disjoint unions?
Note that there is a monotone class which is closed under complements, but not a sigma-algebra: The collection of all unbounded intervals of $\mathbb{R}$.
However, this example is not closed under finite disjoint unions.
No. If $M$ is too small (i.e., empty) or too large (i.e., not a set) then it is not a $\sigma$-algebra. But those are trivial cases. Here is a less pathological counterexample:
Let $X=\{1,2,3,4\}$ and $$M=\big\{\emptyset,\;\;\{1,2\},\;\;\{2,3\},\;\;\{3,4\},\;\;\{1,4\},\;\;\{1,2,3,4\}\big\}$$ Then $M$ is a monotone class closed under complements and finite disjoint unions, but not under intersections.