# Is the assumption of being an algebra crucial in the monotone class theorem?

Let $G$ be an algebra of sets and define $\mathcal{M}(G)$ to be the smallest monotone class containing $G$. Then $\mathcal{M}(G)$ is precisely the $\sigma$-algebra generated by $G$, i.e. $\sigma(G) = \mathcal{M}(G)$.

On proofwiki.org, I've found a proof of a modified version of the statement where $G$ doesn't need to be an algebra, but only closed under complement. I can't find any mistake in their proof.

Since any other textbook contains the Wikipedia version of the statement, I thought being an algebra would be a crucial assumption.

So, is there any mistake in the proofwiki.org proof that I didn't see?

• Watch out: Proofwiki uses an unusual definition of monotone class. The condition on proofwiki is stronger: closure under arbitrary countable unions and intersections, not only monotone unions and intersections. Of course if we require closure under arbitrary countable unions, as mathworker21 said this includes finite unions, so you get an algebra for free. Commented Sep 25, 2020 at 20:17

The proofwiki claim and proof looks correct, but I don't think this is as big of a deal as you seem to think. Clearly if $G$ is an algebra or a collection of subsets closed under complements, then $M(G) \subseteq \sigma(G)$. The bulk of the monotone class theorem is that $M(G)$ is large enough to be all of $\sigma(G)$. So if we only start off with $G$ being a collection of subsets closed under complements, then $M(G)$ in particular will contain all finite unions of sets in $G$ and so will already contain the algebra generated by $G$. A more concise relevant statement is $M(M(G)) = M(G)$, so since we just care about this being large enough, we might as well pretend $G$ is an algebra to start off with.