# Smallest Monotone Class Subset of Smallest Sigma-Algebra?


Here is a restatement of the theorem:

Monotone Class Theorem. Suppose $\mathcal{\A_0}$ is an algebra. Let $\A = \sigma\langle\A_0\rangle$ be the smallest $\sigma$-algebra containing $\A_0$, and $\M$ be the smallest monotone class containing $\A_0$. Then, $\A = \M$.

I am unsure about the logic of the first part of the proof, which reads as follows:

Proof: ($\M \subset \A$). A $\sigma$-algebra is clearly a monotone class, so $\M \subset \A$.

I understand how to show that a $\sigma$-algebra is a monotone class, but cannot see why this necessitates the claim above. It is also easy to see that $|\M| \leq |\A|$:

Proof: ($|\M| \leq |\A|$). Suppose $|\M|>|\A|$. Then, $\M$ is not the smallest monotone class containing $\A_0$, because $\A$ is a monotone class. This is a contradiction!

However, it is unclear to me how we can make any claim about the elements of $A$. Is it not possible for $\M$ to contain an element not in $\A$?

• I suspect that $\mathcal M$ equals the intersection of all monotone classes that contain $\mathcal A_0$. So it is a subset of each monotone class that contains $\mathcal A_0$ and $\mathcal A$ is one of them. – drhab Mar 3 '15 at 18:42
• Oh wow, yes, I completely missed that. Thanks. – Ben Bray Mar 3 '15 at 18:45

Proof: Because the family of monotone classes is closed under intersection, the intersection of all monotone classes containing $$\A_0$$ must be the smallest such class: $$\M = \cap S \qquad S \equiv \{ \mathcal{G} | \mathcal{G} \text{ is a monotone class}, \A_0 \subset \mathcal{G}\}$$ Because $$\A$$ is a monotone class and $$\A_0 \subset \A$$, we know $$\A \in S$$. So, $$\M \cap \A = \M$$, and therefore $$\M \subset \A$$.