Smallest Monotone Class Subset of Smallest Sigma-Algebra?

Question

$\newcommand{\A}{\mathcal{A}}\newcommand{\M}{\mathcal{M}}$ I'm having trouble understanding the first step of the Monotone Class Theorem, which every proof I find seems to claim is obvious.

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 $\A$?

Answer 1

The answer turned out to be quite obvious after all:$\newcommand{\A}{\mathcal{A}}\newcommand{\M}{\mathcal{M}}$

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 $\M \subset \A$.

Thanks to drhab for filling in what I was missing.