Let $X$ be a set and $\mathcal{A} \subset \mathscr{P}(X)$ an algebra and let $m(\mathcal{A})$ and $\sigma(\mathcal{A})$ be the smallest monotone class and the smallest $\sigma$-algebra containing $\mathcal{A})$ respectively.
Why does it hold true that $\sigma(\mathcal{A}) = m(\mathcal{A})$?
Best Answer
Following the (standard) proof from this note, with slightly adapted notation:
It's clear that $m(\mathcal{A}) \subseteq \sigma(\mathcal{A})$ because a $\sigma$-algebra is a monotone class.
To see the reverse, we must show that $m(\mathcal{A})$ is a $\sigma$-algebra. It's clear that $\emptyset \in \mathcal{A} \subseteq m(\mathcal{A})$. Likewise for $X$.
Complementation:
Define $$\mathcal{M}_0=\{A \in m(\mathcal{A}): A^\complement \in m(\mathcal{A})\}$$
and we need to show that $\mathcal{M}_0=m(\mathcal{A})$ to see that the latter is closed under complementation.
First, $\mathcal{A} \subseteq \mathcal{M}_0$ because $\mathcal{A}$ is an algebra and so closed under multiplication (and all its members are in $m(\mathcal{A})$ by definition). Moreover $\mathcal{M}_0$ is a monotone class: let $A_n, n \in \Bbb N$ be an increasing sequence of sets from $\mathcal{M}_0$, with union $A$. Clearly $A \in \mathcal{M}_0$ as it is a monotone class.
Also, for all $n$, $A_n^\complement \in \mathcal{M}_0 (\subseteq m(\mathcal{A}))$ by definition and $A_n$ is a decreasing sequence so $A^\complement = \bigcap_n A_n^\complement \in m(\mathcal{A})$, so we can now conclude that $A \in \mathcal{M}_0$, as required.
With a similar argument we can show that the intersection of a decreasing sequence $\mathcal{M}_0$ is still in that set. So $\mathcal{M}_0$ is a monotone class and from $$\mathcal{A} \subseteq \mathcal{M}_0 \subseteq m(\mathcal{A})$$ it follows that $\mathcal{M}_0 = m(\mathcal{A})$ and this exactly says that $m(\mathcal{A})$ is closed under complementation.
Finite intersections
This is a trickier multistep process. Fix $A \in \mathcal{A}$ for now. The first goal is to show that
$$\forall B \in m(\mathcal{A}): A \cap B \in m(\mathcal{A})\tag{1}$$
Following the idea of the earlier part of the proof we define
$$\mathcal{M}_1 = \{B \in m(\mathcal{A}): A \cap B \in m(\mathcal{A})\}$$
and we note that $\mathcal{A} \subseteq \mathcal{M}_1$ is for free because $\mathcal{A}$ is an algebra and so closed under finite intersections. We will again show that $\mathcal{M}_1$ is a monotone class, so let $B_n$ be an increasing family of members of $\mathcal{M}_1$ with union $B$. Then $A \cap B_n$ is also increasing and by definition in the monotone class $m(\mathcal{A})$ and so $$B\cap A = (\bigcup_n B_n) \cap A = \bigcup_n (B_n \cap A) \in m(\mathcal{A})$$
and it follows that $B \in \mathcal{M}_1$ too.
A similar argument for intersections of decreasing sets $B_n$ using the identity $$(\bigcap_n B_n) \cap A = \bigcap_n (B_n \cap A)$$ shows that $\mathcal{M}_1$ is also closed under decreasing intersections. So $\mathcal{M}_1$ is a monotone class and $$\mathcal{A} \subseteq \mathcal{M}_1 \subseteq m(\mathcal{A})$$ shows that $\mathcal{M}_1 = m(\mathcal{A})$ which implies $(1)$, and as $A \in \mathcal{A}$ was arbitrary, we know
$$\forall A \in \mathcal{A}: \forall B \in m(\mathcal{A}): A \cap B \in m(\mathcal{A})\tag{2}$$
Now for step 2, fix $C \in m(\mathcal{A})$. Define $$\mathcal{M}_2 = \{B \in m(\mathcal{A}): C \cap B \in m(\mathcal{A})\}$$
Note that $(2)$ just says that $\mathcal{A} \subseteq \mathcal{M}_2$
By the exact same argument as we used for $\mathcal{M}_1$ we see that $\mathcal{M}_1$ is a monotone class, and we again get $\mathcal{M}_1 = m(\mathcal{A})$ and this implies that $m(\mathcal{A})$ is closed under intersection by $C$. As $C$ was arbitrary in $m(\mathcal{A})$, $m(\mathcal{A})$ is closed under intersections (and hence unions, using complements, and differences, etc. whatever your basic axioms are you're done showing $\mathcal{A}$ is an algebra).
Countable unions
$m(\mathcal{A})$ is closed under countable unions. This is easy, because a monotone class that is an algebra (which is what the previous points showed) is a $\sigma$-algebra, see my other answer for details. Idea: a countable union can always be seen as an increasing union, using finite unions of the initial parts of the sequence of sets.