> [!theorem] > > Let $X$ be a [[Set|set]] and let $\alg$ be an [[Algebra|algebra]] over $X$. Then the smallest [[Monotone Class|monotone class]] generated by $\alg$ coincides with the smallest [[Sigma Algebra|sigma algebra]] generated by $\alg$. > $ > \cc(\alg) = \cm(\alg) > $ > *Proof*. Denote $\cm = \cm(\alg)$ and $\cc = \cc(\alg)$. Since every $\sigma$-algebra is a monotone class, $\cm \supseteq \cc$. > > Since $\cc \supset \alg$, $\emptyset, X \in \cc$. Let $E \in \cc$, and define > $ > \cc(E) = \bracs{F \in \cc(\alg): E \setminus F, F \setminus E, E \cap F \in \cc} > $ > Then for any increasing sequence $\seq{F_n}$, > $ > \begin{align*} > E \setminus \bigcup_{n \in \nat}F_n &= \bigcap_{n \in \nat}E \setminus F_n \in \cc \\ > \paren{\bigcup_{n \in \nat}F_n} \setminus E &= \bigcup_{n \in \nat}F_n \setminus E \in \cc \\ > E \cap \bigcup_{n \in \nat}F_n &= \bigcup_{n \in \nat}E \cap F_n \in \cc > \end{align*} > $ > their union is in $\cc$. For any decreasing sequence $\seq{G_n}$, > $ > \begin{align*} > E \setminus \bigcap_{n \in \nat}G_n &= \bigcup_{n \in \nat} E \setminus G_n \in \cc \\ > \paren{\bigcap_{n \in \nat}G_n} \setminus E &= \bigcap_{n \in \nat}G_n \setminus E \in \cc \\ > E \cap \bigcap_{n \in \nat}G_n &= \bigcap_{n \in \nat}E \cap G_n \in \cc > \end{align*} > $ > and $\cc(E)$ is a monotone class. > > If $E \in \alg$, then every element of $\alg$ satisfies $E \setminus F, F \setminus E, E \cap F \in \cc$, therefore $\cc(E)$ contains $\alg$ and $\cc(E) = \cc$. Let $F \in \cc$, then $F \in \cc(E)$ for all $E \in \alg$. Since the relation $E \setminus F, F \setminus E, E \cap F \in \cc$ is symmetric, we have $E \in \cc(F)$ for all $E \in \alg$, and $\cc(F) = \cc$. > > Let $E, F \in \cc$, then $E \in \cc(F)$ and $F \in \cc(E)$. Then $E \setminus F$, $F \setminus E$, and $E \cap F$ are all in $\cc$. Using this fact, since $X \in \cc$, $E^c = X \setminus E \in \cc$ for any $E \in \cc$. This way, for any $E, F \in \cc$, > $ > E \cup F = (E^c \cap F^c)^c \in \cc > $ > Therefore $\cc$ is an algebra. > > Let $\seq{E_n} \subset \cc$, then since $\cc$ is closed under countable increasing unions, > $ > \bigcup_{n \in \nat}E_n = \bigcup_{n \in \nat}\bigcup_{i = 1}^{n}E_n \in \cc > $ > it is a $\sigma$-algebra. > > Therefore $\cc \supseteq \cm$.