Generated Sigma-Algebra by Generated Monotone Class

Theorem
Let $X$ be a set, and let $\GG \subseteq \powerset X$ be a nonempty collection of subsets of $X$.

Suppose that $\GG$ satisfies the following condition:


 * $(1):\quad A \in \GG \implies \relcomp X A \in \GG$

that is, $\GG$ is closed under complement in $X$.

Then:


 * $\map {\mathfrak m} \GG = \map \sigma \GG$

where $\mathfrak m$ denotes generated monotone class, and $\sigma$ denotes generated $\sigma$-algebra.

Proof
By Sigma-Algebra is Monotone Class, and the definition of generated monotone class, it follows that:


 * $\map {\mathfrak m} \GG \subseteq \map \sigma \GG$

Next, define $\Sigma$ by:


 * $\Sigma := \set {M \in \map {\mathfrak m} \GG: X \setminus M \in \map {\mathfrak m} \GG}$

By $(1)$, it follows that $\GG \subseteq \Sigma$.

Next, we will show that $\Sigma$ is a $\sigma$-algebra.

To this end, let $G \in \GG$ be arbitrary.

By $(1)$, also $X \setminus G \in \GG$.

Hence from the definition of monotone class:


 * $\O = G \cap \paren {X \setminus G} \in \map {\mathfrak m} \GG$
 * $X = G \cup \paren {X \setminus G} \in \map {\mathfrak m} \GG$

by virtue of Set Difference Intersection with Second Set is Empty Set and Set Difference and Intersection form Partition.

Since $\O = X \setminus X$, it follows that:


 * $X, \O \in \Sigma$

Further, from Set Difference with Set Difference, it follows that:


 * $E \in \Sigma \implies X \setminus E \in \Sigma$

Lastly, for any sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$, the definition of monotone class implies that:


 * $\ds \bigcup_{n \mathop \in \N} E_n \in \map {\mathfrak m} \GG$

Now to ensure that it is in fact in $\Sigma$, compute:

All of the $X \setminus E_n$ are in $\map {\mathfrak m} \GG$ as each $E_n$ is in $\Sigma$.

Hence, from the definition of monotone class, we conclude:


 * $\ds \bigcap_{n \mathop \in \N} \paren {X \setminus E_n} \in \map {\mathfrak m} \GG$

which subsequently implies that:


 * $\ds \bigcup_{n \mathop \in \N} E_n \in \Sigma$

Thus, having verified all three axioms, $\Sigma$ is a $\sigma$-algebra.

Since $\GG \subseteq \Sigma$, this means, by definition of generated $\sigma$-algebra, that:


 * $\map \sigma \GG \subseteq \Sigma \subseteq \map {\mathfrak m} \GG$

By definition of set equality:
 * $\map {\mathfrak m} \GG = \map \sigma \GG$