Generated Sigma-Algebra by Generated Monotone Class

Theorem
Let $X$ be a set, and let $\mathcal G \subseteq \mathcal P \left({X}\right)$ be a nonempty collection of subsets of $X$.

Suppose that $\mathcal G$ satisfies the following condition:


 * $(1):\quad A \in \mathcal G \implies \complement_X \left({A}\right) \in \mathcal G$

that is, $\mathcal G$ is closed under complement in $X$.

Then:


 * $\mathfrak m \left({\mathcal G}\right) = \sigma \left({\mathcal G}\right)$

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:


 * $\mathfrak m \left({\mathcal G}\right) \subseteq \sigma \left({\mathcal G}\right)$

Next, define $\Sigma$ by:


 * $\Sigma := \left\{{M \in \mathfrak m \left({\mathcal G}\right): X \setminus M \in \mathfrak m \left({\mathcal G}\right)}\right\}$

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

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

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

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

Hence from the definition of monotone class:


 * $\varnothing = G \cap \left({X \setminus G}\right) \in \mathfrak m \left({\mathcal G}\right)$
 * $X = G \cup \left({X \setminus G}\right) \in \mathfrak m \left({\mathcal G}\right)$

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

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


 * $X, \varnothing \in \Sigma$

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


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

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


 * $\displaystyle \bigcup_{n \mathop \in \N} E_n \in \mathfrak m \left({\mathcal G}\right)$

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

All of the $X \setminus E_n$ are in $\mathfrak m \left({\mathcal G}\right)$ as each $E_n$ is in $\Sigma$.

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


 * $\displaystyle \bigcap_{n \mathop \in \N} \left({X \setminus E_n}\right) \in \mathfrak m \left({\mathcal G}\right)$

which subsequently implies that:


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

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

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


 * $\sigma \left({\mathcal G}\right) \subseteq \Sigma \subseteq \mathfrak m \left({\mathcal G}\right)$

Thus, by definition of set equality, it follows that:


 * $\mathfrak m \left({\mathcal G}\right) = \sigma \left({\mathcal G}\right)$