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.
Corollary
Let $X$ be a set.
Let $\GG \subseteq \powerset X$ be a non-empty set of subsets of $X$.
Define $\relcomp X \GG$ by:
- $\relcomp X \GG := \set {\relcomp X A: A \in \GG}$
Then:
- $\map \sigma \GG = \map {\mathfrak m} {\GG \cup \relcomp X \GG}$
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:
\(\ds X \setminus \paren {\bigcup_{n \mathop \in \N} E_n}\) | \(=\) | \(\ds \bigcap_{n \mathop \in \N} \paren {X \setminus E_n}\) | De Morgan's Laws: Difference with Union |
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:
- $\map {\mathfrak m} \GG = \map \sigma \GG$
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 3$: Problem $11 \ \text{(ii), (iii)}$