# Monotone class generated by an algebra: Is it a $\sigma$-algebra?

Let $$\mathcal{A}\subseteq\Omega$$ be an algebra and $$\mathcal{M}(\mathcal{A})$$ be the monotone class generated by $$\mathcal{A}$$. Can one show that $$\mathcal{M}(\mathcal{A})$$ is a $$\sigma$$-algebra?

Since $$\mathcal{A}$$ is an algebra, $$\Omega\in\mathcal{A}$$ and thus $$\Omega\in\mathcal{M}(\mathcal{A})$$. And since $$\mathcal{M}(\mathcal{A})$$ is a monotone class, it is closed with respect to countable unions. But how can one show that $$\mathcal{M}(\mathcal{A})$$ is closed with respect to complements? Clearly, if $$A\in\mathcal{A}$$, then $$A^c\in\mathcal{A}\subseteq\mathcal{M}(\mathcal{A})$$, but we would need to start with $$A$$ in the (larger) $$\mathcal{M}(\mathcal{A})$$.

This and others seem to be related, but note the difference: I do not know whether $$\mathcal{A}$$ itself is a monotone class. And monotone classes in general are not closed with respect to complements, see here. The problem appeared when I tried (as claimed here) to find an easy proof for the monotone class theorem when one already has shown Dynkin's $$\pi-\lambda$$ theorem.

## 1 Answer

If $$\mathscr{G}$$ is any family of subsets of $$\Omega$$ containing $$\Omega$$ which is closed under complements (i.e. $$G \in \mathscr{G}\implies G^c\in \mathscr{G}$$), then the minimal monotone class $$m(\mathscr{G})$$ s.t. $$\mathscr{G}\subseteq m(\mathscr{G})$$ is closed under complements. Proof:

(i) $$\mathscr{G}\subseteq \mathscr{F}:=\{F \in m(\mathscr{G}) :F^c\in m(\mathscr{G})\}\subseteq m(\mathscr{G})$$.

(ii) The minimal monotone class of a family of subsets of $$\Omega$$ is monotonic in the sense that $$\mathscr{A}\subseteq \mathscr{B}$$ families of subsets of $$\Omega$$ implies $$m(\mathscr{A})\subseteq m(\mathscr{B})$$. To see this, suppose $$A\in m(\mathscr{A})$$. For any $$\mathscr{D}$$ monotone class containing $$\mathscr{A}$$, then $$A \in \mathscr{D}$$. But $$m(\mathscr{B})$$ is a monotone class containing $$\mathscr{A}$$ since $$\mathscr{A}\subseteq \mathscr{B}\subseteq m(\mathscr{B})$$. So we conclude that $$m(\mathscr{A})\subseteq m(\mathscr{B})$$.

(iii) So we have $$m(\mathscr{F})=m(\mathscr{G})$$ and we need to show that $$\mathscr{F}$$ is a monotone class itself: to see this, first note $$\Omega \in \mathscr{F}$$ since $$\{\Omega,\emptyset\}\subseteq \mathscr{G}$$. Let $$(F_n)_{n \in \mathbb{N}} \subseteq \mathscr{F}$$ increasing; then $$F=\cup_nF_n \in m(\mathscr{G})$$ and $$F^c=\cap_nF_n^c$$. However, $$(F_n^c)_{n \in \mathbb{N}}\subseteq m(\mathscr{G})$$ by the fact that $$(F_n)_{n \in \mathbb{N}} \subseteq \mathscr{F}$$, so $$F^c\in m(\mathscr{G})$$. This implies $$\cup_nF_n=F\in \mathscr{F}$$ by definition of $$\mathscr{F}$$. Let $$(F_n)_{n \in \mathbb{N}} \subseteq \mathscr{F}$$ decreasing. Then $$F=\cap_nF_n\in m(\mathscr{G})$$ and $$F^c=\cup_nF_n^c$$. However, $$(F_n^c)_{n \in \mathbb{N}}\subseteq m(\mathscr{G})$$ by the fact that $$(F_n)_{n \in \mathbb{N}} \subseteq \mathscr{F}$$, so $$F^c\in m(\mathscr{G})$$. This implies $$\cap_nF_n=F\in \mathscr{F}$$ by definition of $$\mathscr{F}$$. But then $$\mathscr{F}$$ is a monotone class, and we conclude that $$\mathscr{F}=m(\mathscr{F})=m(\mathscr{G})$$. This implies that for any $$G\in m(\mathscr{G})\implies G^c \in m(\mathscr{G})$$.