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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.

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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 $G\in m(\mathscr{G})\implies G^c \in m(\mathscr{G})$.

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