# $\pi$-system and monotone class

Assume $P$ is a $\pi$-system (that is, $P$ is closed under finite intersections) and $M$ is monotone class (that is $M$ is a non-empty collection of subsets of $\Omega$ closed under monotone limits).

Show that $P\subset M$ does not imply $\sigma(P)\subset M$.

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This looks like homework. Please read meta.math.stackexchange.com/questions/1803/… and edit accordingly. In the meantime, -1. – Nate Eldredge Dec 15 '11 at 21:20
Isn't $\{\emptyset\}$ both a $\pi$-class and a monotone class? $\{\emptyset\} \subseteq \{\emptyset\}$ but $\sigma(\{\emptyset\}) \subsetneq \{\emptyset\}$. – guy Dec 15 '11 at 21:39

Put $\Omega=\mathbb N$, $P:=\{A_n,n\in\mathbb N\}$, where $A_n=\{k\in\mathbb N\mid k\geq n\}$, and $M:=P\cup\{\emptyset\}$. $P$ is a $\pi$-system since $A_n\cap A_m=A_{\max(m,n)}\in P$, and $M$ is a monotone class since if $\{B_n\}\subset M$ is an increasing sequence, either all the sets are empty hence the union is empty, or for $j$ large enough $B_j=A_{\varphi(j)}$. Then $\bigcup_{n\in\mathbb N}B_n=A_{\min_{n\in\mathbb N}\varphi(n)}\in M$ (indeed $B_n=A_{\varphi(n)}\subset A_{\min_{k\in\mathbb N}\varphi(k)}$ so $\bigcup_{n\in\mathbb N}B_n\subset A_{\min_{n\in\mathbb N}\varphi(n)}$ and since $\varphi$ is a function of integers there is $k_0$ such that $\varphi(k_0)=\min_{n\in\mathbb N}\varphi(n)$ hence $A_{\varphi(k_0)}=B_{k_0}\subset \bigcup_{n\in\mathbb N}B_n$). If $\{B_n\}\subset M$ is an decreasing sequence, then $B_n=A_{\varphi(n)}$ for all $n$ (if $B_{n_0}=\emptyset$ for a $n_0$ then the intersection is empty), and $\bigcup_{n\in\mathbb N}B_n=\begin{cases}\emptyset&\mbox{ if }\sup_n\varphi(n)=+\infty,\\\ B_{\sup_{n\in\mathbb N}\varphi(n)}&\mbox{ otherwise}.\end{cases}$ Since $\sigma(P)=\mathcal P(\mathbb N)$, and $M\neq \mathcal P(\mathbb N)$, we can't have the inclusion $\sigma(P)\subset M$.

However, if $M$ is a $\lambda$-system (that is: contains $\Omega$, is closed under complement and countable disjoint unions), then the last equality is true. This result is useful when we show that two probability measures which agree on a $\pi$-system agree on the the $\sigma$-algebra generated by this $\pi$-system.

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Where are you stuck when you try to prove that $M$ is a monotone class? – Davide Giraudo Dec 16 '11 at 15:22
Let $\Omega$ be any infinite set, and let $\preceq$ be a linear order on $\Omega$. Say that $\Delta\subseteq\Omega$ is downward closed if $\alpha\preceq \beta\in\Delta$ implies that $\alpha\in\Delta$, and let $P=\{\Delta\subseteq\Omega:\Delta\text{ is downward closed}\}$. (Note that $\varnothing$ is vacuously downward closed, so $\varnothing\in P$.) You should have no trouble showing that $P\,$ is itself a monotone class that is not closed under complementation. (In fact this is true as long as $\Omega$ has at least two elements.)