# Monotone classes proof

Homework Statement:
A family ##\mathcal{M} \subset \mathcal{P}(X)## which contains ##X## and is stable under countable unions of increasing sets and countable intersections of decreasing sets
$$(A_n)_{n\in\mathbb{N}}, A_1 \subset A_2 \subset \dots \subset A_n \subset A_{n+1} \uparrow A = \bigcup_{n=1}^{\infty} A_n \Rightarrow A\in \mathcal{M}$$
$$(B_n)_{n\in\mathbb{N}}, B_1 \supset B_2 \supset \dots \supset B_n \supset B_{n+1} \downarrow B = \bigcap_{n=1}^{\infty}B_n \Rightarrow B \in \mathcal{M}$$
is called a ##\textit{monotone class}##. Assume that ##\mathcal{M}## is a monotone class and ##\mathcal{F} \subset \mathcal{P}(X)## any family of sets.

(iii) If ##\mathcal{F}## is ##\cap##-stable, i.e. ##F, G \in \mathcal{F} \Rightarrow F \cap G \in \mathcal{F}##, then so is ##m(\mathcal{F})##.
[Hint: show that the families
$$\sum := \lbrace M \in m(\mathcal{F}) : M \cap F \in m(\mathcal{F}), \forall F \in \mathcal{F} \rbrace$$
$$\sum' := \lbrace M \in m(\mathcal{F}) : M \cap N \in m(\mathcal{F}), \forall N \in m(\mathcal{F}) \rbrace$$
are again monotone classes satisfying ##\mathcal{F} \in \sum, \sum'##.
Relevant Equations:
Here ##m(\mathcal{F})## is the smallest monotone class that contains ##\mathcal{F}##. In other words, if ##L## is a monotone class containing ##\mathcal{F}##, then ##m(\mathcal{F}) \subseteq L##.
My question is how to show ##\mathcal{F} \subset \sum'##. Here is my work for the problem:

Proof of hint: First we'll show ##\sum## is a monotone class. Let ##(A_n)_{n\in\mathbb{N}} \subset \sum## and ##F \in \mathcal{F}##. There are two things to verify. Suppose ##(A_n) \uparrow A = \bigcup_{n=1}^{\infty} A_n##. Then ##(A_n \cap F) \uparrow (A \cap F)##. But ##A_n \cap F \in m(\mathcal{F})## for all ##n##. Hence, ##A \cap F \in m(\mathcal{F})## for all ##F \in \mathcal{F}##. Thus, ##A \in \sum##.

Now, suppose ##(A_n) \downarrow A = \bigcap_{n=1}^{\infty} A_n##. Then ##(A_n \cap F) \downarrow (A \cap F)##. Since ##A_n \cap F \in m(\mathcal{F})## for all ##n##, we have ##A \cap F \in m(\mathcal{F})## for all ##F \in \mathcal{F}##. Thus, ##A \in \sum##. We may conclude ##\sum## is a montone class.

By a similar argument, we can show ##\sum'## is a monotone class. Next, we'd like to show ##\mathcal{F} \subset \sum, \sum'##. Since ##\mathcal{F}## is stable under intersection and ##\mathcal{F} \subset m(\mathcal{F})##, we have ##\mathcal{F} \subset \sum##. I think I need to show ##\sum \subset \sum'## which would give me ##\mathcal{F} \subset m(\mathcal{F}) \subset \sum \subset \sum'##. And clearly ##\sum' \subset m(\mathcal{F})##. This would give ##\sum' = m(\mathcal{F})##, completing the proof.

I tried: Suppose ##A \in \sum##. Let ##N \in m(\mathcal{F})##. We'd like to show ##A \cap N \in m(\mathcal{F})##. If ##N \in \mathcal{F}##, then this is true. Otherwise...

member 587159

member 587159
You have shown that ##\mathcal{F} \subseteq \sum## and ##\sum## is a monotone class. Hence, ##m(\mathcal{F}) \subseteq \sum##.

This allows us to show ##\mathcal{F} \subseteq \sum'##:

If ##A \in \mathcal{F}## and ##N \in m(\mathcal{F})##, then we must show that ##A \cap N \in m(\mathcal{F})##. But now we know that ##N \in \sum##, so since ##A \in \mathcal{F}## we have ##A \cap N \in m(\mathcal{F})##. We conclude ##A \in \sum'##.

fishturtle1
You have shown that ##\mathcal{F} \subseteq \sum## and ##\sum## is a monotone class. Hence, ##m(\mathcal{F}) \subseteq \sum##.

This allows us to show ##\mathcal{F} \subseteq \sum'##:

If ##A \in \mathcal{F}## and ##N \in m(\mathcal{F})##, then we must show that ##A \cap N \in m(\mathcal{F})##. But now we know that ##N \in \sum##, so since ##A \in \mathcal{F}## we have ##A \cap N \in m(\mathcal{F})##. We conclude ##A \in \sum'##.
That makes sense, thank you!

member 587159