Proof of Monotone Classes and Subsets in Measure Theory

[fishturtle1]

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]

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!