My question relates to this previous thread: monotone class theorem, proof
This thread provides a proof for the Monotone Class Theorem, using the following:
Here $\mathcal{A}$ is an algebra and $\mathcal{M}$ is the smallest monotone class containing by $\mathcal{A}$. The proof defines $$ \mathcal{M}(A) = \{F \in \mathcal{M}: A \setminus F, F \setminus A, A \cap F \in \mathcal{M}\} $$ and then it says that "it is easy to verify" that $\mathcal{M}(A)$ is a monotone class. However, this is not clear to me. Why is that the case? I might be missing something completely obvious.
Recall the definition of monotone class: a monotone class is a subset $\mathcal{M} \subset \mathcal{P}(X)$ of the power set of $X$ such that it is closed under taking limit of countable monotone sequence of subsets in $\mathcal{M}$.
Hence if we have to show that $M(A)$ is a monotone class, we have to show that $M(A)$ is closed under union of countable increasing sets and is closed under intersection of countable decreasing sets.
We'll show first that $M(A)$ is closed under taking union of countable increasing sequence of subsets. So let $F_1 \subset F_2 \subset...$ be a countable increasing sequence of subsets in $M(A)$,let their union be $F$, we want to show that $F \in M(A)$.
Namely, we need to show that that the conditions for a set to be in $M(A)$ are satisfied: $F/A \in \mathcal{M}, A/F \in \mathcal{M} , F \cap A \in \mathcal{M}$.
So we first show that $F/A \in \mathcal{M}$. So $F/A=(F_1 \cup F_2 \cup...)/A=(F_1-A)\cup (F_2-A)\cup...$. Here, $(F_i-A)=F_i \cap A^c$ is an increasing sequence of sets and each of them is in $\mathcal{M}$ by definition of $M(A)$. Because $\mathcal{M}$ is a monotone class the union of $F_i \cap A^c$ ,which is $F$, is also in $\mathcal{M}$. So we have just shown that $F/A$ is in $\mathcal{M}$.
Next, we show that $A/F \in \mathcal{M}$. So $A/(F_1 \cup F_2 \cup... )= A \cap (F_1 \cup F_2 \cup... )^c=A \cap (F_1^C \cap F_2^C \cap ...) =(A \cap (F_1)^c \cap (A \cap (F_2)^c) \cap ...$, Now $(A \cap (F_i)^c)$ is a decreasing sets each beloning to $\mathcal{M}$ by definition of $M(A)$. So the their intesection is in $\mathcal{M}$ by the fact that $\mathcal{M}$ is a monotone class. Therefore, we have show that $A/F \in \mathcal{M}$.
Finally we need to show that $A \cap F \in \mathcal{M}$. But this is again because $A \cap F= (A\cap F_1) \cup (A \cap F_2) \cup...$ which is the union of an increasing sequence of sets in the monotone class $\mathcal{M}$ and so their union, which is $F \cap A$, is in $\mathcal{M}$. So we have shown that $A\cap F \in \mathcal{M}$.
Finally, we have to show that $M(A)$ is closed under taking intersection of decreasing sequence of sets. So let $F_1 \supset F_2...$ be a decreasing sequence of sets, and let $F= \bigcap_i F_i$ be their intersection.
So for showing $A/F \in M(A)$, we have $A/F=A \cap(\bigcap_i F_i)^c=A \cap \bigcup_i(F_i)^c = \bigcup_i(A \cap F_i)^c \in M(A)$.
for showing $F/A \in M(A)$, we have $F/A=(\bigcap_i F_i) \cap A^c = \bigcap_i(A^c \cap F_i) \in M(A)$
for showing $F \cap A$, we have that $F \cap A = (\bigcap_i F_i) \cap A=\bigcap_i(A \cap F_i) \in M(A)$.
The reasoning is by similar to the case for showing closed under taking countable union of increasing sequence of subsets.
So we have show the condition for $F$ to be in $M(A)$, i.e $M(A)$ is closed under taking intersection of countable decreasing sequence of subsets.
So having proved that $M(A)$ is closed under taking countable union of increasing sequence of sets and is closed under taking countable intersection of decreasing sequence of sets, we have shown that $M(A)$ is a monotone class.