But both and belong to , since , and is closed under finite intersection, so .
I need help understanding a certain part of the following proof.
Theorem: (Monotone Class Theorem)
Let be a class of subsets of , closed under finite intersections and containing . Let be the smallest class containing which is closed under increasing limits and by difference. Then .
The intersection of classes of sets closed under increasing limits and differences is a class of that type. So by taking the intersection of all such classes, there always exists a smallest class containing which is closed under increasing limits and by differences. For each set , denote
to be the collection of sets such that and .
Clearly is closed under increasing limits and by difference.
Let ; for each one has
and , so .
HERE THE PROBLEM STARTS
Now let . For each we have , and because of the preceeding
, hence , whence , hence .
The rest of the proof is just concluding that is a -algebra.
Back to the problem part. Why must ?
It's easy to see when , because is closed under finite intersections.
You have proved that is closed under increasing limits and difference, and we know that when : hence we have , since is the smallest class of subsets closed under increasing limits and difference containing .
We can apply now the same trick when and conclude the proof.
I think you missed the part where I have trouble. I have trouble understanding why
when . I see how you conclude this when you say .
If , then obviously for each because is closed under finite intersections. This of course implies , whence , because
is the smallest collection of subsets containing that is also closed under increasing limits and differences.
So, if , we could conclude .
If we now assume , then need not be in . And only is closed under finite intersections. The whole is not. How can you then know that for each ?