1. ## Help with proof

I need help understanding a certain part of the following proof.

Theorem: (Monotone Class Theorem)
Let $\mathcal{C}$ be a class of subsets of $\Omega$, closed under finite intersections and containing $\Omega$. Let $\mathcal{B}$ be the smallest class containing $\mathcal{C}$ which is closed under increasing limits and by difference. Then $\mathcal{B} = \sigma(\mathcal{C})$.

Proof:

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 $\mathcal{C}$ which is closed under increasing limits and by differences. For each set $B$, denote
$\mathcal{B}_{B}$ to be the collection of sets $A$ such that $A \in \mathcal{B}$ and $A \cap B \in \mathcal{B}$.

Clearly $\mathcal{B}_{B}$ is closed under increasing limits and by difference.

Let $B \in \mathcal{C}$; for each $C \in \mathcal{C}$ one has
$B \cap C \in \mathcal{C} \subset \mathcal{B}$ and $C \in \mathcal{B}$, so $C \in \mathcal{B}_{B}$.

HERE THE PROBLEM STARTS
Now let $B \in \mathcal{B}$. For each $C \in \mathcal{C}$ we have $B \in \mathcal{B}_{C}$, and because of the preceeding
$B \cap C \in \mathcal{B}$, hence $C \in \mathcal{B}_{B}$, whence $\mathcal{C} \subset \mathcal{B}_{B} \subset \mathcal{B}$, hence $\mathcal{B} = \mathcal{B}_{B}$.

The rest of the proof is just concluding that $\mathcal{B}$ is a $\sigma$-algebra.

Back to the problem part. Why must $B \in \mathcal{B}_{C}$?
It's easy to see when $B \in \mathcal{C}$, because $\mathcal{C}$ is closed under finite intersections.

2. ## Re: Help with proof

Hi,

$B \in \mathcal{B}_C$ iff $B \cap C \in \mathcal{B}$.
But both $B$ and $C$ belong to $\mathcal{B}$, since $\mathcal{C} \subset \mathcal{B}$, and $\mathcal{B}$ is closed under finite intersection, so $B \cap C \in \mathcal{B}$.

3. ## Re: Help with proof

$\mathcal{B}$ is not closed under finite intersection, $\mathcal{C}$ is.

4. ## Re: Help with proof

You have proved that $\mathcal{B}_B$ is closed under increasing limits and difference, and we know that $\mathcal{C} \subset \mathcal{B}_B$ when $B \in \mathcal{C}$ : hence we have $\mathcal{B} \subset \mathcal{B}_B$, since $\mathcal{B}$ is the smallest class of subsets closed under increasing limits and difference containing $\mathcal{C}$.

We can apply now the same trick when $B \in \mathcal{B}$ and conclude the proof.

5. ## Re: Help with proof

Originally Posted by Ahriman
We can apply now the same trick when $B \in \mathcal{B}$ and conclude the proof.
Could you please do this? I don't really see how we can conclude
$B \cap C \in \mathcal{B}_{C}$, for every $C \in \mathcal{C}$.

When you assume $B \in \mathcal{B}$ you can't assume
$\mathcal{B} = \mathcal{B}_{B}$ ?

6. ## Re: Help with proof

This is absolutely the same scheme of proof than before.

We have just showed that for all $B \in \mathcal{B}$ and all $C \in \mathcal{C}$, we have $B \cap C \in \mathcal{B}$.
This means that for every $B \in \mathcal{B}$, we have $\mathcal{C} \subset \mathcal{B}_B$, and so we get $\mathcal{B}_B = \mathcal{B}$ since $\mathcal{B}$ is the smallest bla bla bla ...

7. ## Re: Help with proof

I think you missed the part where I have trouble. I have trouble understanding why
$B \cap C \in \mathcal{B}$ when $B \in \mathcal{B}$. I see how you conclude this when you say $B \in \mathcal{C}$.

If $B \in \mathcal{C}$, then obviously $B \cap C \in \mathcal{C}$ for each $C$ because $\mathcal{C}$ is closed under finite intersections. This of course implies $C \subset \mathcal{B}_{B} \subset \mathcal{B}$, whence $\mathcal{B} = \mathcal{B}_{B}$, because
$\mathcal{B}$ is the smallest collection of subsets containing $\mathcal{C}$ that is also closed under increasing limits and differences.

So, if $B \in \mathcal{C}$, we could conclude $\mathcal{B} = \mathcal{B}_{B}$.

If we now assume $B \in \mathcal{B}$, then $B$ need not be in $\mathcal{C}$. And only $\mathcal{C}$ is closed under finite intersections. The whole $\mathcal{B}$ is not. How can you then know that $B \cap C \in \mathcal{B}$ for each $C \in \mathcal{C}$?

8. ## Re: Help with proof

My second message shows that for $B \in \mathcal{C}$, we have $\mathcal{B}_B = \mathcal{B}$. This exactly says that for every $B \in \mathcal{C}$ and $C \in \mathcal{B}$ we have $B \cap C \in \mathcal{B}$ ! (rename B by C and C by B, and you find what you are expecting)

Next, if $B \in \mathcal{B}$, we have that $\mathcal{C} \subset \mathcal{B}_B$ by the preceding, and so we conclude the proof ...

9. ## Re: Help with proof

Ah, so by the first part (when $B \in \mathcal{C}$ ) we conclude that
$\mathcal{B}_{B} = \mathcal{B}$, so $\mathcal{B}$ actually contains all intersections $B \cap C$, for each $C \in \mathcal{C}$ and $B \in \mathcal{B}$?

10. ## Re: Help with proof

Yes, by definition of $\mathcal{B}_B$.