1. ## Help with proof

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

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

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

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

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

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

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

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

2. ## Re: Help with proof

Hi,

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

3. ## Re: Help with proof

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

4. ## Re: Help with proof

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

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

5. ## Re: Help with proof

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

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

6. ## Re: Help with proof

This is absolutely the same scheme of proof than before.

We have just showed that for all $\displaystyle B \in \mathcal{B}$ and all $\displaystyle C \in \mathcal{C}$, we have $\displaystyle B \cap C \in \mathcal{B}$.
This means that for every $\displaystyle B \in \mathcal{B}$, we have $\displaystyle \mathcal{C} \subset \mathcal{B}_B$, and so we get $\displaystyle \mathcal{B}_B = \mathcal{B}$ since $\displaystyle \mathcal{B}$ 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
$\displaystyle B \cap C \in \mathcal{B}$ when $\displaystyle B \in \mathcal{B}$. I see how you conclude this when you say $\displaystyle B \in \mathcal{C}$.

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

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

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

8. ## Re: Help with proof

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

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

9. ## Re: Help with proof

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

10. ## Re: Help with proof

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