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Thread: Help with proof

  1. #1
    Junior Member TheProphet's Avatar
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    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.
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  2. #2
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    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 $\displaystyle B \cap C \in \mathcal{B}$.
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  3. #3
    Junior Member TheProphet's Avatar
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    Re: Help with proof

    $\displaystyle \mathcal{B}$ is not closed under finite intersection, $\displaystyle \mathcal{C}$ is.
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  4. #4
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    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.
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  5. #5
    Junior Member TheProphet's Avatar
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    Re: Help with proof

    Quote Originally Posted by Ahriman View Post
    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
    $\displaystyle \mathcal{B} = \mathcal{B}_{B}$ ?
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  6. #6
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    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 ...
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  7. #7
    Junior Member TheProphet's Avatar
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    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 $\displaystyle C \in \mathcal{C}$?
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  8. #8
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    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 ...
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  9. #9
    Junior Member TheProphet's Avatar
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    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 $\displaystyle B \in \mathcal{B}$?
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  10. #10
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    Re: Help with proof

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