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I am learning the Monotone Class Theorem from Jacod's Probability Essentials. I don't quite understand the idea of the proof in the book. enter image description here

I don't see the point in the proof at all. What's the use of the "magic" construction $\mathcal{B}_B$? It's only clear to me that

  • $\Omega\in\mathcal{B}$;
  • $\mathcal{B}$ is closed under complement since it's closed by set difference and $\Omega\in\mathcal{B}$;

What's left is to show

  1. $\mathcal{B}$ is closed under countable union;
  2. $\mathcal{B}$ is the smallest $\sigma$-algebra containing $\mathcal{C}$.

However, what's the logical order of the proof? In which part of the proof are these two points shown?


[EDIT:]I think 1 is a direct result of the assumption that $\mathcal{B}$ is closed under increasing limits. For any $\{B_n\}_{n=1}^{\infty}\subset\mathcal{B}$, $\{\cup_{i=1}^{n}B_i\}_{n=1}^{\infty}\subset\mathcal{B}$ is increasing and $$ \cup_{n=1}^{\infty}B_n=\cup_{n=1}^{\infty}{\tilde B}_n, {\tilde B}_n:=\cup_{i=1}^{n}B_i. $$

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$\mathcal{B}$ is closed under finite unions because it is closed under finite intersections and complementations. Hence it is closed under countable unions because it is closed under increasing limits. –  Makoto Kato Sep 1 '12 at 6:04
    
I don't see why one needs $\mathcal{B}_B$ in the proof. –  Jack Sep 2 '12 at 1:06
    
One needs $\mathcal{B}_B$ to prove that $\mathcal{B}$ is closed under finite intersections. –  Makoto Kato Sep 2 '12 at 1:20
    
I don't see why one needs to show $\mathcal{B}$ being closed under finite intersections. As I mentioned in the post, $\mathcal{B}$ can be shown as a $\sigma$-algebra without using $\mathcal{B}_B$, if I am not wrong. –  Jack Sep 2 '12 at 1:31
    
You have to prove that $\mathcal{B}$ is closed under finite unions before you prove that $\mathcal{B}$ is $\sigma$-algebra. –  Makoto Kato Sep 2 '12 at 1:44

1 Answer 1

up vote 1 down vote accepted

Since any $\sigma$-algebra containing $\mathcal{C}$ is closed under increasing limit and set difference, such $\sigma$-algebra must contain $\mathcal{B}$ by definition of $\mathcal{B}$. If $\mathcal{B}$ is a $\sigma$-algebra containing $\mathcal{C}$, then it must be the smallest one. Therefore the main work for the proof is showing that $\mathcal{B}$ is a $\sigma$-algebra, which means

  1. $\Omega\in\mathcal{B}$;
  2. $\mathcal{B}$ is closed under complement;
  3. $\mathcal{B}$ is closed under countable union.

According to OP, it remains to show 3. For any $\{B_n\}_{n=1}^{\infty}\subset\mathcal{B}$, $\{\cup_{i=1}^{n}B_i\}_{n=1}^{\infty}\subset\mathcal{B}$ is increasing and
$$ \cup_{n=1}^{\infty}B_n=\cup_{n=1}^{\infty}{\tilde B}_n, {\tilde B}_n:=\cup_{i=1}^{n}B_i. $$ It suffices to show that ${\tilde B}_n:=\cup_{i=1}^{n}B_i\in\mathcal{B}$. It is for this reason that one needs $\mathcal{B}_B$.

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