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Suppose $(\Omega, \mathcal{\Sigma})$ is a measurable space equipped with two probability measures $P_1$ and $P_2$. $\mathcal{F}$ is a field in $\Omega$ and $\sigma(\mathcal{F})=\mathcal{\Sigma}$. (Added: A field of sets is defined to be closed under finite union and complement and contain $\Omega$.)

I was wondering if $$\lim_{\delta \rightarrow 0} \quad \sup_{B \in \mathcal{F}, P_2(B) < \delta} P_1(B) = 0$$ implies $$\lim_{\delta \rightarrow 0} \quad \sup_{B \in \mathcal{\Sigma}, P_2(B) < \delta} P_1(B) =0$$ and, if yes, what conclusions or theorems can be used to prove it?

Thanks in advance!

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Is a field the same as a $\sigma$-algebra without the $\sigma$ (i.e. closed under finite unions and intersections)? –  Rasmus May 2 '11 at 12:32
    
@Rasmus: Yes, if I understand correctly, that is the only difference between them. –  steveO May 2 '11 at 12:33
1  
@Rasmus: Those are interchangeable terms, field and algebra in this context. I even heard $\sigma$-field being used a few times. –  Asaf Karagila May 2 '11 at 12:49
    
Field is more probabilistic. Algebra is more for analysts. It is the same thing. –  Jonas Teuwen May 2 '11 at 13:08
    
@Rasmus: Sorry I misunderstood your comment. No. A field of sets is closed under finite union and complement and contains $\Omega$$. –  steveO May 2 '11 at 13:30

1 Answer 1

up vote 4 down vote accepted

Yes.

Fix $\epsilon > 0$. By assumption there exists $\delta$ such that for all $B \in \mathcal{F}$ with $P_2(B) < \delta$, we have $P_1(B) \le \epsilon$.

Let $\mathcal{M}$ be the collection of all $B \in \Sigma$ such that either:

  1. $P_2(B) < \delta$ and $P_1(B) \le \epsilon$; or

  2. $P_2(B) \ge \delta$.

One can now apply the monotone class theorem; $\mathcal{M}$ is a monotone class which contains the field $\mathcal{F}$, so $\mathcal{M} = \Sigma$. This implies the desired conclusion.

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Thank you very much! 1) Is $\lim_{\delta \rightarrow 0} \quad \sup_{B \in \mathcal{F}, P_2(B) < \delta} P_1(B) = 0$ a limit in some sense? In your second paragraph, it seems to me it is like a limit of function. 2) Generally speaking, is this way of constructing $\mathcal{M}$ worth memorizing and has it been used in proving some theorems? –  steveO May 2 '11 at 14:47
    
@steveO: 1) Sure, it's the limit as $x \to 0$ of the function $f(x) = \sup_{B \in \mathcal{F}, P_2(B) < x} P_1(B)$ :-) 2) I don't know about this particular $\mathcal{M}$, but monotone class theorem proofs tend to have this form: to show all sets in a $\sigma$-field $\Sigma$ have some property $Q$, show that the collection $\mathcal{M}$ of all sets with property $Q$ form a monotone class, then show there is a field $\mathcal{F}$ which generates $\Sigma$ and such that all sets in $\mathcal{F}$ have property $Q$. –  Nate Eldredge May 2 '11 at 16:04
    
Thanks! How do you go from "for all $B \in \mathcal{F}$ with $P_2(B) < \delta$, we have $P_1(B) \le \epsilon$", to "for all $B \in \mathcal{M}$ with $P_2(B) < \delta$, we have $P_1(B) \le \epsilon$"? –  steveO May 2 '11 at 16:36
    
@steveO: I'm not sure what you're asking. That comes from the way I defined $\mathcal{M}$. Indeed, I specifically chose the definition of $\mathcal{M}$ so that it would be true. –  Nate Eldredge May 2 '11 at 16:56

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