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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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    @Rasmus: Yes. you are right. But being closed under com$p$lement and containing Ω are necessary for its definition.2011-05-02

1 Answers 1

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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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    @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.2011-05-02