# A question involving a field of sets that generates a $\sigma$-algebra

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?

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: 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