I am reading Jun Shao's Mathematical Statistics and I got stuck on the proof of this lemma:
I understood the proof but to extend it from finite measure $\nu$ to a $\sigma$-finite measure $\nu$ is not trivial for me. I have try to decompose $\nu$ as a countable sum of finite measures $\nu_i$. But then, the family $\mathcal{P}$ is not dominated by $\nu_i$. If someone could give me a correct idea, I would be very appreciated.
Note: A measure $\mu$ is dominated by $\nu$ if $\mu$ is absolutely continuous with respect to $\nu$. In other words, $\nu(E)=0$ implies $\mu(E)=0$.
Any sigma finite measure is dominated by a probability measure: let $\nu$ be sigma finite. There exist disjoint sets $A_i, i \geq 1$ such that $\nu (A_i) <\infty$ and $\cup_i A_i$ is the entire space $\Omega$. Let $\nu_0 (E)=\sum\limits_{k=1}^{\infty} \frac {\nu(E\cap A_k)} {2^{k}\nu(A_k)}$. Then $\nu_0$ is a probability measure and $\nu << \nu_0$.
Thus, if your family is dominated by a sigma finite measure then it is also dominated by a finite measure so the first case is applicable.