Jun Shao's Mathematical Statistics - Lemma 2.1 I am trying to understand Jun Shao's proof on Lemma 2.1 of his Mathematical Statistics book.
The statement is that: If a family of population $\mathcal{P}$ is dominated by a $\sigma$-finite measure, then $\mathcal{P}$ is dominated by a probability measure $Q = \Sigma_{i=1}^{\infty}c_i P_i$, where $c_i$'s are non-negative constants with $\Sigma_{i=1}^{\infty}c_i = 1$ and $P_i \in \mathcal{P}$.
Shao only proved the case for finite measure $\nu$. In his proof, he defined $\mathcal{C}$ to be the class of events $C$ for which there exists $P \in \mathcal{P}_0$ such that $P(C) > 0$ and $dP / d\nu > 0$ a.e. $\nu$ on $C$.
I have two questions:

*

*Why does there exist a sequence $\{C_i\} \subset \mathcal{C}$ such that $\nu(C_i) \rightarrow \sup_{C \in \mathcal{C}} \nu(C)$? Does it have anything to do with the finiteness of $\nu$ or the assumption that $dP / d\nu > 0$ a.e. $\nu$ on $C$?


*The assumption that $dP / d\nu > 0$ a.e. $\nu$ on $C$ is not very intuitive to me. Why do we need it to complete the proof? May someone explain to me the motivation behind this assumption?
Thank you in advance.
Here's the original statement and the proof of the lemma:

 A: It is enough to consider $\nu$ finite since any $\sigma$-finite measure $\eta$ is equivalent to a finte measure $\nu$, that is $\eta\ll \nu$, $\nu\ll \eta$. To see this, suppose $\eta$ is an infinite $\sigma$-finite measure. Let $(A_n:n\in\mathbb{N})$ be a pairwise disjoint sequence of measurable sets such that $\Omega=\bigcup_nA_n$ and $0<\eta(A_n)<\infty$ (this is possible since $\eta$ is $\sigma$-finite. Define
$$d\nu = \sum_n \frac{1}{\eta(A_n)}2^{-n}\mathbb{1}_{A_n}\,d\eta$$
Clearly $\nu(\Omega)=1<\infty$. By design, $\nu\ll \eta$ and since $h:=\sum_n \frac{1}{\eta(A_n)}2^{-n}\mathbb{1}_{A_n}>0$, $\eta=\frac{1}{h}\,d\nu$, that is $\eta\ll\nu$.
As for question (1), the finiteness of $\nu$ implies that $\mathcal{C}$ is not empty (it contains the measure $\nu$ that I defined above) and $\{\nu(C): V\in\mathcal{C}\}$ is a bounded set, so the supremum is finite and well defined.
As for question (2), the idea is that one wants to collect a countable sequence of probabilities $P_j$ that are dominated by $\nu$ (i.e. $P\ll \nu)$ and that have positive mass in each of the components $C_j$.
Once $Q$ is defined as $Q=\sum_jc_jP_j$ and $C_0:=\bigcup_jC_j$, you have that
$$\frac{dQ}{d\nu}=\sum_jc_j\frac{dP_j}{d\nu}$$
Notice that $Q( C_0)=\sum_jc_j P_j(C)\geq \sum_jc_jP_i(C_j)>0$ since $0<c_j<1$ and $P(C_j)>0$. Also, if $N_j=\{\omega\in C: \frac{dP_j}{d\nu}(\omega)=0\}$ then by definition of $\mathcal{C}$, $\nu(C_j)=0$; hence $\nu(\bigcup_jC_j)=0$. For $\omega\in C_0\setminus\bigcup_jN_j=C\cap\bigcap_j\Omega\setminus N_j$ we have that $\frac{dP_j}{d\nu}(\omega)>0$ for each $j$. Thus, $\frac{dQ}{d\nu}(\omega)=\sum_jc_j\frac{dP_j}{d\nu}(\omega)>0$.
The remaining of the proof is straight forward.
A: Regarding question (1), it follows from the fact that $P$ is dominated by $\nu$ : Indeed, for any $C$ such that $P(C)>0$, you necessarily have that $\nu(C)>0$ as well, since every $\nu$-null set is $P$-null.
Therefore, the set $\{\nu(C), \ C\in\mathcal C\}$ is a collection of positive numbers, and you can extract a countable collection $C_i$ such that $\nu(C_i)\to\sup_{C\in\mathcal C}\nu(C)$. In case $\nu$ is not finite, it roughly works the same, but with additional $\infty$ symbols to take care of.
Regarding question (2), the assumption on the Radon-Nikodym derivative is necessary to prove that $Q(A\cap C_0) = 0 \implies \nu(A\cap C_0) = 0$. Here are the details :
Since $Q=\sum_i c_iP_i$, the condition $Q(A\cap C_0) = 0$ implies that there exist some (possibly infinitely many) $(P_j)_j$ such that $P_j(A\cap C_0) = 0 $. But, by assumption, we have for every $P_j$ that
$$P_j(A\cap C_0)=\int_{x\in A\cap C_0} \frac{dP_j(x)}{d\nu}\cdot d\nu(x) $$
Notice however that $A\cap C_0\subseteq C_0$ and $C_0\in\mathcal C$, hence the Radon Nikodym derivative $dP_j(x)/d\nu$ is strictly positive for all $x\in A\cap C_0$. But because the integral of a positive function over a positive measure set has to be positive, we conclude that the set $A\cap C_0$ has $\nu$-measure zero (and because $P\ll\nu$, we also conclude that $P(A\cap C_0)=0$).
