Sufficient Statistics: Proof of a lemma by Halmos and Savage I am reading the paper "Application of the Radon-Nikodym Theorem to the Theory of Sufficient Statistics" by Halmos and Savage and have much trouble following the proof of Lemma 7. Below is (my paraphrase of) the statement of the lemma and its proof.
Lemma: Let $\mathcal{P}$ be a family of probability measures on the measurable space $(\Omega, \mathcal{F})$. If $\mathcal{P}$ is dominated by a finite measure $\nu$ (i.e. $\mathbb{P} \ll \nu$ for all $\mathbb{P}\in\mathcal{P}$), then there exists a countable subset $\mathcal{P}_0\subset\mathcal{P}$ such that for every $A\in\mathcal{F}$, $\mathbb{P}_0(A)=0$ for all $\mathbb{P}_0\in\mathcal{P}_0$ implies $\mathbb{P}(A)=0$ for all $\mathbb{P}\in\mathcal{P}$.
The proof provided in the paper is as follows. 
Define the collection of sets $\mathcal{K}$ like so: We say that the $\mathcal{F}$-measurable set $K\in\mathcal{K}$ if and only if there exists some $\mathbb{P}\in\mathcal{P}$ corresponding to this $K$ such that $\mathbb{P}(K)>0$ and $\frac{d\mathbb{P}}{d\nu}>0$ on $K$. Let $\mathcal{C}$ be the collection of all sets of the form $\dot{\bigcup}^\infty_{i=1}K_i$, where $(K_i)^\infty_{i=1}\subset\mathcal{K}$ are disjoint. Then it can be shown that $\mathcal{C}$ is closed under countable unions.
We now construct our countable subset $\mathcal{P}_0$. Choose a sequence $(C_i)^\infty_{i=1}\subset\mathcal{C}$ such that $\nu(C_i)\to\sup_{C\in\mathcal{C}}\nu(C)$ and let $C$ be the union of the $C_i$s. Then since $\mathcal{C}$ is closed under countable unions, $C\in\mathcal{C}$ and there exists $(K_i)^\infty_{i=1}\subset\mathcal{K}$ such that $C=\dot{\bigcup}^\infty_{i=1}K_i$. We then let $\mathcal{P}_0=\{\mathbb{P}_1, \mathbb{P}_2,\cdots\}$, where each $\mathbb{P}_i$ is the probability measure corresponding to $K_i$.
Let $A\in\mathcal{F}$ be such that $\mathbb{P}_i(A)=0$ for each $i$ and $\mathbb{P}$ be any probability measure in $\mathcal{P}$. Denote $B=\{\omega\in\Omega: \frac{d\mathbb{P}}{d\nu}(\omega)>0\}$ Since $\color{red}{\mathbb{P}(A\cap B^c)=0}$, we may assume WLOG that $A\subset B$. It remains to show $\mathbb{P}(A)=0$.
We first show that $\mathbb{P}(A\cap C^c)=0$. Suppose not. Then $\nu(A\cap C^c)>0$ and therefore, because $\color{red}{A\cap C^c\in\mathcal{K}}$, $\color{red}{A\cup C\in\mathcal{C}}$ with $\color{red}{\nu(A\cup C)>\nu(C)}$, contradicting the maximality of $\nu(C)$. Hence $\mathbb{P}(A\cap C^c)=0$.
Finally we show also that $\mathbb{P}(A\cap C)=0$. By $\sigma$-additivity and the fact that $\mathbb{P} \ll \nu$, this follows if we can show that $\nu(A\cap K_i)=0$ for every i. But this is true since $0 = \mathbb{P}_i(A\cap K_i) = \int_{A\cap K_i} \frac{d\mathbb{P}_i}{d\nu} d\nu$ and $\frac{d\mathbb{P}_i}{d\nu}$ is strictly positive on $A\cap K_i$. This completes the proof.
I am most definitely missing many obvious things but I am unable to grasp nor understand why the following claims are true:


*

*$\mathbb{P}(A\cap B^c)=0$

*${A\cap C^c\in\mathcal{K}}$

*$A\cup C\in\mathcal{C}$

*$\nu(A\cup C)$ is strictly greater than $\nu(C)$

*The motivation behind the construction of $\mathcal{P}_0$


I would appreciate it if anyone could enlighten me on this. Thank you!
 A: I'll answer the last item first: $\sigma$-algebras are well-behaved for sequences, and typically need not be well-behaved in terms of anything else. Therefore it is often convenient to reduce some (possibly uncountable) list of conditions to be checked to a countable list of conditions to be checked. 
For the first item, $$\mathbb{P}(A\cap B^c) \leqslant \mathbb{P}(B^c)=\int_{B^c} d\mathbb{P}= \int_{B^c} \frac{d\mathbb{P}}{d\nu} d\nu = \int_{B^c}0 d\nu =0.$$ Here we are using that $\frac{d\mathbb{P}}{d\nu}\equiv 0$ on $B^c$. 
For the next three items, it should be kept in mind that they are proving by contradiction that $\mathbb{P}(A\cap C^c)=0$, so we assume for all three remaining items that $\mathbb{P}(A\cap C^c)>0$. 
Since $A\cap C^c\subset A\subset B$, $\frac{d\mathbb{P}}{d\nu}>0$ on $A\cap C^c$, since the same inequality is true on the larger set $B$. Moreover, we have assumed as our contradiction hypothesis that $\mathbb{P}(A\cap C^c)>0$. These two conditions are what is required for $A\cap C^c$ to be in $\mathcal{K}$. This gives the second item. 
For the third item, we can write $C$ as a countable union $\cup_{i=1}^\infty K_i$ of disjoint members of $\mathcal{K}$. Then $$A\cup C=(A\cap C^c)\cup C= (A\cap C^c)\cup\bigcup_{i=1}^\infty K_i.$$  This is a disjoint union of members of $\mathcal{K}$. We proved $A\cap C^c$ is in $\mathcal{K}$ above. The set $A\cap C^c$ is disjoint from the $K_i$ sets because $A\cap C^c\subset C^c$ and $K_i\subset C$. Since $A\cup C$ is a disjoint, countable union of members of $\mathcal{K}$, it is in $\mathcal{C}$. 
For the last item, since $\nu(A\cap C^c)>0$, $$\nu(A\cup C)=\nu(A\cap C^c)+\nu(C)>\nu(C).$$  
