A countable mixture dominates a sigma-finite dominated family

Choose a countable measurable partition $(E_n{)}_{n\ge 0}$ of $\mathcal{X}$ with $\mu (E_n)<\mathrm{\infty }$ by disjointifying a sigma-finite cover [F2]. Put $c_n=2^{-(n+1)}\mathrm{/}(1+\mu (E_n))$ and $\lambda (A)={\sum }_{n\ge 0}{c}_n\mu (A\cap E_n)$. For disjoint measurable $A_m$, countable additivity of $\mu$ and interchange of nonnegative double series give $\lambda ({\bigcup }_m{A}_m)={\sum }_n{c}_n{\sum }_m\mu (A_m\cap E_n)={\sum }_m\lambda (A_m)$; the interchange is valid because both sides are the supremum of the same finite subsums. Thus $\lambda$ is a measure. Also $\lambda (\mathcal{X})\le {\sum }_{n\ge 0}{2}^{-(n+1)}=1$. Since every $c_n>0$, $\lambda (A)=0$ iff every $\mu (A\cap E_n)=0$, iff $\mu (A)=0$. Thus $\lambda$ is finite and equivalent to $\mu$, so every $P_{\theta }\ll \lambda$.

For each $\theta$, choose the RN density $p_{\theta }=d{P}_{\theta }\mathrm{/}d\lambda$ from [F3], finite outside a $\lambda$-null set, and let $A_{\theta }=\{p_{\theta }>0\}$. Fix ${\theta }_{\ast }$; since $\int p_{{\theta }_{\ast }}d\lambda =1$, $\lambda (A_{{\theta }_{\ast }})>0$. Define $s={\sup }_{F\subseteq \mathrm{\Theta },F\text{ finite}}\lambda ({\bigcup }_{\theta \in F}{A}_{\theta })$, so $0<s\le \lambda (\mathcal{X})<\mathrm{\infty }$. By [A1], choose finite $F_n$ for each $n\ge 0$ with union measure greater than $s-1\mathrm{/}(n+1)$. For all sufficiently large $n$, these unions have positive measure, so the selected index set ${\bigcup }_n{F}_n$ is nonempty. Set $A={\bigcup }_n{\bigcup }_{\theta \in F_n}{A}_{\theta }$. Finite unions from the selected sets have measure at most $s$, so [F4] applied to their indicators gives $\lambda (A)\le s$ by continuity from below; the approximations give $\lambda (A)\ge s$. Hence $\lambda (A)=s$.

Fix $\theta$. If $\lambda (A_{\theta }\setminus A)>0$, the increasing finite unions $({\bigcup }_{n\le N}{\bigcup }_{\varphi \in F_n}{A}_{\varphi })\cup A_{\theta }$ converge to $A\cup A_{\theta }$, whose measure is greater than $s$. [F4] applied to their indicators shows that some finite union then exceeds $s$, contradicting its definition. Therefore $p_{\theta }=0$ $\lambda$-almost everywhere on $A^c$.

Enumerate the nonempty countable collection of selected indices ${\bigcup }_n{F}_n$ as $({\theta }_j)$, allowing repetitions if needed, and choose positive weights summing to one. The mixture $Q={\sum }_j{w}_j{P}_{{\theta }_j}$ has density $q={\sum }_j{w}_j{p}_{{\theta }_j}$ by [F4], so $q>0$ exactly on $A$ up to a $\lambda$-null set. If $Q(C)=0$, then ${\int }_{C}qd\lambda =0$. For each $m\ge 1$, $\lambda (C\cap A\cap \{q\ge 1\mathrm{/}m\})\mathrm{/}m\le {\int }_{C}qd\lambda =0$; these sets cover $C\cap A$ up to the null set where $q$ is not positive on $A$. Hence $\lambda (C\cap A)=0$. Also every $p_{\theta }$ vanishes on $A^c$ by step 3.1, so $P_{\theta }(C)={\int }_C{p}_{\theta }d\lambda =0$. Since $\theta$ was arbitrary, $P_{\theta }\ll Q$ for all $\theta$.