Likelihood tilts are sigma-finite measures

Fix $\theta$ and define $r_{\theta }=p_{\theta }\mathrm{/}{p}_{{\theta }^{\ast }}$ on $\{p_{{\theta }^{\ast }}>0\}$ and $r_{\theta }=0$ elsewhere. The supplied finite-valued nonnegative densities make $r_{\theta }$ nonnegative and finite everywhere. For every real $a\ge 0$, $\{r_{\theta }>a\}=\{p_{{\theta }^{\ast }}>0\}\cap \{p_{\theta }-a{p}_{{\theta }^{\ast }}>0\}\mathrm{.}$ The difference is an everywhere-defined measurable function by [F2], so this intersection is measurable. For $a<0$, the threshold set is $\mathsf{X}$. Thus [F5] proves measurability of $r_{\theta }$, and $Q_{\theta }(A)={\int }_A{r}_{\theta }d{P}_0$ is defined for every $A\in \mathcal{A}$.

By [F3], $Q_{\theta }$ is a measure. For each integer $k\ge 1$, set $E_k=\{x:r_{\theta }(x)\le k\}$. These sets are measurable by step 1.1 and increase to $\mathsf{X}$ because $r_{\theta }$ is finite everywhere. Moreover, $Q_{\theta }(E_k)={\int }_{E_k}{r}_{\theta }d{P}_0\le k{P}_0(E_k)\le k$ by [F4]. Thus $(E_k{)}_{k\ge 1}$ is a countable cover by finite-$Q_{\theta }$-measure sets, and $Q_{\theta }$ is sigma-finite. For the selected minimizer, $r_{{\theta }^{\ast }}=1_{\{p_{{\theta }^{\ast }}>0\}}$, so $Q_{{\theta }^{\ast }}(\mathsf{X})=P_0\{p_{{\theta }^{\ast }}>0\}=1$ by the assumed almost-sure positivity.