Domination and Likelihood Ratios

Choose the singular carrier $N$ from [F1]. For any measurable $E$, the singular part vanishes on $E\setminus N$, so the decomposition gives ${\nu }_{\mathrm{a}\mathrm{c}}(E\setminus N)=\nu (E\setminus N)\ge 0$. Also ${\nu }_{\mathrm{a}\mathrm{c}}(E\cap N)=0$ by [F1] and [F2]. Additivity therefore gives ${\nu }_{\mathrm{a}\mathrm{c}}(E)={\nu }_{\mathrm{a}\mathrm{c}}(E\setminus N)\ge 0$. All displayed component values are finite because the sum is the finite value $\nu$ and the other component is zero on the set being evaluated.

On $E\cap N$, [F2] gives ${\nu }_{\mathrm{a}\mathrm{c}}(E\cap N)=0$, so the decomposition yields ${\nu }_{\mathrm{s}}(E\cap N)=\nu (E\cap N)\ge 0$. The singular part vanishes on $E\setminus N$ by [F1]; hence ${\nu }_{\mathrm{s}}(E)={\nu }_{\mathrm{s}}(E\cap N)\ge 0$. Since $E$ was arbitrary, both components are positive measures.

Choose increasing countable finite-measure covers $(A_i)$ for $\mu$ and $(B_j)$ for $\nu$, using [F1]. Set $X_n={\bigcup }_{i,j\le n}(A_i\cap B_j)$. Then $X_n\uparrow X$ and both $\mu (X_n)$ and $\nu (X_n)$ are finite, since each finite union is contained in finite unions of sets from the corresponding cover. By [F2], $\mathrm{\mid }\nu \mathrm{\mid }(X_n)=\nu (X_n)<\mathrm{\infty }$. Thus [F3] applies. Fix any finite-valued representative $f$ of the nonempty class; it satisfies $\nu (E)={\int }_{E}fd\mu$ for every measurable $E$ and ${\int }_{X_n}\mathrm{\mid }f\mathrm{\mid }d\mu <\mathrm{\infty }$.

For $n,m\ge 1$, let $E_{n,m}=X_n\cap \{f\le -1\mathrm{/}m\}$, which is measurable by [F5]. On $E_{n,m}$, $-f\ge 1\mathrm{/}m$; because $f$ is integrable on $X_n$, [F6] and [F7] give ${\int }_{E_{n,m}}fd\mu \le -\mu (E_{n,m})\mathrm{/}m$. If $\mu (E_{n,m})>0$, the representation in step 1.1 would give $\nu (E_{n,m})<0$, contradicting positivity of $\nu$. Hence every $E_{n,m}$ is $\mu$-null.

The countable union $N={\bigcup }_{n,m\ge 1}{E}_{n,m}$ is measurable and $\mu$-null by [F5]. Since the sets $X_n$ cover $X$, $N=\{f<0\}$.

Define $\tilde{f}=\max \{f,0\}$. By [F7], $\tilde{f}$ is measurable; it is nonnegative, equals $f$ off $N$, and is zero on $N$. For every measurable $E$, absolute continuity gives $\nu (E)=\nu (E\setminus N)$ by [F4], while step 1.1 gives $\nu (E\setminus N)={\int }_{E\setminus N}fd\mu ={\int }_E\tilde{f}d\mu$. Thus $\tilde{f}$ represents the Radon–Nikodym class. The argument also shows that any finite-valued representative is nonnegative outside a null set.

The decomposition $P_1=P_1+0$ has absolutely continuous part $P_1$ and singular part $0$, because the hypothesis gives $P_1\ll P_0$. Comparing it with the decomposition in [F1], uniqueness [F2] gives $P_{1,\mathrm{a}\mathrm{c}}=P_1$ and $P_{1,\mathrm{s}}=0$.

By [F3], $P_{1,\mathrm{a}\mathrm{c}}$ is positive and finite, and $P_0$ is a finite positive measure. Choose a nonnegative representative $\widehat{r}$ supplied by [F4], so $P_{1,\mathrm{a}\mathrm{c}}(A)={\int }_A\widehat{r}d{P}_0$ for every $A\in \mathcal{A}$. By step 1.1, $P_{1,\mathrm{a}\mathrm{c}}=P_1$. The representative $r$ in the statement belongs to this same Radon–Nikodym class, so [F4] gives a measurable $P_0$-null set $N$ outside which $r=\widehat{r}$. By [F5], both $rd{P}_0$ and $\widehat{r}d{P}_0$ are measures; their values on $A\cap N$ are zero by [F8], and their values on $A\setminus N$ agree because the densities agree there. Measure additivity therefore gives ${\int }_{A}rd{P}_0={\int }_A\widehat{r}d{P}_0=P_1(A)$ for every measurable $A$. Thus the density measure $rd{P}_0$ in [F5] equals $P_1$.

For nonnegative measurable $g$, equality of these measures and [F6] give $\int gd{P}_1=\int gd(rd{P}_0)=\int grd{P}_0$. This is the asserted expectation identity in the nonnegative case.

Now let $g$ be real-valued and integrable under $P_1$. By [F7], $g^{+}$ and $g^{-}$ are measurable and integrable. Applying step 3.1 to each gives $\int g^{\pm }rd{P}_0=\int g^{\pm }d{P}_1<\mathrm{\infty }$; applying it to $\mathrm{\mid }g\mathrm{\mid }$ also shows $\int \mathrm{\mid }g\mathrm{\mid }rd{P}_0=\int \mathrm{\mid }g\mathrm{\mid }d{P}_1<\mathrm{\infty }$. Thus $gr\in L^1(P_0)$, and linearity from [F7] yields $\int grd{P}_0=\int g^{+}rd{P}_0-\int g^{-}rd{P}_0=\int gd{P}_1$.

Choose increasing finite-measure covers $(A_i)$ for $\mu$ and $(B_j)$ for $\nu$ by [F1], and set $X_n={\bigcup }_{i,j\le n}(A_i\cap B_j)$. Then $X_n\uparrow \mathcal{X}$, $\mu (X_n)<\mathrm{\infty }$, and $\nu (X_n)<\mathrm{\infty }$. Since $P_{\theta }$ is a probability measure, [F8] gives $\mathrm{\mid }{P}_{\theta }\mathrm{\mid }(X_n)\le 1$. This is a common finite exhaustion for $\mu ,\nu ,P_{\theta }$, so the hypotheses in [F5] hold for every $\theta$, with $(\lambda ,\rho ,\sigma )=(\mu ,\nu ,P_{\theta })$.

By [A1], [F3], and [F4], choose nonnegative representatives $h=d\nu \mathrm{/}d\mu$ and $p_{\theta }^{\mu },p_{\theta }^{\nu }$ for the stated classes. The measurable set $H=\{x:h(x)=0\}$ satisfies $\nu (H)={\int }_{H}hd\mu =0$ by [F3]; then $\mu (H)=0$ by [F6]. Hence $h>0$ $\mu$-almost everywhere.

Fix $\theta$. Applying [F5] on the common exhaustion from step 1.1 gives $p_{\theta }^{\mu }=h{p}_{\theta }^{\nu }$ $\mu$-almost everywhere. For a fixed pair $\theta ,\eta$, remove the union of their two exceptional null sets and the null set $H$ from step 1.2. On the remaining set, both density identities hold and $h>0$, so division by the positive common factor gives $p_{\theta }^{\mu }\mathrm{/}{p}_{\eta }^{\mu }=p_{\theta }^{\nu }\mathrm{/}{p}_{\eta }^{\nu }$ wherever the denominators are positive. The exceptional set may depend on the pair, which proves the stated scope.