Likelihood-ratio change of measure

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 $\int grd{P}_0=\int g^{+}rd{P}_0-\int g^{-}rd{P}_0=\int gd{P}_1$.