Conditional expectation and absolute continuity Let $(\Omega,\mathscr{F})$ be a measurable space and $P$, $Q$ be two probability measures. Assume $Q$ is absolutely continuous with respect to $P$ and $\mathrm{d}Q/\mathrm{d}P=f$. I will use $E^P$ and $E^Q$ to denote the expectations with respect to $P$ and $Q$ respectively.
Assume $X$ is an integrable random variable and $\mathscr{G}\subset\mathscr{F}$ is a sub $\sigma$-algebra. Then I know the following equation $$E^Q(X|\mathscr{G})E^P(f|\mathscr{G})=E^P(Xf|\mathscr{G})$$ holds $Q$-almost surely.
Does this equation also hold $P$-almost surely?
 A: Not generally, no.
It will, however, be true, if the measures $P$ and $Q$ are equivalent measures, meaning that $Q$ is absolutely continuous with respect to $P$, and that $P$ is also absolutely continuous with respect to $Q$.
In the present case, we have only that $Q\ll P$ (denoting the assumption that $Q$ is absolutely continuous with respect to $P$), and this is equivalent to the implication
$$P(A) = 1\Rightarrow Q(A) = 1,$$
but the other direction is not guaranteed. Thus, even though you have a $Q$-almost sure event, it need not be a $P$-almost sure event.
However, if the density $f$ in the expression $\text{d}Q/\text{d}P=f$ is strictly positive, then you will have that $Q$ and $P$ are equivalent.
A: Now I think the above equation also holds $P$ almost surely. 
First of all, I think the statement $$E^Q(X|\mathscr{G})E^P(f|\mathscr{G})=E^P(Xf|\mathscr{G})\quad P-a.s.$$ should be understood in the following way: 
"Let $\xi$ and $\eta$ be any versions of $E^Q(X|\mathscr{G})$ and $E^P(f|\mathscr{G})$ respectively. Then $\xi\eta$ is a version of $E^P(Xf|\mathscr{G})$."
Here is my proof. Because both $\xi$ and $\eta$ are $\mathscr{G}$-measurable, so is $\xi\eta$. Now consider any $A\in\mathscr{G}$. We want to show $$\int_A \xi\eta\mathrm{d}P=\int_A Xf\mathscr{d}P.$$ Notice $$\int_A \xi f\mathrm{d}P=\int_A\xi\mathrm{d}Q=\int_AX\mathrm{d}Q=\int_AXf\mathrm{d}P,$$ where the first and last equalities come from $\mathrm{d}Q/\mathrm{d}P=f$ and the second comes from the assumption that $\xi$ is a version of $E^Q(X|\mathscr{G})$. Hence we only need to show $$\int_A \xi\eta\mathrm{d}P=\int_A \xi f\mathrm{d}P.$$ This is true because $$\int_A \xi \eta\mathrm{d}P=\int_A E^P(\xi f|\mathscr{G})\mathrm{d}P=\int_A \xi f\mathrm{d}P,$$ completing the proof.
In many textbooks, this statement is left as an exercise and it is always written as $$E^Q(X|\mathscr{G})=\frac{E^P(Xf|\mathscr{G})}{E^P(f|\mathscr{G})}\quad Q-a.s.$$ I think the reason that we only get $Q$-a.s 
in this formula is due to the fact that $E^P(f|\mathscr{G})$ might be $0$, i.e., it is possible that $P(E^P(f|\mathscr{G})=0)>0$. However, it is easy to see $Q(E^P(f|\mathscr{G})=0)=0$.
