I am studying Billingsley's section 35 about martingales and I have some difficulties understanding one of the examples. This is the example 35.4.
Suppose we have a measurable space $(\Omega,\mathcal F)$ and let $Q$ and $P$ be probability measures on the latter. Let $\mathcal F_n=\sigma (Y_1,\cdots,Y_n)$ be the sigma algebra generated by the random variables $Y_1,\cdots,Y_n$.
Suppose that under $P$ the distribution of the random vector $(Y_1,\cdots,Y_n)$ has density $p_n(y_1\cdots,y_n)$ with respect to the n-dimensional Lebesgue measure, and under $Q$ it has density $q_n(y_1\cdots,y_n)$. (To avoid technicalities assume that $p_n,q_n$ are everywhere positive.
Then the Radon-Nikodym derivative of $Q$ with respect to $P$ on $\mathcal F_n$ is $$X_n=\frac{q_n(y_1\cdots,y_n)}{p_n(y_1\cdots,y_n)}$$
So far so good, but the author then concludes that $X_n$ is a martigale under $P$ w.r.t. the filtration $\mathcal F_n$.
How can I see this last point,
why is that the following holds?
$$\int_F X_n dP=\int_F X_{n+1} dP, \forall F\in\mathcal F_n$$
Thanks in advance.
EDIT: I may have found a possible solution, but I am not sure about the validity of it.
I can assume that there is a random variable $\Phi\in\mathcal M(\mathcal F)$ such that $$Q(F)=\int_F \Phi dP,\forall F\in\mathcal F_n$$ (One problem might be that then $Q$ is not a probability measure on $\mathcal F$).
Then I can interpret the Radon-Nikodym derivative $X_n$ as the condition expectation $E[\Phi|\mathcal F_n]$. Finally I have that $$E[X_{n+1}|\mathcal F_n]=E[E[\Phi|\mathcal F_{n+1}]|\mathcal F_n]=E[\Phi|\mathcal F_n]=X_n$$
$$E[X_{n+1}]=E[E[X_{n+1}|\mathcal F_n]]=E[X_n]$$
