# Help to understand martingale example from Billingsley

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?

$$\int_F X_n dP=\int_F X_{n+1} dP, \forall F\in\mathcal F_n$$

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 $$E[X_{n+1}|\mathcal F_n]=E[E[\Phi|\mathcal F_{n+1}]|\mathcal F_n]=E[\Phi|\mathcal F_n]=X_n$$
$$E[X_{n+1}]=E[E[X_{n+1}|\mathcal F_n]]=E[X_n]$$