# Levy's Martingale Using Radon Nikodym

Let P and Q be two probability measures on the same space $(\Omega,\mathcal{F},\mathcal{P})$ and let $\mathcal{F_n}$ be filtration. Assume that $Q \ll P$. Let $X_n$ denote the Radon-Nikodym derivative of Q with respect to P on $\mathcal{F_n}$ i.e $X_n$ is $\mathcal{F_n}$-measurable and for any $A \in \mathcal{F_n}$, $Q(A) = \int_AX_ndP$. Show that $(X_n,\mathcal{F_n})$ is a martingale with respect to P, that is $E_P(X_n|\mathcal{F_{n-1}}) = X_{n-1},$ where for any random variable X, $E_P(X) = \int XdP$.

To Prove $(X_n,\mathcal{F_n})$ martingale

i) $X_{n}$ is $\mathcal{F_{n}}$ - measurable.

ii) $E[|X_n|] <\infty$

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

what I don't understand what is the mean of $(X_n,\mathcal{F_n})$ is a martingale with respect to P, that is $E_P(X_n|\mathcal{F_{n-1}}) = X_{n-1},$

Measurability and integrability is a consequence of the Radon-Nikodym theorem. Since $\mathcal{F}_{n-1}\subseteq\mathcal{F}_n$ we have that $$\int_AX_{n-1}\,\mathrm dP=Q(A)=\int X_n\,\mathrm dP$$ for all $A\in\mathcal{F}_{n-1}$ showing that ${\rm E}_P[X_n\mid\mathcal{F}_{n-1}]=X_{n-1}$.