How to justify the martingale condition

By Radon-Nikodym theorem, the conditional expectation of $$X$$ with respect to a $$\sigma$$-algebra $$\mathscr F$$ is a nonnegative random variable denoted by $$\def\E{\mathbf E}\E(X\mid \mathscr F)$$, such that

1. $$\def\E{\mathbf E}\E(X\mid \mathscr F)$$ is $$\mathscr F-measurable$$
2. for all $$A \in \mathscr F\\$$, $$\int_A \def\E{\mathbf E}\E(X\mid \mathscr F)\, dP=\int_A X\, dP$$

In definition of martingale, one condition is $$X_n$$ is $$\mathscr F_n$$-adapted. By definition above, $$\def\E{\mathbf E}\E[X_{n}\mid \mathscr F_n]$$ is well-defined. However, I don't know why $$\def\E{\mathbf E}\E[X_{n+1}\mid \mathscr F_n]$$ is still well-defined in $$X_{n} = \def\E{\mathbf E}\E[X_{n+1}\mid \mathscr F_n]$$.

• You may need to revise your question: it is not clear what you are asking for. – Gordon Dec 17 '18 at 14:46
• The link here explains what adapted means so the same assumption holds for $X_{n+1}$. en.wikipedia.org/wiki/Adapted_process – mark leeds Dec 17 '18 at 15:24
• @Gordon What i am trying to ask is ($X_n$ , $\mathscr F_n$) is martingale if $X_n$ $\in$ $\mathscr F_n$, then we know that $\def\E{\mathbf E}\E[X_{n}\mid \mathscr F_n]$ exist. But it does not say anything about $X_{n+1}$ $\in$ $\mathscr F_n$, then what is $\def\E{\mathbf E}\E[X_{n+1}\mid \mathscr F_n]$? – YellowRiver Dec 17 '18 at 22:54
• $X_n \in\mathscr{F}_n$ means that $\mathbb{E}(X_n\mid \mathscr{F}_n) = X_n$. Here, $\mathbb{E}(X_{n+1}\mid \mathscr{F}_n)$ is the conditional expectation of $X_{n+1}$ with respect to $\mathscr{F}_n$. – Gordon Dec 18 '18 at 14:08