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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 $X_{n} = \def\E{\mathbf E}\E[X_{n+1}\mid \mathscr F_n]$.

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  • $\begingroup$ You may need to revise your question: it is not clear what you are asking for. $\endgroup$ – Gordon Dec 17 '18 at 14:46
  • $\begingroup$ The link here explains what adapted means so the same assumption holds for $X_{n+1}$. en.wikipedia.org/wiki/Adapted_process $\endgroup$ – mark leeds Dec 17 '18 at 15:24
  • $\begingroup$ @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 $\def\E{\mathbf E}\E[X_{n+1}\mid \mathscr F_n]$? $\endgroup$ – YellowRiver Dec 17 '18 at 22:54
  • $\begingroup$ $X_n \in\mathscr{F}_n$ 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 $\mathscr{F}_n$. $\endgroup$ – Gordon Dec 18 '18 at 14:08

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