Seeking for counterexample in closed martingale. In the martingale convergent theorem:
$X_n$ is a martingale with $\sup_n\mathbb E[|X_n|]<\infty$.
Then there exist a $X_\infty\in L^1$ such that $X_n\to X_\infty\text{ a.s.}$
I want to find a example that $\mathbb E[X_\infty\mid\mathcal F_n]=X_n$ for all $n$ not hold.
Thanks in advance.
 A: Counterexample 1: Let $\xi_j$ be independent identically distributed random variables such that $$\mathbb{P}(\xi_j = 0) = \mathbb{P}(\xi_j =2)=\frac{1}{2}.$$ Then $$X_n := \prod_{j=1}^n \xi_j$$ is a non-negative martingale satisfying $\mathbb{E}(X_n)=1$. On the other hand, it is not difficult to see that $X_n \to 0$ almost surely. As $$\mathbb{E}(X_{\infty} \mid \mathcal{F}_n) = 0 \neq X_n$$ we are done.

Counterexample 2: Let $\xi_j$  be independent identically distributed random variables such that $\mathbb{P}(\xi_j = 1)= \mathbb{P}(\xi_j = -1) = 1/2$. Obviously, $X_n := \sum_{j=1}^n \xi_j$ defines a martingale. Define a stopping time $\tau$ by
$$\tau := \inf\{n \geq 0; X_n = -1\}$$
and set $Y_n := 1+ X_{\tau \wedge n}$. By the optional stopping theorem, $(Y_n)_{n \in \mathbb{N}}$ is a martingale. Since $Y_n$ is non-negative we find $$\sup_n \mathbb{E}(|Y_n|) = \sup_n \mathbb{E}(Y_n) = \mathbb{E}Y_0 = 1.$$ It follows from the martingale converence theorem that $$Y_{\infty} = \lim_{n \to \infty} (X_{\tau \wedge n}+1) = X_{\tau}+1 = 0.$$ Again, this implies that $$\mathbb{E}(Y_{\infty} \mid \mathcal{F}_n) \neq Y_n.$$
