From here.
The solution given is that
$$E[X_{n+1}\mid X_n] = 1/2\times 2X_n + 1/2\times 0 = X_n$$
Why exactly? In retrospect, I'm not sure I really got it. I'm trying to think about it in terms of indicator functions over events we must define:
Let $A_n = \{ \omega\mid X_{n+1}(\omega) = 2 X_n(\omega) \}$. Note: $A_n \in \mathscr{F}$ and $P(A_n) = 1/2 \ \forall \ n \in \mathbb{N}$
It looks like $X_{n+1} = 1_{A_n} (2X_n) + 1_{A_n^C}(0)$
$\to E[X_{n+1}\mid X_n] = E[1_{A_n} (2X_n) + 1_{A_n^C}(0)\mid X_n]$
$= E[1_{A_n} (2X_n)\mid X_n]$
$= (2X_n) E[1_{A_n}\mid X_n]$
$= (2X_n) E[1_{A_n}]$ (*)
$= (2X_n) P({A_n}) = X_n$ ?
(*) Is $\sigma(A_n)$ independent of $\sigma(X_n)$ ?
What is $\sigma(X_n)$ in terms of $A_n$ anyway? I guess that:
$\sigma(X_0) = \{ \emptyset, \Omega \}$
$\sigma(X_1) = \sigma(A_0) \because X_1 = 2 \times 1_{A_0}$
$\sigma(X_2) \cdots \subseteq \sigma(\sigma(A_0) \cup \sigma(A_1)) = \sigma(A_0, A_1)$
I think you can save yourself a lot of headache by thinking about it in a more heuristic way like this: \begin{align} \mathbb E[X_{n+1}\mid X_n] &= \mathbb E[X_{n+1} 1_{ \lbrace X_{ n+1} = 2 X_n\rbrace} + X_{n+1} 1_{ \lbrace X_{ n+1} = 0\rbrace} \mid X_n]\\ &=\mathbb E[2 X_{n} 1_{ \lbrace X_{ n+1} = 2 X_n\rbrace} \mid X_n]\\ &= 2X_n \mathbb P(X_{ n+1} = 2 X_n \mid X_n) \end{align} that last probability, by definition, is $1/2$