Ok, I'm really stuck on this one and don't know what to do. Please help me out with some hints.
Here we go;
Consider the sequence of random variables $\lbrace X_n \rbrace_{n\geq 0}$ with values in $\left[0,1\right]$ and $X_0 = a$, where $a \in\left[0,1\right]$. Let the $\sigma$-algebra that's generated by this sequence be $$ \mathcal{F}_n = \sigma\left(X_0,\dots,X_n\right). $$
Let us assume that for $n = 0,1,2,\dots$, $$ \mathbb{P}\left( X_{n+1} = \frac{X_n}{2} \mid \mathcal{F}_n \right) = 1 - X_n, $$ and $$ \mathbb{P}\left( X_{n+1} = \frac{1+X_n}{2} \mid \mathcal{F}_n \right) = X_n. $$
Show that $\lbrace X_n, \mathcal{F}_n \rbrace_{n\geq 0}$ is a Martingale.
Thanks in advance!
Since $0\leqslant X_n\leqslant 1$ a.s. we have $\mathbb E[|X_n|]\leqslant 1$ so that $X_n\in L^1$ for all $n$. If $0<a<1$ then for any $n$ we have \begin{align} \mathbb E[X_{n+1}\mid \mathcal F_n] &= \frac12 X_n\mathbb P\left(X_{n+1}=\frac12 X_n\mid\mathcal F_n\right) + \frac12\left(1+X_n\right)\mathbb P\left(X_{n+1}=\frac12(1+X_n)\mid \mathcal F_n \right)\\ &= \frac12 X_n(1-X_n) + \frac12(1+X_n)X_n\\ &= X_n. \end{align} If $a=0$ then $X_n=0$ a.s. for all $n$, and if $a=1$ then $X_n=1$ a.s. In both of these cases the martingale property holds trivially. We conclude that $\{X_n\}$ is a martingale.