# Limit of a martingale sequence

I see this question on Stochastic Processes by Ross.

Let $$X_n$$ be a Markov process for which $$X_0$$ is uniform on (0, 1) and,

$$X_{n+1}= \begin{cases} \alpha X_n + 1 - \alpha,& \text{w.p. } X_n,\\ \alpha X_n, & \text{w.p. }1-X_n. \end{cases}$$

where $$0<\alpha<1$$. Discuss the limiting properties of the sequence $$\lim_{n\to\infty} X_n$$.

I was able to show $$E[X_{n+1}\mid X_n]=X_n$$, so the sequence is a martingale. Also since it is bounded between 0 and 1, a limit exists by the martingale convergence theorem. Also the mean of this limit is 0.5, as $$E[X_0]=0.5$$ and the mean is constant for a martingale. Can we precisely say what the limit is? Is there anything else we could say about the limit?

Denote $$X_\infty$$ the limit of the process. Let $$x\in(0,1)$$ and define $$A_x$$ as $$A_x\equiv\left\{\omega\in\Omega:\lim_{n\rightarrow\infty}X_n(\omega)=x\right\}$$ Now, fix $$\omega\in A_x$$ and $$\varepsilon>0$$. By definition $$\exists N\in\mathbb{N}, \forall n\geq N:\quad X_n(\omega)\in (x-\varepsilon, x+\varepsilon)\equiv I$$ In particular, for all $$n\geq N$$, $$X_{n+1}(\omega)$$ must lie in the set $$(\alpha(x-\varepsilon),\alpha(x+\varepsilon))\cup(\alpha(x-\varepsilon)+1-\alpha,\alpha(x+\varepsilon)+1-\alpha)\equiv J$$ Notice that if we take $$\varepsilon$$ small enough then $$I\cap J=\emptyset$$, which is impossible since $$X_{n+1}(\omega)\in I\cap J$$. This implies that for all $$x\in(0,1)$$, the set $$A_x$$ is empty. Therefore, $$X_\infty$$ a.s. takes values in $$\left\{0,1\right\}$$, and its distribution can be inferred from the equality $$\mathbb{E}X_\infty=1/2$$.