Martingale convergence proof This is out of Durrett 5.5.7. 
Let $X_n \in [0,1]$ be adapted to $\mathcal{F_n}$. Let $\alpha,\beta > 0$ such that $\alpha + \beta = 1$. Suppose that
$$
 P(X_{n+1} = \alpha + \beta X_n | \mathcal{F_n}) = X_n, \qquad P(X_{n+1}=\beta X_n | \mathcal{F_n} ) = 1 - X_n
$$
Then $X_n \rightarrow X_{\infty}$ a.s., where $X_{\infty} \in \{0,1\}$, and $P(X_{\infty}=1) = \theta$ if $X_0 = \theta$.
Here's my attempt, that I'd like to get some help on. I think the high level idea is right: you show a limit exists a.s. in $[0,1]$, and then you argue the set of sequences which converge to some $\phi \in (0,1)$ must have measure zero. But of course, the formalism is the tricky part.
It's easy enough to check that $(X_n)$ is a martingale w.r.t. $(\mathcal{F_n})$. Therefore, by the martingale convergence theorem, we have $X_n \rightarrow X_{\infty}$ a.s. for some $X_{\infty}$. Clearly $X_{\infty} \in [0,1]$. 
Now, let $\omega \in \Omega$ be such that $X_n(\omega) \rightarrow X_{\infty}(\omega)$, and suppose $X_{\infty}(\omega) = \phi$, where $\phi \in (0,1)$. Clearly, for any $z \in (0,1)$, we have $\beta z < z < \alpha + \beta z$. 
Set 
$$
\epsilon = \min( \phi (1-\beta)/(1+\beta), (\alpha - \phi(1-\beta))/(1+\beta) ).
$$
There exists an $N$ such that for all $n \geq N$, we have $| X_n(\omega) - \phi | \leq \epsilon$. But we have set $\epsilon$ such that
$\alpha + \beta X_n(\omega) > \phi + \epsilon$ and $\beta X_n(\omega) < \phi - \epsilon$. But $X_{n+1}(\omega) \in \{\alpha + \beta X_n(\omega), \beta X_n(\omega)  \}$ (* I can't figure out how to make this argument here more rigorous *). Hence,
$$
  P( \{ \omega : X_{\infty}(\omega) \in (0,1) \}) = 0
$$
which yields the first part of the result. 
Now suppose $X_0 = \theta$. But now we have
$E(X_{\infty}) = P(X_{\infty}=1)$. Since $(X_n)$ is a martingale, $\theta = E(X_0) = E(X_n)$. Since $X_n \rightarrow X_\infty$ a.s., by the dominated convergence theorem we have $E(X_n) \rightarrow E(X_\infty)$, which yields the second result.

Let $Y_n:=X_{n+1}-\beta X_n$. Then taking the expectations in 
$$P(X_{n+1} = \alpha + \beta X_n | \mathcal{F_n}) = X_n, \qquad P(X_{n+1}=\beta X_n | \mathcal{F_n} ) = 1 - X_n$$ 
and using the fact that the expectation of $X_n$ is the same as that of $X_1$ (denoted $p$), we derive that $Y_n$ takes the value $\alpha$ with probability $p$ and zero with probability $1-p$. Since $Y_n\to X_{\infty}-\beta X_{\infty}=\alpha X_\infty$, the possible values of $\alpha X_\infty$ are $\alpha$ or zero.
