How to find the distribution of the random variable $X_{\infty}$
Given $X_0=a$ with $a\in [0,1]$ the distribution
$P(X_{n+1}=\frac{X_n}{2})=1-X_n$ and $P(X_{n+1}=\frac{1+X_n}{2})=X_n$
Then I could show that,
$\texttt{(I)}\ $ $X_n$ is a martingale and converges a.s. and in $L^2$ to some $X_{\infty}$ with $X_n=E(X_\infty|\cal F_n)$, where $\cal{F_n}$$=\sigma(X_0,\dots,X_n)$, so it follows $E(X_\infty)=E(X_\infty|\cal F_0)$$=X_0=a$
$\texttt{(II)}\ $ $E\left((X_{n+1}-X_n)^2\right)=E\left(\frac14X_n(1-X_n)\right)$
it remains to figure out
$E\left(X_\infty(1-X_\infty)\right)$ and the distribution of $X_\infty$
My attempt was the following,
I think from $\texttt{(II)}$ it follows that $0=E\left((X_{\infty}-X_\infty)^2\right)=E\left(\frac14X_\infty(1-X_\infty)\right)$,
Thus $E\left(X_\infty(1-X_\infty)\right)=0$ but this r.v. is nonnegative, so $X_\infty(1-X_\infty)=0$ a.s. And in order that $E(X_\infty)=a$ holds (from $\texttt{(I)}$) one has to define,
$P(X_\infty=0)=1-a$ and $P(X_\infty=1)=a$
Are these steps correct ?