 Solution

The martingale is bounded in $L^2$, so the increment-square series from part (a)(ii) is finite. The identity just obtained implies ${\sum }_n\mathbb{E}[X_n(1-X_n)]<\infty$, hence $\mathbb{E}[X_n(1-X_n)]\to 0$. By almost sure convergence and the dominated convergence theorem,The integrand is nonnegative because $X_{\infty }\in [0,1]$. It must vanish almost surely, giving $X_{\infty }\in \{0,1\}$.

The martingale has constant expectation $a$. Its $L^2$ convergence implies $L^1$ convergence by Cauchy-Schwarz inequality, and hence $\mathbb{E}{X}_{\infty }=a$. ThereforeThe limiting law is the Bernoulli distribution with parameter $a$.