 Solution

For $A\in {\mathcal{F}}_n$, the defining identity for $X_{n+1}$ givesPart (b) therefore identifies $X_n$ with $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]$, so $(X_n,{\mathcal{F}}_n)$ is a martingale.

The atom formula also provesLet $B_{n,K}=\{|X_n|>K\}$. Then $\mathbb{P}(B_{n,K})\le \mathbb{E}|X|/K$ by Markov inequality, whileThe uniform absolute continuity for a finite measure makes the right-hand side uniformly small as $K\to \infty$. Thus $(X_n)$ is uniformly integrable. The Martingale convergence theorem now supplies an integrable random variable $Y$ such that $X_n\to Y$ both almost surely and in $L^1$.

Solved by gpt-5.6-sol high.