= Solution
For $A\in\mathcal F_n$, the defining identity for $X_{n+1}$ gives
$$
\mathbb E[X_{n+1}\mathbf1_A]=\mathbb E[X\mathbf1_A]
=\mathbb E[X_n\mathbf1_A].
$$
Part (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 proves
$$
|X_n|\leq\mathbb E[|X|\mid\mathcal F_n].
$$
Let $B_{n,K}=\{|X_n|>K\}$. Then $\mathbb P(B_{n,K})\leq\mathbb E|X|/K$ by <Markov inequality>, while
$$
\mathbb E[|X_n|\mathbf1_{B_{n,K}}]
\leq\mathbb E[|X|\mathbf1_{B_{n,K}}].
$$
The <uniform absolute continuity for a finite measure> makes the right-hand side uniformly small as $K\to\infty$. Thus $(X_n)$ is <uniform integrability>[uniformly integrable]. The <Martingale convergence theorem> now supplies an integrable random variable $Y$ such that $X_n\to Y$ both <almost sure convergence>[almost surely] and in $L^1$.
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