= Solution
The <martingale convergence theorem> in its $L^1$-bounded form says that a <martingale> with $\sup_n\mathbb E|M_n|<\infty$ has an integrable limit $M_\infty$ and
$$
\boxed{M_n\longrightarrow M_\infty\quad\text{almost surely},\qquad
\mathbb E|M_\infty|\leq\sup_n\mathbb E|M_n|.}
$$
The norm bound on the limit follows from <Fatou lemma>. Boundedness in $L^1$ by itself does not imply <convergence in L1>. The <uniformly integrable martingale convergence theorem> gives the stronger conclusion: if $(M_n)$ is <uniformly integrable>, then $M_n\to M_\infty$ both <almost surely> and in <L1 norm>, and $M_n=\mathbb E[M_\infty\mid\mathcal F_n]$. Conversely, <convergence in L1> implies <uniform integrability>.
For the distinction, the <fair-coin doubling martingale> $M_n=2^n\mathbf1_{\{\text{first }n\text{ tosses are heads}\}}$ has <expectation> one for every $n$ but converges <almost surely> to zero. Its <L1 norm> remains one, so its convergence is not in <L1 norm>. These two formulations specify exactly which hypothesis is needed in part (d).
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