= Solution
Use deterministic times tending to infinity and the <Fatou lemma>:
$$
\boxed{D:=\mathbb E[e^{M_\infty/2}]
\leq\liminf_{n\to\infty}\mathbb E[e^{M_n/2}]
\leq K<\infty.}
$$
Apply the <terminal scaling inequality for stochastic exponentials> to $X=aM$ and $r=1/a$. Part (c) gives $\mathbb E[\mathcal E(aM)_\infty]=1$, so
$$
\mathbb E[\mathcal E(M)_\infty]\geq D^{-2(1/a-1)}.
$$
Letting $a\uparrow1$ shows that the terminal expectation is at least one. The <nonnegative local martingale> $\mathcal E(M)$ starts at one and is a <supermartingale>, so the <Fatou lemma> gives the opposite inequality. Consequently
$$
\boxed{\mathbb E[\mathcal E(M)_\infty]=1.}
$$
The <terminal expectation criterion for a nonnegative local martingale> now closes the argument. Conditional <Fatou lemma> applied to the <supermartingale> at times tending to infinity gives
$$
\mathbb E[\mathcal E(M)_\infty\mid\mathcal F_t]\leq\mathcal E(M)_t.
$$
The left side has expectation one and the right side at most one, so equality holds almost surely. Thus
$$
\boxed{\mathcal E(M)_t
=\mathbb E[\mathcal E(M)_\infty\mid\mathcal F_t],}
$$
and <uniform integrability of conditional expectations> proves that $\mathcal E(M)$ is a <uniformly integrable martingale>. This establishes the required stopped-moment form of the <Kazamaki criterion>.
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