= Solution
For $a<b$, let $U_N[a,b]$ be the number of completed <upcrossings> by the observations $X_1,\ldots,X_N$: buy when the value is at most $a$, sell at the first subsequent value at least $b$, and repeat. With $z^-=\max\{-z,0\}$, one form of the <Doob upcrossing inequality> for a <martingale> is
$$
(b-a)\mathbb E U_N[a,b]\le\mathbb E(X_N-a)^-.
$$
For completeness, hold either zero or one unit according to this rule. The holding at each step is measurable before the next <martingale> increment, so the total gain $G_N$ has <expected value> zero. Completed trades earn at least $b-a$ each; an unfinished trade loses at most $(X_N-a)^-$. Thus $G_N\ge(b-a)U_N[a,b]-(X_N-a)^-$, proving the inequality.
Suppose $C=\sup_n\mathbb E|X_n|<\infty$. For each rational pair $a<b$ the inequality and <monotone convergence> give
$$
\mathbb E U_\infty[a,b]\le\frac{C+|a|}{b-a}<\infty.
$$
Every rational interval therefore has only finitely many <upcrossings>, simultaneously outside a countable union of null sets. If $\liminf X_n<\limsup X_n$, there is a rational interval strictly between them and infinitely many <upcrossings>, a contradiction. Thus the limit $X_\infty$ exists in the extended real line <almost surely>. <Fatou's lemma> now gives
$$
\mathbb E|X_\infty|\le\liminf_n\mathbb E|X_n|\le C.
$$
The limit is finite <almost surely> and integrable. This proves the <martingale convergence theorem> rather than using it as the deduction.
The exact extra condition for <convergence in L1> is \b[<uniform integrability>]:
$$
\boxed{\lim_{K\to\infty}\sup_n\mathbb E\bigl[|X_n|\mathbf1_{\{|X_n|>K\}}\bigr]=0.}
$$
To see sufficiency, first discard the tails of $X_n$ uniformly and the tail of the integrable $X_\infty$. On the set where both absolute values are at most $K$, their difference is bounded by $2K$ and tends to zero <almost surely>; <dominated convergence> finishes the proof. Conversely, if $X_n\to X_\infty$ in $L^1$, then
$$
\mathbb E\bigl[|X_n|\mathbf1_{\{|X_n|>K\}}\bigr]\le2\mathbb E|X_n-X_\infty|+\mathbb E\bigl[|X_\infty|\mathbf1_{\{|X_\infty|>K/2\}}\bigr].
$$
Make the first term uniformly small for all sufficiently large $n$, then control the finitely many remaining integrable variables by increasing $K$. Hence <uniform integrability> is necessary too. In this case the <conditional expectation> identity $X_n=\mathbb E[X_\infty\mid\mathcal F_n]$ follows by taking the $L^1$ limit in $X_n=\mathbb E[X_m\mid\mathcal F_n]$ for $m\ge n$.
The <coin-doubling martingale> shows why an $L^1$ bound is not enough. Let $\xi_1,\xi_2,\ldots$ be independent fair coin indicators and set
$$
X_n=2^n\mathbf1_{\{\xi_1=\cdots=\xi_n=1\}}.
$$
Conditional on the first $n$ coins, the next value is either $2X_n$ or zero with equal <probabilities>, so this is a nonnegative <martingale>. Its <expected value> is always $1$, while a first failure occurs <almost surely>, giving $X_n\to0$ <almost surely>. Yet $\mathbb E|X_n-0|=1$ for every $n$. Indeed, whenever $2^n>K$, its tail <expected value> above $K$ is still $1$. Thus \b[<convergence in L1> fails because <uniform integrability> fails].
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