 Solution

For $a<b$, let $U_N[a,b]$ be the number of completed upcrossings by the observations $X_1,\dots ,X_N$: 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 isFor 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 giveEvery rational interval therefore has only finitely many upcrossings, simultaneously outside a countable union of null sets. If $\mathrm{liminf}{X}_n<\mathrm{limsup}{X}_n$, 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 givesThe 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 uniform integrability: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 $L^1$, thenMake 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 ${\xi }_1,{\xi }_2,\dots$ be independent fair coin indicators and setConditional 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 $X_n\to 0$ 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 convergence in L1 fails because uniform integrability fails.