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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.

The martingale convergence theorem says that a discrete-time martingale $(M_n,{\mathcal{F}}_n)$ satisfyinghas an almost sure convergence limit $M_{\infty }$, finite almost surely and in $L^1$. The theorem asserts that the limit is integrable; it does not assert convergence in L1. More generally, the almost sure submartingale convergence theorem applies to a submartingale with ${\sup }_n\mathbb{E}{M}_n^{+}<\infty$. Uniform integrability is the additional condition that upgrades a martingale's convergence to convergence in L1.

For the requested distinction, let $({\xi }_k)$ be independent fair Bernoulli variables and use their natural filtration. The coin-doubling martingaleis a nonnegative martingale: conditionally on ${\mathcal{F}}_n$, the next factor is $2{\xi }_{n+1}$ with mean one, so $\mathbb{E}(M_{n+1}\mid {\mathcal{F}}_n)=M_n$. Also $\mathbb{E}|M_n|=\mathbb{E}{M}_n=1$ for every $n$, giving the required uniform $L^1$ bound. The probability that all the Bernoulli variables equal one is ${\lim }_n{2}^{-n}=0$. Therefore a zero is eventually encountered almost surely, after which $M_n$ stays zero. ThusThere can be no other $L^1$ limit, since convergence in L1 implies convergence in probability, whose limit must agree with the almost sure limit. This martingale satisfies the almost sure theorem but does not converge in $L^1$.