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A convenient precise version is the almost sure submartingale convergence theorem: if $(S_n)$ is a real submartingale and ${\sup }_n\mathbb{E}{S}_n^{+}<\infty$, then $S_n$ converges almost surely to a finite integrable random variable. In particular the martingale convergence theorem applies to any martingale with ${\sup }_n\mathbb{E}|S_n|<\infty$. It asserts neither convergence in L1 nor preservation of the initial expected value without additional uniform integrability.

Here is a proof including the crossing estimate. Fix real $a<b$, and let $U_n[a,b]$ count completed upcrossings by time $n$. Use a predictable indicator $H_k$ which holds one unit during each crossing attempt: enter when $S_k\le a$, exit when $S_k\ge b$. Witheach completed trade gains at least $b-a$, while the possible unfinished trade loses no more than $(S_n-a{)}^{-}$. Hence $G_n\ge (b-a)U_n[a,b]-(S_n-a{)}^{-}$. The submartingale property and $0\le H_k\le 1$ give $\mathbb{E}{G}_n\le \mathbb{E}(S_n-S_0)$: the complementary predictable gains have nonnegative expected value. ThereforeThe right side is uniformly bounded, since $(S_n-a{)}^{+}\le S_n^{+}+|a|$. By the monotone convergence theorem, every rational interval has finitely many upcrossings almost surely. A real sequence whose lower and upper limits differ crosses some rational interval infinitely often, so the countable intersection of these events gives an extended-real limit. Furthermoreso ${\sup }_n\mathbb{E}|S_n|<\infty$. The Fatou lemma makes the limit finite and integrable, completing the proof. Applying this theorem to $-Z_n$ gives the almost sure supermartingale convergence theorem for nonnegative supermartingales.

For the given adapted multiplicative drift, putThe next denominator $P_{n+1}$ is ${\mathcal{F}}_n$-measurable, and $P_n\ge 1$ makes $Z_n$ integrable. ThusThe nonnegative supermartingale $Z_n$ has a finite almost sure convergence limit $Z_{\infty }$. On the probability-one event where the specified series is finite,Consequently $P_n\uparrow P_{\infty }\in [1,\infty )$ and the multiplicative correction for summable adapted drift givesThe argument does not require a finite expected value for the infinite product or for the total drift.

The almost sure submartingale convergence theorem states that a discrete-time submartingale $Y_n$ satisfying ${\sup }_n\mathbb{E}{Y}_n^{+}<\infty$ converges almost surely to a finite integrable limit. In particular, a martingale with ${\sup }_n\mathbb{E}|M_n|<\infty$ converges almost surely to an integrable $M_{\infty }$, and $\mathbb{E}|M_{\infty }|\le {\sup }_n\mathbb{E}|M_n|$ by Fatou's lemma. This is the martingale convergence theorem. Uniform integrability gives the stronger conclusion of L1 convergence; it is not implied by bounded first moments alone.

A filtration $({\mathcal{F}}_n{)}_{n\ge 0}$ on a probability space is an increasing sequence of sigma-algebras contained in $\mathcal{F}$: ${\mathcal{F}}_n\subseteq {\mathcal{F}}_{n+1}$. A real stochastic process $(X_n)$ is a martingale for this filtration if $X_n$ is ${\mathcal{F}}_n$-measurable, $\mathbb{E}|X_n|<\infty$, andThe first condition is adaptedness; the tower property of conditional expectation then gives $\mathbb{E}[X_n\mid {\mathcal{F}}_m]=X_m$ whenever $m\le n$.

The almost-sure martingale convergence theorem states that if a real martingale satisfies ${\sup }_n\mathbb{E}|X_n|<\infty$, then some integrable random variable $X_{\infty }$ satisfies $X_n\to X_{\infty }$ almost surely. The more general almost sure submartingale convergence theorem needs only ${\sup }_n\mathbb{E}[X_n^{+}]<\infty$ for a submartingale. In particular, a nonnegative martingale converges to a finite integrable limit almost surely, because its expected value is constant. Mere boundedness in the L1 norm does not imply convergence in L1. The stronger uniformly integrable martingale convergence theorem gives both almost-sure convergence and convergence in L1 when the martingale is uniformly integrable.

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 $L^1$.

The almost sure submartingale convergence theorem says that if a submartingale $(Y_n)$ satisfiesthen there is an integrable random variable $Y_{\infty }$ such thatIndeed the positive-part bound also bounds the negative parts, since $\mathbb{E}{Y}_n\ge \mathbb{E}{Y}_0$; the theorem gives a finite limit. In particular, the martingale convergence theorem applies to any martingale with ${\sup }_n\mathbb{E}|M_n|<\infty$.

For a nonnegative martingale, $\mathbb{E}{M}_n=\mathbb{E}{M}_0$ supplies this bound automatically. Its limit obeys $\mathbb{E}{M}_{\infty }\le \mathbb{E}{M}_0$ by the Fatou lemma. Almost sure convergence alone need not preserve the expectation or give convergence in L1. Uniform integrability is the additional condition that supplies convergence in L1 and the identity $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$.