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For a submartingale $(Z_n)$ and $a<b$, let $U_N(a,b)$ count the completed upcrossings of $[a,b]$ by time $N$. The Doob upcrossing inequality givesFor a martingale $(X_n)$, a useful sharper form isHere $x^{+}=\max (x,0)$ and $x^{-}=\max (-x,0)$. To see the martingale form, use a bounded predictable process $H_k\in \{0,1\}$ that holds one unit after an observation at or below $a$ and releases it on the next observation at or above $b$. Its gain $G_N={\sum }_{k=1}^N{H}_k(X_k-X_{k-1})$ is at least $(b-a)U_N-(X_N-a{)}^{-}$: completed trades earn at least $b-a$, and any unfinished trade loses at most $(X_N-a{)}^{-}$. Since $\mathbb{E}{G}_N=0$, the bound follows. For the submartingale version, apply the same strategy to $(Z_n-a{)}^{+}$, itself a submartingale. Each purchase occurs at value zero, so its gain is at least $(b-a)U_N$. The complementary predictable strategy has nonnegative expected gain, bounding this strategy's expected gain by $\mathbb{E}(Z_N-a{)}^{+}-\mathbb{E}(Z_0-a{)}^{+}$, and in particular by the stated right side.

The almost-sure martingale convergence theorem says that if $C={\sup }_n\mathbb{E}|X_n|<\infty$, then $X_n$ converges almost surely to a finite, integrable random variable $Y$, with $\mathbb{E}|Y|\le C$. Indeed, the Doob upcrossing inequality bounds the expected total number of upcrossings of every rational interval by $(C+|a|)/(b-a)$. By monotone convergence, each such total count is finite almost surely; intersect these probability-one events over all rational $a<b$. If the lower and upper limits of a sample path differed, some rational interval strictly between them would be crossed infinitely often. Thus each remaining path has a limit in the extended real line. The Fatou lemma givesso the limit is finite almost surely and integrable. Almost-sure convergence by itself neither preserves expectations nor supplies this uniform first-moment bound.

For the loss of expectation, let $({\eta }_k)$ be independent variables with the fair Bernoulli distribution and setConditional on the past, survival doubles the value with probability $1/2$ and otherwise sets it to zero, so this is a nonnegative martingale. An infinite string of successes has probability zero. The process therefore eventually vanishes almost surely, giving yes: $X_0=1$ and $Y=0$ almost surely are possible. Its expectations remain one at every finite time; the process is not uniformly integrable.

A nonintegrable limit is also possible for a signed martingale. Take independent ${\xi }_k$ withand let $M_0=1$, $M_n=1+{\sum }_{k=1}^n{\xi }_k$. Every increment has mean zero and a finite absolute expectation, so each $M_n$ is integrable and $(M_n)$ is a martingale. Since ${\sum }_k\mathbb{P}({\xi }_k\ne 0)<\infty$, the Borel-Cantelli lemma makes only finitely many increments nonzero almost surely; hence the finite limit $Y=1+{\sum }_k{\xi }_k$ exists almost surely. To compute its absolute expectation, let $A_k$ be the event that only the $k$th increment is nonzero. Independence givesThe infinite product is positive because its tail logarithms are bounded in absolute value by a constant times the summable series ${\sum }_j{2}^{-j}$. The events $A_k$ are disjoint, and on $A_k$, $|Y|\ge 4^k-1$. ThereforeThus yes: a finite almost-sure martingale limit can have infinite absolute expectation, even with deterministic $M_0=1$. This is almost-sure martingale convergence to a nonintegrable limit; the missing hypothesis is the uniform first-moment bound. For a nonnegative martingale, in contrast, the Fatou lemma would force the limit to be integrable.