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An $L^1$-bounded supermartingale has an almost sure convergence limit $X_{\infty }$ in $L^1$. The conditional-expectation martingale $M_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ has uniform integrability and converges to $X_{\infty }$ both almost surely and in $L^1$. To identify its limit, test against every event in every finite-time sigma-algebra and use uniqueness on their generated sigma-algebra. Then $Y=X-M$ is a supermartingale tending to zero almost surely. There is no general nonnegativity conclusion for $Y$, nor general convergence in L1 to zero: the negative of the fair-coin doubling martingale is a counterexample to both stronger assertions.

Let $M_n={\sum }_{k\le n}{a}_k{Z}_k$, and stop on first reaching a fixed positive threshold. If the coefficient squares are summable, uniform integrability gives the stopped mean zero, with the terminal limit used on $T=\infty$. If the coefficient squares diverge, $\mathbb{P}(M_n\ge R)\to 1/2$ for each fixed $R$. Thus $\mathbb{P}({\sup }_n{M}_n=\infty )\ge 1/2$. This event is unaffected by changing finitely many of the independent variables, so the Kolmogorov zero-one law makes its probability one. Hence $T<\infty$ and $M_T$ is at least the positive threshold; its expectation, finite or infinite, cannot be zero.

For $p>1$, ${\sup }_n\mathbb{E}|X_n{|}^p=C<\infty$ gives ${\sup }_n\mathbb{E}[|X_n|1_{|X_n|>M}]\le C{M}^{1-p}\to 0$. Thus an $L^p$ bound implies uniform integrability; a mere $L^1$ bound does not.

If ${\sup }_n\mathbb{E}[X_n^2]<\infty$, then $(X_n)$ is uniformly integrable. Indeed, the Cauchy-Schwarz inequality and Markov inequality give

Stopping a uniformly integrable discrete-time martingale at any stopping time preserves uniform integrability, including when the stopping time can be infinite. For $C={\sup }_n\mathbb{E}|X_n|$, the optional stopping theorem and the Markov inequality giveChoose $R$ first, then $K$, to make this uniformly small.

For one integrable random variable $Y$, the family $Z_{\mathcal{G}}=\mathbb{E}[Y\mid \mathcal{G}]$, over arbitrary sigma-algebras $\mathcal{G}$, has uniform integrability. On $A=\{|Z_{\mathcal{G}}|>K\}$, the defining property of conditional expectation gives $\mathbb{E}[|Z_{\mathcal{G}}|1_A]\le \mathbb{E}[|Y|1_A]$, while $\mathbb{P}(A)\le \mathbb{E}|Y|/K$ by the Markov inequality. The integrable random variable $Y$ has uniformly small absolute integrals over events of sufficiently small probability.

A martingale whose values over its whole time interval form a uniformly integrable family has an integrable terminal limit and is closed by that limit. Conversely, conditional expectations of an integrable terminal variable form a uniformly integrable martingale, by uniform integrability of conditional expectations. For continuous martingales, one may equivalently use uniform integrability of the family stopped at finite stopping times.