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For a continuous function $f$ on $[0,1]$, let $G_n$ be its dyadic slope martingale. Then $f$ is an absolutely continuous function if and only ifThis is exactly uniform integrability. The uniformly integrable martingale convergence theorem gives convergence in L1 $G_n\to g$, while their integrated linear interpolations converge uniformly to $f$. Hence $f(x)=f(0)+{\int }_0^{x}g$. Conversely, if $f$ has density $g\in L^1$, its slopes are $\mathbb{E}[g\mid {\mathcal{F}}_n]$, and the uniform integrability of conditional expectations proves the criterion. The dyadic tail integral is also the sum of the absolute endpoint increments in cells whose slope is at least $\lambda$ in magnitude.

The almost sure supermartingale convergence theorem applies to the $L^1$-bounded supermartingale $X$. Hence $X_n\to X_{\infty }$ almost surely, and the Fatou lemma gives $\mathbb{E}|X_{\infty }|\le {\sup }_n\mathbb{E}|X_n|<\infty$. Take the version of $X_{\infty }$ obtained from the sequence, so it is measurable for ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. SetThe tower property of conditional expectation makes $M$ a martingale, and the uniform integrability of conditional expectations makes it uniformly integrable. Explicitly, with $Z=X_{\infty }$, $C=\mathbb{E}|Z|$, and $R,K>0$, we have $|M_n|\le R+\mathbb{E}[|Z|1_{\{|Z|>R\}}\mid {\mathcal{F}}_n]$, soFirst choose $R$ large, then $K$ large. This proves uniform integrability uniformly in $n$.

The uniformly integrable martingale convergence theorem gives a limit $M_{\infty }$ both almost surely and in $L^1$. For $A\in {\mathcal{F}}_j$ and $n\ge j$, $\mathbb{E}[M_n{1}_A]=\mathbb{E}[X_{\infty }{1}_A]$. Passing to the $L^1$ limit and then extending from the algebra ${\bigcup }_j{\mathcal{F}}_j$ to ${\mathcal{F}}_{\infty }$ identifies $M_{\infty }=X_{\infty }$ almost surely. Finally,Thus $X=M+Y$ with $M$ uniformly integrable and $Y$ a supermartingale tending almost surely to zero. The terminal decomposition of an L1-bounded supermartingale does not require $Y\ge 0$ or $Y_n\to 0$ in $L^1$. Indeed, for $X$ equal to the negative of the fair-coin doubling martingale, $X_{\infty }=0$, $M=0$, and $Y=X$ is negative and has constant absolute mean one.

A family $(Y_i{)}_{i\in I}$ of integrable random variables is uniformly integrable ifThis controls the contributions of large values uniformly over the family, rather than merely bounding its L1 norms.

Put $X_n=\mathbb{E}[Z\mid {\mathcal{F}}_n]$ and $A_n(K)=\{|X_n|>K\}\in {\mathcal{F}}_n$. The conditional Jensen inequality gives $|X_n|\le \mathbb{E}[|Z|\mid {\mathcal{F}}_n]$, hence $\mathbb{E}|X_n|\le \mathbb{E}|Z|$. The Markov inequality therefore givesUsing the measurability of $A_n(K)$ in the defining integral identity for conditional expectation, for every $R>0$ we getThe first term tends to zero as $R\to \infty$ because $Z$ is integrable, by the dominated convergence theorem. Given $\epsilon >0$, first choose $R$ to make it less than $\epsilon /2$, then choose $K$ to make the second term less than $\epsilon /2$. These choices work for every $n$. This proves uniform integrability of conditional expectations:In fact the argument works for an arbitrary family of sigma-algebras; nesting is not needed for this conclusion.

Use deterministic times tending to infinity and the Fatou lemma:Apply the terminal scaling inequality for stochastic exponentials to $X=aM$ and $r=1/a$. Part (c) gives $\mathbb{E}[\mathcal{E}(aM{)}_{\infty }]=1$, soLetting $a\uparrow 1$ shows that the terminal expectation is at least one. The nonnegative local martingale $\mathcal{E}(M)$ starts at one and is a supermartingale, so the Fatou lemma gives the opposite inequality. Consequently

The terminal expectation criterion for a nonnegative local martingale now closes the argument. Conditional Fatou lemma applied to the supermartingale at times tending to infinity givesThe left side has expectation one and the right side at most one, so equality holds almost surely. Thusand uniform integrability of conditional expectations proves that $\mathcal{E}(M)$ is a uniformly integrable martingale. This establishes the required stopped-moment form of the Kazamaki criterion.

The Exponential martingale for Brownian motion is a true martingale: independent increments and the normal distribution identity $\mathbb{E}{e}^{aN}=e^{a^{2}v/2}$ for $N\sim N(0,v)$ give $\mathbb{E}[Z_t\mid {\mathcal{F}}_s]=Z_s$. Its second moment isThe uniform integrability from bounded second moments criterion now applies: for $K>0$,Thus $\{Z_t:0\le t\le T\}$ has uniform integrability. In fact, all values stopped at stopping times bounded by $T$ have uniform integrability, since $Z_{\tau }=\mathbb{E}[Z_T\mid {\mathcal{F}}_{\tau }]$ and uniform integrability of conditional expectations applies. This is a finite-horizon assertion, not an assertion of uniform integrability over all $t\ge 0$.

A sequence of integrable random variables $(X_n)$ has uniform integrability exactly whenThe uniformly integrable martingale convergence theorem states that a uniformly integrable discrete-time martingale $(M_n)$ has an integrable random variable $M_{\infty }$ withIndeed, uniform integrability implies ${\sup }_n\mathbb{E}|M_n|<\infty$, so the martingale convergence theorem gives almost sure convergence. The combination of uniform integrability and almost sure convergence gives convergence in L1. For $m\ge n$, the martingale identity $\mathbb{E}[M_m\mid {\mathcal{F}}_n]=M_n$ passes to the limit by the L1 contraction of conditional expectation.

Conversely, uniform integrability of conditional expectations shows that a martingale of the form $M_n=\mathbb{E}[Y\mid {\mathcal{F}}_n]$ for one integrable random variable $Y$ is uniformly integrable.

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.