Uniform integrability

A family of integrable random variables is uniformly integrable whenAlmost-sure or probabilistic convergence together with uniform integrability upgrades to $L^1$ convergence.

On a finite measure space, a family is uniformly integrable exactly when it admits the displayed bound for some nonnegative increasing convex superlinear function $\Psi$. By the Dunford-Pettis theorem this ensures relative weak compactness in $L^1$. For the sufficient direction, choose $R_A$ so $\Psi (s)\ge As$ for $s>R_A$; then ${\int }_E|v_j|\le R_A\mu (E)+A^{-1}{\sup }_j\int \Psi (|v_j|)$ gives uniform integrability. Conversely choose increasing thresholds $R_m\to \infty$ so ${\sup }_j{\int }_{|v_j|>R_m}|v_j|\le 2^{-m}$. The convex function $\Psi (s)={\sum }_{m\ge 1}(s-R_m{)}_{+}$ is locally a finite sum, has uniformly bounded integrals by summing these tail estimates, and is superlinear because for $s\ge 2R_N$ its first $N$ terms sum to at least $Ns/2$. This proves the criterion in both directions.

Choose increasing $R_m\to \infty$ so $H(t)\ge m^2|t|$ whenever $|t|\ge R_m$. Put $\Psi (s)={\sum }_{m\ge 1}(s-R_m{)}_{+}$ for $s\ge 0$. This is nonnegative, increasing and convex. If $m$ terms are active, then $\Psi (s)\le ms\le H(t)$ for $s=|t|$. For every fixed $N$, $s\ge 2R_N$ gives $\Psi (s)\ge Ns/2$, proving superlinearity. Thus a nonconvex superlinear energy lower bound still supplies the convex function needed in the De la Vallée-Poussin criterion. A continuous superlinear function not initially nonnegative can first have its finite global minimum subtracted.

On a finite-measure space, a bounded subset of $L^1$ is relatively weakly compact exactly when it is uniformly integrable. In particular every uniformly integrable bounded sequence has a weakly convergent subsequence. This is the compactness substitute for reflexivity at exponent one. A superlinear integral bound ${\sup }_j\int H(u_j)<\infty$, with $H\ge 0$ and $H(t)/|t|\to \infty$, gives the required uniform integrability: for $H(t)\ge A|t|$ outside $[-R_A,R_A]$,Choose $A$ large and then the measure of $E$ small. This is the direction used in the direct method in the calculus of variations; a mere bounded $L^1$ norm does not suffice, as concentrating spikes show.

For a uniformly integrable bounded sequence of derivatives on a compact interval and a fixed initial value, its primitives are uniformly bounded and equicontinuous. The increment over any short interval is bounded by the integral of the absolute derivative on that interval. The Arzelà-Ascoli theorem gives relative uniform compactness. If the derivatives converge weakly in $L^1$, tests by interval indicators identify every pointwise limit with $u(x)=\alpha +{\int }_a^{x}v$. Hence every convergent subsequence has this same limit and the primitives converge uniformly. The limit belongs to $W^{1,1}$ and retains any common terminal value. Uniform integrability alone gives no essential-supremum bound on its derivative, so this compactness statement does not assert a Lipschitz limit.

On a finite time horizon, an adapted process is of class D if its values at all stopping times in that horizon form a uniformly integrable family. Deterministic boundedness implies this property. The stopping-time uniformity is stronger than having an integrable value at each fixed time.

An adapted process is of class DL when, for every finite $T$, the family $\{X_{\tau }:\tau \le T\text{ a stopping time}\}$ is uniformly integrable. A local martingale is a true martingale exactly when it has this property. This is a local-in-time condition, not uniform integrability over the whole half-line.

// Target: probability-and-statistics.bigb

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.

For integrable random variables $Z,Y$,Apply this with $Z=X_n$ and $Y=X$ to prove that convergence in L1 implies uniform integrability; control finitely many early terms separately.

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.

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