Uniform integrability is the condition that prevents mass from escaping into rare large values. It is the right replacement for domination in many limit theorems.
Uniform integrability
Uniform integrability controls tails uniformly over a family.
Definition. On a finite measure space, a family \(\mathcal F\subset L^1(\mu)\) is uniformly integrable if \[ \sup_{f\in\mathcal F}\int_{\{|f|>M\}}|f|\,\mathrm{d}\mu\to0 \] as \(M\to\infty\).
Proposition (equivalent finite-measure criterion). On \(\mu(X)<\infty\), \(\mathcal F\) is uniformly integrable if and only if \[ \sup_{f\in\mathcal F}\|f\|_1<\infty \] and for every \(\varepsilon>0\) there is \(\delta>0\) such that \[ \mu(A)<\delta \quad\Longrightarrow\quad \sup_{f\in\mathcal F}\int_A|f|\,\mathrm{d}\mu<\varepsilon. \]
Proof. Tail control at some \(M\) gives \[ \|f\|_1 \le M\mu(X)+\int_{\{|f|>M\}}|f|, \] uniformly in \(f\). Also, \[ \int_A|f| \le M\mu(A)+\int_{\{|f|>M\}}|f|, \] so first choose \(M\) for the tail and then \(\delta\) for the bounded part. Conversely, suppose the two displayed conditions hold. For \(A_f=\{|f|>M\}\), Markov gives \[ \mu(A_f)\le M^{-1}\sup_{g\in\mathcal F}\|g\|_1. \] Choose \(M\) so large that this is below the uniform absolute-continuity threshold; then \(\sup_f\int_{A_f}|f|<\varepsilon\).
Remark (infinite spaces require spatial control). On an infinite measure space, value-tail control alone does not prevent mass from spreading. For \[ f_n=\frac1n\mathbf1_{[0,n]}\quad\text{on }\mathbb R, \] \(f_n\to0\) uniformly and the value tails vanish, but \(\|f_n\|_1=1\). A global Vitali theorem there requires, in addition, uniform tightness of the integrals outside finite-measure sets.
Proposition. If \(|f|\le g\in L^1\) for all \(f\in\mathcal F\), then \(\mathcal F\) is uniformly integrable.
Proof. The tail integral is bounded by \(\int_{\{g>M\}}g\,\mathrm{d}\mu\), which tends to zero. Absolute continuity over small sets follows from the absolute continuity of the integral of \(g\).
de la Vallee-Poussin criterion
Convex superlinear bounds characterize uniform integrability.
Theorem. A family \(\mathcal F\subset L^1\) is uniformly integrable if and only if there is an increasing convex \(\Phi\) with \(\Phi(t)/t\to\infty\) and \(\sup_{f\in\mathcal F}\int\Phi(|f|)\,\mathrm{d}\mu\lt \infty\).
Proof. For convex increasing \(\Phi\) with \(\Phi(0)=0\), the ratio \(\Phi(t)/t\) is increasing. Hence on \(|f|>M\), \[ |f|\le\frac{M}{\Phi(M)}\Phi(|f|), \] and the uniform \(\Phi\)-bound gives tail decay because \(M/\Phi(M)\to0\).
Conversely, choose \(M_k\uparrow\infty\) so that \[ \sup_{f\in\mathcal F}\int_{\{|f|>M_k\}}|f|\,\mathrm{d}\mu\le2^{-k}. \] Choose increasing slopes \(a_k\uparrow\infty\) slowly enough that \[ \sum_k(a_{k+1}-a_k)2^{-k}<\infty, \] and define the convex function \[ \Phi(t)=\int_0^t a(s)\,\mathrm{d}s, \qquad a(s)=a_k\ \text{for }M_k\le s<M_{k+1}. \] Writing \(\Phi(t)\) by summation over its slope increments and using the layer-cake form of the tail bounds yields \(\sup_f\int\Phi(|f|)<\infty\). Since \(a(s)\to\infty\), also \(\Phi(t)/t\to\infty\).
Vitali convergence theorem
Uniform integrability plus convergence in measure is exactly \(L^1\) convergence.
Theorem (Vitali). On a finite measure space, if \(f_n\to f\) in measure and \(\{f_n\}\) is uniformly integrable, then \(f\in L^1\) and \(\|f_n-f\|_1\to0\). Conversely, \(L^1\) convergence implies uniform integrability of \(\{f_n\}\cup\{f\}\).
Proof. Every subsequence has a further subsequence \(f_{n_k}\to f\) almost everywhere. Fatou and the uniform \(L^1\) bound show \(f\in L^1\). The family \(\{f_n-f\}\) is uniformly integrable because \[ \int_A|f_n-f| \le\int_A|f_n|+\int_A|f| \] and the equivalent small-set criterion applies.
Given \(\varepsilon>0\), choose \(\delta\) so that \(\mu(A)<\delta\) implies \[ \sup_n\int_A|f_n-f|<\varepsilon/2. \] Choose \(\eta>0\) with \(\eta\mu(X)<\varepsilon/2\). Convergence in measure gives \(\mu(|f_n-f|>\eta)<\delta\) eventually, and therefore \[ \int|f_n-f| \le \eta\mu(X) +\int_{\{|f_n-f|>\eta\}}|f_n-f| <\varepsilon. \] For the converse, let \(f_n\to f\) in \(L^1\). Given \(\varepsilon\), choose \(N\) so that \(\|f_n-f\|_1<\varepsilon/3\) for \(n\ge N\). The singleton \(\{f\}\) and the finite family \(f_1,\ldots,f_{N-1}\) have uniformly small tails; the \(L^1\) perturbation controls the remaining terms. Hence the whole family is uniformly integrable.
Weak compactness in \(L^1\)
Uniform integrability is also a compactness condition.
Theorem. On a finite measure space, a bounded subset of \(L^1\) is relatively weakly compact if and only if it is uniformly integrable.
Proof. This is the Dunford-Pettis theorem. One direction follows because weak compactness prevents concentration on shrinking sets when tested against bounded indicators. The other direction uses uniform integrability to approximate by bounded functions on finite partitions, then diagonal compactness and duality with \(L^\infty\).
Convergence of expectations
Uniform integrability is the missing condition that prevents rare large observations from surviving a probabilistic limit.
Remark. On a probability space, if \(X_n\to X\) in probability and \(\{X_n\}\) is uniformly integrable, then \[ \mathbb E|X_n-X|\to0, \qquad \mathbb E X_n\to\mathbb E X. \] A convenient sufficient condition is \(\sup_n\mathbb E|X_n|^{1+\delta}<\infty\) for some \(\delta>0\), by de la Vallee-Poussin with \(\Phi(t)=t^{1+\delta}\).