Uniform integrability begins with a failure of ordinary integrability to control a family. A single integrable function has small tails: if $f \in L^1(E, \mathcal E, \mu)$, then the integral of $|f|$ over the set where $|f|$ is large tends to $0$. But when a whole family of functions is moving at once, each member may have a small tail while the location or height of the tail changes from member to member. Uniform integrability asks for one tail estimate that works for the entire family.

[example: A Bounded $L^1$ Family with Moving Spikes] Let $E=(0,1)$ with [Lebesgue measure](/page/Lebesgue%20Measure) $\mathcal L^1$, and define \begin{align*} f_n(x)=n\,\mathbb{1}_{(0,1/n)}(x). \end{align*} For each $n$, the function $f_n$ is measurable, nonnegative, and supported on $(0,1/n)$. Its $L^1$ norm is \begin{align*} \int_E |f_n|\,d\mathcal L^1=\int_0^1 n\,\mathbb{1}_{(0,1/n)}(x)\,d\mathcal L^1(x)=\int_0^{1/n} n\,d\mathcal L^1(x)=n\mathcal L^1((0,1/n))=n\cdot\frac{1}{n}=1. \end{align*} Thus $\sup_n\|f_n\|_{L^1(E)}=1$, so the family is bounded in $L^1(E)$. Now fix $M>0$ and choose an integer $n>M$. Since $f_n(x)=n$ for every $x\in(0,1/n)$ and $f_n(x)=0$ for $x\in(1/n,1)$, the level set is \begin{align*} \{|f_n|>M\}=(0,1/n). \end{align*} Therefore \begin{align*} \int_{\{|f_n|>M\}} |f_n|\,d\mathcal L^1=\int_0^{1/n} n\,d\mathcal L^1(x)=n\mathcal L^1((0,1/n))=1. \end{align*} So no matter how large the threshold $M$ is, some member of the family still has total mass $1$ above that threshold. The mass has not become large in total; instead it has concentrated on smaller and smaller sets, showing that $L^1$ boundedness alone does not prevent concentration. [/example]

The example explains why uniform integrability is a child of integrability, not a replacement for it. Integrability is a property of one function. Uniform integrability is a stability condition for a family: it says that the integrable mass of all members can be controlled uniformly on the parts where the functions are large.

## Definition

The most robust definition should not depend on a special description of the bad set. A family may fail by putting large values on tiny sets, but in applications the tiny sets often come from [convergence in measure](/page/Convergence%20in%20Measure), stopping times, or approximation errors rather than from explicit level sets. Uniform integrability therefore asks for two things at once: bounded total mass, and a uniform version of the absolute continuity of the integral.

[definition: Uniform Integrability] Let $(E, \mathcal E, \mu)$ be a [measure space](/page/Measure%20Space). A family $\mathcal F \subset L^1(E, \mathcal E, \mu)$ is uniformly integrable if \begin{align*} \sup_{f\in\mathcal F}\int_E |f|\,d\mu &< \infty \end{align*} and for every $\varepsilon>0$ there exists $\delta>0$ such that for every $A\in\mathcal E$ with $\mu(A)<\delta$, \begin{align*} \sup_{f\in\mathcal F}\int_A |f|\,d\mu &< \varepsilon. \end{align*} [/definition]

This is a family version of the absolute continuity of the integral. For one integrable function $f$, small measure of $A$ forces $\int_A |f|\,d\mu$ to be small. Uniform integrability demands that the same $\delta$ work for every function in the family, while also excluding unbounded total $L^1$ mass.

The definition is written in terms of arbitrary small sets, but in practice one often detects failure by looking at large values. The spike example failed because the set $\{|f_n|>M\}$ still carried order-one mass no matter how large $M$ was chosen. This raises the first structural question: on the measure spaces where most probability and finite-volume analysis takes place, is it enough to control the large-value tails uniformly? The next theorem answers exactly that question.

[quotetheorem:9925]

For random variables, the same statement is written with expectation. Thus a family $\mathcal X\subset L^1(\Omega,\mathcal F,\mathbb P)$ is uniformly integrable exactly when

\begin{align*} \lim_{K\to\infty}\sup_{X\in\mathcal X}\mathbb E\big[|X|\,\mathbb{1}_{\{|X|>K\}}\big] &= 0. \end{align*}

This form is usually the quickest way to check the condition on probability spaces, while the small-set formulation is the definition that remains stable on general measure spaces.

There is one more convention to record before moving on. On $L^1(\mathbb R^n)$, many analysis texts use "uniform integrability" for a two-part compactness condition: no concentration on small sets and no escape of mass to infinity. Other texts reserve "uniform integrability" for the small-set or large-value condition alone and call the second condition tightness. This page follows the two-part Euclidean convention only when it explicitly says "uniform integrability in $L^1(\mathbb R^n)$" or "Euclidean uniform integrability"; otherwise, uniform integrability means the general-measure definition above.

## Tails, Concentration, and Boundedness

Uniform integrability lives between pointwise integrability and stronger $L^p$ control. It is weaker than having a common $L^p$ bound for some $p>1$, but stronger than boundedness in $L^1$. The difference is exactly whether the family can hide mass in rare large values.

### Higher Integrability

The most common sufficient condition comes from comparing the $L^1$ tail to a higher moment. If the family is uniformly bounded in $L^p$ for some $p>1$, then placing mass far out in the tail becomes increasingly expensive.

[quotetheorem:9926]

The finite-measure hypothesis is doing real work. It turns the estimate on small sets into

\begin{align*} \int_A |f|\,d\mu &\le \|f\|_{L^p(E)}\mu(A)^{1-1/p}, \end{align*}