Uniform integrability

In mathematics, uniform integrability is an important concept in real analysis, functional analysis and measure theory, and plays a vital role in the theory of martingales.

Measure-theoretic definition

Uniform integrability is an extension to the notion of a family of functions being dominated in $L_{1}$ which is central in dominated convergence. Several textbooks on real analysis and measure theory use the following definition:

Definition A: Let $(X,{\mathfrak {M}},\mu )$ be a positive measure space. A set $\Phi \subset L^{1}(\mu )$ is called uniformly integrable if $\sup _{f\in \Phi }\|f\|_{L_{1}(\mu )}<\infty$, and to each $\varepsilon >0$ there corresponds a $\delta >0$ such that

$\int _{E}|f|\,d\mu <\varepsilon$

whenever $f\in \Phi$ and $\mu (E)<\delta .$

Definition A is rather restrictive for infinite measure spaces. A more general definition of uniform integrability that works well in general measure spaces was introduced by G. A. Hunt.

Definition H: Let $(X,{\mathfrak {M}},\mu )$ be a positive measure space. A set $\Phi \subset L^{1}(\mu )$ is called uniformly integrable if and only if

$\inf _{g\in L_{+}^{1}(\mu )}\sup _{f\in \Phi }\int _{\{|f|>g\}}|f|\,d\mu =0$

where $L_{+}^{1}(\mu )=\{g\in L^{1}(\mu ):g\geq 0\}$.

Since Hunt's definition is equivalent to Definition A when the underlying measure space is finite (see Theorem 2 below), Definition H is widely adopted in Mathematics.

The following result provides another equivalent notion to Hunt's. This equivalency is sometimes given as definition for uniform integrability.

Theorem 1: If $(X,{\mathfrak {M}},\mu )$ is a (positive) finite measure space, then a set $\Phi \subset L^{1}(\mu )$ is uniformly integrable if and only if

$\inf _{g\in L_{+}^{1}(\mu )}\sup _{f\in \Phi }\int (|f|-g)^{+}\,d\mu =0$

If in addition $\mu (X)<\infty$, then uniform integrability is equivalent to either of the following conditions

1. $\inf _{a>0}\sup _{f\in \Phi }\int (|f|-a)_{+}\,d\mu =0$.

2. $\inf _{a>0}\sup _{f\in \Phi }\int _{\{|f|>a\}}|f|\,d\mu =0$

When the underlying space $(X,{\mathfrak {M}},\mu )$ is $\sigma$-finite, Hunt's definition is equivalent to the following:

Theorem 2: Let $(X,{\mathfrak {M}},\mu )$ be a $\sigma$-finite measure space, and $h\in L^{1}(\mu )$ be such that $h>0$ almost everywhere. A set $\Phi \subset L^{1}(\mu )$ is uniformly integrable if and only if $\sup _{f\in \Phi }\|f\|_{L_{1}(\mu )}<\infty$, and for any $\varepsilon >0$, there exits $\delta >0$ such that

$\sup _{f\in \Phi }\int _{A}|f|\,d\mu <\varepsilon$

whenever $\int _{A}h\,d\mu <\delta$.

A consequence of Theorems 1 and 2 is that equivalence of Definitions A and H for finite measures follows. Indeed, the statement in Definition A is obtained by taking $h\equiv 1$ in Theorem 2.

Tightness and uniform integrability

Another concept associated with uniform integrability is that of tightness. In this article tightness is taken in a more general setting.

Definition: Suppose measurable space $(X,{\mathfrak {M}},\mu )$ is a measure space. Let ${\mathcal {K}}\subset {\mathfrak {M}}$ be a collection of sets of finite measure. A family $\Phi \subset L_{1}(\mu )$ is tight with respect to ${\mathcal {K}}$ if

$\inf _{K\in {\mathcal {K}}}\sup _{f\in \Phi }\int _{X\setminus K}|f|\,\mu =0$

A tight family with respect to $\Phi ={\mathfrak {M}}\cap L_{1}(\,u)$ is just said to be tight.

When the measure space $(X,{\mathfrak {M}},\mu )$ is a metric space equipped with the Borel $\sigma$ algebra, $\mu$ is a regular measure, and ${\mathcal {K}}$ is the collection of all compact subsets of $X$, the notion of ${\mathcal {K}}$-tightness discussed above coincides with the well known concept of tightness used in the analysis of regular measures in metric spaces

For $\sigma$-finite measure spaces, it can be shown that if a family $\Phi \subset L_{1}(\mu )$ is uniformly integrable, then $\Phi$ is tight. This is capture by the following result which is often used as definition of uniform integrabiliy in the Analysis literature:

Theorem 3: Suppose $(X,{\mathfrak {M}},\mu )$ is a $\sigma$ finite measure space. A family $\Phi \subset L_{1}(\mu )$ is uniformly integrable if and only if

1. $\sup _{f\in \Phi }\|f\|_{1}<\infty$.
2. $\inf _{a>0}\sup _{f\in \Phi }\int _{\{|f|>a\}}|f|\,d\mu =0$
3. $\Phi$ is tight.

When $\mu (X)<\infty$, condition 3 is redundant (see Theorem 1 above).

Condition 2 in Theorem 3 is sometimes replaced by what is called equi-integrability in many books in Analysis : A family ${\mathcal {C}}$ of complex or real valued measurable functions is equi-integrable (or uniformly absolutely continuous with respect to a measure $\mu$) if for any $\varepsilon >0$ there is $\delta >0$ such that $$\sup _{f\in {\mathcal {C}}}\int _{A}|f|\,d\mu \qquad {\text{whenever}}\qquad \mu (A)<\delta$$

Relevant theorems

The following theorem describes a very useful criterion for uniform integrability which is very useful in Probability theory.

de la Vallée-Poussin theorem

Suppose $(X,{\mathfrak {M}},\mu )$ is a finite measure space. The family ${\mathcal {F}}\subset L^{1}(\mu )$ is uniformly integrable if and only if there exists a function $G:[0,\infty )\rightarrow [0,\infty )$ such that $\lim _{t\to \infty }{\frac {G(t)}{t}}=\infty$ and $$\sup _{f\in {\mathcal {F}}}\int _{X}G(|f|)\,d\mu <\infty .$$ The function $G$ can be chosen to be monotone increasing and convex.

Uniform integrability gives a characterization of weak compactness in $L_{1}$.

Dunford–Pettis theorem

Suppose $(X,{\mathfrak {M}},\mu )$ is a $\sigma$-finite measure. A family ${\mathcal {F}}\subset L_{1}(\mu )$ has compact closure in the weak topology $\sigma (L_{1},L_{\infty })$ if and only if ${\mathcal {F}}$ is uniformly integrable.

Probability definition

In probability theory, Definition A or the statement of Theorem 1 are often presented as definitions of uniform integrability using the notation expectation of random variables., that is,

1. A class ${\mathcal {C}}$ of random variables is called uniformly integrable if:

• There exists a finite $M$ such that, for every $X$ in ${\mathcal {C}}$, $\operatorname {E} (|X|)\leq M$ and
• For every $\varepsilon >0$ there exists $\delta >0$ such that, for every measurable $A$ such that $P(A)\leq \delta$ and every $X$ in ${\mathcal {C}}$, $\operatorname {E} (|X|I_{A})\leq \varepsilon$.

or alternatively

2. A class ${\mathcal {C}}$ of random variables is called uniformly integrable (UI) if for every $\varepsilon >0$ there exists $K\in [0,\infty )$ such that $\operatorname {E} (|X|I_{|X|\geq K})\leq \varepsilon \ {\text{ for all }}X\in {\mathcal {C}}$, where $I_{|X|\geq K}$ is the indicator function $I_{|X|\geq K}={\begin{cases}1&{\text{if }}|X|\geq K,\\0&{\text{if }}|X|<K.\end{cases}}$.

Related corollaries

The following results apply to the probabilistic definition.

• Definition 1 could be rewritten by taking the limits as $$\lim _{K\to \infty }\sup _{X\in {\mathcal {C}}}\operatorname {E} (|X|\,I_{|X|\geq K})=0.$$
• A non-UI sequence. Let $\Omega =[0,1]\subset \mathbb {R}$, and define $$X_{n}(\omega )={\begin{cases}n,&\omega \in (0,1/n),\\0,&{\text{otherwise.}}\end{cases}}$$ Clearly $X_{n}\in L^{1}$, and indeed $\operatorname {E} (|X_{n}|)=1\ ,$ for all n. However, $$\operatorname {E} (|X_{n}|I_{\{|X_{n}|\geq K\}})=1\ {\text{ for all }}n\geq K,$$ and comparing with definition 1, it is seen that the sequence is not uniformly integrable.

Non-UI sequence of RVs. The area under the strip is always equal to 1, but $X_{n}\to 0$ pointwise.

• By using Definition 2 in the above example, it can be seen that the first clause is satisfied as $L^{1}$ norm of all $X_{n}$s are 1 i.e., bounded. But the second clause does not hold as given any $\delta$ positive, there is an interval $(0,1/n)$ with measure less than $\delta$ and $E[|X_{m}|:(0,1/n)]=1$ for all $m\geq n$.
• If $X$ is a UI random variable, by splitting $$\operatorname {E} (|X|)=\operatorname {E} (|X|I_{\{|X|\geq K\}})+\operatorname {E} (|X|I_{\{|X|<K\}})$$ and bounding each of the two, it can be seen that a uniformly integrable random variable is always bounded in $L^{1}$.
• If any sequence of random variables $X_{n}$ is dominated by an integrable, non-negative $Y$: that is, for all ω and n, $$|X_{n}(\omega )|\leq Y(\omega ),\ Y(\omega )\geq 0,\ \operatorname {E} (Y)<\infty ,$$ then the class ${\mathcal {C}}$ of random variables $\{X_{n}\}$ is uniformly integrable.
• A class of random variables bounded in $L^{p}$ ($p>1$) is uniformly integrable.

Uniform integrability and stochastic ordering

A family of random variables $\{X_{i}\}_{i\in I}$ is uniformly integrable if and only if there exists a random variable $X$ such that $EX<\infty$ and $|X_{i}|\leq _{\mathrm {icx} }X$ for all $i\in I$, where $\leq _{\mathrm {icx} }$ denotes the increasing convex stochastic order defined by $A\leq _{\mathrm {icx} }B$ if $E\phi (A)\leq E\phi (B)$ for all nondecreasing convex real functions $\phi$.

Relation to convergence of random variables

Main article: Convergence of random variables

A sequence $\{X_{n}\}$ converges to $X$ in the $L_{1}$ norm if and only if it converges in measure to $X$ and it is uniformly integrable. In probability terms, a sequence of random variables converging in probability also converge in the mean if and only if they are uniformly integrable. This is a generalization of Lebesgue's dominated convergence theorem, see Vitali convergence theorem.

References

• Shiryaev, A.N. (1995). Probability (2 ed.). New York: Springer-Verlag. pp. 187–188. ISBN 978-0-387-94549-1.
• Diestel, J. and Uhl, J. (1977). Vector measures, Mathematical Surveys 15, American Mathematical Society, Providence, RI ISBN 978-0-8218-1515-1