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 ${\displaystyle L_{1}}$ which is central in dominated convergence. Several textbooks on real analysis and measure theory use the following definition:^[1]

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

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

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

Definition A is rather restrictive for infinite measure spaces. A more general definition^[2] of uniform integrability that works well in general measure spaces was introduced by G. A. Hunt.

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

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

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

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^[3] provides another equivalent notion to Hunt's. This equivalency is sometimes given as definition for uniform integrability.

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

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

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

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

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

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

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

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

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

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

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

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

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

For ${\displaystyle \sigma }$-finite measure spaces, it can be shown that if a family ${\displaystyle \Phi \subset L_{1}(\mu )}$ is uniformly integrable, then ${\displaystyle \Phi }$ 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 ${\displaystyle (X,{\mathfrak {M}},\mu )}$ is a ${\displaystyle \sigma }$ finite measure space. A family ${\displaystyle \Phi \subset L_{1}(\mu )}$ is uniformly integrable if and only if

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

When ${\displaystyle \mu (X)<\infty }$, 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 ^[4][5][6][7]: A family ${\displaystyle {\mathcal {C}}}$ of complex or real valued measurable functions is equi-integrable (or uniformly absolutely continuous with respect to a measure ${\displaystyle \mu }$) if for any ${\displaystyle \varepsilon >0}$ there is ${\displaystyle \delta >0}$ such that $${\displaystyle \sup _{f\in {\mathcal {C}}}\int _{A}|f|\,d\mu \qquad {\text{whenever}}\qquad \mu (A)<\delta }$$

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^[8][9]

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

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

Dunford–Pettis theorem^[10][11]

Suppose ${\displaystyle (X,{\mathfrak {M}},\mu )}$ is a ${\displaystyle \sigma }$-finite measure. A family ${\displaystyle {\mathcal {F}}\subset L_{1}(\mu )}$ has compact closure in the weak topology ${\displaystyle \sigma (L_{1},L_{\infty })}$ if and only if ${\displaystyle {\mathcal {F}}}$ 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.,^[12][13][14] that is,

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

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

or alternatively

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

Related corollaries

The following results apply to the probabilistic definition.^[15]

• Definition 1 could be rewritten by taking the limits as $${\displaystyle \lim _{K\to \infty }\sup _{X\in {\mathcal {C}}}\operatorname {E} (|X|\,I_{|X|\geq K})=0.}$$
• A non-UI sequence. Let ${\displaystyle \Omega =[0,1]\subset \mathbb {R} }$, and define $${\displaystyle X_{n}(\omega )={\begin{cases}n,&\omega \in (0,1/n),\\0,&{\text{otherwise.}}\end{cases}}}$$ Clearly ${\displaystyle X_{n}\in L^{1}}$, and indeed ${\displaystyle \operatorname {E} (|X_{n}|)=1\ ,}$ for all n. However, $${\displaystyle \operatorname {E} (|X_{n}|I_{\{|X_{n}|\geq K\}})=1\ {\text{ for all }}n\geq K,}$$ 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 ${\displaystyle X_{n}\to 0}$ pointwise.

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

Uniform integrability and stochastic ordering

A family of random variables ${\displaystyle \{X_{i}\}_{i\in I}}$ is uniformly integrable if and only if^[16] there exists a random variable ${\displaystyle X}$ such that ${\displaystyle EX<\infty }$ and ${\displaystyle |X_{i}|\leq _{\mathrm {icx} }X}$ for all ${\displaystyle i\in I}$, where ${\displaystyle \leq _{\mathrm {icx} }}$ denotes the increasing convex stochastic order defined by ${\displaystyle A\leq _{\mathrm {icx} }B}$ if ${\displaystyle E\phi (A)\leq E\phi (B)}$ for all nondecreasing convex real functions ${\displaystyle \phi }$.

Relation to convergence of random variables

Main article: Convergence of random variables

A sequence ${\displaystyle \{X_{n}\}}$ converges to ${\displaystyle X}$ in the ${\displaystyle L_{1}}$ norm if and only if it converges in measure to ${\displaystyle X}$ 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.^[17] This is a generalization of Lebesgue's dominated convergence theorem, see Vitali convergence theorem.