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 $(X,\mathfrak, \mu)$ be a positive measure space. A set $\Phi\subset L^1(\mu)$ is called uniformly integrable if $\sup_\, f\, _<\infty$, 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 slightly more general definition of uniform integrability that works well in general measures spaces was introduced by G. A. Hunt.

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

:$\inf_\sup_\int_, f, \, d\mu=0$

where $L^1_+(\mu)=\$.

For finite measure spaces the following result follows from Definition H:

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

:$\inf_\sup_\int_, f, \, d\mu=0$

Many textbooks in probability present Theorem 1 as the definition of uniform integrability in Probability spaces. When the space $(X,\mathfrak,\mu)$ is $\sigma$-finite, Definition H yields the following equivalency:

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

:$\sup_\int_A, f, \, d\mu <\varepsilon$

whenever $\int_A h\,d\mu <\delta$.

In particular, the equivalence of Definitions A and H for finite measures follows immediately from Theorem 2; for this case, the statement in Definition A is obtained by taking $h\equiv1$ in Theorem 2.

Probability definition

In the theory of probability, 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 $\mathcal$ of random variables is called uniformly integrable if:

* There exists a finite $M$ such that, for every $X$ in $\mathcal$, $\operatorname E(, X, )\leq M$ 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 $\mathcal$, $\operatorname E(, X, I_A)\leq\varepsilon$.

or alternatively

2. A class $\mathcal$ of random variables is called uniformly integrable (UI) if there exists K\in X, I_)\le\varepsilon\ \text X \in \mathcal, where $I_$ is the indicator function $I_ = \begin 1 &\text , X, \geq K, \\ 0 &\text , X, < K. \end$.

Tightness and uniform integrability

One consequence of uniformly integrability of a class $\mathcal$ of random variables is that family of laws or distributions $\$ is Tightness of measures, tight. That is, for each $\delta > 0$, there exists $a > 0$ such that
$P(, X, >a) \leq \delta$
for all $X\in\mathcal$.

This however, does not mean that the family of measures $\mathcal_:=\Big\$ is tight.

There is another notion of uniformity, slightly different than uniform integrability, which also has many applications in Probability and measure theory, and which does not require random variables to have a finite integral

Definition: Suppose $(\Omega,\mathcal,P)$ is a probability space. A classed $\mathcal$ of random variables is uniformly absolutely continuous with respect to $P$ if for any $\varepsilon>0$, there is $\delta>0$ such that
E

Related corollaries

The following results apply to the probabilistic definition.
* Definition 1 could be rewritten by taking the limits as $\lim_ \sup_ \operatorname E(, X, \,I_)=0.$
* A non-UI sequence. Let \Omega = ,1\subset \mathbb, and define $X_n(\omega) = \begin n, & \omega\in (0,1/n), \\ 0 , & \text \end$ Clearly $X_n\in L^1$, and indeed $\operatorname E(, X_n, )=1\ ,$ for all ''n''. However, $\operatorname E(, X_n, , , X_n, \ge K)= 1\ \text n \ge K,$ and comparing with definition 1, it is seen that the sequence is not uniformly integrable.

* 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 \operatorname E(, X, ) = \operatorname E(, X, ,, X, >K)+\operatorname E(, X, ,, X, 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'', $, X_n(\omega), \le Y(\omega),\ Y(\omega)\ge 0,\ \operatorname E(Y) < \infty,$ then the class $\mathcal$ of random variables $\$ is uniformly integrable.
* A class of random variables bounded in $L^p$ ($p > 1$) is uniformly integrable.

Relevant theorems

In the following we use the probabilistic framework, but regardless of the finiteness of the measure, by adding the boundedness condition on the chosen subset of $L^1(\mu)$.

* Nelson Dunford, Dunford– Pettis theoremA class of random variables $X_n \subset L^1(\mu)$ is uniformly integrable if and only if it is relatively compact for the weak topology $\sigma(L^1,L^\infty)$.
* de la Vallée-Poussin theoremThe family $\_ \subset L^1(\mu)$ is uniformly integrable if and only if there exists a non-negative increasing convex function $G(t)$ such that $\lim_ \frac t = \infty \text \sup_\alpha \operatorname E(G(, X_, )) < \infty.$

Relation to convergence of random variables

A sequence $\$ 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.

Citations

References

*
* Diestel, J. and Uhl, J. (1977). ''Vector measures'', Mathematical Surveys 15, American Mathematical Society, Providence, RI {{isbn, 978-0-8218-1515-1

Martingale theory>X, I_A\varepsilon
whenever $P(A)<\delta$.

The term "uniform absolute continuity" is not standard, but is used by some other authors.

Related corollaries

The following results apply to the probabilistic definition.
* Definition 1 could be rewritten by taking the limits as $\lim_ \sup_ \operatorname E(, X, \,I_)=0.$
* A non-UI sequence. Let \Omega = ,1\subset \mathbb, and define $X_n(\omega) = \begin n, & \omega\in (0,1/n), \\ 0 , & \text \end$ Clearly $X_n\in L^1$, and indeed $\operatorname E(, X_n, )=1\ ,$ for all ''n''. However, $\operatorname E(, X_n, , , X_n, \ge K)= 1\ \text n \ge K,$ and comparing with definition 1, it is seen that the sequence is not uniformly integrable.

* 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 \operatorname E(, X, ) = \operatorname E(, X, ,, X, >K)+\operatorname E(, X, ,, X, 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'', $, X_n(\omega), \le Y(\omega),\ Y(\omega)\ge 0,\ \operatorname E(Y) < \infty,$ then the class $\mathcal$ of random variables $\$ is uniformly integrable.
* A class of random variables bounded in $L^p$ ($p > 1$) is uniformly integrable.

Relevant theorems

In the following we use the probabilistic framework, but regardless of the finiteness of the measure, by adding the boundedness condition on the chosen subset of $L^1(\mu)$.

* Nelson Dunford, Dunford– Pettis theoremA class of random variables $X_n \subset L^1(\mu)$ is uniformly integrable if and only if it is relatively compact for the weak topology $\sigma(L^1,L^\infty)$.
* de la Vallée-Poussin theoremThe family $\_ \subset L^1(\mu)$ is uniformly integrable if and only if there exists a non-negative increasing convex function $G(t)$ such that $\lim_ \frac t = \infty \text \sup_\alpha \operatorname E(G(, X_, )) < \infty.$

Relation to convergence of random variables

A sequence $\$ 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.

Citations

References

*
* Diestel, J. and Uhl, J. (1977). ''Vector measures'', Mathematical Surveys 15, American Mathematical Society, Providence, RI {{isbn, 978-0-8218-1515-1

Martingale theory>X_m, : (0, 1/n)=1 for all $m \ge n$.
* If $X$ is a UI random variable, by splitting \operatorname E(, X, ) = \operatorname E(, X, ,, X, >K)+\operatorname E(, X, ,, X, 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'', $, X_n(\omega), \le Y(\omega),\ Y(\omega)\ge 0,\ \operatorname E(Y) < \infty,$ then the class $\mathcal$ of random variables $\$ is uniformly integrable.
* A class of random variables bounded in $L^p$ ($p > 1$) is uniformly integrable.

Relevant theorems

In the following we use the probabilistic framework, but regardless of the finiteness of the measure, by adding the boundedness condition on the chosen subset of $L^1(\mu)$.

* Nelson Dunford, Dunford– Pettis theoremA class of random variables $X_n \subset L^1(\mu)$ is uniformly integrable if and only if it is relatively compact for the weak topology $\sigma(L^1,L^\infty)$.
* de la Vallée-Poussin theoremThe family $\_ \subset L^1(\mu)$ is uniformly integrable if and only if there exists a non-negative increasing convex function $G(t)$ such that $\lim_ \frac t = \infty \text \sup_\alpha \operatorname E(G(, X_, )) < \infty.$

Relation to convergence of random variables

A sequence $\$ 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.

Citations

References

*
* Diestel, J. and Uhl, J. (1977). ''Vector measures'', Mathematical Surveys 15, American Mathematical Society, Providence, RI {{isbn, 978-0-8218-1515-1

Martingale theory>X, I_A\varepsilon
whenever $P(A)<\delta$.

The term "uniform absolute continuity" is not standard, but is used by some other authors.

Related corollaries

The following results apply to the probabilistic definition.
* Definition 1 could be rewritten by taking the limits as $\lim_ \sup_ \operatorname E(, X, \,I_)=0.$
* A non-UI sequence. Let \Omega = ,1\subset \mathbb, and define $X_n(\omega) = \begin n, & \omega\in (0,1/n), \\ 0 , & \text \end$ Clearly $X_n\in L^1$, and indeed $\operatorname E(, X_n, )=1\ ,$ for all ''n''. However, $\operatorname E(, X_n, , , X_n, \ge K)= 1\ \text n \ge K,$ and comparing with definition 1, it is seen that the sequence is not uniformly integrable.

* 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 \operatorname E(, X, ) = \operatorname E(, X, ,, X, >K)+\operatorname E(, X, ,, X, 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'', $, X_n(\omega), \le Y(\omega),\ Y(\omega)\ge 0,\ \operatorname E(Y) < \infty,$ then the class $\mathcal$ of random variables $\$ is uniformly integrable.
* A class of random variables bounded in $L^p$ ($p > 1$) is uniformly integrable.

Relevant theorems

In the following we use the probabilistic framework, but regardless of the finiteness of the measure, by adding the boundedness condition on the chosen subset of $L^1(\mu)$.

* Nelson Dunford, Dunford– Pettis theoremA class of random variables $X_n \subset L^1(\mu)$ is uniformly integrable if and only if it is relatively compact for the weak topology $\sigma(L^1,L^\infty)$.
* de la Vallée-Poussin theoremThe family $\_ \subset L^1(\mu)$ is uniformly integrable if and only if there exists a non-negative increasing convex function $G(t)$ such that $\lim_ \frac t = \infty \text \sup_\alpha \operatorname E(G(, X_, )) < \infty.$

Relation to convergence of random variables

A sequence $\$ 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.

Citations

References

*
* Diestel, J. and Uhl, J. (1977). ''Vector measures'', Mathematical Surveys 15, American Mathematical Society, Providence, RI {{isbn, 978-0-8218-1515-1

Martingale theory>X, I_A\varepsilon
whenever $P(A)<\delta$.

The term "uniform absolute continuity" is not standard, but is used by some other authors.

Related corollaries

The following results apply to the probabilistic definition.
* Definition 1 could be rewritten by taking the limits as $\lim_ \sup_ \operatorname E(, X, \,I_)=0.$
* A non-UI sequence. Let \Omega = ,1\subset \mathbb, and define $X_n(\omega) = \begin n, & \omega\in (0,1/n), \\ 0 , & \text \end$ Clearly $X_n\in L^1$, and indeed $\operatorname E(, X_n, )=1\ ,$ for all ''n''. However, $\operatorname E(, X_n, , , X_n, \ge K)= 1\ \text n \ge K,$ and comparing with definition 1, it is seen that the sequence is not uniformly integrable.

* 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 $m \ge n$.
* If $X$ is a UI random variable, by splitting \operatorname E(, X, ) = \operatorname E(, X, ,, X, >K)+\operatorname E(, X, ,, X, 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'', $, X_n(\omega), \le Y(\omega),\ Y(\omega)\ge 0,\ \operatorname E(Y) < \infty,$ then the class $\mathcal$ of random variables $\$ is uniformly integrable.
* A class of random variables bounded in $L^p$ ($p > 1$) is uniformly integrable.

Relevant theorems

In the following we use the probabilistic framework, but regardless of the finiteness of the measure, by adding the boundedness condition on the chosen subset of $L^1(\mu)$.

* Nelson Dunford, Dunford– Pettis theoremA class of random variables $X_n \subset L^1(\mu)$ is uniformly integrable if and only if it is relatively compact for the weak topology $\sigma(L^1,L^\infty)$.
* de la Vallée-Poussin theoremThe family $\_ \subset L^1(\mu)$ is uniformly integrable if and only if there exists a non-negative increasing convex function $G(t)$ such that $\lim_ \frac t = \infty \text \sup_\alpha \operatorname E(G(, X_, )) < \infty.$

Relation to convergence of random variables

A sequence $\$ 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.

Citations

References

*
* Diestel, J. and Uhl, J. (1977). ''Vector measures'', Mathematical Surveys 15, American Mathematical Society, Providence, RI {{isbn, 978-0-8218-1515-1

Martingale theory>X_m, : (0, 1/n)=1 for all $m \ge n$.
* If $X$ is a UI random variable, by splitting \operatorname E(, X, ) = \operatorname E(, X, ,, X, >K)+\operatorname E(, X, ,, X, 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'', $, X_n(\omega), \le Y(\omega),\ Y(\omega)\ge 0,\ \operatorname E(Y) < \infty,$ then the class $\mathcal$ of random variables $\$ is uniformly integrable.
* A class of random variables bounded in $L^p$ ($p > 1$) is uniformly integrable.

Relevant theorems

In the following we use the probabilistic framework, but regardless of the finiteness of the measure, by adding the boundedness condition on the chosen subset of $L^1(\mu)$.

* Nelson Dunford, Dunford– Pettis theoremA class of random variables $X_n \subset L^1(\mu)$ is uniformly integrable if and only if it is relatively compact for the weak topology $\sigma(L^1,L^\infty)$.
* de la Vallée-Poussin theoremThe family $\_ \subset L^1(\mu)$ is uniformly integrable if and only if there exists a non-negative increasing convex function $G(t)$ such that $\lim_ \frac t = \infty \text \sup_\alpha \operatorname E(G(, X_, )) < \infty.$

Relation to convergence of random variables

A sequence $\$ 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.

Citations

References

*
* Diestel, J. and Uhl, J. (1977). ''Vector measures'', Mathematical Surveys 15, American Mathematical Society, Providence, RI {{isbn, 978-0-8218-1515-1

Martingale theory>X_m, : (0, 1/n)=1 for all $m \ge n$.
* If $X$ is a UI random variable, by splitting \operatorname E(, X, ) = \operatorname E(, X, ,, X, >K)+\operatorname E(, X, ,, X, 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'', $, X_n(\omega), \le Y(\omega),\ Y(\omega)\ge 0,\ \operatorname E(Y) < \infty,$ then the class $\mathcal$ of random variables $\$ is uniformly integrable.
* A class of random variables bounded in $L^p$ ($p > 1$) is uniformly integrable.

Relevant theorems

In the following we use the probabilistic framework, but regardless of the finiteness of the measure, by adding the boundedness condition on the chosen subset of $L^1(\mu)$.

* Nelson Dunford, Dunford– Pettis theoremA class of random variables $X_n \subset L^1(\mu)$ is uniformly integrable if and only if it is relatively compact for the weak topology $\sigma(L^1,L^\infty)$.
* de la Vallée-Poussin theoremThe family $\_ \subset L^1(\mu)$ is uniformly integrable if and only if there exists a non-negative increasing convex function $G(t)$ such that $\lim_ \frac t = \infty \text \sup_\alpha \operatorname E(G(, X_, )) < \infty.$

Relation to convergence of random variables

A sequence $\$ 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.

Citations

References

*
* Diestel, J. and Uhl, J. (1977). ''Vector measures'', Mathematical Surveys 15, American Mathematical Society, Providence, RI {{isbn, 978-0-8218-1515-1

Martingale theory>X_m, : (0, 1/n)=1 for all $m \ge n$.
* If $X$ is a UI random variable, by splitting \operatorname E(, X, ) = \operatorname E(, X, ,, X, >K)+\operatorname E(, X, ,, X, 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'', $, X_n(\omega), \le Y(\omega),\ Y(\omega)\ge 0,\ \operatorname E(Y) < \infty,$ then the class $\mathcal$ of random variables $\$ is uniformly integrable.
* A class of random variables bounded in $L^p$ ($p > 1$) is uniformly integrable.

Relevant theorems

In the following we use the probabilistic framework, but regardless of the finiteness of the measure, by adding the boundedness condition on the chosen subset of $L^1(\mu)$.

* Nelson Dunford, Dunford– Pettis theoremA class of random variables $X_n \subset L^1(\mu)$ is uniformly integrable if and only if it is relatively compact for the weak topology $\sigma(L^1,L^\infty)$.
* de la Vallée-Poussin theoremThe family $\_ \subset L^1(\mu)$ is uniformly integrable if and only if there exists a non-negative increasing convex function $G(t)$ such that $\lim_ \frac t = \infty \text \sup_\alpha \operatorname E(G(, X_, )) < \infty.$

Relation to convergence of random variables

A sequence $\$ 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.

Citations

References

*
* Diestel, J. and Uhl, J. (1977). ''Vector measures'', Mathematical Surveys 15, American Mathematical Society, Providence, RI {{isbn, 978-0-8218-1515-1

Martingale theory>X, I_A\varepsilon
whenever $P(A)<\delta$.

The term "uniform absolute continuity" is not standard, but is used by some other authors.

Related corollaries

The following results apply to the probabilistic definition.
* Definition 1 could be rewritten by taking the limits as $\lim_ \sup_ \operatorname E(, X, \,I_)=0.$
* A non-UI sequence. Let \Omega = ,1\subset \mathbb, and define $X_n(\omega) = \begin n, & \omega\in (0,1/n), \\ 0 , & \text \end$ Clearly $X_n\in L^1$, and indeed $\operatorname E(, X_n, )=1\ ,$ for all ''n''. However, $\operatorname E(, X_n, , , X_n, \ge K)= 1\ \text n \ge K,$ and comparing with definition 1, it is seen that the sequence is not uniformly integrable.

* 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 \operatorname E(, X, ) = \operatorname E(, X, ,, X, >K)+\operatorname E(, X, ,, X, 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'', $, X_n(\omega), \le Y(\omega),\ Y(\omega)\ge 0,\ \operatorname E(Y) < \infty,$ then the class $\mathcal$ of random variables $\$ is uniformly integrable.
* A class of random variables bounded in $L^p$ ($p > 1$) is uniformly integrable.

Relevant theorems

In the following we use the probabilistic framework, but regardless of the finiteness of the measure, by adding the boundedness condition on the chosen subset of $L^1(\mu)$.

* Nelson Dunford, Dunford– Pettis theoremA class of random variables $X_n \subset L^1(\mu)$ is uniformly integrable if and only if it is relatively compact for the weak topology $\sigma(L^1,L^\infty)$.
* de la Vallée-Poussin theoremThe family $\_ \subset L^1(\mu)$ is uniformly integrable if and only if there exists a non-negative increasing convex function $G(t)$ such that $\lim_ \frac t = \infty \text \sup_\alpha \operatorname E(G(, X_, )) < \infty.$

Relation to convergence of random variables

A sequence $\$ 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.

Citations

References

*
* Diestel, J. and Uhl, J. (1977). ''Vector measures'', Mathematical Surveys 15, American Mathematical Society, Providence, RI {{isbn, 978-0-8218-1515-1

Martingale theory>X, I_A\varepsilon
whenever $P(A)<\delta$.

The term "uniform absolute continuity" is not standard, but is used by some other authors.

Related corollaries

The following results apply to the probabilistic definition.
* Definition 1 could be rewritten by taking the limits as $\lim_ \sup_ \operatorname E(, X, \,I_)=0.$
* A non-UI sequence. Let \Omega = ,1\subset \mathbb, and define $X_n(\omega) = \begin n, & \omega\in (0,1/n), \\ 0 , & \text \end$ Clearly $X_n\in L^1$, and indeed $\operatorname E(, X_n, )=1\ ,$ for all ''n''. However, $\operatorname E(, X_n, , , X_n, \ge K)= 1\ \text n \ge K,$ and comparing with definition 1, it is seen that the sequence is not uniformly integrable.

* 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 \operatorname E(, X, ) = \operatorname E(, X, ,, X, >K)+\operatorname E(, X, ,, X, 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'', $, X_n(\omega), \le Y(\omega),\ Y(\omega)\ge 0,\ \operatorname E(Y) < \infty,$ then the class $\mathcal$ of random variables $\$ is uniformly integrable.
* A class of random variables bounded in $L^p$ ($p > 1$) is uniformly integrable.

Relevant theorems

In the following we use the probabilistic framework, but regardless of the finiteness of the measure, by adding the boundedness condition on the chosen subset of $L^1(\mu)$.

* Nelson Dunford, Dunford– Pettis theoremA class of random variables $X_n \subset L^1(\mu)$ is uniformly integrable if and only if it is relatively compact for the weak topology $\sigma(L^1,L^\infty)$.
* de la Vallée-Poussin theoremThe family $\_ \subset L^1(\mu)$ is uniformly integrable if and only if there exists a non-negative increasing convex function $G(t)$ such that $\lim_ \frac t = \infty \text \sup_\alpha \operatorname E(G(, X_, )) < \infty.$

Relation to convergence of random variables

A sequence $\$ 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.

Citations

References

*
* Diestel, J. and Uhl, J. (1977). ''Vector measures'', Mathematical Surveys 15, American Mathematical Society, Providence, RI {{isbn, 978-0-8218-1515-1

Martingale theory>X_m, : (0, 1/n)=1 for all $m \ge n$.
* If $X$ is a UI random variable, by splitting \operatorname E(, X, ) = \operatorname E(, X, ,, X, >K)+\operatorname E(, X, ,, X, 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'', $, X_n(\omega), \le Y(\omega),\ Y(\omega)\ge 0,\ \operatorname E(Y) < \infty,$ then the class $\mathcal$ of random variables $\$ is uniformly integrable.
* A class of random variables bounded in $L^p$ ($p > 1$) is uniformly integrable.

Relevant theorems

In the following we use the probabilistic framework, but regardless of the finiteness of the measure, by adding the boundedness condition on the chosen subset of $L^1(\mu)$.

* Nelson Dunford, Dunford– Pettis theoremA class of random variables $X_n \subset L^1(\mu)$ is uniformly integrable if and only if it is relatively compact for the weak topology $\sigma(L^1,L^\infty)$.
* de la Vallée-Poussin theoremThe family $\_ \subset L^1(\mu)$ is uniformly integrable if and only if there exists a non-negative increasing convex function $G(t)$ such that $\lim_ \frac t = \infty \text \sup_\alpha \operatorname E(G(, X_, )) < \infty.$

Relation to convergence of random variables

A sequence $\$ 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.

Citations

References

*
* Diestel, J. and Uhl, J. (1977). ''Vector measures'', Mathematical Surveys 15, American Mathematical Society, Providence, RI {{isbn, 978-0-8218-1515-1

Martingale theory>X, I_A\varepsilon
whenever $P(A)<\delta$.

The term "uniform absolute continuity" is not standard, but is used by some other authors.

Related corollaries

The following results apply to the probabilistic definition.
* Definition 1 could be rewritten by taking the limits as $\lim_ \sup_ \operatorname E(, X, \,I_)=0.$
* A non-UI sequence. Let \Omega = ,1\subset \mathbb, and define $X_n(\omega) = \begin n, & \omega\in (0,1/n), \\ 0 , & \text \end$ Clearly $X_n\in L^1$, and indeed $\operatorname E(, X_n, )=1\ ,$ for all ''n''. However, $\operatorname E(, X_n, , , X_n, \ge K)= 1\ \text n \ge K,$ and comparing with definition 1, it is seen that the sequence is not uniformly integrable.

* 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 \operatorname E(, X, ) = \operatorname E(, X, ,, X, >K)+\operatorname E(, X, ,, X, 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'', $, X_n(\omega), \le Y(\omega),\ Y(\omega)\ge 0,\ \operatorname E(Y) < \infty,$ then the class $\mathcal$ of random variables $\$ is uniformly integrable.
* A class of random variables bounded in $L^p$ ($p > 1$) is uniformly integrable.

Relevant theorems

In the following we use the probabilistic framework, but regardless of the finiteness of the measure, by adding the boundedness condition on the chosen subset of $L^1(\mu)$.

* Nelson Dunford, Dunford– Pettis theoremA class of random variables $X_n \subset L^1(\mu)$ is uniformly integrable if and only if it is relatively compact for the weak topology $\sigma(L^1,L^\infty)$.
* de la Vallée-Poussin theoremThe family $\_ \subset L^1(\mu)$ is uniformly integrable if and only if there exists a non-negative increasing convex function $G(t)$ such that $\lim_ \frac t = \infty \text \sup_\alpha \operatorname E(G(, X_, )) < \infty.$

Relation to convergence of random variables

A sequence $\$ 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.

Citations

References

*
* Diestel, J. and Uhl, J. (1977). ''Vector measures'', Mathematical Surveys 15, American Mathematical Society, Providence, RI {{isbn, 978-0-8218-1515-1

Martingale theory>X, I_A\varepsilon
whenever $P(A)<\delta$.

The term "uniform absolute continuity" is not standard, but is used by some other authors.

Related corollaries

The following results apply to the probabilistic definition.
* Definition 1 could be rewritten by taking the limits as $\lim_ \sup_ \operatorname E(, X, \,I_)=0.$
* A non-UI sequence. Let \Omega = ,1\subset \mathbb, and define $X_n(\omega) = \begin n, & \omega\in (0,1/n), \\ 0 , & \text \end$ Clearly $X_n\in L^1$, and indeed $\operatorname E(, X_n, )=1\ ,$ for all ''n''. However, $\operatorname E(, X_n, , , X_n, \ge K)= 1\ \text n \ge K,$ and comparing with definition 1, it is seen that the sequence is not uniformly integrable.

* 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 $m \ge n$.
* If $X$ is a UI random variable, by splitting \operatorname E(, X, ) = \operatorname E(, X, ,, X, >K)+\operatorname E(, X, ,, X, 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'', $, X_n(\omega), \le Y(\omega),\ Y(\omega)\ge 0,\ \operatorname E(Y) < \infty,$ then the class $\mathcal$ of random variables $\$ is uniformly integrable.
* A class of random variables bounded in $L^p$ ($p > 1$) is uniformly integrable.

Relevant theorems

In the following we use the probabilistic framework, but regardless of the finiteness of the measure, by adding the boundedness condition on the chosen subset of $L^1(\mu)$.

* Nelson Dunford, Dunford– Pettis theoremA class of random variables $X_n \subset L^1(\mu)$ is uniformly integrable if and only if it is relatively compact for the weak topology $\sigma(L^1,L^\infty)$.
* de la Vallée-Poussin theoremThe family $\_ \subset L^1(\mu)$ is uniformly integrable if and only if there exists a non-negative increasing convex function $G(t)$ such that $\lim_ \frac t = \infty \text \sup_\alpha \operatorname E(G(, X_, )) < \infty.$

Relation to convergence of random variables

A sequence $\$ 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.

Citations

References

*
* Diestel, J. and Uhl, J. (1977). ''Vector measures'', Mathematical Surveys 15, American Mathematical Society, Providence, RI {{isbn, 978-0-8218-1515-1

Martingale theory>X_m, : (0, 1/n)=1 for all $m \ge n$.
* If $X$ is a UI random variable, by splitting \operatorname E(, X, ) = \operatorname E(, X, ,, X, >K)+\operatorname E(, X, ,, X, 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'', $, X_n(\omega), \le Y(\omega),\ Y(\omega)\ge 0,\ \operatorname E(Y) < \infty,$ then the class $\mathcal$ of random variables $\$ is uniformly integrable.
* A class of random variables bounded in $L^p$ ($p > 1$) is uniformly integrable.

Relevant theorems

In the following we use the probabilistic framework, but regardless of the finiteness of the measure, by adding the boundedness condition on the chosen subset of $L^1(\mu)$.

* Nelson Dunford, Dunford– Pettis theoremA class of random variables $X_n \subset L^1(\mu)$ is uniformly integrable if and only if it is relatively compact for the weak topology $\sigma(L^1,L^\infty)$.
* de la Vallée-Poussin theoremThe family $\_ \subset L^1(\mu)$ is uniformly integrable if and only if there exists a non-negative increasing convex function $G(t)$ such that $\lim_ \frac t = \infty \text \sup_\alpha \operatorname E(G(, X_, )) < \infty.$

Relation to convergence of random variables

A sequence $\$ 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.

Citations

References

*
* Diestel, J. and Uhl, J. (1977). ''Vector measures'', Mathematical Surveys 15, American Mathematical Society, Providence, RI {{isbn, 978-0-8218-1515-1

Martingale theory>X_m, : (0, 1/n)=1 for all $m \ge n$.
* If $X$ is a UI random variable, by splitting \operatorname E(, X, ) = \operatorname E(, X, ,, X, >K)+\operatorname E(, X, ,, X, 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'', $, X_n(\omega), \le Y(\omega),\ Y(\omega)\ge 0,\ \operatorname E(Y) < \infty,$ then the class $\mathcal$ of random variables $\$ is uniformly integrable.
* A class of random variables bounded in $L^p$ ($p > 1$) is uniformly integrable.

Relevant theorems

In the following we use the probabilistic framework, but regardless of the finiteness of the measure, by adding the boundedness condition on the chosen subset of $L^1(\mu)$.

* Nelson Dunford, Dunford– Pettis theoremA class of random variables $X_n \subset L^1(\mu)$ is uniformly integrable if and only if it is relatively compact for the weak topology $\sigma(L^1,L^\infty)$.
* de la Vallée-Poussin theoremThe family $\_ \subset L^1(\mu)$ is uniformly integrable if and only if there exists a non-negative increasing convex function $G(t)$ such that $\lim_ \frac t = \infty \text \sup_\alpha \operatorname E(G(, X_, )) < \infty.$

Relation to convergence of random variables

A sequence $\$ 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.

Citations

References

*
* Diestel, J. and Uhl, J. (1977). ''Vector measures'', Mathematical Surveys 15, American Mathematical Society, Providence, RI {{isbn, 978-0-8218-1515-1

Martingale theory>X_m, : (0, 1/n)=1 for all $m \ge n$.
* If $X$ is a UI random variable, by splitting \operatorname E(, X, ) = \operatorname E(, X, ,, X, >K)+\operatorname E(, X, ,, X, 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'', $, X_n(\omega), \le Y(\omega),\ Y(\omega)\ge 0,\ \operatorname E(Y) < \infty,$ then the class $\mathcal$ of random variables $\$ is uniformly integrable.
* A class of random variables bounded in $L^p$ ($p > 1$) is uniformly integrable.

Relevant theorems

In the following we use the probabilistic framework, but regardless of the finiteness of the measure, by adding the boundedness condition on the chosen subset of $L^1(\mu)$.

* Nelson Dunford, Dunford– Pettis theoremA class of random variables $X_n \subset L^1(\mu)$ is uniformly integrable if and only if it is relatively compact for the weak topology $\sigma(L^1,L^\infty)$.
* de la Vallée-Poussin theoremThe family $\_ \subset L^1(\mu)$ is uniformly integrable if and only if there exists a non-negative increasing convex function $G(t)$ such that $\lim_ \frac t = \infty \text \sup_\alpha \operatorname E(G(, X_, )) < \infty.$

Relation to convergence of random variables

A sequence $\$ 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.

Citations

References

*
* Diestel, J. and Uhl, J. (1977). ''Vector measures'', Mathematical Surveys 15, American Mathematical Society, Providence, RI {{isbn, 978-0-8218-1515-1

Martingale theory