Bounded Families and Uniform Integrability

Why a Uniform Bound Controls the Tails

A family of random variables can contain many different distributions and depend on events with very different probabilities. Uniform integrability asks whether their large-value contributions to expectation are nevertheless controlled uniformly. A common bound, even one that holds only almost surely for each member of the family, gives an especially direct answer: sufficiently far into the tail, every member contributes zero.

The important point is that “almost surely” applies separately to each random variable. Uniform integrability does not require choosing representatives that obey a single pointwise bound at every outcome. It requires the tail-expectation condition, which is unchanged by modifications on null sets.

Definition and the Bounded-Family Criterion

Definition: A family $\mathcal{X}$ of integrable random variables is uniformly integrable if $$ \lim_{K\to\infty}\sup_{X\in\mathcal{X}} \mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|>K\}}\right]=0. $$ Here $\mathbf{1}_A$ denotes the indicator of the event $A$. The supremum is taken over the whole family at each threshold $K$.

The order of the supremum and limit matters. For each fixed $X$, integrability ensures that its own tail expectation tends to zero as $K$ increases. Uniform integrability asks for a threshold that works for every member of the family at once.

Theorem: Suppose $\mathcal{X}$ is a family of real-valued random variables and there is a finite constant $C\geq 0$ such that, for every $X\in\mathcal{X}$, $$ \mathbb{P}(|X|\leq C)=1. $$ Then every $X\in\mathcal{X}$ is integrable and $\mathcal{X}$ is uniformly integrable.

Proof. Fix $X\in\mathcal{X}$. Since $|X|\leq C$ almost surely, $$ \mathbb{E}|X|\leq C<\infty, $$ so $X$ is integrable. Now take any $K\geq C$. The event $\{|X|>K\}$ is contained, up to a null set, in $\{|X|>C\}$. Therefore $$ \mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|>K\}}\right]=0. $$ This holds for every $X\in\mathcal{X}$, so the supremum of these tail expectations is zero for every $K\geq C$. In particular, it tends to zero as $K\to\infty$. Thus the family is uniformly integrable. $\square$

The proof gives more than a limit: once the threshold reaches the common bound, every tail expectation vanishes exactly. The bound need not be the smallest possible one. Any finite $C$ that works for every member establishes the conclusion.

Essential Bounds and Pointwise Bounds

A pointwise common bound means that the chosen versions of all random variables satisfy $|X(\omega)|\leq C$ for every $X$ in the family and every outcome $\omega$. The theorem requires less. It is enough that for each $X$, the inequality holds almost surely. The exceptional null set may depend on $X$.

In the language of $L^\infty$, the theorem’s hypothesis is that the essential suprema are uniformly bounded: $$ \sup_{X\in\mathcal{X}}\|X\|_\infty\leq C, $$ where $\|X\|_\infty$ is the essential supremum of $|X|$. Because essential bounds ignore changes on null sets, this condition is about the random variables as almost-sure equivalence classes, rather than about particular pointwise representatives.

Key distinction: A common essential bound is sufficient for uniform integrability. A pointwise common bound is also sufficient, but it is stronger than necessary. It is not necessary to find one exceptional null set that works simultaneously for every member of an arbitrary family.

For a countable family, separate almost-sure bounds can be combined into a common almost-sure bound outside one null set. This fact explains why the distinction can be easy to overlook for sequences.

Proposition: Let $(X_n)_{n\in\mathbb{N}}$ be a countable family and suppose $\mathbb{P}(|X_n|\leq C)=1$ for every $n$, with the same finite $C$. Then there is a single null set outside which $|X_n|\leq C$ for every $n$.

Proof. For each $n$, let $N_n=\{|X_n|>C\}$. The hypothesis gives $\mathbb{P}(N_n)=0$. Set $$ N=\bigcup_{n=1}^{\infty}N_n. $$ By countable subadditivity, $$ \mathbb{P}(N)\leq\sum_{n=1}^{\infty}\mathbb{P}(N_n)=0. $$ Thus $N$ is null. If $\omega\notin N$, then $\omega\notin N_n$ for every $n$, so $|X_n(\omega)|\leq C$ for every $n$. $\square$

Countability is essential to this argument: the probability of a countable union of null sets is zero, but an arbitrary union of null sets need not be null. Uniform integrability itself does not demand a common exceptional set, even for an uncountable family.

Worked Examples

Worked Example: A Family of Bounded Constants

Let $X_t=t$ be the constant random variable with value $t$, where the index ranges over all $t\in[-3,3]$. Then $|X_t|\leq 3$ for every outcome and every index. Each member is integrable, with $\mathbb{E}|X_t|=|t|\leq 3$. For every $K\geq 3$, the event $\{|X_t|>K\}$ is empty for all $t$, and hence $$ \sup_{t\in[-3,3]} \mathbb{E}\!\left[|X_t|\mathbf{1}_{\{|X_t|>K\}}\right]=0. $$ The family is uniformly integrable. This example also shows that the family may be uncountable; no countability of the index set is needed for the bounded-family criterion.

Worked Example: Indicators of Arbitrary Events

Let $(A_\alpha)_{\alpha\in I}$ be any collection of events on a probability space, and set $X_\alpha=\mathbf{1}_{A_\alpha}$. The events need not have equal probabilities or satisfy any independence condition. Each variable takes only the values zero and one, so $|X_\alpha|\leq 1$ everywhere. Thus every $X_\alpha$ is integrable, with $\mathbb{E}|X_\alpha|=\mathbb{P}(A_\alpha)\leq 1$.

If $0\leq K<1$, then $\{|X_\alpha|>K\}=A_\alpha$, and the tail expectation is $\mathbb{P}(A_\alpha)$. If $K\geq 1$, then $\{|X_\alpha|>K\}=\varnothing$, so the tail expectation is zero for every $\alpha$. Therefore $$ \sup_{\alpha\in I} \mathbb{E}\!\left[|X_\alpha|\mathbf{1}_{\{|X_\alpha|>K\}}\right]=0 \qquad\text{for every }K\geq 1, $$ and the family is uniformly integrable.

Worked Example: No Uniform Pointwise Bound Is Needed

On $[0,1]$ with Lebesgue probability measure, define $$ X_n(\omega)=n\mathbf{1}_{\{1/n\}}(\omega), \qquad n\in\mathbb{N}. $$ For each fixed $n$, the singleton $\{1/n\}$ has probability zero. Therefore $X_n=0$ almost surely, and $\|X_n\|_\infty=0$. The family has the common essential bound $C=0$, so the theorem implies that it is uniformly integrable.

Indeed, for each $n$, $|X_n|\mathbf{1}_{\{|X_n|>K\}}$ is zero almost surely for every $K\geq 0$; its expectation is zero, including when $K<n$. Hence the supremum of the tail expectations is zero. Yet there is no finite pointwise common bound: at $\omega=1/n$, $X_n(\omega)=n$, and these values grow without bound as $n$ varies. The example shows why an almost-sure bound must not be confused with a bound holding at every outcome for the chosen versions.

Using the Criterion Carefully

When applying the theorem, identify the bound and check that it is uniform over the family. Bounds that depend on the index do not by themselves give the theorem’s hypothesis: the relevant condition is one finite $C$ that works almost surely for each member. Likewise, knowing that each variable is bounded individually is weaker than knowing that their essential bounds have a finite common upper bound.

For indicators, the bound is immediate because every indicator has absolute value at most one. This is why arbitrary families of indicators are uniformly integrable, regardless of how the events vary. The conclusion concerns the tail expectations; it does not say that the indicators, or their expectations, are all equal.

A common bound is a simple sufficient condition, not a characterization of uniform integrability. The theorem establishes that boundedness in essential supremum implies uniform integrability; it does not assert that every uniformly integrable family has a common essential bound. Keeping this direction of implication clear prevents an overly strong conclusion from a useful criterion.

Check Your Understanding

Use the tail-expectation definition to answer the following questions.

Why does an almost-sure bound $|X|\leq C$ imply that $X$ is integrable?
For a family with a common almost-sure bound $C$, what happens to every tail expectation when $K\geq C$?
What is the difference between a common essential bound and a pointwise common bound?
Why can separate almost-sure bounds for a countable family be combined outside one null set?
For an indicator $\mathbf{1}_A$, what is its tail expectation when $0\leq K<1$, and what is it when $K\geq1$?