Uniform Integrability and Growth Beyond Linear
Uniform integrability controls how much integral a family can place in the large-value tails. The de la Vallée Poussin criterion gives a different, often more usable, way to express that control: apply one suitable convex function to the absolute values, and require the resulting integrals to be bounded uniformly across the family. The key is that the function must grow faster than linearly, so large values carry an increasingly strong penalty.
Throughout, let \((X,\mathcal{M},\mu)\) be a measure space with \(\mu(X)<\infty\). Let \(\mathcal{F}\) be a family of real-valued or complex-valued measurable functions in \(L^1(\mu)\); functions that agree \(\mu\)-a.e. are identified for integrals. We use the tail definition of uniform integrability. The strict inequality in the tail set is part of the convention below.
On a finite measure space this tail definition is equivalent to the small-set criterion: the integrals over measurable sets of sufficiently small measure are uniformly small, together with a uniform \(L^1\) bound. The earlier material on the small-set criterion establishes that equivalence. Here we use tail control directly, because it provides thresholds from which to build the convex function.
The Criterion
The limit condition says that \(\Phi\) grows faster than every fixed linear function: for each \(R>0\), there is a threshold \(K\) such that \(\Phi(t)\ge Rt\) for all \(t>K\). We prove both implications. The construction in the converse direction is important: the criterion asserts the existence of a single function \(\Phi\) that works for the entire family.
Proof. First suppose a function \(\Phi\) as in the theorem exists, and write \(C=\sup_{f\in\mathcal{F}}\int_X\Phi(|f|)\,d\mu<\infty\). For any \(R>0\), the growth condition gives a \(K>0\) such that \(\Phi(t)\ge Rt\) whenever \(t>K\). Since \(\Phi\ge0\), for every \(f\in\mathcal{F}\), $$ \int_{\{|f|>K\}}|f|\,d\mu \le \frac{1}{R}\int_{\{|f|>K\}}\Phi(|f|)\,d\mu \le \frac{C}{R}. $$ Because \(R\) may be arbitrarily large, these tail integrals tend uniformly to zero as \(K\to\infty\): more explicitly, given \(\varepsilon>0\), choose \(R\) with \(C/R<\varepsilon\) if \(C>0\); if \(C=0\), the tail bound is already zero for every such threshold. The family also has uniformly bounded \(L^1\) norms. Fix one threshold \(K\) for which \(\Phi(t)\ge t\) when \(t>K\). Splitting into \(\{|f|\le K\}\) and \(\{|f|>K\}\) gives $$ \int_X|f|\,d\mu \le K\mu(X)+\int_{\{|f|>K\}}|f|\,d\mu \le K\mu(X)+C. $$ Here finiteness of \(\mu(X)\) is used. Thus \(\mathcal{F}\) is uniformly integrable.
For the reverse implication, suppose \(\mathcal{F}\) is uniformly integrable and set \(M=\sup_{f\in\mathcal{F}}\int_X|f|\,d\mu<\infty\). For each integer \(n\ge1\), tail control lets us choose thresholds \(t_n\) so that \(t_n\ge n\), \(t_n\ge t_{n-1}+1\) when \(n\ge2\), and $$ \sup_{f\in\mathcal{F}}\int_{\{|f|>t_n\}}|f|\,d\mu\le\frac{2^{-n}}{n^2}. $$ The thresholds can be made this large because the uniform tail integrals tend to zero. Define, for \(t\ge0\), $$ \Phi(t)=t+\sum_{n=1}^{\infty}n(t-t_n)^+, \qquad (s)^+=\max(s,0). $$ Each summand is nonnegative, nondecreasing and convex. Since \(t_n\ge n\), only finitely many summands are nonzero for any fixed \(t\); consequently \(\Phi(t)\) is finite. It is convex and nondecreasing, and \(\Phi(0)=0\).
We check its growth. Fix an integer \(N\). For \(t>t_N\), the first \(N\) terms give $$ \frac{\Phi(t)}{t}\ge 1+\sum_{n=1}^N n\left(1-\frac{t_n}{t}\right). $$ As \(t\to\infty\), the right-hand side tends to \(1+\sum_{n=1}^N n\). These lower bounds become arbitrarily large as \(N\to\infty\), proving \(\Phi(t)/t\to\infty\).
It remains to check the uniform integral bound. For each \(n\), \((|f|-t_n)^+\le |f|\chi_{\{|f|>t_n\}}\). The partial sums defining \(\Phi(|f|)\) increase to it, so the Monotone Convergence Theorem justifies integrating the series term by term. Therefore $$ \int_X\Phi(|f|)\,d\mu \le \int_X|f|\,d\mu+ \sum_{n=1}^{\infty}n\int_{\{|f|>t_n\}}|f|\,d\mu \le M+\sum_{n=1}^{\infty}\frac{2^{-n}}{n} \le M+1. $$ The bound does not depend on \(f\), so the required supremum is finite. This proves the converse and the criterion. \(\square\)
Proof. For each \(f\), the pointwise inequality \(|f|\le \Phi(|f|)/R\) holds on \(\{|f|>K\}\). Integrating over this set and then bounding its \(\Phi\)-integral by the integral over \(X\) gives the stated inequality. Taking the supremum over \(f\in\mathcal{F}\) completes the proof. \(\square\)
Worked Applications
Worked Example: A Family with an Explicit Superlinear Bound
On \([0,1]\) with Lebesgue measure \(m\), let \(f_n=n\chi_{(0,1/n^2)}\) for integers \(n\ge1\). We verify uniform integrability and find a function \(\Phi\) that certifies it. The endpoint \(1/n^2\) is excluded from the indicated set; changing its membership would not change any integral.
Since \(|f_n|=n\) on an interval of measure \(1/n^2\) and is zero elsewhere, $$ \int_0^1|f_n|\,dm=n\cdot\frac1{n^2}=\frac1n\le1. $$ For a fixed \(K\ge0\), the tail set \(\{|f_n|>K\}\) is \((0,1/n^2)\) if \(n>K\), and is empty if \(n\le K\). Thus $$ \int_{\{|f_n|>K\}}|f_n|\,dm= \begin{cases} 1/n,&n>K,\\ 0,&n\le K. \end{cases} $$ The supremum over integers \(n\ge1\) is \(1/(\lfloor K\rfloor+1)\), which tends to zero. This verifies both conditions in the definition.
Choose \(\Phi(t)=t^{3/2}\). It is nonnegative, convex, nondecreasing, and \(\Phi(0)=0\); moreover \(\Phi(t)/t=\sqrt{t}\to\infty\). Directly, $$ \int_0^1\Phi(|f_n|)\,dm =n^{3/2}\cdot\frac1{n^2} =\frac1{\sqrt n}\le1. $$ At \(n=1\), this integral is \(1\), so the supremum is exactly \(1\). The direct calculation agrees with the criterion.
Worked Example: An Unbounded Function in a Uniformly Integrable Family
On \((0,1)\) with Lebesgue measure, consider the singleton family containing \(f(x)=x^{-1/2}\). This function is unbounded near zero. Its integral is finite: $$ \int_0^1x^{-1/2}\,dx=\left[2\sqrt{x}\right]_0^1=2. $$ For \(K\ge1\), the strict tail condition \(f(x)>K\) is equivalent to \(0<x<K^{-2}\). Therefore $$ \int_{\{f>K\}}f\,dm =\int_0^{K^{-2}}x^{-1/2}\,dx =2\sqrt{K^{-2}}=\frac2K. $$ This tends to zero, so the singleton family is uniformly integrable. For \(0\le K<1\), the tail set is all of \((0,1)\), and its integral is \(2\); this does not affect the limit as \(K\to\infty\).
The function \(\Phi(t)=t^{3/2}\) again has the required convexity and superlinear growth. Its integral is $$ \int_0^1\Phi(f(x))\,dx =\int_0^1x^{-3/4}\,dx =\left[4x^{1/4}\right]_0^1=4. $$ The value is finite, as required for a singleton family. This calculation also checks directly that the unbounded values of \(f\) are controlled by the stronger growth of \(\Phi\).
Worked Example: Bounded Integrals Without Uniform Integrability
On \([0,1]\), set \(g_n=n\chi_{(0,1/n)}\), \(n\ge1\). The integral of each function is $$ \int_0^1|g_n|\,dm=n\cdot\frac1n=1, $$ so the family has uniformly bounded \(L^1\) norms. But for any fixed \(K\ge0\), if \(n>K\), then \(\{|g_n|>K\}=(0,1/n)\) and $$ \int_{\{|g_n|>K\}}|g_n|\,dm=n\cdot\frac1n=1. $$ Hence the supremum of the tail integrals is \(1\) for every \(K\), not a quantity tending to zero. This family is not uniformly integrable.
The criterion also rules out a superlinear convex certificate directly. If \(\Phi(0)=0\), then $$ \int_0^1\Phi(|g_n|)\,dm=\frac{\Phi(n)}{n}. $$ For any \(\Phi\) with \(\Phi(t)/t\to\infty\), these values tend to infinity as \(n\to\infty\); they cannot be uniformly bounded. This confirms the failure predicted by the theorem.
Reading the Criterion Correctly
A common mistake is to infer uniform integrability from a uniform bound on the ordinary integrals. For the family \(g_n\) above, the calculation \(\int|g_n|\,dm=1\) is correct, but it says nothing by itself about how much of that integral lies above a fixed threshold. The incorrect tail calculation would replace the required supremum over \(n\) by a quantity that decreases with \(K\). The correction is to keep \(K\) fixed while taking the supremum: for every \(K\), there are integers \(n>K\), and each such function places its entire integral, equal to \(1\), on \(\{|g_n|>K\}\).
The two directions of the criterion offer complementary ways to use it. If a suitable \(\Phi\) is already available, its growth gives an immediate tail estimate. If uniform integrability is known instead, the thresholds \(t_n\) in the proof encode successively smaller tail integrals; the weighted hinge functions \(n(t-t_n)^+\) combine those controls into one convex function. The finite-measure assumption matters in the first implication when controlling values below a fixed threshold: their integral is at most \(K\mu(X)\).
Check Your Understanding
For each question, pay attention to the uniform quantifier over the family and to the strict tail event.
- In the criterion, why does a uniform bound on \(\int\Phi(|f|)\,d\mu\) control \(\int_{\{|f|>K\}}|f|\,d\mu\) when \(\Phi(t)/t\to\infty\)?
- For \(f_n=n\chi_{(0,1/n^2)}\) on \([0,1]\), compute the tail integral for fixed \(K\ge0\) and identify the supremum over \(n\ge1\).
- Why is \(\mu(X)<\infty\) used when deducing a uniform \(L^1\) bound from a convex-integral bound?
- For \(g_n=n\chi_{(0,1/n)}\), explain why \(\sup_n\int_{\{|g_n|>K\}}|g_n|\,dm\) does not tend to zero.
- In the construction of \(\Phi\), why does choosing \(t_n\to\infty\) ensure that only finitely many terms in its defining sum are nonzero at each fixed \(t\)?