Uniform Integrability in Integrated-Risk Limits

Why Pointwise Limits Need Tail Control

In admissibility arguments, one often compares the risk of a sequence of procedures with the risk of a fixed procedure, then integrates that difference against a prior measure. A pointwise limit of the risk differences is not enough to pass the limit through the integral: contributions from regions of small prior mass can remain substantial if the functions become large there. Uniform integrability rules out precisely this kind of escaping tail contribution.

The central question here is therefore not merely whether the risk differences converge at almost every parameter value, but whether their large values are controlled uniformly across the sequence. With that control, the integral of the difference converges to the integral of its limit. Care is also needed when interpreting this as a difference of integrated risks: that subtraction is valid only when the integrated risks being subtracted are finite.

Uniform Integrability Under a Finite Prior Measure

Definition: Let $\pi$ be a finite measure on a measurable parameter space $(\Theta,\mathcal{A})$. A sequence of measurable, integrable functions $\{g_n\}$ is uniformly integrable with respect to $\pi$ if $$ \lim_{K\to\infty}\sup_n \int_\Theta |g_n(\theta)|\,\mathbf{1}\{|g_n(\theta)|>K\}\,\pi(d\theta)=0. $$ Here $\mathbf{1}\{A\}$ is 1 when condition $A$ holds and 0 otherwise. The condition says that the integral contributed by values above a sufficiently large threshold can be made small uniformly over all $n$.

The functions of interest are often risk differences $D_n(\theta)=R(\theta,\delta_n)-R(\theta,\delta)$, where $\delta_n$ is a sequence of decision rules and $\delta$ is a fixed rule. Uniform integrability concerns this sequence of differences as functions of $\theta$. It is not automatic from pointwise convergence, nor does it follow just from the fact that each $D_n$ is integrable.

Worked Example: A Finite Prior and Bounded Differences

Let $\Theta=\{a,b,c\}$, with prior masses $\pi(\{a\})=1/2$, $\pi(\{b\})=1/3$, and $\pi(\{c\})=1/6$. Suppose risk differences have values $D_n(a)=1+1/n$, $D_n(b)=-1$, and $D_n(c)=0$.

At each parameter value the sequence converges, with limit $D(a)=1$, $D(b)=-1$, and $D(c)=0$. Also $|D_n(\theta)|\leq 2$ for every $n$ and every $\theta$. For $K\geq2$, the tail integral in the definition of uniform integrability is therefore zero. The sequence is uniformly integrable.

The integrals can be computed directly: $$ \int_\Theta D_n\,d\pi =\frac12\left(1+\frac1n\right)-\frac13 =\frac16+\frac{1}{2n}, \qquad \int_\Theta D\,d\pi =\frac12-\frac13 =\frac16. $$ Thus $\int D_n\,d\pi$ tends to $\int D\,d\pi$. On a finite parameter space with a finite measure, this is a straightforward setting in which the limit and integration agree.

The Uniform-Integrability Convergence Theorem

Theorem: Let $\pi$ be a finite measure, and let $g_n$ be measurable, integrable functions on $\Theta$. Suppose $g_n(\theta)\to g(\theta)$ for $\pi$-almost every $\theta$, $g$ is integrable, and $\{g_n\}$ is uniformly integrable with respect to $\pi$. Then $$ \lim_{n\to\infty}\int_\Theta g_n\,d\pi=\int_\Theta g\,d\pi. $$

A proof makes clear where the finite-measure and tail assumptions enter. For $K>0$, define the clipped function $T_K(x)=\max(-K,\min(x,K))$. It agrees with $x$ when $|x|\leq K$, and otherwise replaces $x$ by the nearest endpoint of $[-K,K]$.

Control the limit's tail.
Since $g_n\to g$ almost everywhere, Fatou's lemma applied to $|g_n|\mathbf{1}\{|g_n|>K\}$ gives $$ \int_\Theta |g|\mathbf{1}\{|g|>K\}\,d\pi \leq \liminf_{n\to\infty}\int_\Theta |g_n|\mathbf{1}\{|g_n|>K\}\,d\pi. $$ Indeed, at each point where $g_n\to g$, if $|g|>K$, then eventually $|g_n|>K$ and the displayed integrands converge to $|g|$; if $|g|\leq K$, the left integrand is zero. Uniform integrability therefore makes this tail integral small for large $K$.

Apply bounded convergence to clipped functions.
For fixed $K$, $T_K(g_n)\to T_K(g)$ almost everywhere and $|T_K(g_n)|\leq K$. The constant $K$ is integrable because $\pi(\Theta)<\infty$. The dominated convergence theorem gives $$ \lim_{n\to\infty}\int_\Theta T_K(g_n)\,d\pi =\int_\Theta T_K(g)\,d\pi. $$

Remove the clipping.
For every real $x$, $|x-T_K(x)|\leq |x|\mathbf{1}\{|x|>K\}$. Consequently, the difference between $\int g_n\,d\pi$ and $\int T_K(g_n)\,d\pi$ is bounded in absolute value by the $n$th tail integral. The corresponding bound for $g$ also tends to zero as $K$ increases, by Step 1 and uniform integrability. First take $n\to\infty$ for fixed $K$, using Step 2, and then take $K\to\infty$. The two clipping errors vanish uniformly in $n$, so $\int g_n\,d\pi\to\int g\,d\pi$. $\square$

Worked Example: Pointwise Convergence With a Persistent Tail

On $\Theta=(0,1)$ with Lebesgue measure, set $D_n(\theta)=n\mathbf{1}\{0<\theta<1/n\}$, and let $D(\theta)=0$. For every fixed $\theta>0$, eventually $1/n\leq\theta$, so $D_n(\theta)\to0$. The limiting function is integrable and has integral zero.

Nevertheless, $$ \int_0^1 D_n(\theta)\,d\theta =n\cdot\frac1n=1 $$ for every $n$. The sequence is not uniformly integrable: for any $K>0$, choose $n>K$. Then $$ \int_0^1 |D_n(\theta)|\mathbf{1}\{|D_n(\theta)|>K\}\,d\theta=1. $$ The supremum over $n$ is at least 1 for every $K$, so it does not tend to zero. The mass of the region where $D_n$ is nonzero shrinks, but its height grows enough to preserve the integral. This is the tail contribution pointwise convergence cannot detect.

Using the Theorem for Integrated Risks

Suppose $D_n(\theta)=R(\theta,\delta_n)-R(\theta,\delta)$ converges to $D(\theta)$ for $\pi$-almost every $\theta$, the limit is integrable, and the sequence of differences is uniformly integrable. The theorem gives the unambiguous conclusion $$ \lim_{n\to\infty}\int_\Theta \bigl[R(\theta,\delta_n)-R(\theta,\delta)\bigr]\,\pi(d\theta) =\int_\Theta D(\theta)\,\pi(d\theta). $$ This statement is about the integrals of the differences. If, in addition, both integrated risks are finite for every $n$ and for $\delta$, linearity of integration identifies the left-hand side with the limit of the differences of integrated risks: $$ \int_\Theta [R(\theta,\delta_n)-R(\theta,\delta)]\,\pi(d\theta) =\int_\Theta R(\theta,\delta_n)\,\pi(d\theta) -\int_\Theta R(\theta,\delta)\,\pi(d\theta). $$ The finiteness condition matters. If either integrated risk is infinite, subtraction may be undefined even when the pointwise difference is integrable.

Worked Example: A Shrinking Spike With Uniformly Controlled Tails

Take $\Theta=[0,1]$ with Lebesgue measure and define $$ D_n(\theta)=\theta+n\mathbf{1}\{0<\theta<1/n^2\}. $$ For each fixed $\theta$, the indicator is eventually zero, so $D_n(\theta)\to D(\theta)=\theta$. Each $D_n$ is integrable, and $$ \int_0^1 D_n(\theta)\,d\theta =\int_0^1\theta\,d\theta+n\cdot\frac{1}{n^2} =\frac12+\frac1n. $$ The limit has integral $\int_0^1\theta\,d\theta=1/2$.

To verify uniform integrability, take $K>1$. Outside the spike interval, $D_n(\theta)=\theta\leq1<K$, so no point there contributes to the tail integral. On the spike interval, $D_n(\theta)\leq n+1$. A contribution is possible only when $n+1>K$, and in that case $$ \int_0^1 D_n(\theta)\mathbf{1}\{D_n(\theta)>K\}\,d\theta \leq\frac{n+1}{n^2}. $$ For $K>1$, contributing indices have $n>K-1$; the supremum of $(n+1)/n^2$ over these indices tends to zero as $K\to\infty$. Thus the sequence is uniformly integrable, and its integrals converge to $1/2$, as the theorem predicts.

To see the risk interpretation without any infinite-integral issue, let a fixed procedure have risk $R(\theta,\delta)=2$, and suppose a sequence has risks $R(\theta,\delta_n)=2+D_n(\theta)$. Both risks have finite integrals. The fixed procedure's integrated risk is $2$, while $$ \int_0^1 R(\theta,\delta_n)\,d\theta =2+\frac12+\frac1n. $$ The integrated risk difference is $1/2+1/n$, which converges to $\int_0^1D(\theta)\,d\theta=1/2$.

Common Mistakes and Why the Distinction Matters

Assuming pointwise convergence controls the integral. The shrinking-spike example shows that pointwise convergence can coexist with a nonvanishing integral. Check the tails, not just the pointwise limit.
Checking each function's integrability instead of uniform integrability. In the persistent-tail example, every $D_n$ is integrable, but the tail integrals do not become uniformly small.
Subtracting infinite integrated risks. For example, on $(0,1)$, take $R(\theta,\delta_n)=R(\theta,\delta)=1/\theta$. Then $D_n=0$ is uniformly integrable and converges to zero, but both integrated risks are infinite. Their difference is the undefined expression $\infty-\infty$. The valid conclusion is $\int D_n\,d\pi\to\int D\,d\pi$; identifying this with a difference of integrated risks requires those risks to be finite.

In admissibility arguments, integrated risk differences often provide the quantitative link between a pointwise comparison and a prior-based comparison. Uniform integrability supplies a rigorous way to pass to a limit when the prior measure is fixed and finite. It does not by itself establish a risk comparison at every parameter value; it justifies the stated convergence of integrals under its hypotheses.

Key takeaway: Almost-everywhere convergence plus uniform integrability gives convergence of integrals under a finite measure. For risk functions, state the conclusion first in terms of the integral of the difference; rewrite it as a difference of integrated risks only when both integrated risks are finite.

Check Your Understanding

Use the tail definition and the theorem to answer the following.

What does uniform integrability require of the tail integrals, and how does that differ from requiring each function to be integrable?
For $D_n(\theta)=n\mathbf{1}\{0<\theta<1/n\}$ on $(0,1)$, calculate the integral and explain why the sequence is not uniformly integrable.
Why is finiteness of $\pi(\Theta)$ used when applying dominated convergence to the clipped functions in the proof?
If $D_n\to D$ almost everywhere and the sequence is uniformly integrable, what additional condition lets $\int D_n\,d\pi$ be written as the difference of two integrated risks?
In the shrinking-spike example, what happens to the spike's height, width, and integrated contribution as $n$ increases?