Cancellation Does Not Control Uniform Integrability

Cancellation and Absolute Tails Measure Different Things

A signed integral records the net balance of positive and negative contributions. A complex integral likewise allows contributions with different phases to cancel. Uniform integrability asks a different question: whether the total absolute mass on regions where the function is large becomes uniformly small. Thus, an integral equal to zero does not by itself show that a family has small tails—or even that its absolute integrals are bounded.

Throughout the examples, the measure space is $([0,1],\mathcal{L}_{[0,1]},m)$, where $\mathcal{L}_{[0,1]}$ is the Lebesgue sigma-algebra on $[0,1]$. This is a complete finite measure space, with $m([0,1])=1$. The functions in the first examples are real-valued; the complex example is explicitly identified. All integrals are Lebesgue integrals, and statements about pointwise convergence specify whether they hold everywhere.

Definition: For a family $\mathcal{F}$ of measurable, integrable real- or complex-valued functions on a measure space $(X,\mathcal{M},\mu)$, uniform integrability in the tail-integral sense means $$ \lim_{K\to\infty}\ \sup_{f\in\mathcal{F}}\int_{\{|f|>K\}}|f|\,d\mu=0. $$ Here $|f|$ is absolute value for real-valued $f$ and modulus for complex-valued $f$, and the tail event uses the strict inequality $|f|>K$. This definition tests absolute mass, not the signed or complex integral over the tail.

The cancellation inequality goes in only one direction: for any integrable real- or complex-valued $f$ on $(X,\mathcal{M},\mu)$, $\left|\int_X f\,d\mu\right|\leq\int_X|f|\,d\mu$. Small absolute integral therefore forces a small signed or complex integral, but a small signed or complex integral need not force a small absolute integral. The examples below make the gap exact.

Equal Positive and Negative Spikes

Worked Example: Zero Integral with a Fixed Absolute Tail

For each integer $n\geq1$, define a real-valued function on $[0,1]$ by $$ f_n(x)= \begin{cases} n,&0<x<\dfrac{1}{2n},\\ -n,&\dfrac{1}{2n}<x<\dfrac{1}{n},\\ 0,&\text{otherwise}. \end{cases} $$ Each $f_n$ is measurable and integrable. The two nonzero intervals are disjoint and each has measure $1/(2n)$. Their endpoints are assigned value zero, which does not change any integral.

Compute the signed integral by treating the two intervals separately: $$ \int_{[0,1]} f_n\,dm =n\cdot\frac{1}{2n}+(-n)\cdot\frac{1}{2n} =\frac12-\frac12=0. $$ In contrast, the absolute integral adds the two contributions: $$ \int_{[0,1]}|f_n|\,dm =n\cdot\frac{1}{2n}+n\cdot\frac{1}{2n} =\frac12+\frac12=1. $$

Fix a threshold $K\geq0$. If $n\leq K$, no point satisfies $|f_n|>K$, so the tail integral is zero. If $n>K$, both nonzero intervals satisfy $|f_n|=n>K$, and $$ \int_{\{|f_n|>K\}}|f_n|\,dm=1. $$ For every fixed $K$, there is an integer $n>K$. Consequently $$ \sup_{n\geq1}\int_{\{|f_n|>K\}}|f_n|\,dm=1, $$ so the family $\{f_n:n\geq1\}$ is not uniformly integrable.

Check: adding the two signed contributions gives $1/2-1/2=0$; adding the two absolute contributions gives $1/2+1/2=1$. For the strict tail condition, $n=K$ contributes nothing when $K$ is an integer, but any integer $n>K$ contributes the full value $1$, as required.

This example also has convergence properties that do not repair the failure. For every $x\in[0,1]$, $f_n(x)\to0$: at $x=0$ every value is zero, and for $x>0$, once $n>1/x$, the support $(0,1/n)$ lies below $x$. The measure of the set where $f_n\ne0$ is $1/n$; hence, for every $\varepsilon>0$, $m(\{|f_n|>\varepsilon\})\leq1/n\to0$. Thus $f_n\to0$ everywhere and in measure, while the absolute tails still fail the uniform-integrability test.

An Exact Tail Formula for Balanced Spikes

Proposition: On $([0,1],\mathcal{L}_{[0,1]},m)$, let $a_n>0$, and let $A_n,B_n\in\mathcal{L}_{[0,1]}$ be disjoint sets with $m(A_n)=m(B_n)=\delta_n$. Define the real-valued functions $g_n=a_n\chi_{A_n}-a_n\chi_{B_n}$. For each $K\geq0$, $$ \int_{\{|g_n|>K\}}|g_n|\,dm = \begin{cases} 2a_n\delta_n,&a_n>K,\\ 0,&a_n\leq K. \end{cases} $$ Therefore $\{g_n:n\geq1\}$ is uniformly integrable exactly when $$ \lim_{K\to\infty}\sup_{n\geq1}\bigl(2a_n\delta_n\,\chi_{\{a_n>K\}}\bigr)=0. $$

Proof. On $A_n$, $g_n=a_n$; on $B_n$, $g_n=-a_n$; and outside $A_n\cup B_n$, $g_n=0$. Thus $|g_n|=a_n$ on $A_n\cup B_n$ and is zero elsewhere. If $a_n\leq K$, the strict inequality $|g_n|>K$ holds nowhere. If $a_n>K$, the tail set is $A_n\cup B_n$. Since these sets are disjoint and each has measure $\delta_n$, the tail integral is $a_n(m(A_n)+m(B_n))=2a_n\delta_n$. Taking the supremum over $n$ gives the displayed expression, and its limit is precisely the tail-integral definition of uniform integrability for this family. The formula includes $\delta_n=0$, when both sets have measure zero and the tail integral is zero. $\square$

Worked Example: Larger Amplitudes Hide Unbounded Absolute Mass

Use $A_n=(0,1/(2n))$, $B_n=(1/(2n),1/n)$, and $a_n=n^2$ in the proposition. These sets are disjoint, each has measure $\delta_n=1/(2n)$, and they lie in $[0,1]$. The real-valued function $g_n=n^2\chi_{A_n}-n^2\chi_{B_n}$ is measurable and integrable for each $n$.

The two signed contributions are $$ \int_{A_n}g_n\,dm=n^2\cdot\frac{1}{2n}=\frac n2, \qquad \int_{B_n}g_n\,dm=-n^2\cdot\frac{1}{2n}=-\frac n2. $$ Therefore $\int_{[0,1]}g_n\,dm=n/2-n/2=0$. But the absolute integral is $$ \int_{[0,1]}|g_n|\,dm =n^2\cdot\frac{1}{2n}+n^2\cdot\frac{1}{2n} =\frac n2+\frac n2=n. $$

For a fixed $K\geq0$, whenever $n^2>K$, the strict tail $\{|g_n|>K\}$ contains both intervals, and its absolute integral is $n$. There are integers $n$ with $n^2>K$ and arbitrarily large $n$, so the supremum of the tail integrals is infinite for every fixed $K$. In particular it does not tend to zero; the family is not uniformly integrable. The support has measure $m(A_n\cup B_n)=1/n$, which tends to zero, but the absolute mass on that support is $n$.

Check: substituting $a_n=n^2$ and $\delta_n=1/(2n)$ in the proposition gives $2a_n\delta_n=2n^2/(2n)=n$, matching the direct absolute-integral calculation. The signed integral remains zero because the two contributions are $n/2$ and $-n/2$.

Complex Cancellation Works the Same Way

Worked Example: Three Complex Phases Cancel

Let $\omega=-\tfrac12+i\tfrac{\sqrt3}{2}$. Its square is $\omega^2=-\tfrac12-i\tfrac{\sqrt3}{2}$, so $1+\omega+\omega^2=1-\tfrac12-\tfrac12+i(\tfrac{\sqrt3}{2}-\tfrac{\sqrt3}{2})=0$. Also $|1|=|\omega|=|\omega^2|=1$. Partition $(0,1/n)$ into three consecutive open intervals $I_{n,1},I_{n,2},I_{n,3}$, each of length $1/(3n)$, and define the complex-valued measurable function $$ h_n=n\chi_{I_{n,1}}+n\omega\chi_{I_{n,2}}+n\omega^2\chi_{I_{n,3}}, $$ with value zero elsewhere in $[0,1]$.

Integrating each constant value over its interval gives $$ \int_{[0,1]}h_n\,dm =n\cdot\frac{1}{3n} +n\omega\cdot\frac{1}{3n} +n\omega^2\cdot\frac{1}{3n} =\frac{1+\omega+\omega^2}{3}=0. $$ The modulus on each interval is $n$, since each phase has modulus one. Thus $$ \int_{[0,1]}|h_n|\,dm =n\cdot\frac{1}{3n}+n\cdot\frac{1}{3n}+n\cdot\frac{1}{3n}=1. $$ For $K\geq0$, the strict tail integral is $1$ if $n>K$, and zero if $n\leq K$. As in the real example, its supremum over $n$ is $1$ for every $K$, so the family is not uniformly integrable.

Check: the three interval lengths sum to $3/(3n)=1/n$, and multiplying by the common modulus $n$ gives absolute mass $1$. The complex phases cancel in the integral, but not in the modulus.

What Cancellation Can and Cannot Establish

The essential distinction is between the integral of a function and the integral of its absolute value. For each $n$, the equation $\int f_n\,dm=0$ gives no upper bound on $\int|f_n|\,dm$: the fixed-mass example has absolute integral $1$, while the larger-amplitude example has absolute integral $n$. More importantly for uniform integrability, even a fixed absolute integral is not enough. Its mass must move out of the strict tails uniformly as $K$ increases, and the first and complex examples show that this need not happen.

A common mistake is to calculate the signed tail and treat it as the tail quantity in the definition. For the first example, when $n>K$, the incorrect calculation is $$ \int_{\{|f_n|>K\}}f_n\,dm =\frac12-\frac12=0, $$ followed by the false conclusion that the tail is zero. The definition instead requires $$ \int_{\{|f_n|>K\}}|f_n|\,dm =\frac12+\frac12=1. $$ The same correction applies to complex-valued functions: tail control concerns $|f|$, not the real or complex number $\int_{\{|f|>K\}}f\,dm$.

There is a useful contrast for nonnegative functions. On any measure space $(X,\mathcal{M},\mu)$, if a measurable function $u:X\to[0,\infty]$ has $\int_Xu\,d\mu=0$, then $u=0$ $\mu$-almost everywhere. This statement relies on nonnegativity, which prevents cancellation.

Proof. For each integer $k\geq1$, the measurable set $E_k=\{x:u(x)>1/k\}$ satisfies $(1/k)\chi_{E_k}\leq u$. Monotonicity of the nonnegative Lebesgue integral therefore gives $(1/k)\mu(E_k)\leq\int_Xu\,d\mu=0$, so $\mu(E_k)=0$. Since $u\geq0$, the set $\{u>0\}$ equals $\bigcup_{k\geq1}E_k$: if $u(x)>0$, some integer $k$ has $1/k<u(x)$, and membership in any $E_k$ implies $u(x)>0$. Countable subadditivity gives $\mu(\{u>0\})=0$, which is exactly $u=0$ $\mu$-almost everywhere. $\square$

For signed and complex-valued functions, that argument cannot be applied to the function itself: positive and negative values, or different complex phases, can cancel. When the question is uniform integrability, use the absolute-value tail from the definition and compute it directly.

Check Your Understanding

For each question, distinguish the signed or complex integral from the absolute tail integral.

For the sequence $f_n$ in the first example, compute $\int_{\{|f_n|>K\}}f_n\,dm$ and $\int_{\{|f_n|>K\}}|f_n|\,dm$ when $n>K$. Why do the two answers differ?
Suppose balanced spikes have amplitude $a_n$ and equal lobe measures $\delta_n$. Write the exact tail integral at a fixed threshold $K$, including the case $a_n=K$.
For the complex sequence $h_n$, verify separately that the three phases sum to zero and that each has modulus one. Which calculation determines its uniform-integrability tail?
Can a sequence converge everywhere to zero and in measure while failing to be uniformly integrable? Identify the sequence above and give its tail-integral supremum.
Why does the conclusion that zero integral forces a function to vanish almost everywhere require nonnegativity in the proved contrast?