Reading Uniform Integrability from the Tails
Integrability of each function in a family is an individual condition: for every member, its integral is finite. Uniform integrability asks for control that holds across the whole family at once. The tail formulation measures how much integral can lie where a function has large magnitude, using the same threshold for every member. This makes the order of the quantifiers essential: first choose one threshold \(K\), then take the supremum over the family.
Throughout, \((X,\mathcal{M},\mu)\) is a measure space, not assumed finite, complete, or sigma-finite unless stated. Functions may be real-valued or complex-valued and are \(\mathcal{M}\)-measurable; \(|f|\) denotes absolute value or complex modulus. The space \(L^1(\mu)\) consists of almost-everywhere equivalence classes of such functions with \(\int_X |f|\,d\mu<\infty\), and \(\|f\|_1=\int_X|f|\,d\mu\). Tail integrals do not depend on the chosen representative, since changing a function on a null set does not change the integral.
The uniform \(L^1\) bound is included in the definition on a general measure space. On a finite measure space, however, the tail condition by itself forces this bound, as proved below. Thus on finite measure spaces one often uses the tail condition alone as an equivalent definition. Do not make that simplification on an arbitrary measure space.
Basic Tail Computations
Worked Example: The Unit Ball in \(L^\infty[0,1]\)
Let \(X=[0,1]\), let \(\mathcal{M}\) be the Lebesgue sigma-algebra, and let \(\mu=m\). Take \(\mathcal{F}\) to be the family of all real- or complex-valued measurable functions \(f\) satisfying \(|f|\le1\) almost everywhere. Each belongs to \(L^1(m)\), since \(\int_0^1|f|\,dm\le1\).
If \(K\ge1\), then \(\{|f|>K\}\) is null for every \(f\in\mathcal{F}\). Therefore \(T_{\mathcal{F}}(K)=0\). If \(0\le K<1\), the bound \(|f|\le1\) gives \(\int_{\{|f|>K\}}|f|\,dm\le1\), while the constant function \(f(x)=1\) belongs to the family and has tail integral \(\int_0^1 1\,dm=1\). Hence
Also \(\sup_{f\in\mathcal{F}}\|f\|_1=1\), attained by the constant function \(1\). The family is uniformly integrable. At \(K=1\), the strict inequality in the tail definition matters: points where \(|f|=1\) are not in \(\{|f|>1\}\).
Worked Example: Scalar Multiples of an Indicator
On \((\mathbb{R},\mathcal{L},m)\), let \(E=[0,2]\) and \(\mathcal{F}=\{a\chi_E:-3\le a\le3\}\), a family of real-valued functions in \(L^1(m)\). For \(f=a\chi_E\), its magnitude is \(|a|\) on \(E\) and zero off \(E\). If \(|a|\le K\), the strict superlevel set is empty; if \(|a|>K\), it is \(E\). In the latter case the tail integral is \(|a|m(E)=2|a|\).
For \(0\le K<3\), the supremum over \(-3\le a\le3\) is \(2\cdot3=6\), attained at \(a=3\) (and also at \(a=-3\)). For \(K\ge3\), no parameter satisfies \(|a|>K\), so every tail integral is zero. Consequently, \(T_{\mathcal{F}}(K)=6\) for \(0\le K<3\) and \(T_{\mathcal{F}}(K)=0\) for \(K\ge3\). The norm supremum is \(\sup_{a}2|a|=6\). These computations verify uniform integrability, including the endpoint \(K=3\), where the strict tail is empty.
Check: at \(K=2\), \(a=3\) gives \(\int_{\{|3\chi_E|>2\}}|3\chi_E|\,dm=3m([0,2])=3\cdot2=6\). At \(K=3\), the set \(\{|3\chi_E|>3\}\) is empty, so its integral is \(0\).
What the Tail Condition Does and Does Not Imply
Proof. Since \(T_{\mathcal{F}}(K)\to0\), there is a finite \(K_0\ge0\) such that \(T_{\mathcal{F}}(K_0)\le1\). For each \(f\in\mathcal{F}\), split \(X\) into the disjoint sets \(\{|f|\le K_0\}\) and \(\{|f|>K_0\}\). On the first, \(|f|\le K_0\); on the second, the integral is at most \(T_{\mathcal{F}}(K_0)\). Thus $$ \|f\|_1 =\int_{\{|f|\le K_0\}}|f|\,d\mu+\int_{\{|f|>K_0\}}|f|\,d\mu \le K_0\mu(X)+1. $$ The right side is finite and independent of \(f\), because \(\mu(X)<\infty\). Taking the supremum over \(f\) proves the uniform norm bound. The converse follows directly from the definition of uniform integrability, which includes tail decay. \(\square\)
Worked Example: Vanishing Tails Without a Uniform Norm Bound
Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with counting measure and its full sigma-algebra. For each \(n\ge1\), define the real-valued function \(f_n=\chi_{\{1,\ldots,n\}}\), and let \(\mathcal{F}=\{f_n:n\ge1\}\). Each \(f_n\) is integrable and \(\|f_n\|_1=n\), so \(\sup_n\|f_n\|_1=\infty\).
If \(K\ge1\), then \(f_n\) only takes the values \(0\) and \(1\), so \(\{|f_n|>K\}=\varnothing\) and \(T_{\mathcal{F}}(K)=0\). If \(0\le K<1\), then \(\{|f_n|>K\}=\{1,\ldots,n\}\), whose counting measure is \(n\); the tail integral is \(\sum_{j=1}^n1=n\). Taking the supremum over \(n\) gives \(T_{\mathcal{F}}(K)=\infty\). In particular, \(T_{\mathcal{F}}(K)\to0\), but \(\mathcal{F}\) is not uniformly integrable because its \(L^1\) norms are unbounded.
This also pinpoints a common error: “the tail integral is eventually zero, so the family is uniformly integrable” is not valid on a general measure space. Here the tail is eventually zero, but the mass below the threshold grows without bound. The correction is to check both requirements in the definition: uniform \(L^1\) boundedness and uniform tail decay. This family cannot occur as a counterexample on a finite measure space, by the proposition.
Individual Tails and Finite Families
For one integrable function, the tail integral always tends to zero. This pointwise-in-the-family fact is weaker than the uniform condition: it permits the threshold required for a small tail to depend on the function.
Proof. For positive integers \(n\), define \(g_n=|f|\chi_{\{|f|>n\}}\). These are measurable and satisfy \(0\le g_n\le|f|\), where \(|f|\) is integrable. At every point where \(|f|\) is finite, \(g_n\to0\); an integrable function is finite almost everywhere. The Dominated Convergence Theorem, with integrable dominator \(|f|\), gives \(\int g_n\,d\mu\to0\). For real \(K\to\infty\), choose an integer \(n\le K<n+1\). Then \(\{|f|>K\}\subseteq\{|f|>n\}\), so $$ 0\le\int_{\{|f|>K\}}|f|\,d\mu\le\int_{\{|f|>n\}}|f|\,d\mu. $$ As \(K\to\infty\), the chosen integers \(n\) tend to infinity, and the right side tends to zero. This proves the assertion. \(\square\)
For a finite family, the individual result does become uniform. The finite number of thresholds can be replaced by their maximum; this step is not available for an arbitrary infinite family.
Proof. Write the nonempty finite family as \(\{f_1,\ldots,f_N\}\), allowing repetitions if necessary, and set \(B=\max_{1\le j\le N}\|f_j\|_1<\infty\). Given \(\varepsilon>0\), the preceding proposition gives, for each \(j\), a finite \(K_j\) such that \(\int_{\{|f_j|>K_j\}}|f_j|\,d\mu<\varepsilon\). Set \(K=\max_{1\le j\le N}K_j\). Since \(K\ge K_j\), \(\{|f_j|>K\}\subseteq\{|f_j|>K_j\}\), and hence its tail integral is less than \(\varepsilon\) for every \(j\). Taking the maximum over the finite family gives \(T_{\mathcal{F}}(K)\le\varepsilon\). Because tail integrals are nonincreasing in \(K\), this proves \(T_{\mathcal{F}}(K)\to0\). Together with the bound \(B\), this is uniform integrability. \(\square\)
How to Check a Family
When testing a proposed family, keep the common threshold visible throughout the calculation. For each fixed \(K\), determine the tail integral for a general member, then take the supremum over all members at that same \(K\). Only after that supremum is computed should \(K\) tend to infinity. The following distinction is useful:
| Question | What to verify |
|---|---|
| Are all members integrable? | For every \(f\in\mathcal{F}\), \(\int|f|\,d\mu<\infty\). |
| Are the \(L^1\) norms uniformly bounded? | \(\sup_{f\in\mathcal{F}}\int|f|\,d\mu<\infty\). |
| Do the tails vanish uniformly? | \(\lim_{K\to\infty}\sup_{f\in\mathcal{F}}\int_{\{|f|>K\}}|f|\,d\mu=0\). |
| Is the space finite? | If \(\mu(X)<\infty\), tail decay implies the uniform norm bound; otherwise check it separately. |
The definition concerns the magnitude \(|f|\), so cancellation between positive and negative values, or between complex phases, does not reduce the tail integral. It also concerns integral mass rather than just the measure of the tail set: a small set can still carry a large integral if function values there are large. These distinctions are why uniform integrability is stronger than saying that every member is integrable or that each member has a tail tending to zero on its own.
Check Your Understanding
In each question, keep the strict tail set \(\{|f|>K\}\) and the supremum over the whole family in view.
- On a finite measure space, prove that tail decay implies a uniform \(L^1\) bound by splitting each integral at one fixed threshold.
- For the family \(\{a\chi_{[0,2]}:-3\le a\le3\}\), compute the tail integral at \(K=3\). Why does the answer depend on using \(>\), rather than \(\ge\)?
- For \(f_n=\chi_{\{1,\ldots,n\}}\) under counting measure, compute \(T_{\mathcal{F}}(K)\) separately for \(0\le K<1\) and \(K\ge1\), and state which condition for uniform integrability fails.
- Why does the individual tail limit for each \(f\in L^1(\mu)\) not by itself imply uniform tail decay for an infinite family?
- Show that any finite family of \(L^1(\mu)\) functions has uniformly vanishing tails, and identify where finiteness of the family enters the proof.