Two Ways to Control a Family in \(L^1\)
Uniform integrability can be tested by looking at where functions are large, or by looking at how much integral they can place on sets of small measure. On a finite measure space these viewpoints are equivalent, provided we also require a uniform bound on the \(L^1\) norms. The small-set formulation is useful because it detects concentration even when each individual function is integrable.
Let \((X,\mathcal{M},\mu)\) be a measure space with \(\mu(X)<\infty\), and let \(\mathcal{F}\) be a nonempty family of measurable real- or complex-valued functions in \(L^1(\mu)\). Here \(L^1(\mu)\) means that \(\int_X |f|\,d\mu<\infty\); statements about these integrals depend only on the functions up to equality \(\mu\)-a.e. We use the tail definition of uniform integrability.
The supremum in either condition is over the same family. In particular, the small-set condition is uniform in both the choice of \(f\) and the choice of measurable set \(E\). It controls integrals of absolute values, not signed or complex integrals, so cancellation cannot hide concentration.
The Small-Set Criterion
- \(\mathcal{F}\) is uniformly integrable;
- \(\sup_{f\in\mathcal{F}}\int_X |f|\,d\mu<\infty\), and \(\mathcal{F}\) satisfies the small-set condition.
Proof. Suppose first that \(\mathcal{F}\) is uniformly integrable. By the tail definition, choose \(K\ge 1\) such that $$ \sup_{f\in\mathcal{F}}\int_{\{|f|>K\}}|f|\,d\mu<1. $$ For each \(f\in\mathcal{F}\), split \(X\) into \(\{|f|\le K\}\) and \(\{|f|>K\}\). On the first set \(|f|\le K\), so $$ \int_X |f|\,d\mu \le K\mu(X)+\int_{\{|f|>K\}}|f|\,d\mu < K\mu(X)+1. $$ This bound is independent of \(f\), and \(\mu(X)<\infty\) makes it finite. Thus the \(L^1\) norms are uniformly bounded.
Now fix \(\varepsilon>0\). Uniform integrability gives a \(K\ge1\) such that $$ \sup_{f\in\mathcal{F}}\int_{\{|f|>K\}}|f|\,d\mu<\frac{\varepsilon}{2}. $$ Set \(\delta=\varepsilon/(2K)>0\). For any \(E\in\mathcal{M}\) with \(\mu(E)<\delta\), split the integral over \(E\) according to the same threshold \(K\). The part where \(|f|\le K\) is at most \(K\mu(E)<\varepsilon/2\), and the remaining part is at most the full tail integral. Therefore, for every \(f\in\mathcal{F}\), $$ \int_E |f|\,d\mu \le K\mu(E)+\int_{\{|f|>K\}}|f|\,d\mu <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon. $$ Taking the supremum over \(f\) still gives a value at most the sum of the two uniform bounds, which is strictly less than \(\varepsilon\). This proves the small-set condition.
Conversely, suppose \(C=\sup_{f\in\mathcal{F}}\int_X |f|\,d\mu<\infty\) and the small-set condition holds. If \(C=0\), then every \(\int_X|f|\,d\mu=0\). A nonnegative function with integral zero vanishes \(\mu\)-a.e., so all the tail integrals are zero and the family is uniformly integrable.
Suppose instead that \(C>0\). Fix \(\varepsilon>0\), and use the small-set condition to choose \(\delta>0\) such that \(\mu(E)<\delta\) implies \(\sup_{f\in\mathcal{F}}\int_E|f|\,d\mu<\varepsilon\). Choose \(K>C/\delta\). For each \(f\in\mathcal{F}\), let \(E_f=\{|f|>K\}\), a measurable set. Since \(K\chi_{E_f}\le |f|\), monotonicity of the integral gives $$ K\mu(E_f)\le\int_X |f|\,d\mu\le C, \qquad \mu(E_f)\le\frac{C}{K}<\delta. $$ The small-set condition applies to each \(E_f\), even though that set depends on \(f\). Hence $$ \int_{\{|f|>K\}}|f|\,d\mu =\int_{E_f}|f|\,d\mu <\varepsilon \quad\text{for every }f\in\mathcal{F}. $$ The same \(K\) works for the whole family, so the supremum of the tail integrals is at most \(C/K\) only as a bound on the measures, not on the integrals; the small-set condition gives the required uniform integral bound. Since this argument works for every \(\varepsilon>0\), the tail supremum tends to zero. Thus \(\mathcal{F}\) is uniformly integrable. \(\square\)
The finite-measure hypothesis is used in the first direction when bounding the contribution from \(\{|f|\le K\}\) by \(K\mu(X)\). The reverse direction uses the uniform \(L^1\) bound to make each tail set small, and then uses the small-set condition. These are distinct roles of the two hypotheses.
Worked Applications
Worked Example: A Uniformly Bounded Family
On \([0,1]\) with Lebesgue measure \(m\), let \(\mathcal{F}\) be the family of all measurable real-valued functions satisfying \(|f|\le3\) everywhere. Each is integrable and \(\int_0^1|f|\,dm\le3m([0,1])=3\). For any measurable \(E\subseteq[0,1]\), $$ \int_E|f|\,dm\le 3m(E). $$ Given \(\varepsilon>0\), take \(\delta=\varepsilon/3\). If \(m(E)<\delta\), then \(\sup_{f\in\mathcal{F}}\int_E|f|\,dm\le3m(E)<\varepsilon\). Thus the small-set condition holds. Directly, for every \(K\ge3\), the strict tail set \(\{|f|>K\}\) is empty, so every tail integral is zero. Both formulations confirm uniform integrability.
Check: For \(E=[0,1/12]\), \(m(E)=1/12\). The largest possible integral over \(E\) in this family is \(3(1/12)=1/4\), attained by the constant function \(f=3\). This agrees with the bound \(3m(E)\).
Worked Example: Spikes Have Bounded Norms but Fail Small-Set Control
On \([0,1]\) with Lebesgue measure, define \(f_n=n\chi_{(0,1/n)}\) for integers \(n\ge1\). The interval is open, so its measure is \(1/n\); its endpoints do not affect the integral. Therefore $$ \int_0^1|f_n|\,dm=n\cdot\frac1n=1 $$ for every \(n\), and the \(L^1\) norms are uniformly bounded. But take \(E_n=(0,1/n)\). Then \(m(E_n)=1/n\to0\), while $$ \int_{E_n}|f_n|\,dm=n\cdot\frac1n=1. $$ For example, with \(\varepsilon=1/2\), every proposed \(\delta>0\) admits an integer \(n>1/\delta\), so \(m(E_n)=1/n<\delta\) but \(\int_{E_n}|f_n|\,dm=1\not<1/2\). The small-set condition fails.
The tail test gives the same conclusion. For any \(K>0\), choose an integer \(n>K\). Then \(\{|f_n|>K\}=(0,1/n)\), and its tail integral is \(n(1/n)=1\). Thus the supremum of the tail integrals is at least \(1\) for every \(K\), so it cannot tend to zero. Check: the exact norm computation above gives \(1\), consistent with the constant tail mass on this selected sequence of thresholds.
Worked Example: Unbounded Heights Can Still Be Uniformly Integrable
On \([0,1]\) with Lebesgue measure, set \(g_n=n\chi_{(0,1/n^2)}\) for \(n\ge1\). The interval has measure \(1/n^2\), so $$ \int_0^1|g_n|\,dm=n\cdot\frac1{n^2}=\frac1n. $$ Consequently \(\sup_n\int|g_n|\,dm=1\), attained at \(n=1\). The heights \(n\) are unbounded, but the family is uniformly integrable.
To verify the small-set condition directly, let \(E\subseteq[0,1]\) be measurable and write \(d=m(E)\). Since \(g_n\) is \(n\) on an interval of measure \(1/n^2\) and zero elsewhere, $$ \int_E|g_n|\,dm\le\min\left(nd,\frac1n\right). $$ If \(d=0\), this integral is zero. If \(d>0\) and \(n\le1/\sqrt d\), then \(nd\le\sqrt d\). If \(n>1/\sqrt d\), then \(1/n<\sqrt d\). In either case \(\int_E|g_n|\,dm\le\sqrt d\). Given \(\varepsilon>0\), take \(\delta=\varepsilon^2/4\). Whenever \(m(E)<\delta\), we have \(\sqrt{m(E)}<\varepsilon/2<\varepsilon\), uniformly in \(n\), as required.
For an exact tail check, if \(n\le K\), then \(\{|g_n|>K\}\) is empty. If \(n>K\), it is \((0,1/n^2)\), and its integral is \(1/n\). Hence, for \(K\ge0\), $$ \sup_{n\ge1}\int_{\{|g_n|>K\}}|g_n|\,dm =\frac{1}{\lfloor K\rfloor+1}, $$ where \(\lfloor K\rfloor+1\) is the least integer strictly greater than \(K\). This tends to zero. The example shows why uniform integrability does not require a uniform bound on the pointwise values.
Common Pitfalls and the Role of the Hypotheses
A bounded \(L^1\) norm alone is not enough. The spike family in the previous example has every norm equal to \(1\), but puts integral \(1\) on sets whose measures tend to zero. The small-set condition rules out exactly this concentration. On a finite measure space it also prevents the tail integrals from remaining large, as the theorem proves.
Use absolute values when testing small sets. On \([0,1]\), define $$ h_n=n\chi_{(0,1/(2n))}-n\chi_{(1/(2n),1/n)}. $$ The two intervals are disjoint and each has measure \(1/(2n)\). Thus \(\int_0^1h_n\,dm=n/(2n)-n/(2n)=0\), which might misleadingly suggest there is little mass. But for \(E_n=(0,1/n)\), the signed integral is zero while $$ \int_{E_n}|h_n|\,dm =n\frac1{2n}+n\frac1{2n}=1. $$ Since \(m(E_n)=1/n\to0\), the small-set condition for \(|h_n|\) fails. Uniform integrability concerns the magnitude of the functions, not cancellation in their signed integrals.
Finiteness cannot be omitted from the stated equivalence. On \((\mathbb{R},\mathcal{L},m)\), let \(q_n=\chi_{[0,n]}\). Each \(q_n\) is integrable, and for every \(K\ge1\), the strict tail set \(\{|q_n|>K\}\) is empty. Thus the family satisfies the tail definition, but \(\int_{\mathbb{R}}|q_n|\,dm=n\), which is not uniformly bounded. The finite-measure proof needs \(\mu(X)<\infty\) to control the integral where the functions are below a fixed threshold.
The criterion is therefore a practical two-part test on finite measure spaces: first bound the total \(L^1\) mass uniformly, then ensure that no member of the family can concentrate a substantial amount of that mass on an arbitrarily small measurable set.
Check Your Understanding
For each question, keep the supremum over the family uniform in both the function and the measurable set.
- In the forward implication of the theorem, where exactly is \(\mu(X)<\infty\) used?
- For \(u_n=n^2\chi_{(0,1/n^3)}\) on \([0,1]\), compute \(\int|u_n|\,dm\) and determine whether the tail integrals tend to zero uniformly.
- Why does a uniform \(L^1\) bound imply that the sets \(\{|f|>K\}\) have uniformly small measure when \(K\) is large?
- Give a concrete reason that testing \(\int_E f\,d\mu\), rather than \(\int_E|f|\,d\mu\), can fail to detect concentration.
- Which direction of the small-set criterion can be proved without assuming that the whole measure space has finite measure?