Tail Integrals as a Direct Test
Uniform integrability asks whether the integral carried by large function values becomes uniformly small across a family. For sequences with spikes or truncations, the tail can often be computed exactly, making this definition a practical test rather than an abstract condition. The strict tail set matters: at a threshold \(K\), it is \(\{|f_n|>K\}\), not \(\{|f_n|\ge K\}\).
We work on the finite measure space \(([0,1],\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m([0,1])=1\). The functions below are real-valued, measurable, and integrable. For a family \(\mathcal{F}\) of such functions, uniform integrability means that its tail integrals vanish uniformly:
In the examples, the functions are nonnegative, so the absolute-value signs can be omitted. Each integral is with respect to Lebesgue measure on \([0,1]\). Endpoints of intervals do not affect these integrals because singleton sets have Lebesgue measure zero.
Indicator Spikes
Worked Example: A Spike with Fixed Tail Mass
For each positive integer \(n\), let \(f_n=n\chi_{(0,1/n)}\) on \([0,1]\). The value of \(f_n\) is \(n\) on \((0,1/n)\) and \(0\) elsewhere. Fix \(K>0\). If \(n\le K\), then neither value exceeds \(K\), so the tail set is empty. If \(n>K\), the tail set is \((0,1/n)\). Thus, with the strict threshold,
For every fixed \(K\), there is a positive integer \(n>K\). Therefore \(\sup_n\int_{\{f_n>K\}}f_n\,dm=1\), for every \(K>0\), and this family is not uniformly integrable. Check: for \(K=3\), \(n=3\) gives an empty tail because the function equals \(K\) on its support; \(n=4\) gives tail integral \(4\cdot(1/4)=1\). The supremum is indeed 1.
The calculation illustrates why bounded total integrals do not settle the tail test: here \(\int_0^1 f_n\,dm=1\) for every \(n\), but that mass remains above arbitrarily large thresholds for suitable members of the family. The conclusion follows from the definition directly; no convergence theorem is needed.
Capped Powers and the Domain Endpoint
For \(\alpha>0\) and positive integer \(n\), define \(f_n(x)=\min(n,x^{-\alpha})\) for \(0<x\le1\), and set \(f_n(0)=n\). These are nonnegative measurable functions bounded above by \(n\), hence integrable on \([0,1]\). We derive their tails, taking care that the integration interval cannot extend beyond the domain.
Proof. If \(n\le K\), then \(f_n(x)\le n\le K\) everywhere, so \(\{f_n>K\}\) is empty. Suppose \(n>K\). The capped part of \(f_n\) occurs for \(0\le x\le n^{-1/\alpha}\), where \(f_n=n\). Since \(n>K\), this part lies in the strict tail and contributes \(n\,m((0,n^{-1/\alpha}))=n^{1-1/\alpha}\). For \(x>n^{-1/\alpha}\), \(f_n(x)=x^{-\alpha}\); this exceeds \(K\) precisely when \(x<K^{-1/\alpha}\). Intersecting with the domain \([0,1]\) gives the remaining interval from \(n^{-1/\alpha}\) to \(\min(1,K^{-1/\alpha})\), proving the integral formula. When \(K\ge1\), the upper endpoint is \(K^{-1/\alpha}\). For \(\alpha=1\), the second integral is \(\log(n/K)\), and the first contribution is 1. For \(\alpha\ne1\), put \(\beta=(\alpha-1)/\alpha\). The second integral is \((n^\beta-K^\beta)/(\alpha-1)\), while the first is \(n^\beta\); their sum is \((\alpha n^\beta-K^\beta)/(\alpha-1)\). When \(0<K<1\), the upper endpoint is 1. The second integral is \((1-n^\beta)/(1-\alpha)\), while the first is \(n^\beta\); their sum is \((\alpha n^\beta-1)/(\alpha-1)\). For \(\alpha=1\) this second integral is \(\log n\), giving \(1+\log n\). These cases establish the formulas. \(\square\)
Worked Example: Computing a Capped-Power Tail
Take \(\alpha=\tfrac12\), \(n=16\), and \(K=4\). Then \(f_{16}(x)=\min(16,x^{-1/2})\). The cap changes to the power at \(x=16^{-2}=1/256\), and \(x^{-1/2}>4\) exactly when \(x<1/16\). The strict tail therefore has a capped part \((0,1/256)\) and a power part \((1/256,1/16)\). Their integrals are
The full tail integral is \(\frac1{16}+\frac38=\frac1{16}+\frac6{16}=\frac7{16}\). Check: the general formula gives \(\bigl(\tfrac12\cdot16^{-1}-4^{-1}\bigr)/(-\tfrac12) =(\tfrac1{32}-\tfrac14)/(-\tfrac12)=\frac7{16}\). The value \(\frac38\) is only the second, power-function contribution; omitting the capped part would undercount the tail.
We can now classify the whole family for each exponent, using the same threshold \(K\) for all \(n\).
Proof. First suppose \(0<\alpha<1\). For \(K\ge1\), membership in \(\{f_n>K\}\) implies \(x^{-\alpha}>K\), hence \(x<K^{-1/\alpha}\), and \(f_n(x)\le x^{-\alpha}\) for \(x>0\). Therefore, for every \(n\),
The integral is finite because \(\alpha<1\); its value tends to zero as \(K\to\infty\), since \((\alpha-1)/\alpha<0\). The bound is independent of \(n\), so the family is uniformly integrable.
Now suppose \(\alpha=1\). For any fixed \(K\ge1\), choose integers \(n>K\) and let \(n\to\infty\). The tail formula gives \(1+\log(n/K)\), which tends to infinity. Thus the supremum of the tail integrals is infinite at each such \(K\), and the family is not uniformly integrable. Finally suppose \(\alpha>1\), so \(\beta=(\alpha-1)/\alpha>0\). At any fixed \(K\ge1\), for integers \(n>K\) the formula is \((\alpha n^\beta-K^\beta)/(\alpha-1)\), which tends to infinity as \(n\to\infty\). Again the supremum does not tend to zero. All positive \(\alpha\) have been covered, proving both directions. \(\square\)
Check: for \(\alpha=\tfrac12\), \(K=4\), the uniform bound just proved is \(4^{-1}/(1/2)=1/2\), and the worked tail value \(7/16\) is below it. The bound also decreases to zero as \(K\) increases.
Constant Functions and a Common Endpoint Error
Worked Example: Families of Constants
Let \(g_n(x)=c_n\) on \([0,1]\), where \(c_n\ge0\). If \(c_n\le K\), the strict tail set is empty. If \(c_n>K\), the tail set is all of \([0,1]\) and its integral is \(c_n m([0,1])=c_n\). Thus
For \(c_n=n\), every fixed \(K\) admits integers \(n>K\), and the corresponding tail integrals \(n\) are unbounded. This family is not uniformly integrable. In contrast, for \(c_n=1+1/n\), we have \(c_n\le2\) for every positive integer \(n\); at every \(K\ge2\), all tail sets are empty. This family is uniformly integrable. In fact, for nonnegative constant functions on this unit interval, the family is uniformly integrable exactly when the constants are bounded: a common bound makes all tails empty at larger thresholds, while an unbounded sequence gives unbounded tail integrals at every fixed threshold.
A frequent error in capped-power calculations is to use \(K^{-1/\alpha}\) as the upper endpoint even when it lies outside \([0,1]\). For example, with \(\alpha=1\), \(n=1\), and \(K=\tfrac12\), the function \(f_1(x)=\min(1,x^{-1})\) equals 1 throughout \([0,1]\). Since \(1>\tfrac12\), its tail integral is \(1\). The expression \(1+\log(n/K)=1+\log2\) is wrong here: it integrates the power part past the domain endpoint. The correction is to use \(\min(1,K^{-1/\alpha})\). Since \(K<1\) in this example, the upper endpoint is 1, and the correct formula is \(1+\log n=1\).
These examples show how the tail test separates three mechanisms: a spike can keep a fixed amount of mass above every threshold; a capped power can have tails that vanish, persist, or grow depending on its exponent; and a constant family is governed by whether its values have a common bound. In every case, calculate the tail at one common threshold before taking the supremum over the family.
Check Your Understanding
Use the strict tail set and, whenever relevant, keep the integration endpoint inside \([0,1]\).
- For \(h_n=n\chi_{(0,1/n)}\), calculate \(\int_{\{h_n>5\}}h_n\,dm\) for \(n=5\) and \(n=6\). What does this show about the strict threshold?
- For \(f_n(x)=\min(n,x^{-1/2})\), use the uniform bound to estimate \(\sup_n\int_{\{f_n>9\}}f_n\,dm\).
- For the capped-power family with \(\alpha=1\), compute the tail integral when \(K=\tfrac12\) and \(n=3\), using the correct endpoint.
- Explain why the family of constants \(g_n(x)=\sqrt n\) is not uniformly integrable, even though each \(g_n\) is integrable.
- For which positive values of \(\alpha\) is the family \(\min(n,x^{-\alpha})\) uniformly integrable? Identify the tail estimate that proves the positive case.