Why an \(L^p\) Bound Controls Large Values
A family of functions can have values that become arbitrarily large and still be uniformly integrable, provided the regions where those large values occur are sufficiently small. The useful feature of an \(L^p\) bound with \(p>1\) is that it makes the integral over a large-value tail small, uniformly across the family. The estimate is quantitative: it specifies how fast the tail integral decreases as the threshold increases.
Throughout, let \((X,\mathcal{M},\mu)\) be a finite measure space, so \(\mu(X)<\infty\), and let \(\mathcal{F}\) be a family of measurable real-valued functions \(f:X\to\mathbb{R}\). Completeness of the measure is not required. Integrals of absolute values are with respect to \(\mu\), and all bounds below hold for every member of the family.
For \(1\le p<\infty\), \(f\in L^p(\mu)\) means that \(f\) is measurable and \(\int_X|f|^p\,d\mu<\infty\), with \(\|f\|_p=(\int_X|f|^p\,d\mu)^{1/p}\). The result here requires a uniform bound for some exponent strictly greater than \(1\).
The Tail Estimate
Proof. If \(C=0\), then for every \(f\in\mathcal{F}\), \(\int_X|f|^p\,d\mu=0\). A nonnegative measurable function with integral zero is zero \(\mu\)-almost everywhere, so every tail integral is zero. The conclusion holds.
Suppose now that \(C>0\). Fix \(f\in\mathcal{F}\) and \(K>0\). On the measurable set \(\{|f|>K\}\), the exponent \(1-p\) is negative. Therefore \(|f|^{1-p}\le K^{1-p}\), and hence $$ |f|=|f|^p|f|^{1-p}\le K^{1-p}|f|^p. $$ Integrating this pointwise inequality over the tail set gives $$ \int_{\{|f|>K\}}|f|\,d\mu \le K^{1-p}\int_{\{|f|>K\}}|f|^p\,d\mu \le C K^{1-p}. $$ The final bound does not depend on \(f\), so it remains true after taking the supremum over \(f\in\mathcal{F}\). Since \(p>1\), \(K^{1-p}\to0\) as \(K\to\infty\); thus the family is uniformly integrable. \(\square\)
The bound also gives a direct way to choose a threshold. Given \(\varepsilon>0\), if \(C>0\), take \(K=(C/\varepsilon)^{1/(p-1)}\). Then \(C K^{1-p}=\varepsilon\), so every tail integral is at most \(\varepsilon\) at this same threshold. If a strict bound smaller than \(\varepsilon\) is wanted, choose any larger \(K\).
There is a second quantitative consequence on a finite measure space. For any measurable \(E\in\mathcal{M}\), Hölder's inequality, with conjugate exponent \(p'=p/(p-1)\), gives $$ \int_E |f|\,d\mu\le \left(\int_X|f|^p\,d\mu\right)^{1/p}\mu(E)^{1/p'} \le C^{1/p}\mu(E)^{1/p'}. $$ Thus, when \(C>0\), any \(\delta=(\varepsilon/C^{1/p})^{p'}\) ensures \(\int_E|f|\,d\mu\le\varepsilon\) for every \(f\in\mathcal{F}\) whenever \(\mu(E)\le\delta\). This is the small-set form of uniform control; the tail estimate above gives more specific information about where large values occur.
Worked Example: A Uniform \(L^p\) Bound on a Finite Space
Let \(X=\{a,b\}\) with counting measure, and consider \(f_1(a)=3,\ f_1(b)=0\) and \(f_2(a)=0,\ f_2(b)=4\). For \(p=2\), compute a common \(L^2\) bound and test the tail estimate at \(K=3\).
For \(f_1\), the squared \(L^2\) integral is \(3^2+0^2=9\). For \(f_2\), it is \(0^2+4^2=16\). Hence \(\sup_{f\in\{f_1,f_2\}}\int_X|f|^2\,d\mu=16\), so we may take \(C=16\). At \(K=3\), the strict tail set for \(f_1\) is empty because \(3>3\) is false; its tail integral is \(0\). The strict tail set for \(f_2\) is \(\{b\}\), and its tail integral is \(4\cdot1=4\). The theorem's common bound is \(C K^{1-p}=16\cdot3^{-1}=16/3\), and \(4\le16/3\).
Check: At \(K=4\), neither function has a point with \(|f_j|>4\), so both tail integrals are \(0\). In particular, the value \(4\) itself is excluded by the strict tail condition.
Worked Example: A Truncated Power Function
On \(X=(0,1)\) with Lebesgue measure, let \(f_n(x)=x^{-1/2}\chi_{[1/n,1]}(x)\), for integers \(n\ge2\). Verify a uniform \(L^{3/2}\) bound and calculate the tail for \(K\ge1\).
Since \(f_n\) vanishes outside \([1/n,1]\), $$ \int_0^1|f_n(x)|^{3/2}\,dx =\int_{1/n}^1 x^{-3/4}\,dx =\left[4x^{1/4}\right]_{1/n}^1 =4(1-n^{-1/4})\le4. $$ Thus the theorem applies with \(p=3/2\) and \(C=4\). Its uniform estimate is \(4K^{-1/2}\).
To compute the tail directly, \(x^{-1/2}>K\) holds exactly when \(x<K^{-2}\). The tail set is therefore \([1/n,1]\cap(0,K^{-2})\), up to endpoints of measure zero. If \(n\le K^2\), then \(1/n\ge K^{-2}\), so the tail has measure zero. If \(n>K^2\), integration gives $$ \int_{\{|f_n|>K\}}|f_n(x)|\,dx =\int_{1/n}^{K^{-2}}x^{-1/2}\,dx =2(K^{-1}-n^{-1/2})\le\frac{2}{K}. $$ Consequently the exact calculation gives the uniform bound \(2/K\), which tends to zero and is sharper than the theorem's \(4K^{-1/2}\).
Check: For \(K=4\) and \(n=64\), \(64>4^2=16\), so the tail is nonempty. The direct integral is \(2(1/4-1/8)=1/4\). The theorem's bound at the same threshold is \(4\cdot4^{-1/2}=2\), and \(1/4\le2\). If \(n=16\) at \(K=4\), the lower endpoint is \(1/16=K^{-2}\); equality is not in the strict tail, so the tail integral is \(0\).
Worked Example: Indicator Spikes
On \((0,1)\) with Lebesgue measure, let \(f_n(x)=n\chi_{(0,n^{-2})}(x)\) for integers \(n\ge1\). Find a uniform \(L^2\) bound and the supremum of the tail integrals at a real threshold \(K\ge1\).
The squared function is \(n^2\) on an interval of length \(n^{-2}\), so $$ \int_0^1|f_n|^2\,dx=n^2 n^{-2}=1 $$ for every \(n\). Thus \(C=1\) and \(p=2\); the theorem gives a tail bound \(K^{-1}\).
Because the only nonzero value of \(|f_n|\) is \(n\), the tail set is all of \((0,n^{-2})\) if \(n>K\), and is empty if \(n\le K\). The strict inequality matters when \(n=K\): the value then equals the threshold and is not counted. The tail integral is therefore \(1/n\) when \(n>K\), and \(0\) otherwise. Among integers \(n>K\), \(1/n\) is largest at the smallest such integer, \(\lfloor K\rfloor+1\). Hence $$ \sup_{n\ge1}\int_{\{|f_n|>K\}}|f_n|\,dx =\frac{1}{\lfloor K\rfloor+1}\le\frac1K. $$
Check: At \(K=3\), \(n=3\) contributes \(0\), because \(3>3\) is false. The first nonzero tail occurs at \(n=4\), and its integral is \(4\cdot4^{-2}=1/4\). This equals \(1/(\lfloor3\rfloor+1)\), and \(1/4\le1/3\), as the theorem predicts.
A Common Tail-Estimate Mistake
On the set \(\{|f|>K\}\), the inequality \(|f|\le K\) is false. Replacing \(|f|\) by \(K\) there would incorrectly suggest that \(\int_{\{|f|>K\}}|f|\,d\mu\le K\mu(\{|f|>K\})\). For a concrete failure, take \(X=(0,1)\), \(f(x)=5\chi_{(0,1/25)}(x)\), and \(K=4\). The tail integral is \(5\cdot(1/25)=1/5\), whereas \(K\) times the tail measure is \(4\cdot(1/25)=4/25\); the claimed inequality would say \(1/5\le4/25\), which is false.
The correction is to use the \(L^p\) information, not to bound the function by the threshold on its tail. When \(|f|>K\) and \(p>1\), the factor \(|f|^{1-p}\) is at most \(K^{1-p}\), giving \(|f|\le K^{1-p}|f|^p\). This is the pointwise inequality that makes the uniform tail estimate work. At \(p=1\), its factor becomes \(K^0=1\), which gives no decay as \(K\) increases; this argument therefore depends essentially on an exponent greater than \(1\).
The distinction between a bound for each individual function and a uniform bound for a whole family is also important. The theorem assumes one constant \(C\) works for every \(f\in\mathcal{F}\). With that hypothesis, its tail estimate holds at the same threshold \(K\) for every member, which is exactly the uniformity required in the definition.
Check Your Understanding
For each question, pay attention to the exponent, the common bound, and whether the tail uses a strict inequality.
- If \(\sup_{f\in\mathcal{F}}\int_X|f|^3\,d\mu\le 8\), what tail bound follows at threshold \(K>0\)?
- For \(f_n(x)=x^{-1/3}\chi_{[1/n,1]}(x)\) on \((0,1)\), determine which condition on \(p\) makes \(\sup_n\int_0^1|f_n|^p\,dx\) finite.
- For the spike family \(f_n=n\chi_{(0,n^{-2})}\), what is the tail integral at \(K=5\) when \(n=5\), and when \(n=6\)?
- Why does the tail estimate from a uniform \(L^p\) bound tend to zero when \(p>1\), but not when the same calculation is made with \(p=1\)?
- In the finite-space example, explain why the value \(4\) contributes to the tail at \(K=3\) but not at \(K=4\).