Absolute Integrability Controls Tails
In the previous tutorial, a finite-valued measurable function \(f\) was called integrable when \(\int_X |f|\,d\mu<\infty\). This condition does more than ensure that the signed integral is well-defined: it also limits how much the function can contribute on regions where its magnitude is large, or on sets whose measure is small. These controls are useful when approximating an integral and when deciding whether apparent cancellation is legitimate.
Throughout, \((X,\mathcal{F},\mu)\) is a measure space, and \(f:X\to\mathbb{R}\) is finite-valued and measurable. We use the Monotone Convergence Theorem and the additivity and monotonicity of the nonnegative integral established earlier in this course. For a measurable set \(A\), the notation \(\int_A |f|\,d\mu\) means \(\int_X \mathbf{1}_A|f|\,d\mu\).
A Truncation Criterion
A direct way to examine the size of \(f\) is to cap its magnitude at a fixed height. The truncated function \(|f|\wedge n=\min(|f|,n)\) is bounded, even if \(|f|\) is not. As \(n\) increases, these truncations recover \(|f|\) pointwise. The Monotone Convergence Theorem turns this approximation into a criterion for absolute integrability.
Proof. The functions \(|f|\wedge n\) are nonnegative, measurable, and increase pointwise to \(|f|\). If \(\int_X|f|\,d\mu<\infty\), monotonicity gives $$ \int_X(|f|\wedge n)\,d\mu\leq\int_X|f|\,d\mu<\infty $$ for every \(n\). Their integrals are therefore bounded above uniformly in \(n\).
Conversely, suppose their integrals have a finite upper bound \(C\). The Monotone Convergence Theorem gives $$ \int_X|f|\,d\mu = \lim_{n\to\infty}\int_X(|f|\wedge n)\,d\mu \leq C<\infty. $$ Thus \(f\) is integrable. This criterion is useful when the full integral is difficult to compute but its bounded truncations can be estimated.
Worked Example: A Power Singularity on a Finite Interval
Let \(X=(0,1)\) with Lebesgue measure and \(f(x)=x^{-3/4}\). This function is nonnegative and measurable. For \(0<\delta<1\), direct integration gives $$ \int_\delta^1 x^{-3/4}\,dx = \left[4x^{1/4}\right]_\delta^1 = 4(1-\delta^{1/4}). $$ As \(\delta\) decreases to zero, this expression tends to \(4\). Since the intervals \((\delta,1)\) increase to \((0,1)\) along, for example, \(\delta=1/n\) for \(n\geq2\), the increasing-domain approximation for nonnegative integrals gives $$ \int_0^1 |f(x)|\,dx=4. $$ Hence \(f\) is absolutely integrable, despite becoming unbounded near zero. Its unboundedness alone does not prevent integrability.
For comparison, the related function \(g(x)=x^{-3/2}\) is not absolutely integrable. Indeed, $$ \int_\delta^1 x^{-3/2}\,dx = \left[-2x^{-1/2}\right]_\delta^1 = 2(\delta^{-1/2}-1), $$ which tends to \(+\infty\) as \(\delta\) decreases to zero. Thus \(\int_0^1|g|\,dx=\infty\). The calculation distinguishes two singularities by their accumulated size, not merely by the fact that both functions are unbounded.
Large Values Have Small Total Contribution
For an integrable function, the region where its magnitude exceeds a high threshold has a small integral of \(|f|\). This is a tail estimate: it controls the total contribution from large values, not just the measure of the region on which they occur.
Proof. For each positive integer \(n\), let \(B_n=\{x\in X:|f(x)|\leq n\}\). These sets are measurable, increase with \(n\), and have union \(X\), because \(f\) is finite-valued. The nonnegative measurable functions \(|f|\mathbf{1}_{B_n}\) increase pointwise to \(|f|\). By the Monotone Convergence Theorem, $$ \lim_{n\to\infty}\int_{B_n}|f|\,d\mu=\int_X|f|\,d\mu. $$ The latter integral is finite. Since \(B_n\) and \(X\setminus B_n=\{|f|>n\}\) partition \(X\), additivity of the nonnegative integral gives $$ \int_{\{|f|>n\}}|f|\,d\mu = \int_X|f|\,d\mu-\int_{B_n}|f|\,d\mu. $$ The right-hand side tends to zero. Finally, if \(M\geq n\), then \(\{|f|>M\}\subseteq\{|f|>n\}\), so monotonicity gives $$ 0\leq\int_{\{|f|>M\}}|f|\,d\mu \leq\int_{\{|f|>n\}}|f|\,d\mu. $$ The integer-threshold limit therefore implies the same limit for arbitrary real \(M\to\infty\).
Worked Example: Controlling the Tail on the Real Line
On \(\mathbb{R}\) with Lebesgue measure, consider \(f(x)=(1+|x|)^{-3/2}\). It is nonnegative and measurable. For \(R\geq0\), symmetry and elementary integration give $$ \int_{\{|x|>R\}} f(x)\,dx = 2\int_R^\infty(1+x)^{-3/2}\,dx = \frac{4}{\sqrt{1+R}}. $$ The expression tends to zero as \(R\to\infty\). Taking \(R=0\) also gives \(\int_{\mathbb{R}}|f(x)|\,dx=4\), so \(f\) is integrable.
This calculation controls the contribution outside a large bounded interval. It is not the same as the high-value tail in the theorem: here the omitted region is chosen by the location of \(x\), whereas the theorem uses the size of \(|f(x)|\). Both estimates are useful ways to discard a part of the domain while bounding the resulting error.
Absolute Continuity of the Integral
The high-value tail estimate implies that a set of small measure cannot carry a large integral of \(|f|\). The bound is uniform over all measurable sets of sufficiently small measure; it does not depend on the location or shape of the set.
Proof. Fix \(\varepsilon>0\). By the vanishing high-value tail theorem, choose \(M\geq1\) such that $$ \int_{\{|f|>M\}}|f|\,d\mu<\frac{\varepsilon}{2}. $$ Set \(\delta=\varepsilon/(2M)\), which is positive. For any measurable \(A\), split its contribution between the regions \(\{|f|\leq M\}\) and \(\{|f|>M\}\). On the first region, \(|f|\leq M\); on the second, the integral over \(A\) is at most the full tail integral. Consequently, $$ \int_A|f|\,d\mu \leq M\mu(A)+\int_{\{|f|>M\}}|f|\,d\mu. $$ If \(\mu(A)<\delta\), then \(M\mu(A)<\varepsilon/2\), and the chosen tail integral is also less than \(\varepsilon/2\). Therefore \(\int_A|f|\,d\mu<\varepsilon\), as required. This argument also covers \(\mu(A)=0\), for which the first term is zero.
The signed integral over such a set is controlled as well. The function \(f\mathbf{1}_A\) is integrable because \(|f\mathbf{1}_A|\leq |f|\). Applying the Absolute-Value Bound for an Integrable Function from the previous tutorial yields $$ \left|\int_A f\,d\mu\right| \leq \int_A|f|\,d\mu. $$ Thus the same choice of \(\delta\) ensures that \(\left|\int_A f\,d\mu\right|<\varepsilon\) whenever \(\mu(A)<\delta\). Small sets cannot hide a large signed contribution either.
Worked Example: A Uniform Bound for a Bounded Function
Let \(X=[0,2]\) with Lebesgue measure and \(f(x)=3-x\). Since \(1\leq f(x)\leq3\) on \(X\), for any measurable \(A\subseteq X\), $$ \int_A|f|\,dx\leq3\lambda(A). $$ Given \(\varepsilon>0\), take \(\delta=\varepsilon/3\). If \(\lambda(A)<\delta\), then $$ \int_A|f|\,dx\leq3\lambda(A)<3\delta=\varepsilon. $$ Here a direct bound works because \(f\) is bounded. For an unbounded integrable function, the proof of absolute continuity replaces a global bound by first discarding a high-value tail and then controlling the bounded remainder.
Cancellation Is Not Absolute Integrability
A convergent sequence of signed partial sums does not by itself provide a Lebesgue integral. Under counting measure, the integral of a nonnegative function on \(\mathbb{N}\) is its sum over \(\mathbb{N}\). For a signed function, its positive and negative parts must each have finite integral for the signed integral to be defined as a finite difference.
Worked Example: An Alternating Series on Counting Measure
Give \(\mathbb{N}\) counting measure and set \(f(n)=(-1)^{n+1}/n\). Its positive part is nonzero at odd integers, and its negative part has magnitude \(1/n\) at even integers. The positive-part integral diverges, since for every \(k\geq1\), $$ \frac{1}{2k-1}\geq\frac{1}{2k}, \qquad \sum_{k=1}^{\infty}\frac{1}{2k} = \frac12\sum_{k=1}^{\infty}\frac1k = \infty. $$ The negative-part integral also diverges: $$ \sum_{k=1}^{\infty}\frac{1}{2k} = \infty. $$ Therefore \(\int_{\mathbb{N}}|f|\,d\mu=\infty\), and \(f\) is not integrable. Although the alternating partial sums \(\sum_{n=1}^N(-1)^{n+1}/n\) converge, that cancellation does not make either part finite. The Lebesgue integral of \(f\) is consequently not defined by subtracting its positive and negative integrals.
Why These Estimates Matter
The truncation criterion offers a practical test: uniform control of the integrals of \(|f|\wedge n\) is enough to prove absolute integrability. Once integrability is known, the high-value tail theorem and absolute continuity theorem provide two distinct approximation tools. One controls regions selected by the size of the function; the other controls arbitrary measurable regions selected by their measure.
A common pitfall is to treat a small measure as automatically guaranteeing a small integral. The absolute continuity theorem requires \(f\) to be integrable. For instance, on \((0,1)\), the function \(x^{-3/2}\) has infinite integral near zero, so sets close to zero can have arbitrarily small measure while retaining a large integral. Absolute integrability is precisely what supplies the uniform small-set control proved above.
Check Your Understanding
Use the truncation and tail arguments to answer these questions.
- Why does boundedness of \(\int_X(|f|\wedge n)\,d\mu\) for all positive integers \(n\) imply that \(f\) is integrable?
- In the proof of the vanishing-tail theorem, why do the sets \(\{|f|\leq n\}\) increase to all of \(X\)?
- How does splitting a set \(A\) into low-value and high-value regions prove absolute continuity of the integral?
- Why does the alternating sequence \(f(n)=(-1)^{n+1}/n\) fail to define a Lebesgue integral under counting measure?
- Where does the proof of the absolute continuity theorem use the assumption that \(f\) is integrable?