Equivalent Tests for Lebesgue Integrability

Three Tests for the Same Integrability Condition

A signed function can have positive and negative values, so its integral cannot be defined by subtracting two infinite quantities. The positive and negative parts separate those contributions. This leads to equivalent tests for integrability: the absolute value has finite integral, or, equivalently, both parts have finite integrals. The equivalence matters on spaces of infinite measure, where a formal cancellation may suggest a finite answer even though the Lebesgue integral is undefined.

Unless specified otherwise, let \((X,\mathcal{M},\mu)\) be an arbitrary measure space; it need not be finite, sigma-finite, or complete. Functions in the general statements are measurable and take values in the extended real line \([-\infty,\infty]\). Integrals of nonnegative functions are allowed to take the value \(+\infty\). No convergence theorem is needed here. We use the additivity of the nonnegative Lebesgue integral: for measurable \(u,v:X\to[0,\infty]\), \(\int(u+v)\,d\mu=\int u\,d\mu+\int v\,d\mu\), with extended nonnegative values allowed.

Definition: For a measurable \(f:X\to[-\infty,\infty]\), its positive and negative parts are the nonnegative measurable functions \(f^+=\max(f,0)\) and \(f^-=\max(-f,0)\). Its absolute value satisfies \(|f|=f^++f^-\). The signed Lebesgue integral is defined as \(\int f\,d\mu=\int f^+\,d\mu-\int f^-\,d\mu\) when the two terms are not both \(+\infty\). It is a finite real number exactly when both terms are finite. We call \(f\) integrable if it is finite \(\mu\)-a.e. and its signed integral is finite.

The pointwise identities \(f=f^+-f^-\) and \(|f|=f^++f^-\) hold wherever \(f\) is real-valued; the identities involving \(|f|\) and the parts also hold for extended values, with the signed difference interpreted only when it is defined. The convention “finite \(\mu\)-a.e.” allows an extended-valued function to take an infinite value on a null set. Such values do not affect its integral.

The Equivalence and Its Proof

Proposition: If \(f:X\to[-\infty,\infty]\) is measurable and \(\int |f|\,d\mu<\infty\), then \(f\) is finite \(\mu\)-a.e.

Proof. Let \(N=\{x\in X:|f(x)|=\infty\}\), a measurable set. For each positive integer \(k\), the pointwise inequality \(k\chi_N\le |f|\) holds: on \(N\), the right side is infinite, and off \(N\), the left side is zero. By monotonicity of the nonnegative integral, \(k\mu(N)\le\int |f|\,d\mu<\infty\). Thus \(\mu(N)\le (\int |f|\,d\mu)/k\) for every \(k\). Letting \(k\) increase without bound gives \(\mu(N)=0\), so \(f\) is finite \(\mu\)-a.e. \(\square\)

Theorem (Equivalent Tests for Integrability): For a measurable \(f:X\to[-\infty,\infty]\), the following are equivalent:

\(f\) is integrable;
\(\int |f|\,d\mu<\infty\);
\(\int f^+\,d\mu<\infty\) and \(\int f^-\,d\mu<\infty\).

Whenever these conditions hold, \(\int |f|\,d\mu=\int f^+\,d\mu+\int f^-\,d\mu\) and \(\int f\,d\mu=\int f^+\,d\mu-\int f^-\,d\mu\), with both equalities giving finite real values.

Proof. The identity \(|f|=f^++f^-\) and additivity of the nonnegative integral give, as an equality of extended nonnegative values,

\int |f|\,d\mu=\int f^+\,d\mu+\int f^-\,d\mu.

A sum of two nonnegative extended real numbers is finite exactly when each summand is finite. Therefore condition 2 is equivalent to condition 3. If condition 2 holds, the proposition shows that \(f\) is finite \(\mu\)-a.e.; condition 3 also makes its signed integral finite by the definition above. Hence condition 2 implies condition 1. Conversely, if \(f\) is integrable, its signed integral is finite, which by definition requires both \(\int f^+\,d\mu\) and \(\int f^-\,d\mu\) to be finite. Thus condition 1 implies condition 3. This proves all three equivalences and the stated formulas. \(\square\)

This proof does not require \(\mu(X)<\infty\). In particular, being bounded is not part of the criterion, and finite measure is not a substitute for checking the positive and negative parts when the measure space is infinite.

Worked Example: A Signed Simple Function

On \([0,3]\) with Lebesgue measure \(m\), define \(f=2\) on \([0,1)\), \(f=-3\) on \([1,2)\), and \(f=0\) on \([2,3]\). Determine its positive-part, negative-part, absolute-value, and signed integrals.

The intervals \([0,1)\) and \([1,2)\) each have measure \(1\). On the first interval, \(f^+=2\), \(f^-=0\), and \(|f|=2\). On the second, \(f^+=0\), \(f^-=3\), and \(|f|=3\). On the remaining interval all three functions are zero. Therefore

\int f^+\,dm=2\cdot1=2,\qquad \int f^-\,dm=3\cdot1=3,\qquad \int |f|\,dm=2\cdot1+3\cdot1=5.

Both part integrals are finite, so \(f\) is integrable, and \(\int f\,dm=2-3=-1\). Check: the absolute integral is \(2+3=5\), as the theorem requires, and the signed integral is the contribution \(2\) from the positive interval plus the contribution \(-3\) from the negative interval, namely \(-1\).

Worked Example: Finite Cancellation on the Real Line

With Lebesgue measure \(m\) on \(\mathbb{R}\), define \(f(x)=1/(1+x^2)\) for \(x>0\), \(f(x)=-1/(1+x^2)\) for \(x<0\), and \(f(0)=0\). Compute its parts and its integral.

Here \(f^+(x)=1/(1+x^2)\) for \(x>0\) and is zero otherwise. Similarly, \(f^-(x)=1/(1+x^2)\) for \(x<0\) and is zero otherwise. Using the antiderivative \(\arctan x\),

\int f^+\,dm=\int_0^\infty\frac{dx}{1+x^2} =\lim_{R\to\infty}(\arctan R-\arctan 0)=\frac{\pi}{2},

and substituting \(u=-x\) in the negative half-line gives \(\int f^-\,dm=\int_0^\infty(1+u^2)^{-1}\,du=\pi/2\). Hence \(\int |f|\,dm=\pi/2+\pi/2=\pi<\infty\), so \(f\) is integrable, and \(\int f\,dm=\pi/2-\pi/2=0\). Check: on each symmetric interval \([-R,R]\), the two contributions are \(\arctan R\) and \(-\arctan R\), so the integral there is zero; meanwhile the absolute integral on that interval is \(2\arctan R\), which tends to \(\pi\). The whole-line computations agree with these limits.

When Cancellation Does Not Define an Integral

Worked Example: A Symmetric Improper Integral Is Not Lebesgue Integrable

On \((-1,1)\) with Lebesgue measure, set \(f(x)=1/x\) for \(x\ne0\) and \(f(0)=0\). Determine whether it is integrable, and compare its signed Lebesgue integral with the symmetric improper calculation.

The positive part is \(1/x\) on \((0,1)\), and the negative part is \(1/|x|\) on \((-1,0)\). For \(0<\delta<1\),

\int_\delta^1\frac{dx}{x}=-\ln\delta,\qquad \int_{-1}^{-\delta}\frac{dx}{|x|}=-\ln\delta.

As \(\delta\downarrow0\), both quantities tend to \(+\infty\). Thus \(\int f^+\,dm=\infty\), \(\int f^-\,dm=\infty\), and \(\int |f|\,dm=\infty\). The signed Lebesgue integral is undefined because its definition would require subtracting \(+\infty\) from \(+\infty\). In contrast, for every \(0<\delta<1\), the symmetrically truncated integral is

\int_{-1}^{-\delta}\frac{dx}{x}+\int_\delta^1\frac{dx}{x} =\ln\delta-\ln\delta=0.

Check: each truncated side has magnitude \(-\ln\delta\), which diverges as \(\delta\downarrow0\); only their symmetric sum is zero. The zero therefore describes a particular improper limiting procedure, not a finite Lebesgue integral.

Worked Example: An Infinite Integral Is Not Integrability

On \([0,1]\) with Lebesgue measure, let \(g(x)=1/x\) for \(x>0\) and \(g(0)=0\). Find its positive and negative part integrals and decide whether it is integrable.

The function is nonnegative, so \(g^+=g\) and \(g^-=0\). For \(0<\delta<1\), \(\int_\delta^1g\,dm=-\ln\delta\), which tends to \(+\infty\) as \(\delta\downarrow0\). Consequently \(\int g^+\,dm=\int|g|\,dm=\infty\), while \(\int g^-\,dm=0\). The extended signed integral exists with value \(+\infty\), but \(g\) is not integrable because its integral is not finite. Check: the criterion fails both as a finite absolute integral and as finiteness of both part integrals; the negative part contributes zero, but the positive part is infinite.

A Reliable Checklist

For a measurable real-valued function, or an extended-real-valued function finite almost everywhere, test integrability by computing the two nonnegative integrals \(\int f^+\,d\mu\) and \(\int f^-\,d\mu\), or by computing \(\int|f|\,d\mu\). Either route is valid because the theorem proves their equivalence. If both part integrals are finite, subtraction is legitimate and defines a finite signed integral. If just one is infinite, the signed integral can be \(+\infty\) or \(-\infty\), but the function is still not integrable in the finite-integral sense. If both are infinite, the signed integral is undefined.

Common mistake: the calculation \(\int_{-1}^{-\delta}(1/x)\,dx+\int_\delta^1(1/x)\,dx=0\) for every \(\delta>0\) does not prove that \(1/x\) is Lebesgue integrable on \((-1,1)\). It computes only symmetric truncations. The correction is to check the parts: each has infinite integral, so the signed Lebesgue integral is undefined. In general, a finite value obtained by pairing positive and negative contributions is meaningful only after the absolute-integrability criterion has been checked.

Check Your Understanding

For each question, distinguish finiteness of the signed integral from finiteness of the absolute integral.

State the pointwise relationship between \(|f|\), \(f^+\), and \(f^-\), and use it to prove that \(\int|f|\,d\mu<\infty\) implies both part integrals are finite.
If \(\int f^+\,d\mu=4\) and \(\int f^-\,d\mu=7\), compute \(\int|f|\,d\mu\) and \(\int f\,d\mu\). Is \(f\) integrable?
For the function \(1/x\) on \((-1,1)\), why does the symmetric improper integral not establish Lebesgue integrability?
Can a function have an extended signed integral equal to \(+\infty\) without being integrable? Give the relevant part-integral condition.
Does the equivalence between finite absolute integral and finite positive- and negative-part integrals require \(\mu(X)<\infty\)? Explain.