Common Errors in Comparing Integral Notions

Four Integral Notions, Four Different Questions

When two integration procedures give the same number in a familiar example, it is tempting to regard them as interchangeable. They are not. Proper Riemann integration asks whether a bounded function on a compact interval is Riemann integrable. Improper Riemann integration asks whether specified limits of proper integrals exist. Lebesgue integrability of a real-valued measurable function asks whether the integral of its absolute value is finite. A Cauchy principal value asks whether a particular symmetric limiting procedure exists. The limit in one definition does not automatically answer the question posed by another.

We use Lebesgue measure \(m\) on \([a,b]\), with its Lebesgue sigma-algebra. A real-valued measurable function \(f\) on this interval is Lebesgue integrable when \(\int_{[a,b]}|f|\,dm<\infty\). Equivalently, both \(\int f^+\,dm\) and \(\int f^-\,dm\) are finite, where \(f^+=\max(f,0)\) and \(f^-=\max(-f,0)\). This condition makes \(\int f\,dm=\int f^+\,dm-\int f^-\,dm\) a finite, well-defined number. If both terms are infinite, their difference is not defined.

A proper Riemann integral is defined for a bounded Riemann-integrable function on a compact interval. An improper Riemann integral is a limit of proper Riemann integrals on truncated intervals, with the truncation and limiting procedure specified. For instance, at a left endpoint \(a\), the one-sided improper integral is \(\lim_{t\downarrow a}\int_t^b f(x)\,dx\), if this finite limit exists. For an interior singularity \(c\), the two one-sided limits are tested separately. A principal value at \(c\), in contrast, uses a symmetric truncation around \(c\).

Takeaway: Proper Riemann integrability, convergence of an improper Riemann integral, and Lebesgue integrability are distinct properties. A principal value is a specified limit, not a general substitute for any of them.

Unboundedness and Endpoint Values

Worked Example: An Endpoint Singularity That Is Integrable

Define \(f:[0,1]\to[0,\infty)\) by \(f(0)=0\) and \(f(x)=x^{-1/2}\) for \(0<x\le1\). Determine whether its integral is proper Riemann, improper Riemann, or Lebesgue integrable.

The function is unbounded on \([0,1]\), so it is not eligible for a proper Riemann integral there. Assigning a finite value at \(0\) does not change its behavior near \(0\): for every \(M>0\), \(f(x)>M\) whenever \(0<x<M^{-2}\). For \(0<t\le1\), however, \(f\) is finite and continuous on the compact interval \([t,1]\). Its proper Riemann integral is

\int_t^1x^{-1/2}\,dx =2-2\sqrt{t}.

As \(t\downarrow0\), this tends to \(2\). Thus the one-sided improper Riemann integral converges, and its absolute version converges too, since \(f\ge0\). For the Lebesgue integral, \(f\) is measurable and nonnegative, and the same truncated integrals increase to \(2\). The Monotone Convergence Theorem, applied to \(f\chi_{[t,1]}\) along any sequence \(t\downarrow0\), gives \(\int_{[0,1]}f\,dm=2\). Check: the substitution \(u=\sqrt{x}\), so \(x=u^2\) and \(dx=2u\,du\), gives \(\int_0^1x^{-1/2}\,dx=\int_0^1 2\,du=2\). The endpoint value \(f(0)\) contributes zero to the Lebesgue integral.

This example uses only finite-valued continuous restrictions on the truncated intervals. It does not treat an extended-real-valued function with an interior value \(+\infty\) as continuous or Riemann integrable. If there is an interior singularity, the one-sided behavior on each side must be considered separately.

Worked Example: A Null-Set Change That Destroys Riemann Integrability

On \([0,1]\), let \(f=\chi_{\mathbb{Q}\cap[0,1]}\). Compare it with the zero function.

The rationals in \([0,1]\) form a countable set and hence have Lebesgue measure zero. Therefore \(f=0\) \(m\)-a.e., and its Lebesgue integral is \(0\). To test Riemann integrability, take any partition of \([0,1]\). Every subinterval of positive length contains both rational and irrational points, so the supremum of \(f\) on each such subinterval is \(1\), and its infimum is \(0\). Consequently every upper sum is \(1\) and every lower sum is \(0\). They cannot have a common limit, so \(f\) is not Riemann integrable. Check: the lower and upper sums differ by \(1-0=1\) for every partition, no matter how small the mesh is.

Proposition: On any measure space \((X,\mathcal{M},\mu)\), if real-valued measurable functions \(f\) and \(g\) are equal \(\mu\)-a.e. and \(f\) is Lebesgue integrable, then \(g\) is Lebesgue integrable and \(\int f\,d\mu=\int g\,d\mu\).

Proof. Let \(N=\{x:f(x)\ne g(x)\}\), a measurable null set; if equality a.e. is witnessed by a measurable null set containing the exceptional points, use that set for \(N\). Since \(f=g\) on \(N^c\), the nonnegative functions \(|f|\chi_{N^c}\) and \(|g|\chi_{N^c}\) agree everywhere. Also, the integral of any nonnegative measurable function supported on \(N\) is zero: every nonnegative simple function below it has integral zero because each of its level sets is contained in \(N\). Additivity of the nonnegative integral then gives \(\int |g|\,d\mu=\int |g|\chi_{N^c}\,d\mu=\int |f|\chi_{N^c}\,d\mu=\int|f|\,d\mu<\infty\). Applying the same reasoning to the positive and negative parts shows that the integrals of \(f\) and \(g\) both equal the integral of their common restriction to \(N^c\). Thus their integrals are equal. \(\square\)

The proposition concerns Lebesgue integrability and the Lebesgue integral. It does not say that a null-set change preserves Riemann integrability. The indicator in the example is a concrete counterexample to that broader claim.

Absolute Convergence Is the Bridge

Theorem: Let \(f:(a,b]\to\mathbb{R}\) be finite-valued and continuous, where \(a<b<\infty\). If \(\lim_{t\downarrow a}\int_t^b|f(x)|\,dx\) is finite, then \(f\), extended by any finite value at \(a\), is Lebesgue integrable on \([a,b]\), and its Lebesgue integral equals \(\lim_{t\downarrow a}\int_t^b f(x)\,dx\).

Proof. Choose \(t_n=a+(b-a)/(n+1)\). Then \(t_n\downarrow a\), and the sets \([t_n,b]\) increase to \((a,b]\). The function \(|f|\) is measurable and nonnegative. By the Monotone Convergence Theorem applied to \(|f|\chi_{[t_n,b]}\),

\int_{(a,b]}|f|\,dm =\lim_{n\to\infty}\int_{[t_n,b]}|f|\,dm =\lim_{n\to\infty}\int_{t_n}^b|f(x)|\,dx<\infty.

The middle equality holds because \(f\) is finite and continuous on each compact interval \([t_n,b]\), where its proper Riemann and Lebesgue integrals agree. The value assigned at \(a\) does not affect the integral, because \(\{a\}\) has measure zero. Thus the extension is Lebesgue integrable. Applying the same argument to \(f^+\) and \(f^-\), which are nonnegative and bounded above by \(|f|\), gives finite integrals for both. Their truncated integrals converge separately by the Monotone Convergence Theorem. Subtracting these two finite limits shows that the Lebesgue integral of \(f\) equals the limit of the truncated signed integrals. The absolute-convergence assumption also ensures that the improper signed limit exists: the absolute value of the integral over any tail \([a+s,a+r]\) is at most the integral of \(|f|\) over that tail, which tends to zero as \(s,r\downarrow0\). \(\square\)

The hypotheses matter. The theorem assumes a finite-valued continuous function on the open-at-\(a\) interval, and absolute convergence. It does not establish the same conclusion from conditional convergence alone, and it does not permit an unexamined interior singularity on a truncated interval.

Conditional Convergence and Principal Values

Worked Example: Conditional Convergence Is Not Lebesgue Integrability

Consider \(f(x)=\sin(x)/x\) on \([1,\infty)\), with Lebesgue measure. Its improper integral converges, but its Lebesgue integral is not finite.

For the signed improper integral, Dirichlet's test applies: the partial integrals of \(\sin x\) are bounded, since \(\int_1^R\sin x\,dx=\cos 1-\cos R\), whose absolute value is at most \(2\); and \(1/x\) decreases to zero. Hence \(\lim_{R\to\infty}\int_1^R\sin(x)/x\,dx\) exists.

For absolute integrability, consider the disjoint intervals \(I_k=[k\pi+\pi/6,k\pi+5\pi/6]\), for integers \(k\ge1\). On each interval, \(|\sin x|\ge1/2\), its length is \(2\pi/3\), and \(x\le k\pi+5\pi/6\). Therefore

\int_{I_k}\frac{|\sin x|}{x}\,dx \ge \frac{1}{2}\frac{2\pi/3}{k\pi+5\pi/6} = \frac{\pi/3}{k\pi+5\pi/6}.

The final lower bound is at least \(2/(6k+5)\), since \((\pi/3)/(k\pi+5\pi/6)=2/(6k+5)\). The sum of these bounds diverges by comparison with the harmonic series. Hence \(\int_1^\infty|\sin x|/x\,dx=\infty\), so \(f\) is not Lebesgue integrable. Check: the signed improper integral converges by Dirichlet's test, while the integral of the absolute value diverges; these are different claims, not contradictory conclusions.

Worked Example: A Principal Value Created by Cancellation

For \(f(x)=1/x\) on \((-1,1)\), compare the Cauchy principal value with the two one-sided improper integrals and the Lebesgue integral.

The symmetric principal value is defined by deleting \((-\varepsilon,\varepsilon)\) and taking the limit as \(\varepsilon\downarrow0\). For \(0<\varepsilon<1\),

\int_{-1}^{-\varepsilon}\frac{dx}{x} +\int_{\varepsilon}^{1}\frac{dx}{x} = (\ln\varepsilon-0)+(0-\ln\varepsilon)=0.

Thus the principal value is \(0\). But the one-sided integrals do not converge to finite values: \(\int_{-1}^{-\varepsilon}dx/x=\ln\varepsilon\to-\infty\), whereas \(\int_{\varepsilon}^{1}dx/x=-\ln\varepsilon\to+\infty\). Also \(\int_{-1}^1|1/x|\,dx=\infty\). The Lebesgue integral is therefore not defined as a finite integral: its positive and negative parts both have infinite integrals. Check: symmetric truncation cancels \(\ln\varepsilon\) against \(-\ln\varepsilon\), but the two one-sided limits separately diverge. Cancellation in the principal-value procedure does not produce Lebesgue integrability.

A Reliable Comparison Checklist

Before identifying an integral or comparing two values, check what the definition requires:

Proper Riemann: Is the function bounded and Riemann integrable on the entire compact interval? An endpoint assignment does not repair unboundedness nearby.
Improper Riemann: Which endpoint, infinite tail, or interior singularity is truncated? For an interior singularity, are the one-sided limits finite separately?
Lebesgue: Is the function measurable, and is \(\int|f|\,dm\) finite? A convergent signed improper integral does not establish this.
Principal value: Is the specified symmetric limit being computed? Do not identify it with separate one-sided integrals unless those limits also exist.
Null sets: Equality a.e. preserves the Lebesgue integral for integrable functions, but does not guarantee Riemann integrability.

A frequent error is to write \(\int_{-1}^1(1/x)\,dx=0\) because the integrand is odd. The calculation that gives zero is the symmetric principal-value calculation, not an ordinary improper integral: the two one-sided limits are \(-\infty\) and \(+\infty\), and the Lebesgue integral is undefined. Another frequent error is to assume that changing function values on a null set never affects integrability in any sense. The rational-indicator example shows exactly why that statement must be restricted to the Lebesgue integral.

Check Your Understanding

For each question, identify the integral notion and state which hypothesis or limiting procedure matters.

For \(f(x)=x^{-2/3}\) on \((0,1]\), with any finite assigned value at \(0\), is the integral proper Riemann, improper Riemann, and Lebesgue integrable?
Why does changing \(0\) to \(\chi_{\mathbb{Q}\cap[0,1]}\) preserve the Lebesgue integral but not preserve Riemann integrability?
What extra condition, beyond convergence of an improper signed integral, is used in the theorem that gives a finite Lebesgue integral?
For \(1/x\) on \((-1,1)\), which limiting procedure gives zero, and why does that not define either the ordinary improper integral or the Lebesgue integral?
For \(\sin(x)/x\) on \([1,\infty)\), what conclusion follows from Dirichlet's test, and what separate calculation rules out Lebesgue integrability?