Cancellation in an Improper Integral
An improper Riemann integral can converge even when the accumulated area of the absolute value is infinite. The convergence then depends on positive and negative contributions cancelling as the endpoint moves outward. Lebesgue integration treats those contributions separately: a finite Lebesgue integral of a real-valued function requires both its positive and negative parts to have finite integrals. Thus conditional convergence of an improper integral is not, by itself, Lebesgue integrability.
We work on \(([1,\infty),\mathcal{L},m)\), the Lebesgue measure space obtained by restricting Lebesgue measure on \(\mathbb{R}\). It is complete and sigma-finite, but not finite. All functions below are real-valued and measurable. For a real-valued function \(f\), define \(f^+(x)=\max(f(x),0)\) and \(f^-(x)=\max(-f(x),0)\). Then \(f=f^+-f^-\) and \(|f|=f^++f^-\) pointwise. A finite Lebesgue integral means \(\int f^+\,dm<\infty\) and \(\int f^-\,dm<\infty\); equivalently, \(\int |f|\,dm<\infty\). These conditions ensure that \(\int f\,dm=\int f^+\,dm-\int f^-\,dm\) is a finite real number.
The key issue is the order of operations. The improper integral takes the limit of the signed truncations, allowing cancellation at each finite endpoint \(R\). The Lebesgue integral of a signed function is instead defined through its positive and negative parts. If both parts have infinite integral, their difference is not defined as a Lebesgue integral, even if the signed truncations have a finite limit.
What Conditional Convergence Implies for the Parts
Proof. For each positive integer \(n\), set \(P_n=\int_1^n f^+\,dm\) and \(N_n=\int_1^n f^-\,dm\). The pointwise identities \(f=f^+-f^-\) and \(|f|=f^++f^-\), together with additivity of the integral on the bounded interval, give \(P_n-N_n=\int_1^n f\,dm\) and \(P_n+N_n=\int_1^n|f|\,dm\). Hence \(P_n-N_n\to I\), so the differences are bounded, while \(P_n+N_n\to+\infty\). Both \(P_n\) and \(N_n\) are nondecreasing sequences of nonnegative numbers. If \(P_n\) were bounded, then \(N_n=P_n-(P_n-N_n)\) would also be bounded, because \(P_n-N_n\) is bounded; this would contradict \(P_n+N_n\to+\infty\). Thus \(P_n\to+\infty\). If \(N_n\) were bounded, the identity \(P_n=N_n+(P_n-N_n)\) would make \(P_n\) bounded, which has just been ruled out. Thus \(N_n\to+\infty\) as well. By the Monotone Convergence Theorem applied to \(f^+\chi_{[1,n]}\) and \(f^-\chi_{[1,n]}\), whose nonnegative measurable sequences increase pointwise to \(f^+\) and \(f^-\), respectively, their integrals on \([1,\infty)\) are both infinite. Therefore the signed Lebesgue integral would require subtracting \(+\infty\) from \(+\infty\), which is undefined. \(\square\)
The proposition separates two claims that are sometimes conflated. Conditional convergence is a statement about a limit of signed truncations. Lebesgue integrability is a statement about the total size of the positive and negative contributions. The following examples verify both features directly.
Worked Example: The Integral of sin x Divided by x
Consider the continuous real-valued function \(f(x)=\sin x/x\) on \([1,\infty)\). First, for \(1\le A<B\), integration by parts gives
Since \(|\cos x|\le1\), the absolute value of this tail is at most
This bound holds for every \(B>A\). In particular, given any \(\varepsilon>0\), choosing \(A>2/\varepsilon\) makes every tail from \(A\) to \(B>A\) have absolute value less than \(\varepsilon\). The Cauchy criterion for a finite limit therefore shows that \(\int_1^R\sin x/x\,dx\) converges as \(R\to\infty\).
Now consider absolute values. For every integer \(k\ge1\), take the interval \([k\pi+\pi/6,\ k\pi+5\pi/6]\). Its length is \(2\pi/3\), and \(|\sin x|\ge1/2\) throughout it. Also, \(x\le(k+1)\pi\) there, since \(5\pi/6<\pi\). Consequently,
These intervals are disjoint. The sum of the lower bounds diverges: grouping \(1/(k+1)\) over blocks with \(2^j\le k+1<2^{j+1}\) gives at least \(2^j/2^{j+1}=1/2\) from each block. Thus the absolute truncation integrals tend to infinity. The proposition now gives \(\int f^+\,dm=\int f^-\,dm=+\infty\), so \(f\) does not have a Lebesgue integral, despite its convergent ordinary improper integral. Check: the bound \(2/A\) tends to zero as \(A\to\infty\), while the lower bounds for disjoint absolute-value intervals have a divergent sum; these establish convergence and absolute divergence by separate estimates.
Changing the Oscillation or the Endpoint
Worked Example: A Quadratic Oscillatory Phase
On \([1,\infty)\), let \(g(x)=\sin(x^2)/x\). For every \(R>1\), the substitution \(t=x^2\), with \(dt=2x\,dx\), gives \(dx/x=dt/(2t)\), and hence
The right-hand side converges as \(R\to\infty\), because the integral of \(\sin t/t\) converges by the tail estimate in the preceding example and \(R^2\to\infty\). Applying the same substitution to the nonnegative absolute-value integral yields
The right-hand side tends to infinity by the disjoint-interval lower bound already established. Thus the improper integral of \(g\) is conditionally convergent, and \(g\) is not Lebesgue integrable. Check: the factor \(1/2\) comes from \(dx/x=dt/(2t)\); multiplying a convergent limit by \(1/2\) preserves finiteness, and multiplying a quantity tending to infinity by \(1/2\) still gives divergence.
Worked Example: Oscillation Near a Singular Endpoint
Consider \(h(x)=\sin(1/x)/x\) on \((0,1]\), with Lebesgue measure restricted to that interval. This is real-valued, measurable and continuous at every point of \((0,1]\), though it has no finite limit at \(0\). For \(0<a<1\), the substitution \(t=1/x\) gives \(dx=-dt/t^2\) and \(dx/x=-dt/t\). Therefore the one-sided truncation is
As \(a\downarrow0\), the upper endpoint \(1/a\to\infty\), so these signed truncations converge by the result for \(\sin t/t\). For the absolute values the same substitution gives
which tends to infinity. Hence the ordinary one-sided improper integral at \(0\) converges conditionally, but \(h\) has no finite Lebesgue integral on \((0,1]\). Check: the reversal of endpoints and the minus sign in \(dx=-dt/t^2\) produce the limits \(1\) and \(1/a\) in increasing order; the absolute-value substitution has the same limits and no cancellation.
Why the Cancellation Does Not Define a Lebesgue Integral
A common mistake is to write “the improper integral converges, so the Lebesgue integral has the same value.” For \(f(x)=\sin x/x\), this reasoning confuses a limit of signed integrals with integrability. The interval-by-interval calculation shows that the absolute integral is infinite, and the proposition shows that both \(\int f^+\,dm\) and \(\int f^-\,dm\) are infinite. The proposed Lebesgue expression would therefore be \(+\infty-(+\infty)\), which is undefined—not the finite limit of the signed truncations.
Equivalently, finite Lebesgue integrability of a real-valued \(f\) requires \(\int |f|\,dm<\infty\). The absolute convergence condition rules out reliance on cancellation and is exactly what permits the positive and negative contributions to be integrated separately as finite quantities. Conditional improper convergence remains meaningful and useful, but it is a different notion. In the examples above, the cancellation is tied to the order of the truncation endpoints; it does not produce a finite Lebesgue integral on the stated measure space.
Check Your Understanding
For each question, distinguish the signed truncation limit from the integral of the absolute value.
- State the definition of conditional convergence for an improper integral on \([1,\infty)\).
- Why does convergence of \(\int_1^R f\,dm\) together with divergence of \(\int_1^R|f|\,dm\) force both \(\int f^+\,dm\) and \(\int f^-\,dm\) to be infinite?
- For \(\sin x/x\), identify the explicit intervals used to bound the absolute integral below, and explain why their lower bounds have divergent sum.
- Use \(t=x^2\) to express the signed truncation integral of \(\sin(x^2)/x\) in terms of \(\sin t/t\).
- For \(\sin(1/x)/x\) on \((0,1]\), what happens to the transformed upper endpoint as \(a\downarrow0\), and what does this imply for the signed and absolute truncations?