TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11

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Uniform oscillatory tail mass forces failure of absolute convergence

Statement

Let $g \geq 0$ be monotone on $[a, \infty)$ and suppose $\int_{a}^{\infty} g$ diverges. Let $a \leq x_{0} < x_{1} < \dots$ tend to infinity and satisfy $x_{j + 1} - x_{j} \leq M$ for some $M > 0$ , and suppose a locally integrable $f$ satisfies $\int_{x_{j}}^{x_{j + 1}} ∣ f (x) ∣ d x \geq δ > 0$ for every $j$ . Then $\int_{a}^{\infty} ∣ f (x) g (x) ∣ d x$ diverges, so $\int_{a}^{\infty} f g$ cannot converge absolutely.

Facts & Assumptions

Given: The nonnegative monotone $g$ , bounded-gap sequence, and uniform block mass in the statement.

[L1]

Proper integration preserves order and is additive on adjacent intervals (If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ ).

[L2]

A nonnegative improper integral converges exactly when its truncations are bounded (A nonnegative improper integral converges iff its truncated integrals are bounded).

[L3]

Removing finitely many terms does not affect divergence of a nonnegative series (A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail).

Proof

technique · direct

1.1

Suppose first that $g$ is nonincreasing. On the $j$ th block, $g (x) \geq g (x_{j + 1})$ , hence [L1, L2, L3] $\int_{x_{j}}^{x_{j + 1}} ∣ f ∣ g \geq δ g (x_{j + 1}) .$ Also $\int_{x_{j}}^{x_{j + 1}} g \leq M g (x_{j})$ . Since $\int g$ diverges, additivity and [L2] force $\sum_{j} g (x_{j})$ , and hence its shifted tail, to diverge. Thus the block lower bounds for $\int ∣ f ∣ g$ have unbounded partial sums.

L1L2L3

1.2

If $g$ is nondecreasing, divergence of its integral implies it is positive at some point; thereafter $g (x_{j})$ is bounded below by a positive constant. Now $\int_{x_{j}}^{x_{j + 1}} ∣ f ∣ g \geq δ g (x_{j})$ , whose partial sums diverge.

given

2.1

In either monotonicity case the nonnegative truncations of $∣ f g ∣$ are unbounded, so [L2] proves divergence and the definition rules out absolute convergence.

L2∎

Depends on

Used by

The sine integral converges conditionally but not absolutely Counterexample
The sine integral is improperly Riemann integrable and not Lebesgue integrable Counterexample

Dependency tree · two levels

41 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

William F. Trench, Introduction to Real Analysis, exercises following Section 3.4 (standard reference, not scraped)