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

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A nonnegative improper integral converges iff its truncated integrals are bounded

Statement

Let $f \geq 0$ be Riemann integrable on every compact subinterval of $[a, \infty)$ . Then $\int_{a}^{\infty} f$ converges if and only if the set ${\int_{a}^{R} f : R > a}$ is bounded above. In the convergent case its supremum is the value of the improper integral. The analogous assertion holds at either finite singular endpoint and at $- \infty$ , with truncations directed toward that endpoint.

Facts & Assumptions

Given: A nonnegative, locally Riemann-integrable $f$ at one singular end.

[L1]

Monotonicity of the proper integral makes integrals over nonnegative functions nonnegative (If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ).

[L2]

A bounded monotone real sequence converges to its supremum or infimum (A monotone sequence converges if and only if it is bounded).

[L3]

Finite truncations may be moved without changing convergence (Improper convergence is independent of finite truncations and split points).

Proof

technique · direct

1.1

At $+ \infty$ , $F (R) = \int_{a}^{R} f$ is nondecreasing by [L1]. If $F (R)$ converges, its range is bounded. Conversely, if its range is bounded above, the integer sequence $F (n)$ is bounded and nondecreasing, so [L2] gives $F (n) \to S = \sup_{n} F (n)$ .

L1L2

2.1

For $n \leq R \leq n + 1$ , monotonicity gives $F (n) \leq F (R) \leq F (n + 1)$ . Hence $F (R) \to S$ . Every real truncation lies below a later integer truncation, so $S$ is also the supremum of the full truncation range.

step 1.1L1

3.1

Reciprocal truncations and the same squeeze prove the finite-endpoint forms; reversing orientation proves the $- \infty$ form. Moving the initial finite endpoint is harmless by [L3].

L3∎

Depends on

Used by

A bounded truncation function need not have an improper limit Counterexample
Comparison tests for improper integrals Theorem
Uniform oscillatory tail mass forces failure of absolute convergence Theorem

Dependency tree · two levels

26 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, Theorem 3.4.5 (standard reference, not scraped)