Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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A nonnegative improper integral converges iff its truncated integrals are bounded

Statement

Let f≥0f\ge0 be Riemann integrable on every compact subinterval of [a,∞)[a,\infty). Then ∫a∞f\int_a^\infty f converges if and only if the set {∫aRf:R>a}\left\{\int_a^R f:R>a\right\} 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 ff at one singular end.

[L1]

Monotonicity of the proper integral makes integrals over nonnegative functions nonnegative (If f≤gf \le g on [a,b][a,b] and both are integrable then ∫abf≤∫abg\int_a^b f \le \int_a^b g; and m(b−a)≤∫abf≤M(b−a)m(b-a) \le \int_a^b f \le 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)=∫aRfF(R)=\int_a^R f is nondecreasing by [L1]. If F(R)F(R) converges, its range is bounded. Conversely, if its range is bounded above, the integer sequence F(n)F(n) is bounded and nondecreasing, so [L2] gives F(n)→S=sup⁡nF(n)F(n)\to S=\sup_nF(n).

L1L2
2.1

For n≤R≤n+1n\le R\le n+1, monotonicity gives F(n)≤F(R)≤F(n+1)F(n)\le F(R)\le F(n+1). Hence F(R)→SF(R)\to S. Every real truncation lies below a later integer truncation, so SS 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∎

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