TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Comparison and absolute comparison tests for improper multiple integrals

Statement

For locally integrable $0 \leq f \leq g$ , comparison on compact subsets gives the same inequality for improper integrals:

$0 \leq \int_{D} f \leq \int_{D} g .$

If $∣ f ∣ \leq g$ and $\int_{D} g < + \infty$ , then $f$ is absolutely improperly integrable.

Facts & Assumptions

Given: An open set $D$ and locally Riemann-integrable functions with the pointwise inequalities in the Statement.

[L1]

Every compact Jordan exhaustion computes a nonnegative improper integral (Every Jordan exhaustion computes a nonnegative improper multiple integral).

[L2]

A signed function is improperly integrable precisely when the nonnegative improper integral of its absolute value is finite (Improper multiple integrals and absolute convergence on open sets).

[L3]

Proper Riemann integrals on a nondegenerate rectangle preserve pointwise inequalities (Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $R^{m}$ ).

[L4]

The integral over a bounded Jordan set is the bounding-rectangle integral of the zero extension (The Riemann integral of a bounded function over a bounded Jordan measurable set).

Proof

technique · direct

1.1

On every compact Jordan $K \subseteq D$ , extend $f ∣_{K}$ and $g ∣_{K}$ by zero to one bounding rectangle. Their zero extensions satisfy the same pointwise inequalities, so [L3] and [L4] give $0 \leq \int_{K} f \leq \int_{K} g$ ; taking the defining suprema, equivalently using [L1] on any exhaustion, gives $0 \leq \int_{D} f \leq \int_{D} g$ .

L1L3L4

2.1

If $∣ f ∣ \leq g$ and $\int_{D} g$ is finite, step 1.1 applied to $∣ f ∣$ gives $\int_{D} ∣ f ∣ \leq \int_{D} g < + \infty$ , so [L2] gives absolute improper integrability of $f$ .

step 1.1L2∎

Depends on

Used by

Dependency tree · two levels

22 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

V. Guillemin, MIT 18.101 Analysis II Lecture Notes, §§3.7–3.8 (standard reference, not scraped)