LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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An integrable dominator gives uniform tail control on every compact parameter set

Statement

An integrable dominator gives one compact Jordan core outside which every dominated slice has uniformly small integral.

Precisely, let $D \subseteq R^{n}$ be open, let $I \subseteq R$ be an interval, and let $f : D \times I \to R$ have locally Riemann-integrable slices $f_{t}$ . Let $C \subseteq I$ be compact and suppose $∣ f (x, t) ∣ \leq g (x)$ for $x \in D$ and $t \in C$ , where $g \geq 0$ is locally Riemann integrable and $\int_{D} g < + \infty$ . Then every $f_{t}$ with $t \in C$ , and every difference $f_{t} - f_{s}$ , is absolutely improperly integrable. For every $ε > 0$ there is a compact Jordan $K \subseteq D$ such that for every $s, t \in C$ ,

$∣ \int_{D} f_{t} - \int_{K} f_{t} ∣ < ε, ∣ \int_{D} (f_{t} - f_{s}) - \int_{K} (f_{t} - f_{s}) ∣ < 2 ε .$

Facts & Assumptions

Given: The functions, compact parameter set, dominator, and $ε > 0$ of the Statement.

[L1]

Every compact Jordan exhaustion computes the nonnegative improper integral, independently of the exhaustion (Every Jordan exhaustion computes a nonnegative improper multiple integral).

[L2]

For locally integrable $0 \leq u \leq v$ , one has $\int_{D} u \leq \int_{D} v$ ; if $∣ u ∣ \leq v$ and $\int_{D} v < + \infty$ , then $u$ is absolutely improperly integrable (Comparison and absolute comparison tests for improper multiple integrals).

[L3]

Proper multidimensional integrals are linear, monotone, and satisfy $∣ \int u ∣ \leq \int ∣ u ∣$ (Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $R^{m}$ ).

[L4]

Every open subset of $R^{n}$ has a compact Jordan exhaustion (Every open subset of $R^{n}$ admits a compact Jordan exhaustion).

[L5]

For an absolutely improperly integrable function, proper integrals along every compact Jordan exhaustion converge to its improper integral (Absolute convergence makes signed improper multiple integrals independent of exhaustion).

Proof

technique · direct

1.1L1L4choose

Choose a compact Jordan exhaustion $(K_{j})$ by [L4]. Then [L1] gives $\int_{K_{j}} g \to \int_{D} g < + \infty$ , so choose $K = K_{j}$ with $0 \leq \int_{D} g - \int_{K} g < ε$ .

2.1step 1.1L1L2L3L5

For every $t \in C$ , [L2] first makes $f_{t}$ absolutely improperly integrable. For every later exhaustion member $K_{i} \supseteq K$ , [L3] applied to the zero extensions gives $∣ \int_{K_{i}} f_{t} - \int_{K} f_{t} ∣ \leq \int_{K_{i}} g - \int_{K} g$ . Passing to the exhaustion limits by [L1] and [L5] gives $∣ \int_{D} f_{t} - \int_{K} f_{t} ∣ \leq \int_{D} g - \int_{K} g < ε$ .

3.1step 2.1L1L2L3L5algebra∎

Since $∣ f_{t} - f_{s} ∣ \leq ∣ f_{t} ∣ + ∣ f_{s} ∣ \leq 2 g$ , [L2] makes every difference absolutely improperly integrable, and the same [L1], [L3], and [L5] argument gives the second estimate with $2 ε$ , uniformly for $s, t \in C$ .

Depends on

Used by

Dependency tree · two levels

30 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

W. F. Trench, Functions Defined by Improper Integrals, §7 (standard reference, not scraped)