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TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21
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Locally dominated parameter-dependent improper multiple integrals are continuous

Statement

A locally dominated parameter-dependent improper multiple integral is continuous in the parameter.

More precisely, let f:D×IR be continuous and suppose it is locally dominated near each parameter in the sense of Parameter-dependent improper multiple integrals. Then F(t)=Df(x,t)dx is continuous on I in the relative topology.

Facts & Assumptions

Given: The continuous integrand f, parameter interval I, and local domination in the Statement; fix t0I.

[L1]

If locally integrable slices ft on an open D satisfy ftg on a compact parameter set, where g0 and Dg<+, then for every η>0 one compact Jordan KD satisfies DftKft<η for every such t (An integrable dominator gives uniform tail control on every compact parameter set).

[L2]

A continuous map from a compact metric space is uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).

[L3]

Proper multidimensional Riemann integrals are monotone and satisfy the absolute-value estimate (Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in Rm).

[L5]

An integrable nonnegative dominator makes every locally integrable dominated slice absolutely improperly integrable (Comparison and absolute comparison tests for improper multiple integrals).

[L6]

Every continuous real function on a compact Jordan set is Riemann integrable there (A continuous real function on a compact Jordan measurable set is Riemann integrable over that set).

Proof

technique · direct
1.1

Choose a compact relative parameter neighborhood CI of t0 and an integrable g dominating ft there. Continuity gives local integrability and [L5] gives absolute improper integrability of the slices. Given ε>0, [L1] supplies a compact Jordan core KD on which both tail errors are below ε/3.

L1L5choose
1.2

The restriction of f to the compact set K×C is uniformly continuous by [L2], while [L6] supplies all proper core integrals. Hence, for tC sufficiently close to t0, f(x,t)f(x,t0)<ε/(3(1+contK)) for every xK, and [L3] with [L4] makes the compact-core integral difference smaller than ε/3.

L2L3L4L6
2.1

Split F(t)F(t0) into the two tail errors and the core-integral difference. Steps 1.1 and 1.2 make its absolute value smaller than ε, proving relative continuity at t0, including a one-sided parameter endpoint.

step 1.1step 1.2algebra

Depends on

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Sources