LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-29

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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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The published Riemann change-of-variables theorem already gives the Lebesgue formula for continuous compactly supported integrands

Statement

Let $U, V \subseteq R^{n}$ be open and let $T : U \to V$ be a $C^{1}$ diffeomorphism. If $f : V \to R$ is continuous and compactly supported, then $\int_{V} f (y) d λ_{n} (y) = \int_{U} f (T (x)) ∣ \det D T (x) ∣ d λ_{n} (x) .$

Facts & Assumptions

Given: Open sets $U, V \subseteq R^{n}$ , a $C^{1}$ diffeomorphism $T : U \to V$ , and a continuous compactly supported function $f : V \to R$ .

[L1]

The published Riemann change-of-variables theorem holds for compactly supported integrands. (A compactly supported Riemann integrand admits the global change-of-variables formula from a diffeomorphism near the relevant compact preimage)

[L2]

On Jordan measurable compact sets, the Riemann and Lebesgue integrals agree. (Lebesgue outer measure is at most Jordan outer content, and a bounded Jordan measurable set is Lebesgue measurable with Lebesgue measure equal to its Jordan content)

Proof

technique · direct

1.1L1L2

Because $f$ has compact support inside $V$ , there is a compact Jordan set $K \subseteq V$ containing $supp (f)$ . Then $f$ vanishes on $V ∖ K$ , so both integrals in the statement reduce to integrals over $K$ and $T^{- 1} (K)$ .

2.1L1L2step 1.1∎

The Riemann theorem [L1] applies on these compact sets, and [L2] identifies those Riemann integrals with the corresponding Lebesgue integrals. Therefore the displayed Lebesgue change-of-variables formula holds.

Depends on

Used by

Dependency tree · two levels

31 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

Gerald B. Folland, Real Analysis, 2nd ed., Theorem 2.47 (standard reference, not scraped)