CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09

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A smooth positive density with infinite mass

Statement refuted

The assertion that every positive smooth density measure has finite total mass fails for the density $∣ d x ∣$ on $R$ .

Facts & Assumptions

Given: Assume ${A C}_{ω}$ . Manifolds are Hausdorff, second countable and smooth, with boundary allowed; $n = 0$ is allowed unless excluded. Densities are pointwise Borel, $0 \cdot \infty = 0$ , and $λ_{0} (R^{0}) = 1$ . Independent B-page witness: compact-finite Euclidean density with total mass infinity.

[F1]

Positive smooth densities give Radon volume: The measure of a positive finite smooth density is locally finite and compact-finite.

[F2]

Intrinsic density measure and its chart restriction: The identity-chart density one integrates to Lebesgue measure.

[F3]

A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included: A closed interval has its length as measure.

Counterexample

1.1

On $R$ the coefficient $r_{x} = 1$ is positive and smooth. Its measure is locally finite, and $μ_{r} ([- N, N]) = \int_{- N}^{N} 1 d x = 2 N$ for every integer $N \geq 1$ . Each compact K is bounded, hence contained in some $[- N, N]$ and has finite measure.

F1F2F3

2.1

Given any finite $L > 0$ , an integer $N > L / 2$ yields $μ_{r} (R) \geq 2 N > L$ , proving infinite total mass. Thus the hypothesis holds and the asserted finite-total-mass conclusion fails. The same chart computation gives $μ_{r} (\emptyset) = μ_{r} ({0}) = 0$ and $μ_{r} ([0, 1]) = 1$ .

F2F3step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Folland, Real Analysis, second edition, §11.4 pp.361–363; Theorems 2.14–2.15 pp.50–51 (standard reference, not scraped)