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

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False: locally finite volume has finite total mass

Statement

False assertion: every positive smooth density whose Borel measure is locally finite has finite total mass.

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 $λ_{0} (R^{0}) = 1$ . Witness R with unit density; explicit arbitrarily large finite-interval masses.

[F1]

Positive smooth densities give Radon volume: A finite-valued positive smooth density has locally finite, compact-finite measure.

[F2]

Intrinsic density measure and its chart restriction: A chart restriction computes its measure by the coordinate integral.

[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: An interval has its length as Lebesgue measure.

Refutation

1.1

Take $M = R$ with its identity chart and $r = ∣ d x ∣$ , coefficient one. This is a finite-valued positive smooth density, so its measure is locally finite and finite on compact sets. For every integer $N \geq 1$ , $μ_{r} ([- N, N]) = \int_{- N}^{N} 1 d x = 2 N$ .

F1F2F3

2.1

For any finite proposed bound $L$ , choose an integer $N > L / 2$ . Monotonicity gives $μ_{r} (R) \geq 2 N > L$ , so the total mass is infinite. The empty set and singleton have mass zero by the same length formula, and intervals of length one have mass one; none of these local values bounds the total.

F3step 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)