CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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.
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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A linear bijection need not preserve Jordan content

Statement refuted

Every linear bijection of $R^{2}$ preserves Jordan content.

Facts & Assumptions

Given: The unit square $Q = [0, 1]^{2}$ and the linear map $T (x, y) = (2 x, y)$ .

[L1]

If $E$ is bounded and Jordan measurable, a linear map with matrix $A$ sends $E$ to a bounded Jordan measurable set of content $∣ \det A ∣ cont (E)$ (A linear endomorphism of $R^{n}$ sends bounded Jordan sets to bounded Jordan sets and scales their content by the absolute determinant).

[L2]

A rectangle has volume equal to the product of its nonnegative side lengths (Axis-parallel rectangles in $R^{m}$ and their volume).

[L3]

Jordan inner and outer contents are obtained from finite rectangular inner families and outer covers, with every inner sum at most every outer sum (Jordan inner and outer content and Jordan measurable bounded sets in $R^{m}$ ).

Counterexample

technique · direct

1.1

The matrix of $T$ is $diag (2, 1)$ , with inverse $diag (1 / 2, 1)$ and determinant $2$ , and $T (Q) = [0, 2] \times [0, 1]$ .

L1algebra

2.1

Each rectangle itself is both a one-rectangle inner family and outer cover, so [L2] and [L3] give $cont (Q) = 1$ and $cont (T (Q)) = 2$ ; equivalently [L1] gives the same scaling. Thus the bijection $T$ does not preserve Jordan content.

step 1.1L1L2L3∎

Depends on

Used by

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Dependency tree · two levels

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