CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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Without sigma-finiteness, the rectangle formula need not determine a unique product measure

Statement refuted

The rectangle formula determines at most one product measure even without sigma-finiteness.

Counterexample

technique · direct

Let $μ$ be Lebesgue measure on $[0, 1]$ , let $ν$ be counting measure on $[0, 1]$ , let $D : = {(x, y) \in [0, 1]^{2} : x = y}$ , and let $ρ, τ$ be the two measures on the product sigma-algebra supplied by the standard non-sigma-finite Lebesgue/counting construction in the listed Tao source.

Facts & Assumptions

Given: Lebesgue measure $μ$ on $[0, 1]$ , counting measure $ν$ on $[0, 1]$ , and the diagonal $D : = {(x, y) \in [0, 1]^{2} : x = y}$ .

[L1]

The listed Tao source's standard non-sigma-finite Lebesgue/counting construction yields measures $ρ, τ$ on the product sigma-algebra such that $ρ (A \times B) = μ (A) ν (B) = τ (A \times B)$ on measurable rectangles and $ρ (D) = 1 \neq 0 = τ (D)$ .

Verification

1.1

By [L1], $ρ$ and $τ$ agree on every measurable rectangle.

2.1

The same fact [L1, step 1.1] gives $ρ (D) = 1 \neq 0 = τ (D)$ , so the two measures are distinct. Therefore the rectangle formula does not determine a unique product measure without sigma-finiteness.

L1step 1.1∎

Used by

Nothing in the library uses this result yet.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Terence Tao, An Introduction to Measure Theory, Remark 1.7.12 (standard reference, not scraped)