False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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FALSE: Tonelli's theorem still holds without any sigma-finiteness hypothesis

Statement

For arbitrary measure spaces, every nonnegative product-measurable function satisfies Tonelli's theorem.

Facts & Assumptions

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

[L1]

Tonelli's theorem holds on sigma-finite product spaces. (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product)

[A1]

Counting measure on the uncountable set $[0, 1]$ is not sigma-finite in the sense of Finite, sigma-finite, and semifinite measures.

[A2]

For every $x, y \in [0, 1]$ , the diagonal sections are $D_{x} = {x}$ and $D^{y} = {y}$ .

Refutation

technique · direct

1.1A1algebra

The diagonal $D$ is closed in $[0, 1]^{2}$ , so $f = 1_{D}$ is a nonnegative measurable function; by [A1], it lives on a product space outside the sigma-finite scope of [L1].

1.2A2algebra

For fixed $x \in [0, 1]$ , [A2] gives $ν (D_{x}) = 1$ , so $\int_{[0, 1]} ν (D_{x}) d μ (x) = \int_{0}^{1} 1 d x = 1.$

1.3A2algebra

For fixed $y \in [0, 1]$ , [A2] gives $μ (D^{y}) = 0$ , so $\int_{[0, 1]} μ (D^{y}) d ν (y) = \int_{[0, 1]} 0 d ν = 0.$

2.1L1step 1.1step 1.2step 1.3∎

Steps 1.2 and 1.3 give unequal iterated integrals for the same nonnegative measurable function on a non-sigma-finite product space. Hence the displayed universal claim is false, and [L1] cannot be extended by simply deleting sigma-finiteness.

Depends on

Used by

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Dependency tree · two levels

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Sources

Gerald B. Folland, Real Analysis, 2nd ed., remark after Theorem 2.37 (standard reference, not scraped)