False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-29

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FALSE: the product Lebesgue sigma-algebra is the full Euclidean Lebesgue sigma-algebra

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). For all $m, n \geq 1$ , $L (R^{m}) \otimes L (R^{n}) = L (R^{m + n}) .$

Facts & Assumptions

Given: The Axiom of Countable Choice, Lebesgue measure $λ$ on $R$ , a non-Lebesgue-measurable set $N \subseteq R$ , the set $E : = {0} \times N \subseteq R^{2}$ , and the line $Z : = {0} \times R$ .

[L1]

The Euclidean Lebesgue measure is the completion of the product measure. (The Euclidean Lebesgue measure is the completion of the product of the factor Lebesgue measures)

[L2]

Every section of a product-measurable set is measurable. (Every section of a product-measurable set is measurable)

[L3]

On measurable rectangles, the product measure satisfies $(λ \times λ) (A \times B) = λ (A) λ (B)$ . (For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique)

Refutation

technique · direct

1.1L3algebra

For each $n \geq 1$ , the rectangle $Z_{n} : = {0} \times [- n, n]$ satisfies $(λ \times λ) (Z_{n}) = λ ({0}) λ ([- n, n]) = 0$ by [L3], so $Z = ⋃_{n \geq 1} Z_{n}$ is $(λ \times λ)$ -null.

2.1L1step 1.1

Because $E \subseteq Z$ , [L1] makes $E$ Lebesgue measurable in $R^{2}$ .

3.1L1L2step 2.1∎

If the displayed equality were true, then $E$ would belong to $L (R) \otimes L (R)$ . But then [L2] would force the horizontal section $E_{0} = N$ to be Lebesgue measurable, a contradiction. Therefore the product sigma-algebra is strictly smaller than the full Euclidean Lebesgue sigma-algebra. The correct statement is the completion statement of [L1].

Depends on

Used by

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Dependency tree · two levels

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Sources

John K. Hunter, Measure Theory, Example 5.20 (standard reference, not scraped)