CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-09-08 (Codex)

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A nonmeasurable subset of a null line shows that the product of complete measures need not be complete

Statement refuted

Assuming the Axiom of Countable Choice, the product of two complete measure spaces is always complete.

Counterexample

technique · direct

Assume AC. Take a Vitali set $V \subseteq [0, 1]$ and put $E : = {0} \times V \subseteq R^{2} .$ The two factors are complete Lebesgue measure spaces, but their product is not complete. The verification supplies the witness using AC, not countable choice alone.

Facts & Assumptions

Given: The Axiom of Choice (The Axiom of Choice). Write $λ$ for Lebesgue measure on $L (R)$ .

[L1]

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

[L2]

Assuming countable choice, Lebesgue measure is sigma-finite. (Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure)

[L3]

For sigma-finite factors, the product measure satisfies the rectangle formula. (For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique)

[L4]

Assuming countable choice, the intervals ${0} = [0, 0]$ and $[0, 1]$ are Lebesgue measurable with measures $0$ and $1$ respectively. (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)

[L5]

Assuming countable choice, Euclidean Lebesgue measure is complete. (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume)

[L6]

Under AC a Vitali set on $[0, 1]$ exists (Assuming choice on the cosets of $Q$ in $R$ , a Vitali set in $[0, 1]$ exists, Vitali set on $[0, 1]$ ), and every such set is non-Lebesgue-measurable (Assuming the Axiom of Choice, a Vitali set is not Lebesgue measurable).

[L7]

Completeness means that every subset of a measurable null set is measurable (Complete measure spaces).

[L8]

Countable choice asks for a choice function on every sequence of nonempty sets (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Verification

1.1givenL6L8construct

AC implies [L8]: apply AC to the family ${X_{n} : n \in N}$ of a sequence of nonempty sets and compose its choice function with $n \mapsto X_{n}$ . This supplies countable choice for [L2], [L4] and [L5]. AC also supplies the selector in [L6]; fix that Vitali set $V \subseteq [0, 1]$ , which is non-Lebesgue-measurable, and set $E = {0} \times V$ .

2.1step 1.1L2L3L4algebra

Both factors are sigma-finite by [L2]. The rectangle $Z = {0} \times [0, 1]$ is product-measurable and [L3] and [L4] give $(λ \times λ) (Z) = λ ({0}) λ ([0, 1]) = 0 \cdot 1 = 0$ . Since $V \subseteq [0, 1]$ , we have $E \subseteq Z$ .

3.1step 1.1step 2.1L1L5L7∎

If $E$ belonged to $L (R) \otimes L (R)$ , then [L1] would make $E_{0} = {y : (0, y) \in E} = V$ Lebesgue measurable, contrary to step 1.1. Thus the subset $E$ of the measurable null set $Z$ is not product-measurable. By [L7] the product is not complete, although [L5] makes both factors complete.

Depends on

Used by

A completed-product measurable set can have a nonmeasurable exceptional section Counterexample

Dependency tree · two levels

62 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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