CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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Two disjoint nonmeasurable subsets of $[0, 1]$ can have the measurable union $[0, 1]$

Statement refuted

Refuted claim: if $E$ and $F$ are disjoint subsets of $[0, 1]$ and $E \cup F$ is Lebesgue measurable, then both $E$ and $F$ are Lebesgue measurable.

Facts & Assumptions

Given: The Axiom of Choice.

[L1]

Assuming the Axiom of Choice, a Vitali set in $[0, 1]$ exists (Assuming choice on the cosets of $Q$ in $R$ , a Vitali set in $[0, 1]$ exists).

[L2]

Every Vitali set is not Lebesgue measurable (Assuming the Axiom of Choice, a Vitali set is not Lebesgue measurable).

[L3]

The interval $[0, 1]$ is Lebesgue measurable with measure $1$ (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).

Counterexample

technique · direct

1.1L1L3construct

Choose a Vitali set $V \subseteq [0, 1]$ by [L1], and put $W : = [0, 1] ∖ V$ . Then $V$ and $W$ are disjoint and $V \cup W = [0, 1]$ , which is measurable by [L3].

2.1step 1.1L2L3∎

The set $V$ is not measurable by [L2]. If $W$ were measurable, then $V = [0, 1] ∖ W$ would also be measurable because $[0, 1]$ is measurable, contradiction. So $W$ is not measurable either, and the pair $(V, W)$ refutes the claim.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 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

Vitali set (Wikipedia) (standard reference, not scraped)

Two disjoint nonmeasurable subsets of [0,1][0,1] can have the measurable union [0,1][0,1]