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Assuming the Axiom of Choice, a Vitali set is not Lebesgue measurable

Statement

Assume the Axiom of Choice. Let $V \subseteq [0, 1]$ be a Vitali set. Then $V$ is not Lebesgue measurable.

Facts & Assumptions

Given: The Axiom of Choice and a Vitali set $V \subseteq [0, 1]$ .

[L2]

Assuming countable choice, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

[L3]

Assuming countable choice, $[0, 1]$ and $[- 1, 2]$ are Lebesgue measurable with measures $1$ and $3$ 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).

[L4]

$Q$ is countably infinite ( $Q$ is countably infinite).

[L5]

A measure vanishes on $\emptyset$ and is countably additive on pairwise disjoint measurable sequences (Measures on sigma-algebras).

[L6]

For every real $M$ there is a natural number $n \geq 1$ with $M < n$ (Every complete ordered field is Archimedean).

Proof

technique · contradiction

1.1

Enumerate $Q \cap [- 1, 1]$ as $(q_{k})_{k \in N}$ by [L4]. The translates $V + q_{k}$ are pairwise disjoint: if $v_{1} + q_{i} = v_{2} + q_{j}$ then $v_{1} - v_{2} = q_{j} - q_{i} \in Q$ , so the selector property of $V$ forces $v_{1} = v_{2}$ and then $q_{i} = q_{j}$ . They cover $[0, 1]$ , because every $x \in [0, 1]$ differs by a rational in $[- 1, 1]$ from the unique $v \in V$ in its equivalence class, and each lies inside $[- 1, 2]$ because $0 \leq v \leq 1$ and $- 1 \leq q_{k} \leq 1$ .

givenL4algebra

2.1

Suppose, for contradiction, that $V$ is Lebesgue measurable with $λ (V) = 0$ . Then every translate $V + q_{k}$ is measurable with measure $0$ by [L2], and countable additivity applied to the pairwise disjoint family of step 1.1 gives $1 = λ ([0, 1]) \leq λ ⁣ (⋃_{k \in N} (V + q_{k})) = \sum_{k = 0}^{\infty} λ (V + q_{k}) = 0$ , contradicting [L3].

step 1.1L2L3L5assume-contra

2.2

Suppose instead that $V$ is Lebesgue measurable with $λ (V) > 0$ . By [L6] choose a natural number $m \geq 1$ with $3 < m λ (V)$ . The first $m$ translates of step 1.1 are pairwise disjoint and all lie inside $[- 1, 2]$ , so [L2], [L3] and finite additivity from [L5] give $3 = λ ([- 1, 2]) \geq \sum_{k < m} λ (V + q_{k}) = m λ (V)$ , contradicting the choice of $m$ .

step 1.1L2L3L5L6assume-contra

3.1

Steps 2.1 and 2.2 rule out both possible values of the measure of a measurable Vitali set, so $V$ is not Lebesgue measurable.

step 2.1step 2.2discharge-contradiction∎

Remarks

The proof uses only countably many rational translates after the initial selector has been fixed. The countable part is not where the choice cost sits.

Depends on

Used by

A Vitali set shows that not every subset of ℝ is Lebesgue measurable Counterexample
Two disjoint nonmeasurable subsets of [0,1] can have the measurable union [0,1] Counterexample
FALSE: assuming the Axiom of Choice, every subset of ℝ is Lebesgue measurable False statement

Dependency tree · two levels

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Sources

T. Tao, An Introduction to Measure Theory (GSM 126), Section 1.2 (standard reference, not scraped)
Jacek Cichoń, Aleksander Kharazishvili, and Bogdan Węglorz, Subsets of the Real Line, Theorem 8.2 (standard reference, not scraped)