TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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Assuming the Axiom of Choice, no translation-invariant measure on $P (R)$ is both finite and nonzero on $[0, 1]$

Statement

Assume the Axiom of Choice. There is no measure $μ$ on $P (R)$ such that

$μ (E + q) = μ (E)$ for every subset $E \subseteq R$ and every rational $q$ ;
$0 < μ ([0, 1]) < + \infty$ .

Facts & Assumptions

Given: The Axiom of Choice, a measure $μ$ on $P (R)$ , rational-translation invariance of $μ$ , and $0 < μ ([0, 1]) < + \infty$ .

[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]

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

[L3]

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

[L4]

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

By [L1] fix a Vitali set $V \subseteq [0, 1]$ , and enumerate $Q \cap [- 1, 1]$ as $(q_{k})_{k \in N}$ by [L2]. Exactly as in the Vitali argument, the translates $V + q_{k}$ are pairwise disjoint, they cover $[0, 1]$ , and they all lie inside $[- 1, 2]$ .

L1L2construct

1.2

Since $[- 1, 2] \subseteq [- 1, 0] \cup [0, 1] \cup [1, 2]$ , translation invariance and finite subadditivity from [L3] give $μ ([- 1, 2]) \leq 3 μ ([0, 1]) < + \infty$ .

givenL3algebra

2.1

If $μ (V) = 0$ , then every translate $V + q_{k}$ has measure $0$ and countable additivity on the disjoint family of step 1.1 gives $μ ([0, 1]) = 0$ , contradicting the hypotheses.

step 1.1L3assume-contra

2.2

If $μ (V) > 0$ , choose a natural number $m \geq 1$ with $μ ([- 1, 2]) < m μ (V)$ by [L4]. The first $m$ translates from step 1.1 are pairwise disjoint subsets of $[- 1, 2]$ , so [L3] gives $μ ([- 1, 2]) \geq \sum_{k < m} μ (V + q_{k}) = m μ (V)$ , contradicting the choice of $m$ .

step 1.1step 1.2L3L4assume-contra

3.1

Steps 2.1 and 2.2 rule out both possibilities for $μ (V)$ , so no such translation-invariant measure $μ$ exists.

step 2.1step 2.2discharge-contradiction∎

Depends on

Used by

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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, Chapter 8 (standard reference, not scraped)

Assuming the Axiom of Choice, no translation-invariant measure on P(R)P(R) is both finite and nonzero on [0,1][0,1]