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Every subset of $R$ of positive Lebesgue outer measure contains a nonmeasurable subset

Statement

Assume the Axiom of Choice. Let $A \subseteq R$ satisfy $λ^{*} (A) > 0$ . Then $A$ contains a subset that is not Lebesgue measurable.

Facts & Assumptions

Given: The Axiom of Choice and a subset $A \subseteq R$ with $λ^{*} (A) > 0$ .

[L0]

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

[L2]

Assuming countable choice, if a Lebesgue measurable subset of $R$ has positive measure then its difference set contains an open ball about $0$ (If a Lebesgue measurable subset of $R^{n}$ has positive measure, its difference set contains an open ball about the origin).

[L3]

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

[L4]

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

[L5]

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

Proof

technique · contradiction

1.1

By [L0] fix a Vitali set $V \subseteq [0, 1]$ . For each rational $q$ put $A_{q} : = A \cap (V + q)$ . The sets $A_{q}$ are pairwise disjoint because the translates $V + q$ are, and they cover $A$ because the rational translates of a Vitali selector cover all of $R$ .

givenL0L3construct

2.1

Suppose some $A_{q}$ were Lebesgue measurable with positive measure. Then [L2] would give an open interval about $0$ inside $A_{q} - A_{q}$ . But if $x, y \in A_{q} \subseteq V + q$ , then $x - y \in Q$ implies $(x - q) - (y - q) \in Q$ with both terms in $V$ , so the selector property forces $x = y$ ; thus $A_{q} - A_{q}$ contains no nonzero rational. Every open interval about $0$ contains a nonzero rational, contradiction. Hence every measurable $A_{q}$ has measure $0$ .

step 1.1L2assume-contra

3.1

Suppose, for contradiction, that every $A_{q}$ is Lebesgue measurable. Then step 2.1 gives $λ (A_{q}) = 0$ for every rational $q$ , and by [L4] each $A_{q}$ is measurable with outer measure $0$ . Since the family is pairwise disjoint and covers $A$ , countable additivity from [L5] gives $λ (A) = 0$ , hence $λ^{*} (A) = 0$ , contradicting the hypothesis. Therefore at least one set $A_{q}$ is not Lebesgue measurable, and it is a subset of $A$ .

step 1.1step 2.1L3L4L5discharge-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)
Steinhaus theorem (Wikipedia) (standard reference, not scraped)

Every subset of RR of positive Lebesgue outer measure contains a nonmeasurable subset