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

Statement

Assume the Axiom of Countable Choice. Let $B \subseteq R$ be a Bernstein set. Then $B$ is not Lebesgue measurable.

Facts & Assumptions

Given: The Axiom of Countable Choice and a Bernstein set $B \subseteq R$ .

[L1]

A Bernstein set has inner measure $0$ , and in every nondegenerate bounded interval its intersection has full outer measure (A Bernstein set has inner measure $0$ , and in every nondegenerate interval its intersection has full outer measure).

[L2]

For bounded subsets of $R$ , Lebesgue measurability is equivalent to equality of inner and outer measure (For bounded subsets of $R$ , Lebesgue measurability is equivalent to equality of inner and outer measure).

Proof

technique · direct

1.1

Let $I = [0, 1]$ . Step [L1] gives $λ_{*} (B \cap I) = 0$ and $λ^{*} (B \cap I) = λ (I) = 1$ .

2.1

The bounded set $B \cap I$ therefore has unequal inner and outer measure, so [L2] says it is not Lebesgue measurable. If $B$ itself were Lebesgue measurable, then its intersection with the measurable interval $I$ would be too, contradiction. Hence $B$ is not Lebesgue measurable.

step 1.1L2∎

Depends on

Used by

Assuming countable choice and a well-ordering of the real line, a Bernstein set is dense, has inner measure 0, and is not Lebesgue measurable Counterexample

Dependency tree · two levels

21 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

Jacek Cichoń, Aleksander Kharazishvili, and Bogdan Węglorz, Subsets of the Real Line, Chapter 8 (standard reference, not scraped)