False statementConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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FALSE: some measurable set has density one half in every interval

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

There is a Lebesgue measurable set $E \subseteq R$ such that $\frac{λ (E \cap I)}{λ (I)} = \frac{1}{2}$ for every nondegenerate bounded interval $I$ .

Facts & Assumptions

Given: The Axiom of Countable Choice, and assume there is a measurable set $E \subseteq R$ with half-density in every nondegenerate bounded interval.

[L1]

Almost every point of a measurable set is a density-one point of that set. (Lebesgue density theorem)

[L2]

The interval $[0, 1]$ has 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)

Refutation

technique · direct

1.1

Applying the hypothesis to $I = [0, 1]$ and [L2], one gets [L2, given, algebra] $λ (E \cap [0, 1]) = \frac{1}{2} .$ So $E$ has positive measure.

L2givenalgebra

2.1

By [L1], almost every point of $E$ is a density-one point of $E$ . Choose [L1, step 1.1, given, choose, contradiction: density at x, discharge-contradiction] such a point $x \in E$ . But the hypothesis applied to every interval $(x - r, x + r)$ gives $\frac{λ (E \cap (x - r, x + r))}{2 r} = \frac{1}{2} (r > 0),$ so the density of $E$ at $x$ is $1 / 2$ , not $1$ . This contradiction refutes the claim.

L1step 1.1givenchoosecontradiction: density at xdischarge-contradiction∎

Depends on

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Sources

Terence Tao, An Introduction to Measure Theory, Exercise 1.6.26(ii) (standard reference, not scraped)