False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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FALSE: a dense subset of $R$ of outer measure zero and a dense subset of full inner measure cannot both meet every open interval

Statement

A dense subset of $R$ of outer measure zero and a dense subset of full inner measure cannot both meet every open interval.

Facts & Assumptions

Given: The rational reals $Q_{R} \subseteq R$ .

[L1]

$Q_{R}$ is dense in $R$ , its complement is dense, and every nonempty open subset of $R$ is uncountable (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable).

[L2]

Every at most countable subset of $R$ is Lebesgue null; in particular $λ (Q_{R}) = 0$ (Every at most countable subset of $R^{n}$ is Lebesgue null; in particular $λ_{1} (Q) = 0$ ).

[L3]

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).

[L4]

$λ_{*} (E) = \sup {λ (K) : K \subseteq E compact}$ (Lebesgue inner measure on the real line).

[L5]

Every bounded nondegenerate interval is Lebesgue measurable with its usual length (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

The set $Q_{R}$ is dense by [L1], and it has outer measure $0$ because it is countably infinite and therefore Lebesgue null by [L2].

L1L2

2.1

Let $I \subseteq R$ be a bounded nondegenerate interval. Then $I ∖ Q_{R}$ is measurable, because $I$ is measurable by [L5] and $Q_{R}$ has measure $0$ by step 1.1. Also $λ (I ∖ Q_{R}) = λ (I)$ , so [L3] gives $λ_{*} (I ∖ Q_{R}) = λ (I)$ .

step 1.1L3L4L5algebra

3.1

Every nonempty open interval contains both a rational and an irrational by [L1]. So $Q_{R}$ is dense, and $R ∖ Q_{R}$ is also dense. Step 2.1 shows that inside every bounded nondegenerate interval the latter has full inner measure. These two dense sets therefore coexist, and the statement is false.

step 2.1L1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

38 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

Dense set (Wikipedia) (standard reference, not scraped)
Null set (Wikipedia) (standard reference, not scraped)

FALSE: a dense subset of RR of outer measure zero and a dense subset of full inner measure cannot both meet every open interval