CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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$\mathbb{Q} \cap [0,1]$ has measure zero and not content zero, although it is bounded

Statement refuted

Refuted claim: every set of measure zero has content zero (FALSE: every set of measure zero has content zero).

The witness is $E := \mathbb{Q}_{\mathbb{R}} \cap [0,1]$ , the rationals of the unit interval (The rationals embed densely in the reals, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length). It is at most countable, hence null; it is bounded; and every finite family of intervals covering it has total length at least $1$ , because the union of finitely many closed intervals is closed and contains the closure of $E$ , which is all of $[0,1]$ . The refutation is carried out in full in FALSE: every set of measure zero has content zero; this item records the witness and says what makes it work.

Facts & Assumptions

Given: The set $E = \mathbb{Q}_{\mathbb{R}} \cap [0,1]$ .

[A1]

The refuted claim: every subset of $\mathbb{R}$ of measure zero has content zero.

[L1]

$E$ is at most countable, being a subset of the countable set $\mathbb{Q}_{\mathbb{R}}$ , and therefore null ( $\mathbb{Q}$ is countably infinite, Every subset of an at most countable set is at most countable, Finite, countably infinite, countable, uncountable, The rationals embed densely in the reals, Every at most countable subset of $\mathbb{R}$ has measure zero).

[L2]

Every point of $[0,1]$ is adherent to $E$ , so any closed set containing $E$ contains $[0,1]$ ; and a finite union of closed intervals is closed (FALSE: every set of measure zero has content zero, Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$ , and every nonempty open subset of $\mathbb{R}$ is uncountable, The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points, Arbitrary unions and finite intersections of open subsets of $\mathbb{R}$ are open, and dually for closed sets, Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L3]

A finite family of closed intervals covering $[0,1]$ has total length at least $1$ (If finitely many intervals cover a closed bounded interval $[a,b]$ , the sum of their lengths is at least $b - a$ ).

[L4]

Content zero means a finite cover of total length below every positive $\varepsilon$ ; on compact sets content zero and measure zero coincide (Measure zero (a countable cover by intervals of total length below every $\varepsilon$ ) and content zero (a finite such cover), For a compact subset of $\mathbb{R}$ , measure zero and content zero coincide).

Counterexample

technique · direct

1.1

$E$ has measure zero by [L1], and $E \subseteq [0,1]$ is bounded.

1.2

Any finite family of closed intervals covering $E$ has total length at least $1$ : its union is closed by [L2] and contains $E$ , hence contains $[0,1]$ by [L2], and [L3] applies.

L2L3

2.1

So $E$ does not have content zero, since a witness at $\varepsilon = 2^{-1}$ would give a finite cover of total length at most $2^{-1} < 1$ ; $E$ therefore witnesses the failure of [A1].

step 1.1step 1.2A1L4∎

Remarks

Boundedness is not what is missing. $E$ is bounded, and its failure is as large as it can be: no finite cover does better than the trivial cover of $[0,1]$ by itself. What $E$ lacks is closedness, and hence compactness; adding it repairs the implication completely (For a compact subset of $\mathbb{R}$ , measure zero and content zero coincide).
The closure is the whole obstruction. Content zero is insensitive to passing to the closure, since a finite union of closed intervals is closed, whereas measure zero is not: $E$ is null and $\overline{E} = [0,1]$ is not (A sequence of intervals covering $[a,b]$ has total length at least $b - a$ , so no interval of positive length has measure zero). That single asymmetry is the entire difference between the two notions.
Compare the compact case. The Cantor set is uncountable and null, and being compact it also has content zero (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points). So the failure here is not about cardinality: a much larger null set can have content zero, and a countable one need not.

Depends on

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Sources

Jordan measure (Wikipedia) (standard reference, not scraped)
Null set (Wikipedia) (standard reference, not scraped)
MIT 18.125, Homework 2: Measure-zero sets (standard reference, not scraped)
UAF Math 641, Measure Theory notes (standard reference, not scraped)

Q∩[0,1]\mathbb{Q} \cap [0,1]Q∩[0,1] has measure zero and not content zero, although it is bounded