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

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FALSE: every set of measure zero has content zero

Statement

False claim: every set of measure zero has content zero (Measure zero (a countable cover by intervals of total length below every $\varepsilon$ ) and content zero (a finite such cover)).

The converse is true and is A set of content zero has measure zero; the two notions do coincide for compact sets (For a compact subset of $\mathbb{R}$ , measure zero and content zero coincide). The claim above drops the compactness, and boundedness alone is not a substitute: the witness below is a bounded set of measure zero with no finite cover by intervals of total length less than $1$ .

Facts & Assumptions

Given: The set $E := \mathbb{Q}_{\mathbb{R}} \cap [0,1]$ , where $\mathbb{Q}_{\mathbb{R}}$ is the image of $\mathbb{Q}$ in $\mathbb{R}$ (The rationals embed densely in the reals).

[A1]

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

[L1]

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

[L2]

$\mathbb{Q}_{\mathbb{R}}$ is dense in $\mathbb{R}$ : strictly between any two reals lies a rational (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 rationals embed densely in the reals).

[L3]

$[c,d]$ is a closed set, a finite union of closed sets is closed, and $\overline{A}$ is the set of points every neighbourhood of which meets $A$ , so a closed set containing $A$ contains $\overline{A}$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, Arbitrary unions and finite intersections of open subsets of $\mathbb{R}$ are open, and dually for closed sets, 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, The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L4]

If $[a,b] \subseteq \bigcup_{j \le n}[c_j,d_j]$ with $a \le b$ and $c_j \le d_j$ , then $\sum_{j \le n}(d_j - c_j) \ge b - a$ (If finitely many intervals cover a closed bounded interval $[a,b]$ , the sum of their lengths is at least $b - a$ ).

[L5]

$A$ has content zero when for every real $\varepsilon > 0$ it has a finite cover by closed intervals of total length at most $\varepsilon$ (Measure zero (a countable cover by intervals of total length below every $\varepsilon$ ) and content zero (a finite such cover)).

[L6]

Every nonempty finite set of reals has a maximum and a minimum (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L7]

Ordered-field arithmetic: $0 < 1$ , so $2 > 0$ and $2^{-1} > 0$ and $2^{-1} < 1$ ; adding a constant and multiplying by a positive preserve an inequality; the order is total and transitive (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Refutation

technique · direct

1.1

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

1.2

Every $x \in [0,1]$ is adherent to $E$ : given a real $\varepsilon > 0$ , put $p := \max\{0,\ x - \varepsilon\}$ and $q := \min\{1,\ x + \varepsilon\}$ , which exist by [L6]. Then $p < q$ : indeed $p \le x \le q$ by [L7] and $0 \le x \le 1$ , while $p = x$ would need $x \le 0$ hence $x = 0 < \min\{1,\varepsilon\} = q$ , and $q = x$ would need $x \ge 1$ hence $x = 1 > \max\{0, 1-\varepsilon\} = p$ , and otherwise $p < x < q$ . By [L2] there is a rational strictly between $p$ and $q$ ; it lies in $[0,1]$ because $0 \le p$ and $q \le 1$ , and within $\varepsilon$ of $x$ because $x - \varepsilon \le p$ and $q \le x + \varepsilon$ . So $N_\varepsilon(x) \cap E \ne \varnothing$ .

L2L6L7

2.1

Let $n \in \mathbb{N}$ and $c_0 \le d_0, \dots, c_n \le d_n$ be any finite family of closed intervals with $E \subseteq \bigcup_{j \le n}[c_j,d_j]$ . The union $\bigcup_{j\le n}[c_j,d_j]$ is a closed set by [L3], and it contains $E$ , hence contains $\overline{E}$ by [L3]; by step 1.2 every point of $[0,1]$ lies in $\overline{E}$ , so $[0,1] \subseteq \bigcup_{j \le n}[c_j,d_j]$ and [L4] gives $\sum_{j \le n}(d_j - c_j) \ge 1$ .

step 1.2L3L4

3.1

So no finite family of closed intervals covers $E$ with total length at most $2^{-1} < 1$ , and $E$ does not have content zero by [L5] and [L7]; yet $E$ has measure zero by step 1.1. The claim [A1] therefore fails at $E$ and is false.

step 1.1step 2.1A1L5L7∎

Remarks

Boundedness is not the missing hypothesis, closedness is. $E$ is bounded and its failure is total: no finite cover does better than total length $1$ , the same bound as for all of $[0,1]$ . What $E$ lacks is closedness, and with it compactness; For a compact subset of $\mathbb{R}$ , measure zero and content zero coincide shows that supplying it repairs the implication completely.
The gap between the two notions is a quantifier, not a constant. Given $\varepsilon$ , the countable cover of $E$ from Every at most countable subset of $\mathbb{R}$ has measure zero uses intervals whose lengths shrink geometrically; no finite initial segment of it covers $E$ , because the rationals left over are still dense in $[0,1]$ . Compactness is exactly what turns a countable cover into a finite one, and that is the whole content of the repair.
The named witness is $\mathbb{Q} \cap [0,1]$ has measure zero and not content zero, although it is bounded ↗.

Depends on

Used by

ℚ ∩ [0,1] has measure zero and not content zero, although it is bounded Counterexample

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 135 results over 35 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

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)