TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)

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For a compact subset of $\mathbb{R}$ , measure zero and content zero coincide

Statement

Let $K \subseteq \mathbb{R}$ be compact (Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset), equivalently closed and bounded (A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded). Then

$K \text{ has measure zero} \quad \Longleftrightarrow \quad K \text{ has content zero}$

(Measure zero (a countable cover by intervals of total length below every $\varepsilon$ ) and content zero (a finite such cover)).

The implication from content zero to measure zero is A set of content zero has measure zero and needs no hypothesis on $K$ . The other direction is the one that uses compactness, and it uses it exactly as A sequence of intervals covering $[a,b]$ has total length at least $b - a$ , so no interval of positive length has measure zero does: a countable cover is enlarged to an open cover at an arbitrarily small cost in total length, and compactness reduces the open cover to a finite one.

Facts & Assumptions

Given: A compact set $K \subseteq \mathbb{R}$ and a real $\varepsilon > 0$ . Throughout, $\theta := 2^{-1}$ .

[L1]

$A$ is null when for every real $\eta > 0$ there are sequences $(a_k)$ , $(b_k)$ with $a_k \le b_k$ , $A \subseteq \bigcup_k[a_k,b_k]$ and $\sum_{k<i}(b_k - a_k) \le \eta$ for every $i$ ; $A$ has content zero when the same holds with a finite list (Measure zero (a countable cover by intervals of total length below every $\varepsilon$ ) and content zero (a finite such cover)).

[L2]

A set of content zero is null (A set of content zero has measure zero).

[L3]

$[c,d]$ has length $d - c \ge 0$ for $c \le d$ ; $(c,d)$ is the open interval with the same endpoints and is contained in $[c,d]$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L4]

Every open interval is an open set (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).

[L5]

$K$ is compact: from every family of open sets whose union contains $K$ , either $K = \varnothing$ and the empty subfamily covers it, or there are $m \in \mathbb{N}$ and members $U_0, \dots, U_m$ of the family whose union contains $K$ ; compactness is equivalent to being closed and bounded (Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset, A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded).

[L6]

Powers and the geometric series: $\theta^0 = 1$ , $\theta^{k+1} = \theta^k\theta$ , $\theta^k > 0$ , and $\sum_{k=0}^{\infty}\theta^k = 2$ for $\theta = 2^{-1}$ ; a series of nonnegative terms has all its partial sums at most its sum (Integer powers $a^m$ , For $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ the series diverges, Series, partial sums, convergence and the sum, divergence, and the tail series, A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum).

[L7]

Finite sums: additivity, scaling, splitting and monotonicity in the terms (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L8]

Every finite list of naturals has an upper bound in $\mathbb{N}$ , by induction on its length and the totality of the order of $\mathbb{N}$ (The principle of mathematical induction, Trichotomy of the order on $\mathbb{N}$ , Order on the natural numbers).

[L9]

Ordered-field arithmetic: $0 < 1$ , so $2 > 0$ , $4 > 0$ , $8 > 0$ and $t \cdot 8^{-1} > 0$ for $t > 0$ ; adding a constant and multiplying by a positive preserve an inequality (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.

Proof

technique · direct

1.1

One direction is immediate: if $K$ has content zero then $K$ is null by [L2], with no hypothesis on $K$ used. It remains to prove the converse for compact $K$ .

L2suffices: only the forward direction remains

1.2

If $K = \varnothing$ , then for every real $\varepsilon > 0$ the single interval $[0,0]$ covers $K$ and has total length $0 \le \varepsilon$ , so $K$ has content zero by [L1]. Hence suppose $K \ne \varnothing$ for the rest of the proof.

L1cases

2.1

Assume $K$ is null and let the real $\varepsilon > 0$ be given. By [L1] applied with $\eta := \varepsilon \cdot 2^{-1} > 0$ fix sequences $(a_k)$ , $(b_k)$ with $a_k \le b_k$ , $K \subseteq \bigcup_k [a_k,b_k]$ and $\sum_{k<i}(b_k - a_k) \le \varepsilon \cdot 2^{-1}$ for every $i \in \mathbb{N}$ .

step 1.1givenL1L9choose

3.1

Put $\delta_k := \varepsilon \cdot 8^{-1} \cdot \theta^{k}$ , a positive real by [L6] and [L9], and $J_k := (a_k - \delta_k,\ b_k + \delta_k)$ , an open set by [L4] containing $[a_k,b_k]$ by [L3] and [L9]. Hence $\{\, J_k : k \in \mathbb{N} \,\}$ is a family of open sets whose union contains $K$ , and the closed interval $[a_k - \delta_k,\ b_k + \delta_k]$ has length $(b_k - a_k) + 2\delta_k = (b_k - a_k) + \varepsilon \cdot 4^{-1} \cdot \theta^{k}$ by [L3] and [L9].

step 2.1L3L4L6L9

4.1

By [L5] there are $m \in \mathbb{N}$ and members $J_{k_0}, \dots, J_{k_m}$ of that family covering $K$ , and by [L8] there is $N \in \mathbb{N}$ with $k_t \le N$ for every $t \le m$ ; then $K \subseteq \bigcup_{k \le N} J_k \subseteq \bigcup_{k \le N}[a_k - \delta_k,\ b_k + \delta_k]$ by [L3].

step 1.2step 3.1L3L5L8choose

5.1

The total length of that finite list is $\sum_{k \le N}\big((b_k - a_k) + \varepsilon \cdot 4^{-1}\theta^{k}\big) = \sum_{k<N+1}(b_k - a_k) + \varepsilon \cdot 4^{-1}\sum_{k<N+1}\theta^{k} \le \varepsilon \cdot 2^{-1} + \varepsilon \cdot 4^{-1} \cdot 2 = \varepsilon$ , by [L7], step 2.1, [L6] and [L9].

step 2.1step 3.1step 4.1L6L7L9

6.1

So for every real $\varepsilon > 0$ the finite list of step 4.1 covers $K$ with total length at most $\varepsilon$ , which by [L1] is exactly the statement that $K$ has content zero; together with step 1.1 the two notions coincide on compact sets.

step 1.1step 1.2step 4.1step 5.1L1∎

Remarks

Compactness, not boundedness, is what does the work. $\mathbb{Q} \cap [0,1]$ is bounded and null and does not have content zero ( $\mathbb{Q} \cap [0,1]$ has measure zero and not content zero, although it is bounded ↗); it fails to be closed, and the finite subcover step is exactly what it cannot supply.
The theorem is what makes content zero usable at all. Every set to which content zero is applied on this page is compact: the Cantor set (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) and the Smith-Volterra-Cantor set (The Smith-Volterra-Cantor set: the same construction removing, at stage $n \ge 1$ , an open middle interval of length $4^{-n}$ from each of the $2^{n-1}$ remaining intervals) are both closed and bounded, so for them the two notions may be used interchangeably, and the finite form is the one that combines with If finitely many intervals cover a closed bounded interval $[a,b]$ , the sum of their lengths is at least $b - a$ .
The cost of opening up the cover is $\varepsilon \cdot 2^{-1}$ , chosen in advance. Splitting the budget in half before the enlargement, rather than after, is what keeps the final total at $\varepsilon$ exactly; the same bookkeeping appears in A sequence of intervals covering $[a,b]$ has total length at least $b - a$ , so no interval of positive length has measure zero and in A countable union of measure-zero sets has measure zero, by countable choice.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 116 results over 29 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)
Heine-Borel theorem (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 11 (standard reference, not scraped)
MIT 18.125, Homework 2: Measure-zero sets (standard reference, not scraped)

For a compact subset of R\mathbb{R}R, measure zero and content zero coincide