LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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If finitely many intervals cover a closed bounded interval $[a,b]$ , the sum of their lengths is at least $b - a$

Statement

Let $a, b \in \mathbb{R}$ with $a \le b$ , let $n \in \mathbb{N}$ , and let $c_0 \le d_0, \ \dots, \ c_n \le d_n$ be reals such that

$[a,b] \;\subseteq\; \bigcup_{j \le n} [c_j, d_j] ,$

the intervals being those of Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length. Then

$\sum_{j=0}^{n} (d_j - c_j) \;\ge\; b - a .$

The same bound holds for a cover by bounded intervals of any of the four bounded forms, since an interval with endpoints $c \le d$ is contained in $[c,d]$ and has the same length $d - c$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length); replacing each covering interval by the closed interval on its endpoints changes no length and only enlarges the union. In particular a finite family of intervals of total length strictly below $b - a$ cannot cover $[a,b]$ , which is the form in which this lemma is used throughout the page.

This is the one quantitative fact underlying everything about measure zero here. Without it nothing forbids a set of measure zero from being all of $[0,1]$ . Four items on this page rest on it: A sequence of intervals covering $[a,b]$ has total length at least $b - a$ , so no interval of positive length has measure zero directly, and through that lemma 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, The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero and FALSE: every set of measure zero has content zero. Two of the worked items on the companion page rest on it as well.

Facts & Assumptions

Given: For $n \in \mathbb{N}$ let $P(n)$ be the assertion: for all reals $a \le b$ and all reals $c_0 \le d_0, \dots, c_n \le d_n$ with $[a,b] \subseteq \bigcup_{j \le n}[c_j,d_j]$ , one has $\sum_{j \le n}(d_j - c_j) \ge b - a$ . The lemma is that $P(n)$ holds for every $n \in \mathbb{N}$ .

[L1]

$[c,d] = \{\, x : c \le x \le d \,\}$ , its length is $d - c \ge 0$ when $c \le d$ , and $[a,b]$ is nonempty exactly when $a \le b$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L2]

Finite sums: $\sum_{j \le n} t_j = \sum_{j < n+1} t_j$ with $\sum_{j<0} t_j = 0$ and $\sum_{j<m+1} t_j = \sum_{j<m} t_j + t_m$ ; sums split as $\sum_{j<m} t_j = \sum_{j<i} t_j + \sum_{j=i}^{m-1} t_j$ for $i \le m$ , where $\sum_{j=i}^{m-1} t_j = \sum_{l < m-i} t_{i+l}$ ; a sum of nonnegative terms is nonnegative, and each single term is at most the whole sum (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L3]

Induction on $\mathbb{N}$ (The principle of mathematical induction).

[L4]

Ordered-field arithmetic: $0 < 1$ , so $2 := 1 + 1 > 0$ and $0 < t \cdot 2^{-1} < t$ for $t > 0$ ; 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.

Proof

technique · induction

1.1

The assertion to be proved is $P(n)$ for every $n \in \mathbb{N}$ , with $P$ as in the Given, and the argument is an induction on $n$ using [L3].

givenL3induction

1.2

Base, $n = 0$ . Let $a \le b$ and $[a,b] \subseteq [c_0,d_0]$ with $c_0 \le d_0$ . Then $a \in [a,b]$ and $b \in [a,b]$ by [L1], so $c_0 \le a$ and $b \le d_0$ , whence $d_0 - c_0 \ge b - a$ by [L4]; and $\sum_{j \le 0}(d_j - c_j) = d_0 - c_0$ by [L2]. So $P(0)$ holds.

baseL1L2L4

1.3

Induction hypothesis. Fix $n \in \mathbb{N}$ and assume $P(n)$ .

ihgiven

2.1

The induction step: the two easy cases. Let $a \le b$ and let $c_0 \le d_0, \dots, c_{n+1} \le d_{n+1}$ satisfy $[a,b] \subseteq \bigcup_{j \le n+1}[c_j,d_j]$ ; write $S := \sum_{j \le n+1}(d_j - c_j)$ , a sum of nonnegative terms by [L1]. If $a = b$ then $b - a = 0 \le S$ by [L2]. Otherwise $a < b$ ; then $a \in [a,b]$ by [L1], so there is $i \le n+1$ with $a \in [c_i,d_i]$ , that is $c_i \le a \le d_i$ , and we fix one such $i$ . If $d_i \ge b$ then $d_i - c_i \ge b - a$ by [L4], and $d_i - c_i \le S$ by [L2], so $S \ge b - a$ . There remains the case $a < b$ and $d_i < b$ .

step 1.1L1L2L4choose

3.1

The induction step: the remaining case, where the $i$ -th interval is deleted. Assume $a < b$ and $d_i < b$ , and define $n+1$ pairs by $(c'_l, d'_l) := (c_l, d_l)$ for $l < i$ and $(c'_l, d'_l) := (c_{l+1}, d_{l+1})$ for $i \le l \le n$ ; by the splitting law and the index-shift convention of [L2], $S' := \sum_{l \le n}(d'_l - c'_l) = S - (d_i - c_i)$ . Let $\eta$ be any real with $0 < \eta \le b - d_i$ and put $c := d_i + \eta$ , so $d_i < c \le b$ . Every $x \in [c,b]$ satisfies $x \ge c > d_i$ , hence $x \notin [c_i,d_i]$ by [L1], and satisfies $a \le d_i < x \le b$ , hence $x \in [a,b]$ ; so $x$ lies in some $[c_j,d_j]$ with $j \ne i$ , that is in some $[c'_l,d'_l]$ . Thus $[c,b] \subseteq \bigcup_{l \le n}[c'_l,d'_l]$ with $c \le b$ , and step 1.3 gives $S' \ge b - c = b - d_i - \eta$ .

step 1.3step 2.1L1L2L4

4.1

Passing to the limiting value of $\eta$ , and the conclusion. In the case of step 3.1 one has $S' \ge b - d_i$ : were $S' < b - d_i$ , the real $\eta_0 := (b - d_i - S') \cdot 2^{-1}$ would satisfy $0 < \eta_0 < b - d_i$ by [L4], so step 3.1 would give $S' \ge b - d_i - \eta_0 = (b - d_i + S') \cdot 2^{-1} > S'$ , which is impossible. Hence $S = (d_i - c_i) + S' \ge (d_i - a) + (b - d_i) = b - a$ by [L4], using $c_i \le a$ from step 2.1. Together with the cases settled in step 2.1 this proves $P(n+1)$ , so by [L3] $P(n)$ holds for every $n \in \mathbb{N}$ .

step 2.1step 3.1L2L3L4discharge-induction∎

Remarks

Why the argument does not simply take $[d_i, b]$ . The point $d_i$ itself may be covered by the deleted interval and by nothing else, so the remaining intervals need not cover $[d_i,b]$ . They do cover $[d_i + \eta,\, b]$ for every positive $\eta$ , and that is enough: the bound $b - d_i - \eta$ holds for all such $\eta$ , and step 4.1 removes the $\eta$ . Every attempt to shortcut this step by taking a closed left endpoint at $d_i$ is false as stated.
Degenerate covering intervals are allowed and cost nothing. A pair with $c_j = d_j$ contributes the single point $c_j$ and the length $0$ , so a list may always be padded to a longer one, which is what A set of content zero has measure zero does.
The bound is sharp. The single interval $[a,b]$ covers $[a,b]$ with total length exactly $b - a$ , and no cover does better.
This is not the Heine-Borel theorem, and it does not use it. The lemma is a statement about finitely many intervals and is proved by counting alone; compactness enters only when a countable cover has to be reduced to a finite one, which is what A sequence of intervals covering $[a,b]$ has total length at least $b - a$ , so no interval of positive length has measure zero does with it.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 35 results over 11 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

Heine-Borel theorem (Wikipedia) (standard reference, not scraped)
Null set (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 11 (standard reference, not scraped)
UAF Math 641, Measure Theory notes (standard reference, not scraped)

If finitely many intervals cover a closed bounded interval [a,b][a,b][a,b], the sum of their lengths is at least b−ab - ab−a