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

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Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
AI-adapted — a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A sequence of intervals covering $[a,b]$ has total length at least $b - a$ , so no interval of positive length has measure zero

Statement

Let $a, b \in \mathbb{R}$ with $a \le b$ , let $(a_k)_{k \in \mathbb{N}}$ and $(b_k)_{k \in \mathbb{N}}$ be sequences of reals with $a_k \le b_k$ for every $k$ , and suppose

$[a,b] \;\subseteq\; \bigcup_{k \in \mathbb{N}} [a_k, b_k] .$

If $M \in \mathbb{R}$ satisfies $\sum_{k < n} (b_k - a_k) \le M$ for every $n \in \mathbb{N}$ , then

$M \;\ge\; b - a .$

Consequently, if $a < b$ then no subset of $\mathbb{R}$ containing $[a,b]$ has measure zero (Measure zero (a countable cover by intervals of total length below every $\varepsilon$ ) and content zero (a finite such cover)); in particular none of the four bounded intervals $[a,b]$ , $(a,b)$ , $[a,b)$ , $(a,b]$ with $a < b$ has measure zero, so measure zero is not a vacuous notion.

This is the countable strengthening of If finitely many intervals cover a closed bounded interval $[a,b]$ , the sum of their lengths is at least $b - a$ , and it is what compactness is spent on: the countable cover is enlarged to an open one at an arbitrarily small cost in total length, and A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded reduces it to a finite cover, where the finite lemma applies.

Facts & Assumptions

Given: Reals $a \le b$ , sequences $(a_k)$ and $(b_k)$ with $a_k \le b_k$ for every $k$ and $[a,b] \subseteq \bigcup_k [a_k,b_k]$ , and a real $M$ with $\sum_{k<n}(b_k - a_k) \le M$ for every $n \in \mathbb{N}$ . Throughout, $\theta := 2^{-1}$ .

[L1]

Measure zero: $A$ is null when for every real $\varepsilon > 0$ there is a sequence of closed intervals covering $A$ all of whose partial total lengths are $\le \varepsilon$ ; a subset of a null set is null (Measure zero (a countable cover by intervals of total length below every $\varepsilon$ ) and content zero (a finite such cover)).

[L2]

$[c,d] = \{\, x : c \le x \le d \,\}$ has length $d - c \ge 0$ when $c \le d$ ; $(c,d)$ is the open interval; a closed bounded interval is bounded (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, Lower bound, bounded below, bounded set).

[L3]

Every open interval $(c,d)$ is an open set and every interval $[c,d]$ is a closed 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).

[L4]

A subset of $\mathbb{R}$ is compact exactly when it is closed and bounded (A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded); from every family of open sets whose union contains a compact set, either the set is empty and the empty subfamily covers it, or one can extract $m \in \mathbb{N}$ and members $U_0, \dots, U_m$ of the family whose union already contains it (Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset).

[L5]

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

[L6]

Powers and the geometric series: $\theta^0 = 1$ and $\theta^{k+1} = \theta^k \theta$ , all $\theta^k > 0$ for $\theta > 0$ , and $\sum_{k=0}^{\infty} \theta^k = 1/(1-\theta) = 2$ 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 by a constant, splitting, and monotonicity in the terms (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L8]

Every finite list $k_0, \dots, k_m$ of naturals has an upper bound $K \in \mathbb{N}$ : by induction on $m$ , taking $K = 0$ for the empty case and replacing $K$ by whichever of $K$ and $k_{m+1}$ is the larger, the order of $\mathbb{N}$ being total (The principle of mathematical induction, Trichotomy of the order on $\mathbb{N}$ , Order on the natural numbers).

[L9]

Ordered-field arithmetic: $0 < 1$ , so $2 > 0$ 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 · contradiction

1.1

Suppose, for contradiction, that $M < b - a$ . Since $\sum_{k<0}(b_k - a_k) = 0$ by [L7], we have $M \ge 0$ , so $b - a > 0$ and $a < b$ . Put $\varepsilon := (b - a - M) \cdot 2^{-1}$ , a positive real by [L9].

assume-contragivenL7L9

2.1

For $k \in \mathbb{N}$ put $\delta_k := \varepsilon \cdot 4^{-1} \cdot \theta^{k}$ , a positive real by [L6] and [L9], and $J_k := (a_k - \delta_k,\ b_k + \delta_k)$ . Each $J_k$ is an open set by [L3], and $[a_k,b_k] \subseteq J_k$ because $a_k - \delta_k < a_k \le x \le b_k < b_k + \delta_k$ for $x \in [a_k,b_k]$ , by [L2] and [L9]. Hence $[a,b] \subseteq \bigcup_k [a_k,b_k] \subseteq \bigcup_k J_k$ , so $\{\, J_k : k \in \mathbb{N} \,\}$ is a family of open sets whose union contains $[a,b]$ . The length of the interval with endpoints $a_k - \delta_k$ and $b_k + \delta_k$ is $(b_k - a_k) + 2\delta_k = (b_k - a_k) + \varepsilon \cdot 2^{-1} \cdot \theta^{k}$ , by [L2] and [L9].

step 1.1givenL2L3L6L9

3.1

$[a,b]$ is closed and bounded by [L2] and [L3], hence compact by [L4]; so there are $m \in \mathbb{N}$ and members $J_{k_0}, \dots, J_{k_m}$ of the family with $[a,b] \subseteq J_{k_0} \cup \dots \cup J_{k_m}$ . By [L8] fix $K \in \mathbb{N}$ with $k_t \le K$ for every $t \le m$ ; then every $J_{k_t}$ occurs among $J_0, \dots, J_K$ , so $[a,b] \subseteq \bigcup_{k \le K} J_k$ .

step 2.1L2L3L4L8choose

4.1

By [L5], applied to the $K+1$ intervals $J_k$ with endpoints $a_k - \delta_k \le b_k + \delta_k$ , one gets $\sum_{k \le K} \big( (b_k - a_k) + \varepsilon \cdot 2^{-1} \cdot \theta^{k} \big) \ge b - a$ .

step 2.1step 3.1L5

5.1

The left-hand side is at most $M + \varepsilon$ : by [L7] it splits as $\sum_{k < K+1}(b_k - a_k) + \varepsilon \cdot 2^{-1} \sum_{k < K+1} \theta^{k}$ , the first sum is $\le M$ by hypothesis, and the second is $\le \varepsilon \cdot 2^{-1} \cdot 2 = \varepsilon$ by [L6]. So $b - a \le M + \varepsilon = (b - a + M) \cdot 2^{-1} < b - a$ by [L9], which is impossible; the assumption of step 1.1 is untenable and $M \ge b - a$ . For the consequence, let $a < b$ and let $A \supseteq [a,b]$ be null; taking $\varepsilon_1 := (b-a) \cdot 2^{-1} > 0$ in [L1] gives a sequence of closed intervals covering $A$ , hence covering $[a,b]$ , with every partial total length $\le \varepsilon_1$ , so what has just been proved gives $(b-a) \cdot 2^{-1} \ge b - a$ and hence $b - a \le 0$ by [L9], contradicting $a < b$ . Finally each of $(a,b)$ , $[a,b)$ , $(a,b]$ and $[a,b]$ with $a < b$ contains $[a', b']$ for $a' := a + (b-a) \cdot 4^{-1}$ and $b' := b - (b-a) \cdot 4^{-1}$ , which satisfy $a < a' < b' < b$ by [L9], so none of them is null.

step 1.1step 2.1step 4.1givenL1L6L7L9discharge-contradiction∎

Remarks

What the hypothesis $\sum_{k<n}(b_k - a_k) \le M$ says. It is the working form of "the total length is at most $M$ " recorded in Measure zero (a countable cover by intervals of total length below every $\varepsilon$ ) and content zero (a finite such cover): for nonnegative terms, having all partial sums below $M$ is the same as convergence with sum below $M$ . Stating the lemma with partial sums avoids assuming convergence, and the conclusion is therefore also the statement that a cover of $[a,b]$ whose total length diverges is no counterexample.
The $\varepsilon$ is spent on making the cover open, not on the estimate. Enlarging $[a_k,b_k]$ to $(a_k - \delta_k, b_k + \delta_k)$ adds $2\delta_k$ to the $k$ -th length, and the geometric choice $\delta_k = \varepsilon \theta^k/4$ makes the whole added amount at most $\varepsilon$ , however many intervals are used. This is the standard device and it recurs in For a compact subset of $\mathbb{R}$ , measure zero and content zero coincide.
Compactness is not optional here. Without it the finite lemma cannot be reached, and the countable statement is genuinely stronger than the finite one: $\mathbb{Q} \cap [0,1]$ is covered by countably many intervals of total length below any $\varepsilon$ , and by no finite family of total length below $1$ ( $\mathbb{Q} \cap [0,1]$ has measure zero and not content zero, although it is bounded ↗).

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 116 results over 28 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

Null set (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)
UAF Math 641, Measure Theory notes (standard reference, not scraped)

A sequence of intervals covering [a,b][a,b][a,b] has total length at least b−ab - ab−a, so no interval of positive length has measure zero