TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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A countable union of measure-zero sets has measure zero, by countable choice

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $(A_{n})_{n \in N}$ be a sequence of subsets of $R$ , each of measure zero (Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover)). Then

$⋃_{n \in N} A_{n} has measure zero.$

By the padding convention of Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover) and Finite, countably infinite, countable, uncountable the same conclusion covers the union of an at most countable family of null sets, a finite family being extended by copies of $\emptyset$ .

The hypothesis ${A C}_{ω}$ is spent at exactly one step, step 2.1 below, where one covering sequence is selected for every $A_{n}$ at once. Each $A_{n}$ has many such covers and nullity provides no rule for singling one out. Nothing else in the proof selects anything: the diagonal enumeration and the estimate are formulas.

Facts & Assumptions

Given: A sequence $(A_{n})_{n \in N}$ of null subsets of $R$ and a real $ε > 0$ . Throughout, $θ : = 2^{- 1}$ .

[A1]

The Axiom of Countable Choice: every family $(X_{n})_{n \in N}$ of nonempty sets has a function $f$ on $N$ with $f (n) \in X_{n}$ for every $n$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L1]

$A$ is null when for every real $η > 0$ there are sequences $(a_{k})$ , $(b_{k})$ with $a_{k} \leq b_{k}$ , $A \subseteq ⋃_{k} [a_{k}, b_{k}]$ and $\sum_{k < n} (b_{k} - a_{k}) \leq η$ for every $n$ (Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover)).

[L2]

There is a bijection $J : N \times N \to N$ , with inverse $J^{- 1}$ ( $N \times N \approx N$ , Injection, surjection, bijection).

[L3]

Powers and the geometric series: $θ^{0} = 1$ , $θ^{m + 1} = θ^{m} θ$ , $θ^{m} > 0$ , and $\sum_{m = 0}^{\infty} θ^{m} = 2$ for $θ = 2^{- 1}$ ; a series of nonnegative terms has all its partial sums at most its sum (Integer powers $a^{m}$ , For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ 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).

[L4]

Finite sums: additivity, scaling, splitting and monotonicity in the terms; a sum of nonnegative terms is nonnegative and does not decrease when further nonnegative terms are adjoined, so a sum of finitely many nonnegative terms indexed injectively inside a finite rectangle is at most the sum over the whole rectangle (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L5]

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

[L6]

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

Let the real $ε > 0$ be given and put $ε_{n} : = ε \cdot θ^{n + 1}$ for $n \in N$ , a positive real by [L3] and [L6]. Let $X_{n}$ be the set of all pairs of sequences $((a_{k}), (b_{k}))$ with $a_{k} \leq b_{k}$ for every $k$ , $A_{n} \subseteq ⋃_{k} [a_{k}, b_{k}]$ and $\sum_{k < i} (b_{k} - a_{k}) \leq ε_{n}$ for every $i \in N$ . Each $A_{n}$ is null, so each $X_{n}$ is nonempty by [L1].

givenL1L3L6

2.1

By [A1] fix $f$ with $f (n) \in X_{n}$ for every $n$ , and write $f (n) = ((a_{k}^{n})_{k}, (b_{k}^{n})_{k})$ . This is the one and only application of countable choice in the proof.

step 1.1A1choose

3.1

By [L2] fix a bijection $J : N \times N \to N$ and define sequences $(c_{j})$ and $(d_{j})$ by $c_{J (m, k)} : = a_{k}^{m}$ and $d_{J (m, k)} : = b_{k}^{m}$ , which is a total definition because $J$ is a bijection; then $c_{j} \leq d_{j}$ for every $j$ . Every $x \in ⋃_{n} A_{n}$ lies in some $A_{m}$ , hence in some $[a_{k}^{m}, b_{k}^{m}] = [c_{J (m, k)}, d_{J (m, k)}]$ by step 2.1, so $⋃_{n} A_{n} \subseteq ⋃_{j} [c_{j}, d_{j}]$ .

step 2.1L2

4.1

Fix $i \in N$ . The pairs $J^{- 1} (j)$ for $j < i$ are finitely many and pairwise distinct, so by [L5] there is $N \in N$ with both coordinates of each of them at most $N$ ; since all the terms $d_{j} - c_{j}$ are nonnegative, [L4] gives $\sum_{j < i} (d_{j} - c_{j}) \leq \sum_{m \leq N} (\sum_{k \leq N} (b_{k}^{m} - a_{k}^{m}))$ . For each $m \leq N$ the inner sum is $\sum_{k < N + 1} (b_{k}^{m} - a_{k}^{m}) \leq ε_{m}$ by step 2.1, so the whole is at most $\sum_{m \leq N} ε \cdot θ^{m + 1} = ε \cdot θ \sum_{m < N + 1} θ^{m} \leq ε \cdot 2^{- 1} \cdot 2 = ε$ , by [L3], [L4] and [L6].

step 3.1L3L4L5L6

5.1

Steps 3.1 and 4.1 exhibit, for the given $ε > 0$ , sequences of closed intervals covering $⋃_{n} A_{n}$ with every partial total length at most $ε$ ; since $ε > 0$ was arbitrary, [L1] gives that $⋃_{n} A_{n}$ has measure zero.

step 1.1step 3.1step 4.1L1∎

Remarks

Why the slack is geometric. The $n$ -th set is covered to within $ε \cdot 2^{- n - 1}$ and the budgets sum to $ε$ , exactly as in Every at most countable subset of $R$ has measure zero, of which this theorem is the abstract form: applying it to the singletons ${x_{n}}$ of a listing recovers that lemma, at the cost of an appeal to ${A C}_{ω}$ that the direct proof avoids. The expenditure is the same one, and made for the same reason, as in Countable unions of at most countable sets, assuming ${A C}_{ω}$ .
No rearrangement theorem is used, and none is available here. The estimate is made on finite partial sums only, and every finite partial sum of the doubly-indexed family is compared with a sum over a finite rectangle, which is a finite rearrangement. The theory of rearranging infinite series is not in the reading order at this point, and the proof is arranged so as not to need it.
The bound is on the total length, not on the number of intervals. The combined cover is countable even when each $A_{n}$ is covered by infinitely many intervals, which is exactly what $N \times N \approx N$ supplies. Nothing analogous holds for content zero: a countable union of sets of content zero need not have content zero, since $Q \cap [0, 1]$ is such a union ( $Q \cap [0, 1]$ has measure zero and not content zero, although it is bounded ↗).
This is where the two smallness notions of the page separate cleanly. A countable union of null sets is null, whereas a countable union of nowhere dense sets is meager and, by Baire category in $R$ , by nested intervals with canonically chosen rational endpoints: a countable intersection of dense open sets is dense, so $R$ is not a countable union of nowhere dense sets, never all of $R$ ; and yet $R$ is the union of a meager set and a null set ( $R$ is the union of a meager set and a set of measure zero, so smallness of category and smallness of measure are independent notions ↗).

Depends on

Used by

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Sources

Null set (Wikipedia) (standard reference, not scraped)
Axiom of countable choice (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)