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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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Elementary volume is a sigma-finite premeasure on the algebra of elementary sets

Statement

Let n≥1, let En be the algebra of elementary subsets of Rn (The elementary sets form an algebra of subsets of Rn containing every half-open box) and let μ0 be elementary volume (The sum of the volumes of a disjoint box decomposition of an elementary set does not depend on the decomposition). Then μ0 is a sigma-finite premeasure on En (Premeasures on algebras of sets): μ0(∅)=0; whenever (Ak)k∈N is a pairwise disjoint sequence in En whose union A again lies in En,

μ0(A)  =  ∑k=0∞μ0(Ak);

and Rn=⋃k∈N(−k,k]n with μ0((−k,k]n)<+∞ for every k.

No choice principle is used. The one place where a textbook proof selects countably many objects is the enlargement of each Ak, and here the enlarged set is the canonical Ak+1/(m+1) of Every elementary set is squeezed in volume between a compact subset and an elementary set whose interior contains it with m the least natural number that works, which is a definition rather than a selection. The compact inner set and the finite subcover are each a single instantiation of an existential statement.

Facts & Assumptions

Given: A natural number n≥1, the algebra En with elementary volume μ0, and a pairwise disjoint sequence (Ak)k∈N in En whose union A lies in En.

[L1]

En is an algebra of subsets of Rn, it contains every half-open box, and it is closed under intersection of two members and under difference (The elementary sets form an algebra of subsets of Rn containing every half-open box).

[L2]

For every n≥1, there is exactly one function μ0:En→[0,+∞] whose value at A is the sum of the volumes of the members of any presentation of A by a finite list of pairwise disjoint half-open boxes; it satisfies μ0(∅)=0 and μ0(B)=vol⁡(B) for every half-open box B (The sum of the volumes of a disjoint box decomposition of an elementary set does not depend on the decomposition).

[L3]

Elementary volume is finitely additive on pairwise disjoint elementary sets, monotone, and finitely subadditive (Elementary volume is finitely additive, monotone and finitely subadditive on the elementary algebra).

[L4]

A+δ is an elementary set, it is determined by A and δ alone, it contains A, and every point of A is an interior point of A+δ in (Rn,d2) (Every elementary set is squeezed in volume between a compact subset and an elementary set whose interior contains it, claim 1).

[L5]

For every real ε>0 there is m∈N with μ0(A+1/(m+1))≤μ0(A)+ε (Every elementary set is squeezed in volume between a compact subset and an elementary set whose interior contains it, claim 2).

[L6]

If μ0(A)<+∞, then for every real ε>0 there are an elementary set A′ and a compact set K⊆Rn with A′⊆K⊆A and μ0(A)≤μ0(A′)+ε (Every elementary set is squeezed in volume between a compact subset and an elementary set whose interior contains it, claim 3).

[L7]

For a nonempty box vol⁡(B):=+∞ when ai=−∞ or bi=+∞ for some i<n, and vol⁡(B):=∏i<n(bi−ai) when every ai and every bi is real; B(a,b):={ x∈Rn:ai<xi≤bi  for every i<n }; and (u,v]n:=B(u,v) (Half-open boxes in Rn and their volume).

[L8]

A subset E⊆Rn is an elementary set when there are a natural number m and a list B0,…,Bm−1 of half-open boxes with E=⋃j<mBj (Elementary sets: the finite unions of half-open boxes in Rn).

[F1]

A premeasure on an algebra A0 vanishes at the empty set and is countably additive whenever a disjoint sequence in A0 has its union in A0; it is sigma-finite if there is a sequence (Pn) in A0 with X=⋃nPn and μ0(Pn)<+∞ for every n (Premeasures on algebras of sets).

[F2]

The nonnegative extended sum of a sequence in [0,+∞] is ∑k=0∞ak:=sup⁡n∈Nsn, the supremum of its nondecreasing partial sums (Series in the nonnegative extended real line).

[F3]

A is a compact subset of X if and only if for every set I and every family (Ui)i∈I of open subsets of X with A⊆⋃i∈IUi there are n∈N and indices i0,…,in∈I with A⊆Ui0∪⋯∪Uin, or else A=∅ (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, claim 3; Open cover, subcover, compact metric space, and compact subset of a metric space).

[F5]

Every nonempty subset S⊆N has a least element (The well-ordering principle).

[F6]

If ∣r∣<1 then ∑k=0∞rk=1/(1−r); in particular ∑k=0∞2−k=2 (For ∣r∣<1, ∑k≥0rk=1/(1−r), and for ∣r∣≥1 the series diverges).

[F7]

Every complete ordered field F is Archimedean: for every x∈F there is a natural number n≥1 with x<n⋅1F (Every complete ordered field is Archimedean).

[F8]

For sequences of reals, ∑k<n(ak+bk)=∑k<nak+∑k<nbk; if ak≤bk for all k<n then ∑k<nak≤∑k<nbk; and if ak≥0 for all k<n then ∏k<nak≥0, with ∏k<nak>0 when every ak>0 (Laws of finite sums and finite products, claims 1, 4 and 6).

[F9]

Finite sums and finite products of a sequence of reals are defined by the recursions Σ0=0, Σσ(n)=Σn+an and Π0=1, Πσ(n)=Πn⋅an (Finite sums and finite products, by recursion).

[F10]

For a,b∈R‾, a+b:=+∞ when a=+∞ and b≠−∞, or b=+∞ and a≠−∞ (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined).

Proof

technique · direct
1.1L1L2F1

μ0 is a function on the algebra En with values in [0,+∞] and μ0(∅)=0, which is the first premeasure clause.

1.2L1L2L7L8F7

Each cube (−k,k]n is a half-open box, hence elementary, with μ0((−k,k]n)=(2k)n<+∞ for k≥1 and μ0((−0,0]n)=0; and ⋃k∈N(−k,k]n=Rn, because for x∈Rn the Archimedean property supplies a natural k≥1 above each of the finitely many reals ∣xi∣.

1.3L3F2

For every N, finite additivity gives ∑k<Nμ0(Ak)=μ0(⋃k<NAk) and monotonicity gives μ0(⋃k<NAk)≤μ0(A), so every partial sum is at most μ0(A) and therefore ∑k=0∞μ0(Ak)≤μ0(A), that supremum being the nonnegative extended sum.

1.4L2L3L4L5L6F2F3F4F5F6F8

Suppose μ0(A)<+∞ and let ε>0 be real. Fix an elementary A′ and a compact K with A′⊆K⊆A and μ0(A)≤μ0(A′)+ε; for each k let mk be the least natural number with μ0(Ak+1/(mk+1))≤μ0(Ak)+ε2−k, which exists because the set of such naturals is nonempty and N is well ordered, and put Uk:=int⁡(Ak+1/(mk+1)), an open set containing Ak. Since K⊆A=⋃kAk⊆⋃kUk, compactness yields finitely many indices covering K, hence a natural N with K⊆⋃k<NUk, so that A′⊆⋃k<NAk+1/(mk+1); monotonicity, finite subadditivity and the geometric series then give μ0(A′)≤∑k<Nμ0(Ak+1/(mk+1))≤∑k<Nμ0(Ak)+ε∑k<N2−k≤∑k=0∞μ0(Ak)+2ε, whence μ0(A)≤∑k=0∞μ0(Ak)+3ε; as ε was an arbitrary positive real and μ0(A) is finite, μ0(A)≤∑k=0∞μ0(Ak).

2.1step 1.4L1L2L3L7L8F2F7F8F9F10

Suppose instead μ0(A)=+∞ and put T:=∑k=0∞μ0(Ak); if T=+∞ then μ0(A)≤T holds, and if T<+∞ a contradiction follows. Fix a disjoint box presentation A=⋃j<qBj; some Bj0=B(a,b) has infinite volume, hence is nonempty with ai0=−∞ or bi0=+∞ for some i0<n. Take y∈Bj0 and put αi:=yi−1 when ai=−∞ and αi:=(ai+yi)/2 otherwise, so that ai<αi<yi and c:=∏i<nui>0, where ui:=yi−αi for i≠i0 and ui0:=1. For a real R exceeding every ∣αi∣ and every ∣yi∣, the box DR with parameter pairs (αi,yi] for i≠i0 and (αi0,R], respectively (−R,yi0], in coordinate i0 according as bi0=+∞ or ai0=−∞, satisfies DR⊆Bj0∩(−R,R]n⊆A∩(−R,R]n and has volume at least (R−∣αi0∣−∣yi0∣)c. On the other hand A∩(−R,R]n is elementary of finite volume and is the disjoint union of the elementary sets Ak∩(−R,R]n, so step 1.4 and monotonicity give μ0(A∩(−R,R]n)≤∑k=0∞μ0(Ak∩(−R,R]n)≤T; taking R above (T/c)+∣αi0∣+∣yi0∣+1 by the Archimedean property contradicts this.

3.1step 1.1step 1.2step 1.3step 1.4step 2.1F1∎

Steps 1.3, 1.4 and 2.1 give μ0(A)=∑k=0∞μ0(Ak) in every case, which with steps 1.1 and 1.2 makes μ0 a sigma-finite premeasure on En.

Depends on

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Sources