Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generated
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • 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.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Finiteness of the moments of a bounded finite Borel measure on R3

Statement

Assume the Axiom of Countable Choice ACω (The Axiom of Countable Choice (ACω)). Let ν be a measure on (R3,B(R3)) (Mass distributions on R3: total mass, first moment and inertia matrix) that is finite and bounded: ν(R3)<+∞ and ν(R3∖B‾R(0))=0 for some R>0 (Open ball, closed ball and sphere in a metric space). Then:

  1. for all j,k∈{1,2,3} the functions ρ↦ρj and ρ↦ρjρk are Borel measurable and ν-integrable, with ∫∣ρj∣ dν≤R ν(R3) and ∫∣ρjρk∣ dν≤R2 ν(R3);
  2. consequently m:=ν(R3) and the scalar integrals ∫ρj dν and ∫(∥ρ∥22δjk−ρjρk) dν are finite, with ∣∫ρj dν∣≤Rm and ∣∫(∥ρ∥22δjk−ρjρk) dν∣≤2R2m. When m>0, these integrals are precisely the components of the first moment μ(ν) and inertia matrix I(ν) of Mass distributions on R3: total mass, first moment and inertia matrix. When m=0, all these integrals vanish; the zero measure is not called a mass distribution.

Facts & Assumptions

Given: A finite Borel measure ν on R3 and a real R>0 with ν(R3∖B‾R(0))=0.

[L1]

Every continuous map R3→R is Borel measurable, under ACω Continuous functions on Euclidean spaces are Borel measurable, Borel measurable and Lebesgue measurable functions on Rn; and sums, scalar multiples and pointwise products of measurable real-valued functions are measurable Arithmetic and lattice operations preserve measurability whenever they are defined.

[L2]

For a nonnegative simple function the nonnegative Lebesgue integral equals the simple integral; in particular the constant function R on R3 has ∫R dν=R ν(R3) The nonnegative Lebesgue integral, The integral of a nonnegative simple function.

[L3]

The nonnegative integral is monotone and homogeneous for nonnegative scalars: if 0≤f≤g then ∫f dν≤∫g dν, and ∫cf dν=c∫f dν for c≥0 Monotonicity and nonnegative homogeneity of the nonnegative integral.

[L4]

A real measurable function f with ∫∣f∣ dν<+∞ is integrable, and then ∣∫f dν∣≤∫∣f∣ dν Integrable real and complex functions, and their integrals, The modulus of an integral is bounded by the integral of the modulus.

[L5]

ρ=ρ1e1+ρ2e2+ρ3e3 satisfies ∣ρj∣≤∥ρ∥2 for each j and ∥ρ∥2≤R exactly on B‾R(0) (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, Open ball, closed ball and sphere in a metric space).

Proof

technique · direct
1.1givenL1

Each coordinate function ρ↦ρj is continuous, hence Borel measurable by [L1]; the products ρ↦ρjρk are then measurable by the product clause of [L1].

2.1step 1.1givenL1L2L3L5L6algebra

Let N:=R3∖B‾R(0), a Borel set by [L6], with ν(N)=0. We justify the almost-everywhere comparison from the definition of the nonnegative integral: if a nonnegative measurable f satisfies f≤B on Nc for a finite constant B≥0, every nonnegative simple minorant s≤f has the same simple integral as s1Nc. Indeed, in a finite disjoint level-set representation of s, the contribution of each intersection with N is zero, while ν(A)=ν(A∖N) for every measurable level set A. Pointwise s1Nc≤B, so [L2] and [L3] give ∫s dν≤Bν(R3); taking the supremum over simple minorants gives ∫f dν≤Bν(R3) by The nonnegative Lebesgue integral. Apply this with f=∣ρj∣, B=R, and with f=∣ρjρk∣, B=R2: measurability follows from step 1.1 and [L1], and [L5] gives the bounds on Nc. Thus ∫∣ρj∣ dν≤Rν(R3) and ∫∣ρjρk∣ dν≤R2ν(R3).

3.1step 2.1givenL4

The bounds of step 2.1 are finite because ν is finite, so each of the functions ρj and ρjρk has finite integral of its modulus, hence is ν-integrable by [L4].

4.1step 2.13.1givenL1L4L5algebra∎

Each coordinate integrand is integrable by step 3.1. Each inertia integrand gjk(ρ):=∥ρ∥22δjk−ρjρk is measurable by [L1], and on Nc its modulus is at most 2R2 by [L5]. The simple-minorant comparison proved in step 2.1 gives ∫∣gjk∣ dν≤2R2m<∞, so [L4] makes gjk integrable and gives ∣∫gjk dν∣≤2R2m. Likewise [L4] and step 2.1 give ∣∫ρj dν∣≤Rm. If m>0, ν is a mass distribution and these are its moment and inertia components by Mass distributions on R3: total mass, first moment and inertia matrix. If m=0, the same bounds force every displayed integral to be zero; this case is excluded from the definition's mass-distribution terminology. Thus clause 2 holds for every measure in the hypotheses.

Depends on

Dependency tree · 2 levels

Sources