TheoremBase

The Integral over the Unit Cell of a Product of One-Variable Functions

For bounded Borel functions of one variable, the integral over the torus of the product of their coordinate evaluations is the product of their integrals over the unit interval.

Statement

We work in the setting of The Flat Torus: Standing Notation, used here with a natural number nn satisfying 1≤n1\le n; the initial segments [n][n], Euclidean space Rn\mathbb{R}^{n}, the half-open unit cell QQ, the measure space (Q,BQ,λQ)(Q,\mathcal{B}_{Q},\lambda_{Q}) together with the notation ∫Tnv dx\int_{\mathbb{T}^{n}}v\,dx, and the integral and the notion of an integrable map are the ones fixed there. Let B(R)\mathcal{B}(\mathbb{R}) be the Borel σ\sigma-algebra of the real line and let λ\lambda be Lebesgue measure on it. Finite products of real numbers are the finite products of that definition, formed in the field of real numbers.

Put

J={t∈R:0≤t<1}.J=\{t\in\mathbb{R}:0\le t<1\}.

Then JJ is an interval, since 0≤y0\le y and y<1y<1 whenever x≤y≤zx\le y\le z with x,z∈Jx,z\in J; hence J∈B(R)J\in\mathcal{B}(\mathbb{R}) by Borel Sigma-Algebra on the Real Line. The closed intervals [0,1][0,1] and [1,1]={1}[1,1]=\{1\} are intervals too, hence also lie in B(R)\mathcal{B}(\mathbb{R}), and claim 4 of Lebesgue Measure on the Real Line: Existence, Lengths of Intervals and Uniqueness gives λ([0,1])=1\lambda([0,1])=1 and λ({1})=0\lambda(\{1\})=0. Since {1}⊆[0,1]\{1\}\subseteq[0,1], λ([0,1])\lambda([0,1]) is finite and J=[0,1]∖{1}J=[0,1]\setminus\{1\}, claim 3 of Basic Properties of a Measure gives λ(J)=1−0=1\lambda(J)=1-0=1. Let (J,BJ,λJ)(J,\mathcal{B}_{J},\lambda_{J}) denote the restriction of (R,B(R),λ)(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda) to JJ furnished by claim 1 of that lemma, so that BJ={A∈B(R):A⊆J}\mathcal{B}_{J}=\{A\in\mathcal{B}(\mathbb{R}):A\subseteq J\} and λJ(A)=λ(A)\lambda_{J}(A)=\lambda(A). By the description of the cell recorded in The Half-Open Unit Cell Tiles Euclidean Space (i), a point x∈Rnx\in\mathbb{R}^{n} lies in QQ if and only if xi∈Jx_{i}\in J for every i∈[n]i\in[n].

Let gi:J→Rg_{i}:J\to\mathbb{R} be measurable with respect to BJ\mathcal{B}_{J}, for every i∈[n]i\in[n]. Then the following hold.

1. (The product function) Setting

G(x)=∏i=1ngi(xi)(x∈Q)G(x)=\prod_{i=1}^{n}g_{i}(x_{i})\qquad(x\in Q)

defines a map G:Q→RG:Q\to\mathbb{R}, and GG is measurable with respect to BQ\mathcal{B}_{Q}.

2. (The integral of the product) Suppose in addition that for every i∈[n]i\in[n] there is a real number MiM_{i} with ∣gi(t)∣≤Mi|g_{i}(t)|\le M_{i} for every t∈Jt\in J, where ∣ ⋅ ∣|\,\cdot\,| is the absolute value. Then each gig_{i} is integrable with respect to λJ\lambda_{J}, the map GG is integrable with respect to λQ\lambda_{Q}, and

∫TnG dx=∏i=1n∫Jgi dλJ.\int_{\mathbb{T}^{n}}G\,dx=\prod_{i=1}^{n}\int_{J}g_{i}\,d\lambda_{J}.

Proofs

Log in to submit a proof.

Loading...

Citations

Loading…

Dependencies

Loading…

Related

0 relations

Curated associations between results. These are editable and subjective — they do not replace the dependency graph, which is derived from the references in the text.

No relations recorded yet.

Comments

Log in to comment.

Loading…