TheoremBase

Borel Sets of ℝ^{m+n} and the Product σ-Algebra of ℝ^m and ℝ^n

Concatenation of an m-tuple and an n-tuple is measurable from the product of the Borel σ-algebras of ℝ^m and ℝ^n to the Borel σ-algebra of ℝ^{m+n}, with measurable inverse, so a set of pairs is product-measurable exactly when its concatenated image is Borel; likewise ℝ and ℝ^1 are identified measurably.

Statement

In the setting of Measurable Functions and the Lebesgue Integral: Notation, let m,n∈Nm,n\in\mathbb{N}. Let ι:Rm×Rn→Rm+n\iota:\mathbb{R}^{m}\times\mathbb{R}^{n}\to\mathbb{R}^{m+n} be the concatenation function, a bijection by Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space (iii), and let B(Rm)⊗B(Rn)\mathcal{B}(\mathbb{R}^{m})\otimes\mathcal{B}(\mathbb{R}^{n}) be the product σ\sigma-algebra on Rm×Rn\mathbb{R}^{m}\times\mathbb{R}^{n} of the Borel σ\sigma-algebras of Rm\mathbb{R}^{m} and Rn\mathbb{R}^{n}.

Concatenation is measurable. ι\iota is (B(Rm)⊗B(Rn))/B(Rm+n)(\mathcal{B}(\mathbb{R}^{m})\otimes\mathcal{B}(\mathbb{R}^{n}))/\mathcal{B}(\mathbb{R}^{m+n})-measurable.

Its inverse is measurable. The inverse ι−1:Rm+n→Rm×Rn\iota^{-1}:\mathbb{R}^{m+n}\to\mathbb{R}^{m}\times\mathbb{R}^{n} is B(Rm+n)/(B(Rm)⊗B(Rn))\mathcal{B}(\mathbb{R}^{m+n})/(\mathcal{B}(\mathbb{R}^{m})\otimes\mathcal{B}(\mathbb{R}^{n}))-measurable. Consequently, for every E⊆Rm×RnE\subseteq\mathbb{R}^{m}\times\mathbb{R}^{n},

E∈B(Rm)⊗B(Rn)  ⟺  ι[E]∈B(Rm+n).E\in\mathcal{B}(\mathbb{R}^{m})\otimes\mathcal{B}(\mathbb{R}^{n})\iff\iota[E]\in\mathcal{B}(\mathbb{R}^{m+n}).

The real line and R1\mathbb{R}^{1}. For t∈Rt\in\mathbb{R}, let τ(t)∈R1\tau(t)\in\mathbb{R}^{1} be the 11-tuple with component τ(t)1=t\tau(t)_{1}=t, that is, the constant function on [1][1] with value tt. The function τ:R→R1\tau:\mathbb{R}\to\mathbb{R}^{1} is a bijection whose inverse is x↦x1x\mapsto x_{1}; τ\tau is B(R)/B(R1)\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R}^{1})-measurable, and its inverse is B(R1)/B(R)\mathcal{B}(\mathbb{R}^{1})/\mathcal{B}(\mathbb{R})-measurable. Consequently, for every B⊆RB\subseteq\mathbb{R},

B∈B(R)  ⟺  τ[B]∈B(R1).B\in\mathcal{B}(\mathbb{R})\iff\tau[B]\in\mathcal{B}(\mathbb{R}^{1}).

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