Let d∈N+d \in \mathbb{N}^+ be a positive integer, and x=(x1,…,xn)x = (x_1, \ldots, x_n) be a generic point in Rd\mathbb{R}^d.

Definition (Rectangle). Let aj,bj∈Ra_j, b_j \in \mathbb{R} with aj<bja_j < b_j, for j∈[d]j \in [d]. The following set of points is called a rectangle in Rd\mathbb{R}^d:

×i∈[d](ai,bi]≐{x∈Rd:xi∈(ai,bi],∀i∈[d]}. \times_{i \in [d]} (a_i, b_i] \doteq \{ x \in \mathbb{R}^d : x_i \in (a_i, b_i], \forall i \in [d] \}.

Let S0\mathcal{S}_0 be the collection of all possible rectangles.

Definition (Borel Sets). Let Rd=σ(S0)\mathcal{R}^d = \sigma(\mathcal{S}_0) be the σ\sigma-algebra generated by the rectangles. Elements in Rd\mathcal{R}^d are called Borel sets.

Properties

Theorem. Rd\mathcal{R}^d contains the open sets.

Proof. Let GG be an open set in Rd\mathbb{R}^d. For each x∈Gx \in G, there exits a rectangle Rx=×i∈[d](ai,bi]R_x = \times_{i \in [d]} (a_i, b_i] such that

  1. x∈Rx⊂Gx \in R_x \subset G.
  2. ai,bia_i, b_i are rational numbers, for each i∈[d]i \in [d].

Therefore, G=∪x∈GRxG = \cup_{x \in G} R_x. However, there are countable number of rectangles whose endpoints are rational numbers. Hence G∈RdG \in \mathcal{R}^d.

□\square

Corollary. Rd\mathcal{R}^d contains the closed sets.

Theorem. Let

S1≐{×i∈[d][ai,bi):aj<bj,∀j∈[d]}S2≐{×i∈[d](ai,bi):aj<bj,∀j∈[d]}S3≐{×i∈[d][ai,bi]:aj≤bj,∀j∈[d]} \begin{aligned} \mathcal{S}_1 &\doteq \Big\{ \times_{i \in [d]} [a_i, b_i) : a_j < b_j , \forall j \in [d] \Big\} \\ \mathcal{S}_2 &\doteq \Big\{ \times_{i \in [d]} (a_i, b_i) : a_j < b_j , \forall j \in [d] \Big\} \\ \mathcal{S}_3 &\doteq \Big\{ \times_{i \in [d]} [a_i, b_i] : a_j \le b_j , \forall j \in [d] \Big\} \end{aligned}

Then σ(S0)=σ(S1)=σ(S2)=σ(S3)\sigma(\mathcal{S}_0) = \sigma(\mathcal{S}_1) = \sigma(\mathcal{S}_2) = \sigma(\mathcal{S}_3).

Proof. We just prove that σ(S0)=σ(S2)\sigma(\mathcal{S}_0) = \sigma(\mathcal{S}_2).

  1. ×i∈[d](ai,bi]=∩n≥1(×i∈[d](ai,bi+1/n])\times_{i \in [d] } (a_i, b_i] = \cap_{n \ge 1} \Big( \times_{i \in [d]} (a_i, b_i + 1 / n] \Big), therefore ×i∈[d](ai,bi]∈σ(S2)\times_{i \in [d] } (a_i, b_i] \in \sigma(\mathcal{S}_2). It follows that σ(S0)⊂σ(S2)\sigma(\mathcal{S}_0) \subset \sigma(\mathcal{S}_2)

  2. ×i∈[d](ai,bi)=∪n≥1(×i∈[d](ai,bi−1/n])\times_{i \in [d] } (a_i, b_i) = \cup_{n \ge 1} \Big( \times_{i \in [d]} (a_i, b_i - 1 / n] \Big), therefore ×i∈[d](ai,bi)∈σ(S0)\times_{i \in [d] } (a_i, b_i) \in \sigma(\mathcal{S}_0). It follows that σ(S2)⊂σ(S0)\sigma(\mathcal{S}_2) \subset \sigma(\mathcal{S}_0)

Via similar argument, we can prove that σ(S0)=σ(S3)\sigma(\mathcal{S}_0) = \sigma(\mathcal{S}_3), σ(S1)=σ(S2)\sigma(\mathcal{S}_1) = \sigma(\mathcal{S}_2) and σ(S1)=σ(S3)\sigma(\mathcal{S}_1) = \sigma(\mathcal{S}_3).

□\square