semi-ring (in measure theory)

Prerequisites

$E_{i}$ 为 $Ω_{i}$ 上的半环，则 $S = E_{1} \times E_{2} \times \dots \times E_{n}$ 为半环.

证明：

S = {i = 1 \prod n A_{i} ∣ A_{i} \in E_{i}}

(1) $\emptyset \in S$
显然
(2) 若 $A, B \in S$ ，则 $A \cap B \in S$

$A = \prod_{i = 1}^{n} A_{i}$
$B = \prod_{i = 1}^{n} B_{i}$

$A \cap B = \prod_{i = 1}^{n} A_{i} \cap B_{i} \in S$

(3) $A, B \in S$ ,则 $A ∖ B = ⋃_{i = 1}^{n} C_{i}$ ，其中 $C_{i} \in S$

$A = \prod_{i = 1}^{n} A_{i}$
$B = \prod_{i = 1}^{n} B_{i}$
设 $A_{i} ∖ B_{i} = ⨄_{j = 1}^{m_{i}} D_{ij}$

A ∖ B = A \cap B^{c} = i = 1 \prod n A_{i} \cap (i = 1 \prod n B_{i})^{c} = i = 1 \prod n A_{i} \cap j = 1 ⋃ n (i < j \prod Ω_{i} \times B_{j}^{c} \times i > j \prod Ω_{i}) = j = 1 ⋃ n (i < j \prod A_{i} \times (A_{j} ∖ B_{j}) \times i > j \prod A_{i}) = j = 1 ⋃ n (i < j \prod (A_{i} \cap B_{i}) \times (A_{j} ∖ B_{j}) \times i > j \prod A_{i}) = j = 1 ⋃ n (i < j \prod (A_{i} \cap B_{i}) \times (k = 1 ⨄ m_{j} D_{j k}) \times i > j \prod A_{i}) = j = 1 ⋃ n k = 1 ⨄ m_{j} (i < j \prod (A_{i} \cap B_{i}) \times (D_{j k}) \times i > j \prod A_{i}) = j = 1 ⨄ n k = 1 ⨄ m_{j} (i < j \prod (A_{i} \cap B_{i}) \times (D_{j k}) \times i > j \prod A_{i})

$E = {[a, b) ∣ a, b \in R}$ 是半环

证明：

(1) $\emptyset \in E$

$\emptyset = [a, a) \in E$

(2) 若 $A, B \in E$ ，则 $A \cap B \in E$

$A = [a, b)$
$B = [c, d)$

$A \cap B = [a, b) \cap [c, d) = [max (a, c), min (b, d))$

(3) $A, B \in E$ ,则 $A ∖ B = ⋃_{i = 1}^{n} C_{i}$ ，其中 $C_{i} \in E$

$A = [a, b)$
$B = [c, d)$

$B^{c} = (- \infty, c) \cup [d, \infty)$

A ∖ B = A \cap B^{c} = [a, b) \cap ((- \infty, c) \cup [d, \infty)) = [a, min (b, c)) \cup [max (a, d), b)

所以 ${[a, b) ∣ a, b \in R}$ 是半环