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Finite disjoint unions of measurable rectangles form an algebra generating the product sigma-algebra
Statement
Let and be measurable spaces. The family of finite disjoint unions of measurable rectangles in is an algebra of subsets of , and it generates .
Facts & Assumptions
Given: Measurable spaces and .
A measurable rectangle has the form with and . (Measurable rectangles in a product of measurable spaces)
The product sigma-algebra is the sigma-algebra generated by the measurable rectangles. (The product sigma-algebra and its finite iterates)
For rectangles, and
If for , take the nonempty Boolean atoms generated by in and by in . The products of an -atom and a -atom are finitely many pairwise disjoint measurable rectangles partitioning , and each , hence also , is the union of a subfamily of these product atoms.
Proof
By [L1] and [A1], the intersection of two measurable rectangles is again a measurable rectangle, and the complement of a measurable rectangle is a finite union of measurable rectangles.
Let be the family of finite disjoint unions of measurable rectangles. It contains and . If , then [A2] disjointifies the finite union into finitely many pairwise disjoint measurable rectangles, so . Likewise belongs to by step 1.1. Thus is an algebra.
Every measurable rectangle belongs to , so [L2] gives The reverse inclusion holds because every member of is a finite union of measurable rectangles and hence lies in . Therefore , and is an algebra generating the product sigma-algebra.
Depends on
- Measurable rectangles in a product of measurable spaces
- The product sigma-algebra and its finite iterates
- Algebras of subsets
- $A \times (B \cup C) = (A \times B) \cup (A \times C)$, $A \times (B \cap C) = (A \times B) \cap (A \times C)$, $A \times (B \setminus C) = (A \times B) \setminus (A \times C)$, $(A \cap B) \times (C \cap D) = (A \times C) \cap (B \times D)$; $A \times B = \varnothing$ if and only if $A = \varnothing$ or $B = \varnothing$; and for nonempty $A$ and $B$, $A \times B \subseteq C \times D$ if and only if $A \subseteq C$ and $B \subseteq D$
Used by
Dependency tree · two levels
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Sources
- John K. Hunter, Measure Theory, paragraph before Definition 5.10 (standard reference, not scraped)