Sigma-Algebra Closed under Union
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Theorem
Let $X$ be a set, and let $\Sigma$ be a $\sigma$-algebra on $X$.
Let $A, B \in \Sigma$ be measurable sets.
Then $A \cup B \in \Sigma$, where $\cup$ denotes set union.
Corollary
Let $A_1, \ldots, A_n \in \Sigma$.
Then $\ds \bigcup_{k \mathop = 1}^n A_k \in \Sigma$.
Proof
Define $A_1 = A, A_2 = B$, and for $n \in \N, n \ge 2: A_n = \O$.
Then by Sigma-Algebra Contains Empty Set, axiom $(3)$ of a $\sigma$-algebra applies.
Hence:
- $\ds \bigcup_{n \mathop \in \N} A_n = A \cup B \in \Sigma$
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $3.2 \ \text{(ii)}$