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 $\displaystyle \bigcup_{k \mathop = 1}^n A_k \in \Sigma$.


Proof

Define $A_1 = A, A_2 = B$, and for $n \in \N, n \ge 2: A_n = \varnothing$.

Then by Sigma-Algebra Contains Empty Set, axiom $(3)$ of a $\sigma$-algebra applies.

Hence $\displaystyle \bigcup_{n \mathop \in \N} A_n = A \cup B \in \Sigma$.

$\blacksquare$


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