Sigma-Algebra Closed under Union

Theorem
Let $X$ be a set, and let $\mathcal A$ be a $\sigma$-algebra on $X$.

Let $A, B \in \mathcal A$ be measurable sets.

Then $A \cup B \in \mathcal A$, where $\cup$ denotes set union.

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 $\displaystyle \bigcup_{n \in \N} A_n = A \cup B \in \mathcal A$.