# Sigma-Algebra Closed under Union

## 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:

$\ds \bigcup_{n \mathop \in \N} A_n = A \cup B \in \Sigma$

$\blacksquare$