σ-Algebras of Sets

Recall from the Algebras of Sets page that if $X$ is a set then an algebra on $X$ is a collection $\mathcal A$ of subsets of $X$ with the following properties:

• If $A_1, A_2 \in \mathcal A$ then $A_1 \cup A_2 \in \mathcal A$.

• If $A \in \mathcal A$ then $A^c \in \mathcal A$.

We will now define a new type of algebra on a set.

Definition: Let $X$ be a set. A $\sigma$-Algebra on $X$ is a nonempty collection $\mathcal A$ of subsets of $X$ with the following properties: 1) If $A \in \mathcal A$ then $A^c \in \mathcal A$. 2) If $(A_n)_{n=1}^{\infty} \subseteq \mathcal A$ then $\bigcup_{n=1}^{\infty} A_n \in \mathcal A$.

Proposition 1: Let $X$ be a set and let $\mathcal A$ be a $\sigma$-algebra on $X$. If $(A_n)_{n=1}^{\infty} \subseteq \mathcal A$ then $\bigcap_{n=1}^{\infty} A_n \in \mathcal A$.

• Proof: Since $(A_n)_{n=1}^{\infty} \subseteq \mathcal A$, we have that $A_n^c \in \mathcal A$ for each $n \in \mathbb{N}$ by property (1). Since $(A_n^c)_{n=1}^{\infty} \subseteq \mathcal A$ we have by property (2) that:

(1)

\begin{align} \quad \bigcup_{n=1}^{\infty} A_n^c = \left ( \bigcap_{n=1}^{\infty} A_n \right )^c \in \mathcal A \end{align}

• So by property (1) again, $\bigcap_{n=1}^{\infty} A_n \in \mathcal A$.