σ-Algebras of Sets

# 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) For every sequence $(A_n)_{n=1}^{\infty}$ in $\mathcal A$ we have that $\displaystyle{\bigcup_{i=1}^{\infty} A_i \in \mathcal A}$. 2) For every set $A \in \mathcal A$ we have that the complement, $A^c \in \mathcal A$.

Note that the definitions of an algebra and a $\sigma$-algebra on a set $X$ are similar. The only difference is that an algebra on a set $X$ is closed under finite unions, while an algebra on a set $X$ is closed under countably infinite unions.//

The following proposition tells us that every $\sigma$-algebra on $X$ is an algebra on $X$.

 Proposition: Let $\mathcal A$ be a $\sigma$-algebra on $X$. Then $\mathcal A$ is an algebra on $X$.
• Proof: Let $\mathcal A$ be a $\sigma$-algebra on $X$. Let $A_1, A_2 \in \mathcal A$. Consider the following sequence:
(1)
\begin{align} \quad (A_n^*)_{n=1}^{\infty} = (A_1, A_2, A_2, ...) \end{align}
• Since $\mathcal A$ is a $\sigma$-algebra on $X$ we have that:
(2)
\begin{align} \quad A_1 \cup A_2 = A_1 \cup A_2 \cup A_2 \cup ... = \bigcup_{i=1}^{\infty} A_n^* \in \mathcal A \end{align}
• Furthermore, for every $A \in \mathcal A$ we have that $A^c \in \mathcal A$. So $\mathcal A$ is an algebra on $X$. $\blacksquare$