> [!definition] > > Let $X$ be a [[Set|set]]. Then $\cm \subseteq \pow{X}$ is a $\sigma$-algebra if it is [[Closure|closed]] under complement and [[Cardinality|countable]] unions. > $ > A \in \cm \Rightarrow A^c \in \cm \quad > \bracs{A}_{1}^{\infty} \in \cm \Rightarrow \bigcup_{i \in \nat}A_i \in \cm > $ > > $\sigma$-algebras are also closed under countable intersections. > > In other words, $\cm$ is an [[Algebra|algebra]] that is also closed under *countable* unions. > > $\bracs{}, X$ belong to every $\sigma$-algebra. > [!definition] > > Let $X$ be an [[Cardinality|uncountable]] set, then > $ > \mathcal{A} = \bracs{E \subset X: |E| < \aleph_1 \text{ or } |E^c| < \aleph_1} > $ > is the **$\sigma$-algebra of countable/co-countable sets**. > [!definition] > > Let $X$ be a set and $\ce \subseteq \pow{X}$, then the smallest $\sigma$-algebra containing $\ce$, $\agb{\ce}$, is the $\sigma$-algebra **generated** by $\ce$. > [!theorem] > > Let $X$ be a set and $\ce, \cf \subseteq \pow{X}$, then > $ > \ce \subseteq \agb{\cf} \Rightarrow \agb{\ce} \subseteq \agb{\cf} > $ > > *Proof*. By closure.