Intersection of Sigma-Algebras

Theorem
Let $X$ be a set.

Let $\left({\mathcal{A}_i}\right)_{i \in I}$ be a collection of $\sigma$-algebras on $X$.

Then $\mathcal A := \displaystyle \bigcap_{i \in I} \mathcal{A}_i$ is also a $\sigma$-algebra on $X$.

Here, $\displaystyle \bigcap$ denotes set intersection.

Proof
Verify the axioms for a $\sigma$-algebra in turn:

Axiom $(1)$
As all the $\mathcal{A}_i$ are $\sigma$-algebras, $X \in \mathcal{A}_i$ for all $i \in I$.

Hence $X \in \mathcal A$ by definition of intersection.

Axiom $(2)$
Let $A \in \mathcal A$.

Then for all $i \in I$, $A \in \mathcal{A}_i$, by definition of intersection.

As all the $\mathcal{A}_i$ are $\sigma$-algebras, also, for all $i \in I$:


 * $X \setminus A \in \mathcal{A}_i$

Hence $X \setminus A \in \mathcal A$ by definition of intersection.

Axiom $(3)$
Let $\left({A_n}\right)_{n \in \N}$ be a sequence in $\mathcal A$.

Then it is also a sequence in every $\mathcal{A}_i$.

Hence, for all $i \in I$, as the $\mathcal{A}_i$ all are $\sigma$-algebras:


 * $\displaystyle \bigcup_{n \in \N} A_n \in \mathcal{A}_i$

By definition of intersection, it follows that:


 * $\displaystyle \bigcup_{n \in \N} A_n \in \mathcal A$