Intersection of Sigma-Algebras

Theorem

Let $X$ be a set.

Let $\family {\Sigma_i}_{i \mathop \in I}$ be a indexed set of $\sigma$-algebras on $X$.

Then $\Sigma := \ds \bigcap_{i \mathop \in I} \Sigma_i$ is also a $\sigma$-algebra on $X$.

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

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

As all the $\Sigma_i$ are $\sigma$-algebras, $X \in \Sigma_i$ for all $i \in I$.

Hence $X \in \Sigma$ by definition of intersection.

$\Box$

Let $E \in \Sigma$.

Then for all $i \in I$, $E \in \Sigma_i$, by definition of intersection.

As all the $\Sigma_i$ are $\sigma$-algebras, also, for all $i \in I$:

$X \setminus E \in \Sigma_i$

Hence $X \setminus E \in \Sigma$ by definition of intersection.

$\Box$

Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence in $\Sigma$.

Then it is also a sequence in every $\Sigma_i$.

Hence, for all $i \in I$, as the $\Sigma_i$ are all $\sigma$-algebras:

$\ds \bigcup_{n \mathop \in \N} E_n \in \Sigma_i$

By definition of intersection, it follows that:

$\ds \bigcup_{n \mathop \in \N} E_n \in \Sigma$

$\blacksquare$