Sigma-Algebra Closed under Countable Intersection

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Theorem

Let $X$ be a set, and let $\Sigma$ be a $\sigma$-algebra on $X$.

Suppose that $\left({E_n}\right)_{n \in \N} \in \Sigma$ is a collection of measurable sets.


Then $\displaystyle \bigcap_{n \mathop \in \N} E_n \in \Sigma$, where $\displaystyle \bigcap$ denotes set intersection.


Proof

\(\displaystyle \forall n \in \N: \ \ \) \(\displaystyle E_n \in \Sigma\) \(\implies\) \(\displaystyle X \setminus E_n \in \Sigma\) Axiom $(2)$ for $\sigma$-algebras
\(\displaystyle \) \(\implies\) \(\displaystyle \bigcup_{n \mathop \in \N} \left({X \setminus E_n}\right) \in \Sigma\) Axiom $(3)$ for $\sigma$-algebras
\(\displaystyle \) \(\implies\) \(\displaystyle X \setminus \left({\bigcup_{n \mathop \in \N} \left({X \setminus E_n}\right) }\right) \in \Sigma\) Axiom $(2)$ for $\sigma$-algebras

From De Morgan's laws: Complement of Intersection:

$\displaystyle \bigcup_{n \mathop \in \N} \left({X \setminus E_n}\right) = X \setminus \left({\bigcap_{n \mathop \in \N} E_n}\right)$

Also, by Set Difference with Set Difference and Set Union Preserves Subsets:

$\displaystyle X \setminus \left({X \setminus \left({\bigcap_{n \mathop \in \N} E_n}\right) }\right) = \bigcap_{n \mathop \in \N} E_n$

Combining the previous equalities, it follows that:

$\displaystyle \bigcap_{n \mathop \in \N} E_n \in \Sigma$

$\blacksquare$


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