Sigma-Algebra is Delta-Algebra

From ProofWiki
Jump to navigation Jump to search

Theorem

A $\sigma$-algebra is also a $\delta$-algebra.


Proof

Let $\SS$ be a $\sigma$-algebra whose unit is $\mathbb U$.

Let $A_1, A_2, \ldots$ be a countably infinite collection of elements of $\SS$.


Then:

\(\ds \forall i: \, \) \(\ds \mathbb U \setminus A_i\) \(\in\) \(\ds \SS\) $\SS$ is closed under relative complement with $\mathbb U$
\(\ds \leadsto \ \ \) \(\ds \bigcup_{i \mathop = 1}^\infty \paren {\mathbb U \setminus A_i}\) \(\in\) \(\ds \SS\) $\SS$ is closed under countable unions
\(\ds \leadsto \ \ \) \(\ds \mathbb U \setminus \bigcap_{i \mathop = 1}^\infty A_i\) \(\in\) \(\ds \SS\) De Morgan's Laws
\(\ds \leadsto \ \ \) \(\ds \bigcap_{i \mathop = 1}^\infty A_i\) \(\in\) \(\ds \SS\) $\SS$ is closed under relative complement with $\mathbb U$


Thus $\SS$ is a $\delta$-algebra.

$\blacksquare$


Also see