Sigma-Algebra is Delta-Algebra

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

• Delta-Algebra is Sigma-Algebra