Delta-Algebra is Sigma-Algebra

Theorem

Every $\delta$-algebra is a $\sigma$-algebra.

Proof

Let $\DD$ be a $\delta$-algebra whose unit is $\mathbb U$.

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

Then:

\(\ds \forall i: \, \)

\(\ds \mathbb U \setminus A_i\)

\(\in\)

\(\ds \DD\)

$\DD$ is closed under relative complement with $\mathbb U$

\(\ds \leadsto \ \ \)

\(\ds \bigcap_{i \mathop = 1}^\infty \paren {\mathbb U \setminus A_i}\)

\(\in\)

\(\ds \DD\)

$\DD$ is closed under countable intersections

\(\ds \leadsto \ \ \)

\(\ds \mathbb U \setminus \bigcup_{i \mathop = 1}^\infty A_i\)

\(\in\)

\(\ds \DD\)

De Morgan's Laws: Complement of Union

\(\ds \leadsto \ \ \)

\(\ds \bigcup_{i \mathop = 1}^\infty A_i\)

\(\in\)

\(\ds \DD\)

$\DD$ is closed under relative complement with $\mathbb U$

Thus $\DD$ is also a $\sigma$-algebra.

$\blacksquare$

Also see

• Sigma-Algebra is Delta-Algebra