Equivalence of Definitions of Sigma-Algebra

Definition 1 implies Definition 3
Let $\Sigma$ be a system of sets on a set $X$ such that:
 * $(1): \quad X \in \Sigma$
 * $(2): \quad \forall A, B \in \Sigma: \relcomp X A \in \Sigma$
 * $(3): \quad \ds \forall A_n \in \Sigma: n = 1, 2, \ldots: \bigcup_{n \mathop = 1}^\infty A_n \in \Sigma$

Let $A, B \in \Sigma$.

From the definition:
 * $\forall A \in \Sigma: A \subseteq X$.

Hence from Intersection with Subset is Subset:
 * $\forall A \in \Sigma: A \cap X = A$

Hence $X$ is the unit of $\Sigma$.

So by definition 2 of $\sigma$-ring it follows that $\Sigma$ is a $\sigma$-ring with a unit.

Thus $\Sigma$ is a $\sigma$-algebra by definition 3.

Definition 3 implies Definition 1
Let $\Sigma$ be a $\sigma$-ring with a unit $X$.

By definition, $X \in \Sigma$.

From definition 2 of $\sigma$-ring, $\Sigma$ is:
 * $(1) \quad$ closed under set difference.
 * $(2) \quad$ closed under countable union

From Unit of System of Sets is Unique, we have that:
 * $\forall A \in \Sigma: A \subseteq X$

from which we have that $X \setminus A = \relcomp X A$.

So $\Sigma$ is a $\sigma$-algebra by definition 1.

Definition 1 implies Definition 2
First, (SA1') is the same as (SA1).

Secondly, (SA3') is a special case of (SA3).

It remains to show (SA2').

Let $A,B\in\Sigma$ be arbitrary.

Observe:

By (SA2):
 * $\relcomp X A\in\Sigma$

Furthermore by (SA3):
 * $\relcomp X A \cup B\in\Sigma$

Finally, by (SA2):
 * $A\setminus B = \relcomp X {\relcomp X A \cup B}\in\Sigma$

Hence (SA2') is verified.

Definition 2 implies Definition 1
First, (SA1) is the same as (SA1').

Secondly, (SA2) is a special case of (SA2'), since for all $A\subseteq X$:
 * $\relcomp X A = X \setminus A$

by definition.

It remains to show (SA3).

Let $A_n\in\Sigma$ be arbitrary for $n=1,2,\ldots$.

Let $F_1 := A_1$

For all $n\geq 2$, let recursively:
 * $F_n := A_n\setminus \paren { F_1 \sqcup \cdots \sqcup F_{n-1}}$

Recursively applying (SA2') and (SA3'), we obtain:
 * $F_n\in\Sigma$

for all $n=1,2,\ldots$.

By construction, $\family F_n$ is disjoint and:
 * $\bigcup _{n=1} ^\infty A_n = \bigsqcup _{n=1} ^\infty F_n$

where the belongs to $\Sigma$ by (SA3').

Hence (SA3) is verified.

Definition 1 implies Definition 4
Immediate from the definition of algebra along with the added condition of closure under countable unions.

Definition 4 implies Definition 1
By definition $1$ of algebra of sets, an algebra has the properties:

Replacing $(\text {AS} 2)$ with closure under countable unions immediately yields the first definition.