Equivalence of Definitions of Sigma-Algebra

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
=== 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, $(\text {SA} 1')$ is the same as $(\text {SA} 1)$.
Secondly, $(\text {SA} 3')$ is a special case of $(\text {SA} 3)$.
It remains to show $(\text {SA} 2')$.
Let $A, B \in \Sigma$ be arbitrary.
Observe:
By $(\text {SA} 2)$:
:$\relcomp X A \in \Sigma$
Furthermore by $(\text {SA} 3)$:
:$\relcomp X A \cup B \in \Sigma$
Finally, by $(\text {SA} 2)$:
:$A \setminus B = \relcomp X {\relcomp X A \cup B} \in \Sigma$
Hence $(\text {SA} 2')$ is verified.
=== Definition 2 implies Definition 1 ===
First, $(\text {SA} 1)$ is the same as $(\text {SA} 1')$.
Secondly, $(\text {SA} 2)$ is a special case of $(\text {SA} 2')$, since for all $A \subseteq X$:
:$\relcomp X A = X \setminus A$
by definition of relative complement.
It remains to show $(\text {SA} 3)$.
Let $A_n \in \Sigma$ be arbitrary for $n = 1, 2, \ldots$.
Let $F_1 := A_1$.
For all $n \ge 2$, let recursively:
:$F_n := A_n \setminus \paren {F_1 \sqcup \cdots \sqcup F_{n - 1} }$
==== Sublemma ====
For all $n = 1, 2, \ldots$ we have:
:$F_n \in \Sigma$
and:
:$A_1 \cup \cdots \cup A_n = F_1 \sqcup \cdots \sqcup F_n$
==== Proof of Sublemma ====
We shall prove by induction.
For $n = 1$ the claim is trivial, as $F_1 = A_1$.
Suppose that the claim is true for $n = k - 1$.
In particular:
:$F_1 \sqcup \cdots \sqcup F_{k - 1} \in \Sigma$
Therefore by $(\text {SA} 2')$:
:$F_k = A_k \setminus \paren {F_1 \sqcup \cdots \sqcup F_{k - 1} } \in\Sigma$
Furthermore:
Therefore:
:$\ds \bigcup_{n \mathop = 1}^\infty A_n = \bigsqcup_{n \mathop = 1}^\infty F_n$
where the belongs to $\Sigma$ by $(\text {SA} 3')$.
Hence $(\text {SA} 3)$ 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.
== Sources ==
* : $\S 1.1$: Rings and Algebras
* : $\S 1.1$: Rings and Algebras: Theorem $1.1.1$
* : $\S 1.2$
Category:Sigma-Algebras