Equivalence of Definitions of Sigma-Algebra

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Theorem

The following definitions of the concept of Sigma-Algebra are equivalent:

Definition 1

Let $X$ be a set.

Let $\Sigma$ be a system of subsets of $X$.


$\Sigma$ is a $\sigma$-algebra over $X$ if and only if $\Sigma$ satisfies the sigma-algebra axioms:

\((\text {SA} 1)\)   $:$   Unit:    \(\ds X \in \Sigma \)      
\((\text {SA} 2)\)   $:$   Closure under Complement:      \(\ds \forall A \in \Sigma:\) \(\ds \relcomp X A \in \Sigma \)      
\((\text {SA} 3)\)   $:$   Closure under Countable Unions:      \(\ds \forall A_n \in \Sigma: n = 1, 2, \ldots:\) \(\ds \bigcup_{n \mathop = 1}^\infty A_n \in \Sigma \)      


Definition 2

Let $X$ be a set.

Let $\Sigma$ be a system of subsets of $X$.


$\Sigma$ is a $\sigma$-algebra over $X$ if and only if $\Sigma$ satisfies the sigma-algebra axioms:

\((\text {SA} 1')\)   $:$   Unit:    \(\ds X \in \Sigma \)      
\((\text {SA} 2')\)   $:$   Closure under Set Difference:      \(\ds \forall A, B \in \Sigma:\) \(\ds A \setminus B \in \Sigma \)      
\((\text {SA} 3')\)   $:$   Closure under Countable Disjoint Unions:      \(\ds \forall A_n \in \Sigma: n = 1, 2, \ldots:\) \(\ds \bigsqcup_{n \mathop = 1}^\infty A_n \in \Sigma \)      


Definition 3

A $\sigma$-algebra $\Sigma$ is a $\sigma$-ring with a unit.

Definition 4

Let $X$ be a set.

A $\sigma$-algebra $\Sigma$ over $X$ is an algebra of sets which is closed under countable unions.


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.

$\Box$


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.

$\Box$


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:

\(\ds A \setminus B\) \(=\) \(\ds A \cap \relcomp X B\) Set Difference as Intersection with Relative Complement
\(\ds \) \(=\) \(\ds \relcomp X {\relcomp X A} \cap \relcomp X B\) Relative Complement of Relative Complement
\(\ds \) \(=\) \(\ds \relcomp X {\relcomp X A \cup B}\) De Morgan's Law

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.

$\Box$


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:

\(\ds A_1 \cup \cdots \cup A_k\) \(=\) \(\ds \paren {A_1 \cup \cdots \cup A_{k - 1} } \cup A_k\)
\(\ds \) \(=\) \(\ds \paren {F_1 \sqcup \cdots \sqcup F_{k - 1} } \cup A_k\) induction assumption
\(\ds \) \(=\) \(\ds \paren {F_1 \sqcup \cdots \sqcup F_{k - 1} } \sqcup \paren {A_k \setminus \paren {F_1 \sqcup \cdots \sqcup F_{k - 1} } }\)
\(\ds \) \(=\) \(\ds \paren {F_1 \sqcup \cdots \sqcup F_{k - 1} } \sqcup F_k\)

$\Box$


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Therefore:

$\ds \bigcup_{n \mathop = 1}^\infty A_n = \bigsqcup_{n \mathop = 1}^\infty F_n$

where the right hand side belongs to $\Sigma$ by $(\text {SA} 3')$.

Hence $(\text {SA} 3)$ is verified.

$\Box$


Definition 1 implies Definition 4

Immediate from the definition of algebra along with the added condition of closure under countable unions.

$\Box$


Definition 4 implies Definition 1

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

\((\text {AS} 1)\)   $:$   Unit:    \(\ds X \in \Sigma \)      
\((\text {AS} 2)\)   $:$   Closure under Union:      \(\ds \forall A, B \in \Sigma:\) \(\ds A \cup B \in \Sigma \)      
\((\text {AS} 3)\)   $:$   Closure under Complement Relative to $X$:      \(\ds \forall A \in \Sigma:\) \(\ds \relcomp X A \in \Sigma \)      

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

$\blacksquare$


Sources

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