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.

A $\sigma$-algebra $\Sigma$ over $X$ is a system of subsets of $X$ with the following properties:

\((\text {SA} 1)\)   $:$   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.

A $\sigma$-algebra $\Sigma$ over $X$ is a system of subsets of $X$ with the following properties:

\((\text {SA} 1')\)   $:$   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 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

Follows directly from the definitions, as a disjoint union is a type of union.

$\Box$


Definition 2 implies Definition 1

Let $\Sigma$ be a system of sets on a set $X$ such that:

$(1): \quad X \in \Sigma$
$(2): \quad \forall A, B \in \Sigma: \relcomp X A \in \Sigma$
$(3): \quad \ds \forall A_n \in \Sigma: n = 1, 2, \ldots: \bigsqcup_{n \mathop = 1}^\infty A_n \in \Sigma$

Conditions $(1)$ and $(2)$ in definition 2 are identical to that of conditions $(1)$ and $(2)$ in definition 1.

Let $\family {E_n}_{n \mathop \in \N}$ be a countable indexed family of sets in $\Sigma$.

By Union of Indexed Family of Sets Equal to Union of Disjoint Sets:

$\ds \bigsqcup_{n \mathop \in \N} F_n = \bigcup_{n \mathop \in \N} E_n$

for an appropriately constructed countable indexed family of disjoint sets in $\Sigma$.


By the hypotheses of definition $2$:

$\ds \bigsqcup_{k \mathop \in \N}^\infty F_k$

is measurable.


Thus $\ds \bigcup_{n \mathop \in \N} E_n$ is also measurable by $(\text {SA} 2)$ of definition $1$.

$\Box$


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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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