# Equivalence of Definitions of Sigma-Algebra

## Theorem

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

### Definition 1

Let $X$ be a set.

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

 $(SA \, 1)$ $:$ Unit: $\displaystyle X \in \mathcal R$ $(SA \, 2)$ $:$ Closure under Complement: $\displaystyle \forall A \in \mathcal R:$ $\displaystyle \complement_X \left({A}\right) \in \mathcal R$ $(SA \, 3)$ $:$ Closure under Countable Unions: $\displaystyle \forall A_n \in \mathcal R: n = 1, 2, \ldots:$ $\displaystyle \bigcup_{n \mathop = 1}^\infty A_n \in \mathcal R$

### Definition 2

Let $X$ be a set.

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

 $(SA \, 1')$ $:$ Unit: $\displaystyle X \in \mathcal R$ $(SA \, 2')$ $:$ Closure under Complement: $\displaystyle \forall A \in \mathcal R:$ $\displaystyle \complement_X \left({A}\right) \in \mathcal R$ $(SA \, 3')$ $:$ Closure under Countable Disjoint Unions: $\displaystyle \forall A_n \in \mathcal R: n = 1, 2, \ldots:$ $\displaystyle \bigsqcup_{n \mathop = 1}^\infty A_n \in \mathcal R$

### Definition 3

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

### Definition 4

Let $X$ be a set.

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

## Proof

### Definition 1 implies Definition 3

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

$(1): \quad X \in \mathcal R$
$(2): \quad \forall A, B \in \mathcal R: \complement_X \left({A}\right) \in \mathcal R$
$(3): \quad \displaystyle \forall A_n \in \mathcal R: n = 1, 2, \ldots: \bigcup_{n \mathop = 1}^\infty A_n \in \mathcal R$

Let $A, B \in \mathcal R$.

From the definition:

$\forall A \in \mathcal R: A \subseteq X$.

Hence from Intersection with Subset is Subset:

$\forall A \in \mathcal R: A \cap X = A$

Hence $X$ is the unit of $\mathcal R$.

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

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

$\Box$

### Definition 3 implies Definition 1

Let $\mathcal R$ be a $\sigma$-ring with a unit $X$.

By definition, $X \in \mathcal R$.

From definition 2 of $\sigma$-ring, $\mathcal R$ is:

$(1) \quad$ closed under set difference.
$(2) \quad$ closed under countable union

From Unit of System of Sets is Unique, we have that:

$\forall A \in \mathcal R: A \subseteq X$

from which we have that $X \setminus A = \complement_X \left({A}\right)$.

So $\mathcal R$ 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 $\mathcal R$ be a system of sets on a set $X$ such that:

$(1): \quad X \in \mathcal R$
$(2): \quad \forall A, B \in \mathcal R: \complement_X \left({A}\right) \in \mathcal R$
$(3): \quad \displaystyle \forall A_n \in \mathcal R: n = 1, 2, \ldots: \bigsqcup_{n \mathop = 1}^\infty A_n \in \mathcal R$

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

Let $\left\{ {E_n}\right\}_{n \mathop \in \N}$ be a countable indexed family of sets in $\mathcal R$.

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

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

By the hypotheses of definition 2:

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

is measurable.

Thus $\displaystyle \bigcup_{n \mathop \in \N} E_n$ is also measurable by condition 2 of definition 1.

$\Box$

### 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 of an algebra of sets, an algebra has the properties:

 $(AS \, 1)$ $:$ Unit: $\displaystyle X \in \mathcal R$ $(AS \, 2)$ $:$ Closure under Union: $\displaystyle \forall A, B \in \mathcal R:$ $\displaystyle A \cup B \in \mathcal R$ $(AS \, 3)$ $:$ Closure under Complement Relative to $X$: $\displaystyle \forall A \in \mathcal R:$ $\displaystyle \complement_X \left({A}\right) \in \mathcal R$

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

$\blacksquare$