Equivalence of Definitions of Sigma-Algebra

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.

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.

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.

For condition $(3)$, let $\left\{ { E_j} \right \}_{j \mathop = 0}^\infty$ be a countable set of sets in $\mathcal R$ indexed by $\N$.

Define another set $\displaystyle \left\{ { F_k } \right \}_{j \mathop = 0}^\infty$ by:


 * $\displaystyle F_k = E_k \setminus \left({ \bigcup_{j \mathop = 0}^{k \mathop - 1} E_j }\right)$

Consider distinct $F_m,F_n$

let $m < n$

Then the construction of $F_k$ implies that:

... which means that $\displaystyle \left\{ { F_k } \right \}_{j \mathop = 0}^\infty$ is a family of pairwise disjoint sets.

By the hypotheses of definition 2:


 * $\displaystyle \bigsqcup_{n \mathop = 0}^\infty F_k$

...is measurable.

By De Morgan's:


 * $\displaystyle F_k = \bigcap_{j \mathop = 0}^{k \mathop - 1} \left({E_k \setminus E_j}\right)$

Taking unions over all $k$:

Hence the equivalence, as $\left\{ { E_j} \right \}_{j \mathop = 0}^\infty$ was an arbitrary countable set of sets in $\mathcal R$.

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

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