Equivalence of Definitions of Sigma-Ring

Definition 1 implies Definition 2
Let $\text {SR}$ be a ring of sets which is closed under countable unions.

We have:

which are exactly $\text {SR} 1$ and $\text {SR} 2$.

Then as $\text {SR}$ is closed under countable unions:


 * $\ds A_1, A_2, \ldots \in \text {SR} \implies \bigcup_{n \mathop = 1}^\infty A_n \in \text {SR}$

and so $\text {SR} 3$ is fulfilled.

Definition 2 implies Definition 1
Let $\text {SR}$ be a system of sets such that:

As noted above, $\text {SR} 1$ and $\text {SR} 2$ are exactly $\text {RS} 1_2$ and $\text {RS} 2_2$.

Let $A, B \in \text {SR}$.

Let $A_1 = A, A_2 = B$ and $A_n = \O$ for all $n = 3, 4, \ldots$

Then:
 * $\ds \forall A_n \in \text {SR}: n = 1, 2, \ldots: \bigcup_{n \mathop = 1}^\infty A_n = A \cup B \in \text {SR}$

Thus criterion $(\text {RS} 3_2)$ is fulfilled.

So $\text {SR}$ is a ring of sets which is closed under countable unions.

Definition 2 implies Definition 3
$\text {SR} 1$ is exactly $\text {SR} 1'$ and $\text {SR} 2$ is exactly $\text {SR} 2'$.

$\text {SR} 3$ allows arbitrary countable unions, and therefore a fortiori implies $\text {SR} 3'$.

Definition 3 implies Definition 2
As noted above, $\text {SR} 1$ and $\text {SR} 2$ are exactly $\text {SR} 1'$ and $\text {SR} 2'$.

Let $A_1, A_2, \ldots \in \Sigma$.

We have to show that $\ds \bigcup_{n \mathop = 1}^\infty A_n \in \Sigma$.

Let $B_n = A_n \setminus \ds\bigcup_{i \mathop = 1}^{n-1}A_i$ for all $n\in\N$.

Because:
 * $B_n = \paren {\cdots \paren {\paren {A_n \setminus A_{n - 1} } \setminus A_{n - 2} } \setminus \cdots \setminus A_1}$

and $\text {SR} 2'$, we have:
 * $\forall n \in \N: B_n \in \Sigma$

As $B_n \subseteq A_n$ for all $n \in \N$, it follows that:
 * $\ds \bigcup_{n \mathop = 1}^\infty B_n \subseteq \bigcup_{n \mathop = 1}^\infty A_n$

Let:
 * $x \in \ds\bigcup_{n \mathop = 1}^\infty A_n$

and:
 * $k = \min \set {n \in \N: x \in A_n}$

Then:
 * $x \in B_k$

Therefore:
 * $x \in \ds \bigcup_{n \mathop = 1}^\infty B_n$

Hence:
 * $\ds \bigcup_{n \mathop = 1}^\infty A_n \subseteq \bigcup_{n \mathop = 1}^\infty B_n$

Together with
 * $\ds \bigcup_{n \mathop = 1}^\infty B_n \subseteq \bigcup_{n \mathop = 1}^\infty A_n$

it follows that:
 * $\ds \bigcup_{n \mathop = 1}^\infty B_n = \bigcup_{n \mathop = 1}^\infty A_n$

Let $i, j \in \N, i < j$.

Then:
 * $B_i \cap B_j \subseteq A_i \cap B_j = A_i \cap \paren {A_j \setminus \ds \bigcup_{k \mathop = 1}^{j - 1} A_k} = \O$

So $B_1, B_2, \ldots$ is pairwise disjoint.

Hence $\ds\bigcup_{n \mathop = 1}^\infty A_n = \ds \bigsqcup_{n \mathop = 1}^\infty B_n \in \Sigma$, as $B_n \in \Sigma$ for all $n \in \N$ and $\text {SR} 3'$.