# Sigma-ring

In mathematics, a nonempty collection of sets is called a σ-ring (pronounced sigma-ring) if it is closed under countable union and relative complementation.

## Formal definition

Let ${\displaystyle {\mathcal {R}}}$  be a nonempty collection of sets. Then ${\displaystyle {\mathcal {R}}}$  is a σ-ring if:

1. ${\displaystyle \bigcup _{n=1}^{\infty }A_{n}\in {\mathcal {R}}}$  if ${\displaystyle A_{n}\in {\mathcal {R}}}$  for all ${\displaystyle n\in \mathbb {N} }$
2. ${\displaystyle A\setminus B\in {\mathcal {R}}}$  if ${\displaystyle A,B\in {\mathcal {R}}}$

## Properties

These two properties imply:

${\displaystyle \bigcap _{n=1}^{\infty }A_{n}\in {\mathcal {R}}}$  if ${\displaystyle A_{1},A_{2},\ldots }$  are elements of ${\displaystyle {\mathcal {R}}.}$

This is because

${\displaystyle \bigcap _{n=1}^{\infty }A_{n}=A_{1}\setminus \bigcup _{n=2}^{\infty }\left(A_{1}\setminus A_{n}\right).}$

Every σ-ring is a δ-ring but there exist δ-rings that are not σ-rings.

## Similar concepts

If the first property is weakened to closure under finite union (i.e., ${\displaystyle A\cup B\in {\mathcal {R}}}$  whenever ${\displaystyle A,B\in {\mathcal {R}}}$ ) but not countable union, then ${\displaystyle {\mathcal {R}}}$  is a ring but not a σ-ring.

## Uses

σ-rings can be used instead of σ-fields (σ-algebras) in the development of measure and integration theory, if one does not wish to require that the universal set be measurable. Every σ-field is also a σ-ring, but a σ-ring need not be a σ-field.

A σ-ring ${\displaystyle {\mathcal {R}}}$  that is a collection of subsets of ${\displaystyle X}$  induces a σ-field for ${\displaystyle X}$ . Define ${\displaystyle {\mathcal {A}}=\{A\cup B^{c}|A,B\in {\mathcal {R}}\}}$ . Then ${\displaystyle {\mathcal {A}}}$  is a σ-field over the set ${\displaystyle X}$  - to check closure under countable union, recall a ${\displaystyle \sigma }$ -ring is closed under countable intersections. In fact ${\displaystyle {\mathcal {A}}}$  is the minimal σ-field containing ${\displaystyle {\mathcal {R}}}$  since it must be contained in every σ-field containing ${\displaystyle {\mathcal {R}}}$ .