From: Frank Rothenhäusler <uzspbl@uni-bonn.de>
Subject: Re: Sigma-Algebra
Date: Tue, 21 Sep 1999 00:18:23 +0200
Newsgroups: sci.math
To: "Kenneth A. Osmond" <osmond@execpc.com>
Keywords: Is a sigma-algebra also an algebra?

"Kenneth A. Osmond" wrote:

> Is there a good proof out there for proving:  If A is a sigma-algebra, A
> is also an algebra?

By definition, a system A of subsets of a set M is a sigma-algebra in M, if
(1) M is an element of A,
(2) if U is in A then so is CU, its compliment with respect to M,
(3) the union of  any countable system of sets in A also lies in A.

A is an algebra, if
(1) A contains the empty set,
(2) with U,V in A the sets U\V an UuV are in A,
(3) M is in A

("u" denoting union of sets)

Conditions (1),(2) mean that A is a ring. One can easily prove that A is a
ring
in this sense iff A is a commutative ring in the algebraic sense with
respect to the operations "intersection of sets" as multiplication and
"symmetric difference" as addition. The latter is defined as follows:

symmetric difference (U,V) := (U\V)u(V\U)

Adding condition (3) only adds the neutral element of the multiplication, so
we
get a commutative ring with unity.

The implication "A sigma-algebra -> A algebra" directly follows from the
identities
U\V = C(CUuV) and CM=empty set.

Bye,

Frank