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I know that $\sigma$ in $\sigma$-algebra stands for the closure under countable union property. What about "algebra"? Surely it cannot algebra over a field or a ring as defined in algebra textbooks. Also why is $\sigma$-algebra also called $\sigma$-field and what is meant by "field"?

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"Algebra" = Boolean algebra (of sets). I don't know why the term "field" came to be used for this. –  GEdgar Dec 27 '12 at 13:20
    
From tomasz reply, I think "field" stands for "field of subsets" which is usually defined by the three axioms above. Since any field of subsets is a Boolean algebra, this probably results in the term $\sigma$-algebra, where "algebra" stands for "Boolean algebra". I am not sure historically which comes first. –  learn_maths Dec 27 '12 at 13:38
    
But why was the word "field" chosen for that? "Algebra" at least previously existed in connection with Boole's work. –  GEdgar Dec 27 '12 at 13:40
    
That I am not sure. The properties of $\sigma$-fields are not the same as those of commutative division rings in general. –  learn_maths Dec 27 '12 at 14:07
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up vote 1 down vote accepted

A field of sets is a family $\mathcal F$ of subsets of a given set $X$ satisfying the axioms:

  1. $X\in \mathcal F$
  2. For any $A,B\in \mathcal F$ we have $A\cup B\in \mathcal F$.
  3. For any $A\in \mathcal F$ we have $X\setminus A\in \mathcal F$.

In other words, it's a boolean algebra of sets with the usual operations. Algebra, in this context, is actually synonymous to field. A $\sigma$-field (-algebra) corresponds to a $\sigma$-complete boolean algebra.

Worth mentioning, it actually is quite naturally a ring in the usual algebraic sense (like any boolean algebra). You're right that it can't be a field except the most trivial two-element case (as zero divisors abound).

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Thanks, I think I got it now. –  learn_maths Dec 27 '12 at 13:30
    
@learn_maths: In that case, you should accept the answer (there's a "tick" symbol on the left of the answer, under the voting arrows). –  tomasz Dec 27 '12 at 16:47
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