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Introduction
When defining a measure for a set we usually cannot hope to make every subset of measurable. Instead we must usually restrict our attention to a specific collection of subsets of , requiring that this collection be closed under operations that we would expect to preserve measurability. A -algebra is such a collection.
Definition
Given a set , a -algebra in is a collection of subsets of such that:
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.
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Any union of countably many elements of is an element of .
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The complement of any element of in is an element of .
Notes
It follows from the definition that any -algebra in also satisfies the properties:
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.
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Any intersection of countably many elements of is an element of .
Note that a -algebra is a field of sets that is closed under countable unions and countable intersections (rather than just finite unions and finite intersections).
Given any collection of subsets of , the -algebra generated by is defined to be the smallest -algebra in such that . This is well-defined, as the intersection of any non-empty collection of -algebras in is also a -algebra in .
Examples
For any set , the power set is a -algebra in , as is the set .
A more interesting example is the Borel -algebra in , which is the -algebra generated by the open subsets of , or, equivalently, the -algebra generated by the compact subsets of .
Mathematics Subject Classification
28A60 Measures on Boolean rings, measure algebras- Forums
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Comments
question about \sigma-algebra
In relation to Djao's entry \sigma-algebra, \emptyset \in \mathcal{B}(E) appears like a premise. As far as I know, \emptyset \in A \forall A. So, why this one is not redundant?
BTW, in my browser IE6 the entry title looks like [red cross]$\sigma$-algebra, missing the greek letter sigma.
Re: question about \sigma-algebra
You seem to be confusing membership with inclusion (being a subset of). The empty set is a subset of every set, but it is not, for example, a member of itself.
Re: question about \sigma-algebra
Hi ratboy,
Yes! Now it is clearer for me why E \in \mathcal{B(E)}.
Thank you!
Pedro