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- From: ara@zurich.ai.mit.edu (Allan Adler)
- Newsgroups: sci.math.research
- Subject: Re: Set-free measure theory
- Message-ID: <ARA.93Jan26233454@camelot.ai.mit.edu>
- Date: 27 Jan 93 04:34:54 GMT
- Article-I.D.: camelot.ARA.93Jan26233454
- References: <C1GuEp.M99@sugar.neosoft.com>
- Sender: Daniel Grayson <dan@math.uiuc.edu>
- Organization: M.I.T. Artificial Intelligence Lab.
- Lines: 28
- Approved: Daniel Grayson <dan@math.uiuc.edu>
- Originator: dan@symcom.math.uiuc.edu
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- In-Reply-To: claird@NeoSoft.com's message of 26 Jan 93 14:47:10 GMT
-
- In article <C1GuEp.M99@sugar.neosoft.com> claird@NeoSoft.com (Cameron Laird) writes:
-
- Conjecture: the literature already includes a proof
- that there is nothing novel about "set-free measures",
- that is, they all have realizations as (extended)
- measures. Does anyone know?
-
- Apparently, you don't require the boolean algebra to be a sigma algebra.
- In that case, the Stone representation theorem seems to answer your
- question: every boolean algebra is isomorphic to the algebra of
- clopen (i.e. simultaneously open and closed) subsets of a totally
- disconnected compact Hausdorff space.
-
- If you want to work with sigma algebras and want the representation
- to be a sigma homomorphism, the Stone representation doesn't work anymore.
- I'm sure a lot of work has been done on the representation of sigma
- algebras as sigma algebras of subsets, but I don't know the results
- on it.
-
- If your Boolean algebra is atomic, you might get a suitable representation
- of it as subsets of the set of its atoms.
-
- If you look at Borel sets modulo sets of measure zero, you don't have
- sets any more. This is not an atomic boolean algebra in general.
-
-
- Allan Adler
- ara@altdorf.ai.mit.edu
-