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- From: kosowsky@minerva.harvard.edu (Jeffrey J. Kosowsky)
- Newsgroups: sci.math
- Subject: Borel product measures
- Message-ID: <KOSOWSKY.92Nov17115347@minerva.harvard.edu>
- Date: 17 Nov 92 16:53:47 GMT
- Sender: usenet@das.harvard.edu (Network News)
- Organization: Harvard Robotics Lab, Harvard University
- Lines: 32
-
-
- Suppose X and Y are topological spaces. Let E and F be their
- respective Borel sigma-algebras. I am interested in knowing when the
- product sigma-algebra ExF is itself Borel.
- I know the following facts:
-
- 1] If the topologies on X and Y are second countable, then clearly ExF
- is Borel.
-
- 2] In general ExF is not Borel.
-
- 3] Let u and v be Borel measures on (X,E) and (Y,F) respectively. Then
- the completion (ExF)* with respect to uxv is not in general Borel.
-
- 3] If (X, E, u) and (Y, F, v) are locally compact, sigma-compact,
- sigma-finite Borel regular measure spaces, the (XxY, (ExF)*, uxv) is a
- locally compact, sigma-compact, sigma-finite *Borel* regular measure
- space.
- (This follows from a construction using the Riesz representation
- xtheorem)
-
- So I am interested in knowing under what conditions the completion
- (ExF)* with respect to uxv contains all Borel sets.
- Now, (3) shows that locally compact, sigma-compact, sigma-finite Borel
- regular measures are sufficient to insure a Borel product
- sigma-algebra. Can these conditions be relaxed. e.g. are sigma-finite
- Borel regular measures sufficient?
-
- Please provide a proof and/or reference for your suggested sufficient
- conditions. Thank you!
-
- Jeff Kosowsky
-