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I am very interested in counterexamples when cylindrical sigma-algebra is not equal to the borel. I read link text and link text.

Especially interessted in l-infinity space. I've read Talgat and Fremlin's article about Gaussian measure on l-infinitive, but if there is another method to prove that, without measures? And there is another idea of that proof by Vakhania "Probability Distributions on Banach Spaces" et al, p. 23 - 24, but i cant get how to prove it exactly.

And what about [0,1]^[0,1] space. It's a separable metric space as i could prove, but what about sigma-algebras.

Would be great if you could give me any links of literature about cylindrical sigma-algebra relationship with Borel or Baire, examples and counterexamples.

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    $\begingroup$ $[0,1]^{[0,1]}$ (with the product topology) is not metrizable. $\endgroup$ – Nate Eldredge Mar 21 '11 at 17:00
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    $\begingroup$ And $[0,1]^{[0,1]}$ is interesting for the sigma-algebra, too. The product sigma-algebra (= the Baire sigma-algebra) has the property that every set in it depends on only countably many coordinates. In particular, a singleton is NOT in this sigma-algebra. So the Baire sigma-algebra is not the same as the Borel sigma-algebra. $\endgroup$ – Gerald Edgar Mar 21 '11 at 18:29
  • $\begingroup$ Ok, they are not equal. It would be great if there is example of Borel, not Baire(Cylindrical) set. I think of [0,1/2]^{[0,1]}, but didnt prove it yet. Maybe you can give me other examples? $\endgroup$ – Ravil Mudarisov Mar 21 '11 at 19:42
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    $\begingroup$ Correct, a singleton is a closed set, therefore Borel. Also, the more complicated set $[0,1/2]^{[0,1]}$ is closed, therefore Borel. $\endgroup$ – Gerald Edgar Mar 21 '11 at 23:54
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    $\begingroup$ Set $[0,1/2]^{[0,1]}$ does not depend on only countably many coordinates, so it is not Baire. These are comments rather than answers, since they are not about $l^\infty$. $\endgroup$ – Gerald Edgar Mar 22 '11 at 13:39

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