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I see that there are two differen concepts for Sigma Algebras on cartesian products over the real numbers. The first one is the Borel Sigma Algebra created by the product topology. The other one is the cylindrical sigma algebra. Actually, when I read the definition of cylinder sets via projections see Wikipedia, I first thought that these concepts would be the same. Somehow, this seems to be wrong. So how are they related to each other. Is one of them coarser than the other or are they in general incomparable and why is the cylindrical one used in probability theory?

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    $\begingroup$ Where do vector spaces come in? By "cylindrical $\sigma$-algebra", you mean the $\sigma$-algebra generated by rectangles? $\endgroup$ Commented May 23, 2014 at 19:15
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    $\begingroup$ See math.stackexchange.com/questions/248032/… $\endgroup$ Commented May 23, 2014 at 19:32
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    $\begingroup$ they coincide in the countable case if each topology has a countable basis. $\endgroup$ Commented May 23, 2014 at 20:14
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    $\begingroup$ so why would one introduce this product of sigma algebras, instead of the more natural one from topology? (by the way: Thanks for your help so far) $\endgroup$
    – user66906
    Commented May 23, 2014 at 20:17
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    $\begingroup$ In general, you do not want too many sets to be measurable. To specify a measure, you have to specify its value for each measurable set and that is easier if there are less measurable sets. $\endgroup$ Commented May 23, 2014 at 21:02

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The article to which you linked actually says that cylindrical $\sigma$-algebra is contained in Borel $\sigma$-algebra. The answer by Michael Greinecker shows the reverse inclusion may fail. Concerning

why would one introduce this product of sigma algebras, instead of the more natural one from topology?

I'd say that from the measure theory point of view the cylindrical $\sigma$ algebra is more natural. They both begin with the same basis of sets, finitely restricted rectangles. From here we either

  1. take the smallest $\sigma$-algebra containing the rectangles, or
  2. take the smallest topology containing the rectangles, and then take the smallest $\sigma$-algebra containing that topology.

Clearly, the first approach (which gives cylindrical $\sigma$-algebra) is more direct. The second approach (which gives the Boreal $\sigma$-algebra) involves taking uncountable unions of basis sets in the process of generating topology. Uncountable unions do not play well with measures.


In "small" spaces like $\mathbb R^n$, where the $\sigma$-algebras are the same, we usually think in Borel terms. This is mostly because for analysis it is important to have close connection to the topology of Euclidean space. In huge uncountable products, there is not much that topology can do for us.

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  • $\begingroup$ just out of curiosity: But the product of the borel sigma-algebras is always equal to the cylindrical sigma algebra? or is the product of sigma-algebras another concept? $\endgroup$
    – user66906
    Commented May 24, 2014 at 14:51
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    $\begingroup$ @user108886 If the number of factors is finite, it's the same as $\sigma$-algebra. If it's infinite, you need to clarify what you take as a basis: arbitrary boxes, or finitely restricted boxes. The latter gives cylindrical $\sigma$-algebra. The former gives something larger. $\endgroup$
    – user147263
    Commented May 24, 2014 at 16:23
  • $\begingroup$ What are the bad consequences of using Borel $\sigma$-algebra in uncountable product space? Is it because it is hard to handle ("uncountable unions do not play well with measures") or does it bring in some unwanted measurable sets? And why is cylindrical $\sigma$-algebra sufficient for usage? Or is it so that cylindrical $\sigma$-algebra is always what we want, but we just choose to call it Borel $\sigma$-algebra in the finite dimensional case? $\endgroup$
    – Ziyuan
    Commented Dec 1, 2017 at 23:24

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