The Unapologetic Mathematician

Mathematics for the interested outsider

Measurable Spaces, Measure Spaces, and Measurable Functions

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April 26, 2010 - Posted by | Analysis, Measure Theory

14 Comments »

  1. is it enough [tex] mathcal{S} [/tex] be a [tex] \sigma ring [/tex] a probability book i’m studying says the space needs a [tex] \sigma – Field[/tex]
    Could you clarify please

    thanks

    Comment by cappa | April 27, 2010 | Reply

  2. A \sigma-field is a different name for a \sigma-algebra. As I point out above, in many cases \mathcal{S} actually is a \sigma-algebra (or -field). In this case, X itself is in \mathcal{S}, and so it’s clear that every point in X is in some measurable set.

    However, we’re going to allow \mathcal{S} to just be a \sigma-ring — X itself might not be in \mathcal{S} — so long as every point in X is still in some measurable set.

    Comment by John Armstrong | April 27, 2010 | Reply

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  11. Is every measurable space is topological space

    Comment by M. A. Hossain | September 27, 2011 | Reply

  12. Not in any natural way. The obvious thing would be to try to use the \sigma-algebra of measurable sets as the collection of open sets in the topology, but it’s not necessarily closed under arbitrary unions.

    Comment by John Armstrong | September 27, 2011 | Reply

  13. Actually my query is as follows:
    If T is a sigma-algebra In X, is T also a topology in X?

    Comment by M. A. Hossain | September 28, 2011 | Reply

  14. No, as I said; T is not necessarily closed under arbitrary (uncountable) unions.

    Comment by John Armstrong | September 28, 2011 | Reply


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