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Is there a measure for every Borel σ-algebra?

  1. Feb 11, 2012 #1

    Fredrik

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    The Borel σ-algebra can be defined on every topological space. Does that mean that every topological space can be turned into a measure space?
     
  2. jcsd
  3. Feb 11, 2012 #2
    Yes, in a trivial way. Just define [itex]\mu(S)=0[/itex] for all S. Or take the counting measure. Or a point measure.

    A better answer is actually Riesz representation theorem. This says that all regular measures on a locally compact Hausdorff space coincide with positive functionals

    [tex]T:C_c(X)\rightarrow \mathbb{R}[/tex]

    (with [itex]C_c(X)[/itex] the functions to [itex]\mathbb{R}[/itex] with compact support). Such a functional exist because of the Hahn-Banach theorem.

    The same thing can be done with [itex]C_0(X)[/itex] actually. In that case: the measures are the positive functionals, the probability measures are the states and the point measures represent the pure states on X.

    In general, the existence of non-trivial measures is not a simple problem and requires set theory (see for example measurable cardinals)
     
    Last edited: Feb 11, 2012
  4. Feb 11, 2012 #3

    lavinia

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    I wonder if the space has a cardinality greater than the continuum whether there are any measures other than the trivial one.
     
  5. Feb 12, 2012 #4
    The first thing to do is to define the word "the trivial one".

    This is quite an interesting theory. In general, this is studied in Boolean algebras and the so-called measure algebras.
     
  6. Feb 12, 2012 #5

    Fredrik

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    Excellent answer as always. Thanks. I've been meaning to study the Riesz representation theorem. It will be the first thing I do when I'm done with the basics of integration theory.
     
    Last edited: Feb 12, 2012
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