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Measure theory in R^n and in abstract spaces

  1. Mar 17, 2009 #1

    WWGD

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    Hi:

    I am trying to review the way L^p spaces are treated differently

    in Royden. In Ch.6, he treats them under "Classical Banach Spaces",

    and then again, in his Ch.11 , under "Abstract Spaces".

    This is what I understand: (Please comment/correct)

    In the case of abstract spaces, we do not deal with R^n,

    or subsets of R^n, where the idea of measure is "natural" , and

    where we start by defining the measure of an open "box" ( a

    subset bounded by open intervals (a1,b1),(a2,b2),..,(an,bn))

    which we define to have measure (a1-b1)(a2-b2)...(an-bn)

    and build our measure around this, by defining the measure of a set

    using the infimum of coverings by these boxes.



    In abstract spaces, instead of R^n, we choose any topological space,

    then define a sigma algebra , like, e.g., the Borel algebra generated by

    open sets, or one generated by the compact subsets, etc. , which we

    define to be the collection of measurable sets.

    Then we define measurability in terms of being a member of

    the sigma algebra, and , given abstract spaces X= (X,s_X,mu_X), Y=(Y,s_Y,mu_Y) , with

    s_X,s_Y sigma algebras and mu_X,mu_Y measures (satisfying the basic

    axioms ;subadditivity, monotonicity eyc.) , we then define f:X-->Y ; X,Y as above

    is measurable, if for U_y in s_Y (the sigma -algebra in Y ), f^-1(U_Y) is in s_X (the

    sigma-algebra of X).



    What I am not clear on, is the issue of integrability: in abstract

    spaces, we integrate against a measure. How is this different from

    Lebesgue integration, where we do a sum :Sumf(y)( mu_X (x):f(x)=y)

    i.e., multiply any value f(y) in the range by the measure of its preimage?.

    It seems to be the same?.

    Question:

    Is any of these abstract spaces , or measure triples (X, s_X,mu_X) a Hilbert space,

    or are they all Banach spaces?. Are there also non-measurable subsets in these

    abstract spaces?.

    Thanks for Comments/Corrections/Examples/Refs
     
  2. jcsd
  3. Mar 17, 2009 #2
    A measure space (a set with sigma algebra and measure) does not need to be a topological space (elementary measure theory requires no topology).

    The integral of real or complex valued functions can be defined for any measure space, in the case of the Borel sigma algebra on R^n it is Lebesgue's integral. For this reason the general version of this integral for arbitrary measure spaces is also often called the Lebesgue integral.

    I think you are confusing measure spaces with L^p spaces. For any measure space X one can define L^p(X), these are Banach spaces and L^2(X) is also a Hilbert space.

    Yes, all subsets not in the sigma-algebra (by definition).
     
  4. Mar 17, 2009 #3

    WWGD

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    Thanks For Your Comments, YYAT:


    Do we ever work with P(X) as the sigma-algebra on X, to have all subsets be measurable?

    I know we cannot do this with the Lebesgue measure.


    Thanks again.
     
  5. Mar 17, 2009 #4
    This formula only works for "simple functions" (linear combinations of indicator functions of measurable sets). For general measurable functions one uses approximation by simple functions to define the integral, as explained in http://en.wikipedia.org/wiki/Lebesgue_integral#Integration".

    Take, for example, the counting measure on Z. Note that the measure is really a set theoretic concept in this case (cardinality).
     
    Last edited by a moderator: Apr 24, 2017
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