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Interesting math theorem in measure theory

  1. Oct 1, 2006 #1
    Sorry if this is kind of vague, but the other day, one of my math profs told me about a theorem which he thought was particularly interesting. I might be missing or getting a condition wrong, but here goes:

    Suppose [tex]I(f, d)[/tex] is a real-valued function, where f is a real-valued function always defined on a region d, and that region d. if:

    1. [tex]I(f_1 + f_2, d) = I(f_1, d) + I(f_2, d)[/tex]
    2. If [tex]d = d_1 \cup d_2[/tex], and [tex]d_1 \cap d_2 = \emptyset[/tex], then [tex]I(f, d) = I(f, d_1) + I(f, d_2)[/tex]
    3. If [itex]a[/itex] is a real constant, then [tex]I(af, d) = aI(f, d)[/tex]

    Then, [tex]I(f, d) = \int_d f[/tex]

    I asked about where I could find this in a book or online and he said a graduate-level book on measure theory (which unfortunately I don't have any). I did look through Spivak's calculus on manifolds, but nothing came close even.

    Does anyone recognize this?
     
    Last edited: Oct 1, 2006
  2. jcsd
  3. Oct 1, 2006 #2

    matt grime

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    As stated it cannot be true: the functional I(f,d) = k*int_d d for instance satisfies the same criterion as above, as do many others: you need some kind of normalization process.

    What is true is that you're looking at the Riesz Representation theorem. Note I am very unsure of the spelling: it could be Riezche, Rietzche, or who knows what. It says that every linear functional (yes, that is the key word) is given by integrating with respect to something.

    http://en.wikipedia.org/wiki/Riesz_representation_theorem

    There are two related eaiser to understand ideas. The latter is the riesz theorem for hilbert spaces.

    1. Given the space of polynomials on one variable, R[x], define an operator D on R[x] that is linear, sends x to 1, and some other things and you can prove that D must be differentiation. Can't recall the exact statement.

    2. If H is a hilbert space (possibly with some conditions), and if f is a linaer functional then there is a y such that f(x) =<x,y> for all x.
     
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