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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.


    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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