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Simplifying Radon-Nykodim App?

  1. Jul 19, 2012 #1

    Bacle2

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    Hi, I'm trying to work this out. Just a small question:

    Let (X, m, μ) be a measure-triple (X a set; m a sigma-algebra, μ

    a measure) , and let,

    f:X-->ℝ be measurable. For B a Borel set, define :

    v(B):=μ(f-1(B)). Then show:

    a)v is a measure in the Borel subsets of R , and

    b)If g: ℝ-->[0 , oo) is any Borel-measurable function on R,

    ∫ℝ (gdv) = ∫X (gof)dμ

    First one is easy, but my method for b seems too long b). I

    am trying to work this with Radon-Nykodim theorem. I'm pretty sure we can pull back

    the measure on R to have the absolute continuity condition satisfied, and same thing

    for the sigma-finiteness, i.e., by pulling back the Borel measure on R.

    I'm just curious as to whether there is a faster way of doing this problem.

    TIA
     
  2. jcsd
  3. Jul 19, 2012 #2

    micromass

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    It's very easy to verify it the usual way:
    1) First verify it for step functions
    2) Use monotone convergence to verify it for positive functions
    3) Verify it for general functions using [itex]f=f^+-f^-[/itex].
     
  4. Jul 20, 2012 #3

    Bacle2

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    Right; thanks, but I don't know to what extent this simplifies the necessary

    calculations in Radon-Nykodim. R-N works, but it just seems long. I thought maybe

    there was some corollary to help give it a short proof. Thanks, tho.
     
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