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Measure Theory Question

  1. Feb 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex](X,\mathcal{B},\mu)[/itex] be a measure space and [itex]g[/itex] be a nonnegative measurable function on [itex]X[/itex]. Set [itex]\nu (E)=\int_{E}g\,d\mu[/itex]. Prove that [itex]
    \nu[/itex] is a measure and [tex]\int f\, d \nu =\int fg\,d\mu[/tex] for all nonnegative measurable functions [itex]f[/itex] on [itex]X[/itex].



    3. The attempt at a solution
    I'm basically at a total loss on how to start this. I'll keep working on it tonight, and will add to it as I go.
     
  2. jcsd
  3. Feb 20, 2012 #2

    morphism

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    Have you tried applying the definition of "measure" to [itex]\nu[/itex]? You'll have to use some basic results, like the monotone convergence theorem, to get stuff to work out.
     
  4. Feb 29, 2012 #3
    I've shown for non-negative simple functions that [tex]\int \phi\,d\nu=\int \phi g \,d\mu.[/tex] Now I wish to show it in general for non-negative measurable functions. So I say let [itex]f[/itex] be a non-negative measurable function on X. Fix [itex]\phi[/itex] as a simple function such that [itex]0\le\phi \le f[/itex]. Hence we have [itex]\int \phi\,d\nu=\int \phi g \,d\mu[/itex], and thus [tex]\int f\, d\nu=\sup_{\phi} \int \phi \, d\,\nu=\sup_{\phi} \int g\phi\, d\,\mu \overset{?}{=} \int gf\, d\mu.[/tex]

    Am I allowed to make that last equality?
     
  5. Feb 29, 2012 #4
    Should I do some inequalities for simple functions above and below? I feel like that last equality should be an inequality. Hrmf.
     
  6. Feb 29, 2012 #5

    Ray Vickson

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    Try it first for step functions f, of the form
    [tex] f = \sum_{i=1}^n c_i \chi(I_i), [/tex] where [itex] I_1, I_2,..., I_n [/itex] is a measurable partition of [itex] R [/itex] and [itex] \chi(A) [/itex] is the characteristic function of a set [itex] A \subset R: \: \chi(A)(x) = 0 \text{ if } x \not\in A, \text{ and } \chi(A)(x) = 1 \text{ if } x \in A.[/itex]

    RGV
     
  7. Feb 29, 2012 #6
    Hey, thanks for the reply!

    I've already done this for simple functions, I'm just stuck on how to show that any non-negative measurable function satisfies the equality. Should i take approximations from above and below? or should what I gave in my first reply be sufficient?
     
  8. Feb 29, 2012 #7

    Ray Vickson

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    I don't know how your textbook defines the integral for general functions, but many treatments define it as the limit of integrals of step functions.

    RGV
     
  9. Feb 29, 2012 #8
    I'm using Royden, where the integral of a non-negative function is defined as
    [tex]\int f\, d\,\mu =\sup \{\int \phi \, d\mu :0\le\phi\le f,\,\phi \, simple\}[/tex]
     
  10. Feb 29, 2012 #9
    So my question is if I'm allowed to say this:
    [tex]\int f\, d\nu = \sup \{\int \phi\, d\nu :0\le\phi\le f,\, \phi \,simple\} =\sup \{\int \phi g\,d\mu :0\le\phi\le f,\,\phi\,simple\} = \int fg\,d\mu?[/tex]
     
  11. Mar 1, 2012 #10
    Nevermind, I'll just use Monotone Convergence on a sequence of simple functions.
     
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