A question on proof of Riesz Representation Theorem when p=1

  1. Jun 23, 2010 #1
    This question comes from the proof of Riesz Representation Theorem in Bartle's "The Elements of Integration and Lebesgue Measure", page 90-91, as the image below shows.

    The equation (8.10) is [tex]G(f)=\int fgd\mu[/tex].

    The definition of [tex]L^\infty[/tex] space is as follows:

    My question is: why the g determined by Radon-Nikodym Theorem is in [tex]L^\infty[/tex]? I can only prove that it is Lebesgue integrable, that is, belongs to [tex]L^1[/tex] space, but the proof mentions no word on why it is in [tex]L^\infty[/tex], that is, bounded a.e.. Could you please tell me how to prove this? Thanks!
  2. jcsd
  3. Jun 25, 2010 #2


    User Avatar
    Science Advisor

    The L^1 property already comes from Radon-Nikodym, right? Radon-Nikodym says such a g in L^1 exists. To prove it is also in L^\infty, what about this:

    Suppose g is not a.e. bounded, then for every n we can find An with [itex]0<\mu(A_n)<\infty[/itex] such that for all [itex]x\in A_n[/itex] we have [itex]|g(x)|>n[/itex]. Now take


    [tex]\|G\|=\sup\frac{|Gf|}{\|f\|}\geq \frac{|Gf_0|}{\|f_0\|}=\frac{1}{\mu(A_n)}\left|\int fg\right|=\frac{1}{\mu(A_n)}\int_{A_n} |g|>n,[/tex]

    in contradiction with G being bounded.
  4. Jun 25, 2010 #3
    Yes! this is the proof! Thank you for the ingenious construction, I got it.

Have something to add?
Similar Discussions: A question on proof of Riesz Representation Theorem when p=1