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

by zzzhhh
Tags: proof, representation, riesz, theorem
zzzhhh is offline
Jun23-10, 12:20 AM
P: 39
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!
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Landau is offline
Jun25-10, 06:46 AM
Sci Advisor
P: 905
Quote Quote by zzzhhh View Post
I can only prove that it is Lebesgue integrable, that is, belongs to [tex]L^1[/tex] space
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.
zzzhhh is offline
Jun25-10, 08:27 PM
P: 39
Yes! this is the proof! Thank you for the ingenious construction, I got it.

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