A question on proof of Riesz Representation Theorem when p=1by zzzhhh Tags: proof, representation, riesz, theorem 

#1
Jun2310, 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 9091, 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 RadonNikodym 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
Jun2510, 06:46 AM

Sci Advisor
P: 905

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]f_0:=\frac{1_{A_n}g}{g}.[/tex] Then [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. 



#3
Jun2510, 08:27 PM

P: 39

Yes! this is the proof! Thank you for the ingenious construction, I got it.



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