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

by zzzhhh
Tags: proof, representation, riesz, theorem
 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 $$G(f)=\int fgd\mu$$. The definition of $$L^\infty$$ space is as follows: My question is: why the g determined by Radon-Nikodym Theorem is in $$L^\infty$$? I can only prove that it is Lebesgue integrable, that is, belongs to $$L^1$$ space, but the proof mentions no word on why it is in $$L^\infty$$, that is, bounded a.e.. Could you please tell me how to prove this? Thanks!
P: 905
 Quote by zzzhhh I can only prove that it is Lebesgue integrable, that is, belongs to $$L^1$$ 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 $0<\mu(A_n)<\infty$ such that for all $x\in A_n$ we have $|g(x)|>n$. Now take
$$f_0:=\frac{1_{A_n}|g|}{g}.$$

Then

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

in contradiction with G being bounded.
 P: 39 Yes! this is the proof! Thank you for the ingenious construction, I got it.

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