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

[zzzhhh]

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.
[URL]http://i3.6.cn/cvbnm/ac/9a/a3/3d06837bc78f74ba103b6d242a78e3a1.png[/URL]

The equation (8.10) is $$G(f)=\int fgd\mu$$.

The definition of $$L^\infty$$ space is as follows:
[URL]http://i3.6.cn/cvbnm/a5/0e/22/259193b92d8d2ef4878532eefec4d900.png[/URL]

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!

 

[Landau]

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.

 

[zzzhhh]

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