616 CHAPTER 20. REPRESENTATION THEOREMS

Proof: Define a new measure µ̃ , according to the rule

µ̃ (E)≡∫

Erdµ. (20.3.13)

Thus µ̃ is a finite measure on S . Now define a mapping, η : Lp(Ω,µ)→ Lp(Ω, µ̃) by

η f = r−1p f .

Then||η f ||pLp(µ̃)

=∫ ∣∣∣r− 1

p f∣∣∣p rdµ = || f ||pLp(µ)

and so η is one to one and in fact preserves norms. I claim that also η is onto. To see this,let g ∈ Lp(Ω, µ̃) and consider the function, r

1p g. Then∫ ∣∣∣r 1

p g∣∣∣p dµ =

∫|g|p rdµ =

∫|g|p dµ̃ < ∞

Thus r1p g ∈ Lp (Ω,µ) and η

(r

1p g)= g showing that η is onto as claimed. Thus η is one

to one, onto, and preserves norms. Consider the diagram below which is descriptive of thesituation in which η∗ must be one to one and onto.

h,Lp′ (µ̃) Lp (µ̃)′ , Λ̃

η∗

→ Lp (µ)′ ,Λ

Lp (µ̃)η

← Lp (µ)

Then for Λ∈ Lp (µ)′ , there exists a unique Λ̃∈ Lp (µ̃)′ such that η∗Λ̃=Λ,∣∣∣∣∣∣Λ̃∣∣∣∣∣∣= ||Λ|| . By

the Riesz representation theorem for finite measure spaces, there exists a unique h∈ Lp′ (µ̃)

which represents Λ̃ in the manner described in the Riesz representation theorem. Thus||h||Lp′ (µ̃) =

∣∣∣∣∣∣Λ̃∣∣∣∣∣∣= ||Λ|| and for all f ∈ Lp (µ) ,

Λ( f ) = η∗Λ̃( f )≡ Λ̃(η f ) =

∫h(η f )dµ̃ =

∫rh(

f−1p f)

dµ

=∫

r1p′ h f dµ.

Now ∫ ∣∣∣∣r 1p′ h∣∣∣∣p′ dµ =

∫|h|p

′rdµ = ||h||p

′

Lp′ (µ̃)< ∞.

Thus∣∣∣∣∣∣∣∣r 1

p′ h∣∣∣∣∣∣∣∣

Lp′ (µ)= ||h||Lp′ (µ̃) =

∣∣∣∣∣∣Λ̃∣∣∣∣∣∣= ||Λ|| and represents Λ in the appropriate way. If

p = 1, then 1/p′ ≡ 0. This proves the Lemma.A situation in which the conditions of the lemma are satisfied is the case where the

measure space is σ finite. In fact, you should show this is the only case in which theconditions of the above lemma hold.