638 CHAPTER 20. REPRESENTATION THEOREMS

Therefore, picking E ∈S1, you can define a complex measure,

λ (E)≡ limn→∞

∫E

gndµ

Then λ ≪ µ and so by Corollary 20.2.9 on Page 609 and the fact shown above that Ω1 isσ finite there exists a unique S1 measurable g ∈ L1 (Ω1,µ) such that

λ (E) =∫

Egdµ ≡ lim

n→∞

∫E

gndµ.

Extend g to equal 0 outside Ω1.It remains to show {gn} converges weakly. It has just been shown that for every s a

simple function measurable with respect to S1∫Ω

gnsdµ =∫

Ω1

gnsdµ →∫

Ω1

gsdµ =∫

Ω

gsdµ

Now let f ∈ L∞ (Ω1,S1,µ) and pick a uniformly bounded representative of this function.Then by Theorem 11.3.9 on Page 241 there exists a sequence of simple functions converg-ing uniformly to f and so {∫

Ω

gn f dµ

}converges because∣∣∣∣∫

Ω

gn f dµ−∫

Ω

g f dµ

∣∣∣∣ ≤ ∣∣∣∣∫Ω

gnsdµ−∫

Ω

gsdµ

∣∣∣∣+∫

Ω

|gn|εdµ +∫

Ω

|g|εdµ

≤ Cε + ||g||1 ε +

∣∣∣∣∫Ω

gnsdµ−∫

Ω

gsdµ

∣∣∣∣for suitable simple s satisfying supω∈Ω1

|s(ω)− f (ω)| < ε and the last term converges to0 as n→ ∞.(

L1 (Ω,S ,µ))′ is a space I don’t know much about due to a possible lack of σ finite-

ness of Ω. However, it does follow that for i the inclusion map of L1 (Ω1,S1,µ) intoL1 (Ω,S ,µ) which merely extends the function as 0 off Ω1 and f ∈

(L1 (Ω,S ,µ)

)′, there

exists h ∈ L∞ (Ω1) such that for all g ∈ L1 (Ω1,S1,µ)

i∗ f (g) =∫

Ω1

hgdµ.

This is because i∗ f ∈(L1 (Ω1,S1,µ)

)′ and Ω1 is σ finite and so the Riesz representationtheorem applies to get a unique such h ∈ L∞ (Ω1) . Then since all the gn equal 0 off Ω1,

f (gn) = i∗ f (gn) =∫

Ω1

hgndµ