636 CHAPTER 20. REPRESENTATION THEOREMS

Lemma 20.9.2 Let { fn} be a sequence of functions in L1 (Ω,S ,µ). Then there exists a σ

finite set of S , Ω1, and a σ algebra of subsets of Ω1,S1, such that S1 ⊆S , fn = 0 offΩ1, fn ∈ L1 (Ω1,S1,µ), and S1 = σ (A ), the σ algebra generated by A , for some A acountable algebra.

Proof: Let En denote the sets which are of the form{f−1n (B(z,r)) : z ∈Q+ iQ,r > 0,r ∈Q, and 0 /∈ B(z,r)

}Since each En is countable, so is

E ≡ ∪∞n=1En

Now let Ω1 ≡ ∪E . I claim Ω1 is σ finite. To see this, let

Wn =

{ω ∈Ω : | fk (ω)|> 1

nfor some k = 1,2, · · · ,n

}Thus if ω ∈Wn, it follows that for some r ∈Q, z ∈Q+ iQ sufficiently close to fk (ω)

ω ∈ f−1k (B(z,r)) ∈ Ek

and so ω ∈ ∪nk=1Ek and consequently, Wn ∈ ∪n

k=1Ek. Also

µ (Wn)1n≤∫

Wn

n

∑k=1| fk (ω)|dµ < ∞.

Now if ω ∈Ω1, then for some k,ω is contained in a set of Ek. Therefore, for that k,

fk (ω) ∈ B(z,r)

where r is a positive rational number and z ∈Q+ iQ and B(z,r) does not contain 0. There-fore, fk (ω) is at a positive distance from 0 and so for large enough n,ω ∈Wn. Take n solarge that 1/n is less than the distance from B(z,r) to 0 and also larger than k.

By Lemma 20.9.1 there exists a countable algebra of sets A which contains E . LetS1 ≡ σ (A ). It remains to show fn (ω) = 0 off Ω1 for all n. Let ω /∈ Ω1. Then ω is notcontained in any set of Ek and so fk (ω) cannot be nonzero. Hence fk (ω) = 0. This provesthe lemma.

The following Theorem is the main result on sequential compactness in L1 (Ω,S ,µ).

Theorem 20.9.3 Let K ⊆ L1 (Ω,S ,µ) be such that for some C > 0 and all f ∈ K,

|| f ||L1 ≤C (20.9.23)

and K also satisfies the property that if {En} is a decreasing sequence of measurable setssuch that ∩∞

n=1En = /0, then for all ε > 0 there exists nε such that if n≥ nε , then∣∣∣∣∫En

f dµ

∣∣∣∣< ε (20.9.24)

for all f ∈ K. Then every sequence of functions of K has an L1 (Ω,µ) weakly convergentsubsequence.