640 CHAPTER 20. REPRESENTATION THEOREMS

Then since µ is finite, there exists nε such that if n≥ nε , then µ (En)≤ ε/2(1+λ ε) . Thenletting f ∈ K, ∫

En

| f |dµ =∫

En∩[| f |≥λ ε ]| f |dµ +

∫En∩[| f |<λ ε ]

| f |dµ

≤ ε/2+∫

En

λ ε dµ < ε/2+ ε/2 = ε

This proves 20.9.26

Corollary 20.9.6 Let (Ω,S ,µ) be a measure space in which µ (Ω) < ∞ and let K ⊆L1 (Ω,S ,µ) be equi integrable. Then every sequence from K has a weakly convergentsubsequence.

Proof: From Lemma 20.9.5 the hypotheses of Theorem 20.9.3 are satisfied.It is also convenient to consider the following proposition.

Proposition 20.9.7 Let (Ω,S ,µ) be a measure space in which µ (Ω) < ∞. Then K ⊆L1 (Ω,S ,µ) is equi integrable if and only if K is uniformly integrable and there exists aconstant, M such that for all f ∈ K, || f ||L1 ≤M.

Proof: First suppose K is equi integrable. Then pick λ such that for all f ∈ K,∫[| f |≥λ ]

| f |dµ < 1.

Then for f ∈ K ∫Ω

| f |dµ =∫[| f |≥λ ]

| f |dµ +∫[| f |<λ ]

| f |dµ

≤ 1+λ µ (Ω)≡M.

Also, if ε > 0, pick λ so large that for all f ∈ K∫[| f |≥λ ]

| f |dµ <ε

2.

Then letting A ∈S ,∫A| f |dµ =

∫A∩[| f |≥λ ]

| f |dµ +∫

A∩[| f |<λ ]| f |dµ

<ε

2+λP(A)

and so if P(A) is sufficiently small, this is less than ε. Thus K is uniformly integrable.Now suppose || f ||1 ≤M for all f ∈ K and K is uniformly integrable. Then∫

[| f |≥λ ]| f |dµ ≥ λP([| f | ≥ λ ])