23.4. WEAK COMPACTNESS 631

of disjoint elements of S having the property that 1 < µ(Ωn) < ∞, ∪∞n=1Ωn = Ω. Define

r(x) = ∑∞n=1

1n2 XΩn(x) µ(Ωn)

−1, µ̃(E) =∫

E rdµ . Thus∫

Ωrdµ = µ̃(Ω) = ∑

∞n=1

1n2 < ∞ so

µ̃ is a finite measure. The above lemma gives the existence part of the conclusion of thetheorem. Uniqueness is done as before. ■

With the Riesz representation theorem, it is easy to show that Lp(Ω), p> 1 is a reflexiveBanach space. Recall Definition 21.2.14 on Page 546 for the definition. From this, oneobtains a weak compactness result.

23.4 Weak CompactnessTheorem 23.4.1 For (Ω,S ,µ) a σ finite measure space and p > 1, Lp(Ω) is re-flexive. Thus every bounded sequence has a weakly convergent subsequence.

Proof: Let δ r : (Lr(Ω))′ → Lr′(Ω) be defined for 1r +

1r′= 1 by

∫(δ rΛ)g dµ = Λg

for all g ∈ Lr(Ω). From Theorem 23.3.6 δ r is one to one, onto, continuous and linear.By the open map theorem, δ

−1r is also one to one, onto, and continuous (δ rΛ equals the

representor of Λ). Thus δ∗r is also one to one, onto, and continuous by Corollary 21.2.11.

Now observe that J = δ∗p ◦δ

−1q . To see this, let z∗ ∈ (Lq)′, y∗ ∈ (Lp)′,

δ∗p ◦δ

−1q (δ qz∗)(y∗) = (δ ∗pz∗)(y∗) = z∗(δ py∗) =

∫(δ qz∗)(δ py∗)dµ,

J(δ qz∗)(y∗) = y∗(δ qz∗) =∫(δ py∗)(δ qz∗)dµ .

Therefore δ∗p ◦δ

−1q = J on δ q(Lq)′ = Lp. But the two δ maps are onto and so J is also onto.

■What about weak compactness in L1 (Ω)? I will give a simple sufficient condition in the

case of a finite measure space. More can be said. See for example Dunford and Schwartz[16]. I have this in my Topics in Analysis book also. Recall Proposition 10.9.6 on Page293 which says equi-integrable is the same as bounded and uniformly integrable. Thus inthe following, you can replace equi-integrable with bounded and uniformly integrable.

Theorem 23.4.2 Let (Ω,F ,µ) be a finite measure space and let { fn} be a se-quence in L1 (Ω) which is equi-integrable. Then there exists a subsequence which con-verges weakly in L1 (Ω) to some function f .

Proof: Let

En ≡{

f−1n (B(z,r)) : r is a positive rational and z ∈Q+ iQ

}.

Let E = ∪∞n=1En. Thus E and En are countable. Also, every open set is the countable union

of these sets B(z,r). Now let K be all finite intersections of sets of E and include /0and Ω in K . Then σ (K ) contains inverse images of Borel sets for each fn. Thus eachfn is measurable with respect to σ (K ). Also K is countable. Then, using a Cantordiagonal argument, we can have

∫E gndµ converges for all E ∈K . Let G be those sets

G ∈ σ (K ) such that∫

G gndµ converges. Suppose Gk are disjoint and each in G . Then,

since Ω has finite measure, limn→∞ µ

(∪∞

j=kG j

)= 0 because ∑k µ (Gk) converges. Let G=

∪∞k=1Gk and so

∫G gndµ−

∫G gmdµ =

∫G ∑

∞k=1 XGk (gn−gm)dµ =∑

Nk=1

∫Gk

(gn−gm)dµ+