Kenneth Kuttler

216 CHAPTER 9. THE LEBESGUE INTEGRAL

Then the relation between this and uniform integrability is as follows.

Proposition 9.9.6 In the context of the above definition, S is equi-integrable if andonly if it is a bounded subset of L1 (Ω) which is also uniformly integrable.

Proof: ⇒ I need to show S is bounded and uniformly integrable. First considerbounded. Choose λ to work for ε = 1. Then for all f ∈S,∫

| f |dµ =∫[| f |>λ ]

| f |dµ +∫[| f |≤λ ]

| f | ≤ 1+λ µ (Ω)

Thus it is bounded. Now let E be a measurable subset of Ω. Let λ go with ε/2 in thedefinition of equi-integrable. Then for all f ∈S,∫

E| f |dµ ≤

∫[| f |>λ ]

| f |dµ +∫

E∩[| f |≤λ ]| f |dµ ≤ ε

2+λ µ (E)

Then let µ (E) be small enough that λ µ (E)< ε/2 and this shows uniform integrability.⇐ I need to verify equi-integrable from bounded and uniformly integrable. Let δ be

such that if µ (E) < δ , then∫

E | f |dµ < ε for all f ∈ S. If not, then there exists fn ∈ Swith [| fn|> n]> δ . Thus

∫| fn|dµ ≥

∫[| fn|>n] | fn|dµ ≥ nµ ([| fn|> n])> nδ and so S is not

bounded after all. ■The following theorem is Vitali’s convergence theorem.

Theorem 9.9.7 Let { fn} be a uniformly integrable set of complex valued functions,µ(Ω)< ∞, and fn(x)→ f (x) a.e. where f is a measurable complex valued function. Thenf ∈ L1 (Ω) and limn→∞

∫Ω| fn− f |dµ = 0.

Proof: First it will be shown that f ∈ L1 (Ω). By uniform integrability, there existsδ > 0 such that if µ (E) < δ , then

∫E | fn|dµ < 1 for all n. By Egoroff’s theorem, there

exists a set E of measure less than δ such that on EC, { fn} converges uniformly. There-fore, for p large enough, and n > p,

∫EC

∣∣ fp− fn∣∣dµ < 1 which implies

∫EC | fn|dµ <

1+∫

Ω

∣∣ fp∣∣dµ.Then since there are only finitely many functions, fn with n≤ p, there exists

a constant, M1 such that for all n,∫

EC | fn|dµ < M1. But also,∫Ω

| fm|dµ =∫

EC| fm|dµ +

∫E| fm| ≤M1 +1≡M.

Therefore, by Fatou’s lemma,∫

Ω| f |dµ ≤ liminfn→∞

∫| fn|dµ ≤ M, showing that f ∈ L1

as hoped.Now S∪{ f} is uniformly integrable so there exists δ 1 > 0 such that if µ (E) < δ 1,

then∫

E |g|dµ < ε/3 for all g ∈ S∪{ f}.By Egoroff’s theorem, there exists a set, F with µ (F) < δ 1 such that fn converges

uniformly to f on FC. Therefore, there exists m such that if n>m, then∫

FC | f − fn|dµ < ε

3 .It follows that for n > m,∫

Ω

| f − fn|dµ ≤∫

FC| f − fn|dµ +

∫F| f |dµ +

∫F| fn|dµ <

3= ε,

which verifies the claim of the theorem. ■

216 CHAPTER 9. THE LEBESGUE INTEGRALThen the relation between this and uniform integrability is as follows.Proposition 9.9.6 In the context of the above definition, © is equi-integrable if andonly if it is a bounded subset of L! (Q) which is also uniformly integrable.Proof: = I need to show G is bounded and uniformly integrable. First considerbounded. Choose A to work for € = 1. Then for all f € G,J flan = Doon vide fo |f| <1+Ap(Q)Thus it is bounded. Now let E be a measurable subset of Q. Let A go with €/2 in thedefinition of equi-integrable. Then for all f € G,EJifldus [ ifique fe ifaw <5 + Am (E)E (f|>4] EN(|f|<A] 2Then let u (E) be small enough that Au (E) < €/2 and this shows uniform integrability.< I need to verify equi-integrable from bounded and uniformly integrable. Let 6 besuch that if u(£) < 6, then f, |f|du < e for all f € G. If not, then there exists f,, € Gwith || fn] >] > 6. Thus f|fnldu > fiip,)snj\fnl du = ne ([|fn| > 2]) > 26 and so G is notbounded after all. HfThe following theorem is Vitali’s convergence theorem.Theorem 9.9.7 Lez {fn} be a uniformly integrable set of complex valued functions,L(Q) <9, and f,(x) > f(x) ae. where f is a measurable complex valued function. Thenf EL! (Q) and lity fo |fn — f|du = 0.Proof: First it will be shown that f € L'(Q). By uniform integrability, there exists6 > 0 such that if u(E) < 6, then f,|f,|du < 1 for all n. By Egoroff’s theorem, thereexists a set E of measure less than 6 such that on E©, { f,} converges uniformly. There-fore, for p large enough, and n > p, {pc | fp — fn|du < 1 which implies fic |fnldp <1+ fo | fp| dp.Then since there are only finitely many functions, f, with n < p, there existsaconstant, M, such that for all n, fic | fn| du < M1. But also,[i vfnlaw = | binkdie+ | \fml S Mit 1M,Q EC ETherefore, by Fatou’s lemma, fo |,f|du <liminf,. f |fnl|du <M, showing that f € L!as hoped.Now GU {f} is uniformly integrable so there exists 6; > 0 such that if u(E) < 61,then J, |g|du < €/3 for all g e GU{f}.By Egoroff’s theorem, there exists a set, F with u(F) < 6, such that f, convergesuniformly to f on F©. Therefore, there exists m such that if n > m, then fre |f — fnld < §.It follows that for n > m,—€ €— + —E[it-tlaws [op tldus [itldur [ Uildu<5+5+5 =e,which verifies the claim of the theorem.