292 CHAPTER 10. THE ABSTRACT LEBESGUE INTEGRAL

Now let η > 0 be given and pick m0 such that m−10 < η . If ω ∈FC, then ω ∈

∞⋂m=1

ECk(m)m.

Hence ω ∈ ECk(m0)m0

so | fn(ω)− f (ω)| < 1/m0 < η for all n > k(m0). This holds for all

ω ∈ FCand so fn converges uniformly to f on FC.Now if E ̸= /0, consider {XEC fn}∞

n=1. Each XEC fn has real and imaginary parts mea-surable and the sequence converges pointwise to XE f everywhere. Therefore, from thefirst part, there exists a set of measure less than ε,F such that on FC,{XEC fn} convergesuniformly to XEC f . Therefore, on (E ∪F)C , { fn} converges uniformly to f . This provesthe theorem. ■

10.9.2 The Vitali Convergence TheoremThe Vitali convergence theorem is a convergence theorem which in the case of a finitemeasure space is superior to the dominated convergence theorem.

Definition 10.9.2 Let (Ω,F ,µ) be a measure space and let S ⊆ L1(Ω). S isuniformly integrable if for every ε > 0 there exists δ > 0 such that for all f ∈S

|∫

Ef dµ|< ε whenever µ(E)< δ .

Lemma 10.9.3 If S is uniformly integrable, then |S| ≡ {| f | : f ∈S} is uniformly inte-grable. Also S is uniformly integrable if S is finite.

Proof: Let ε > 0 be given and suppose S is uniformly integrable. First suppose thefunctions are real valued. Let δ be such that if µ (E)< δ , then |

∫E f dµ|< ε

2 for all f ∈S.Let µ (E)< δ . Then if f ∈S,∫

E| f |dµ ≤

∫E∩[ f≤0]

(− f )dµ +∫

E∩[ f>0]f dµ =

∣∣∣∣∫E∩[ f≤0]f dµ

∣∣∣∣+ ∣∣∣∣∫E∩[ f>0]f dµ

∣∣∣∣<

ε

2+

ε

2= ε.

In general, if S is a uniformly integrable set of complex valued functions, the inequalities,∣∣∣∣∫ERe f dµ

∣∣∣∣≤ ∣∣∣∣∫Ef dµ

∣∣∣∣ , ∣∣∣∣∫EIm f dµ

∣∣∣∣≤ ∣∣∣∣∫Ef dµ

∣∣∣∣ ,imply ReS ≡ {Re f : f ∈S} and ImS ≡ {Im f : f ∈S} are also uniformly integrable.Therefore, applying the above result for real valued functions to these sets of functions, itfollows |S| is uniformly integrable also.

For the last part, is suffices to verify a single function in L1 (Ω) is uniformly integrable.To do so, note that from the dominated convergence theorem, limR→∞

∫[| f |>R] | f |dµ = 0.

Let ε > 0 be given and choose R large enough that∫[| f |>R] | f |dµ < ε

2 . Now let µ (E)< ε

2R .Then ∫

E| f |dµ =

∫E∩[| f |≤R]

| f |dµ +∫

E∩[| f |>R]| f |dµ

< Rµ (E)+ε

2<

ε

2+

ε

2= ε.

This proves the lemma. ■The following gives a nice way to identify a uniformly integrable set of functions.