176 CHAPTER 7. THE ABSTRACT LEBESGUE INTEGRAL

Lemma 7.10.3 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.

Lemma 7.10.4 Let S be a subset of L1 (Ω,µ) where µ (Ω) < ∞. Let t → h(t) be acontinuous function which satisfies

limt→∞

h(t)t

= ∞

Then S is uniformly integrable and bounded in L1 (Ω) if

sup{∫

Ω

h(| f |)dµ : f ∈S

}= N < ∞.