250 CHAPTER 11. ABSTRACT MEASURE AND INTEGRATION

where ck ∈ C and µ (Ek)< ∞. For s a complex simple function as above, define

I (s)≡n

∑k=1

ckµ (Ek) .

Lemma 11.4.3 The definition, 11.4.2 is well defined. Furthermore, I is linear on the vectorspace of complex simple functions. Also the triangle inequality holds,

|I (s)| ≤ I (|s|) .

Proof: Suppose ∑nk=1 ckXEk (ω) = 0. Does it follow that ∑k ckµ (Ek) = 0? The suppo-

sition impliesn

∑k=1

ReckXEk (ω) = 0,n

∑k=1

ImckXEk (ω) = 0. (11.4.22)

Choose λ large and positive so that λ +Reck ≥ 0. Then adding ∑k λXEk to both sides ofthe first equation above,

n

∑k=1

(λ +Reck)XEk (ω) =n

∑k=1

λXEk

and by Lemma 11.3.8 on Page 241, it follows upon taking∫

of both sides thatn

∑k=1

(λ +Reck)µ (Ek) =n

∑k=1

λ µ (Ek)

which implies ∑nk=1 Reckµ (Ek) = 0. Similarly, ∑

nk=1 Imckµ (Ek) = 0 so

n

∑k=1

ckµ (Ek) = 0

Thus if∑

jc jXE j = ∑

kdkXFk

then ∑ j c jXE j +∑k (−dk)XFk = 0 and so the result just established verifies ∑ j c jµ (E j)−∑k dkµ (Fk) = 0 which proves I is well defined.

That I is linear is now obvious. It only remains to verify the triangle inequality.Let s be a simple function,

s = ∑j

c jXE j

Then pick θ ∈ C such that θ I (s) = |I (s)| and |θ | = 1. Then from the triangle inequalityfor sums of complex numbers,

|I (s)| = θ I (s) = I (θs) = ∑j

θc jµ (E j)

=

∣∣∣∣∣∑jθc jµ (E j)

∣∣∣∣∣≤∑j

∣∣θc j∣∣µ (E j) = I (|s|) .

This proves the lemma.With this lemma, the following is the definition of L1 (Ω) .