20.2. VECTOR MEASURES 605

Then 20.2.5 implies that whenever the Ei are disjoint, |µ|(∪nj=1E j)≥∑

nj=1 |µ|(E j). There-

fore,∞

∑j=1|µ|(E j)≥ |µ|(∪∞

j=1E j)≥ |µ|(∪nj=1E j)≥

n

∑j=1|µ|(E j).

Since n is arbitrary,

|µ|(∪∞j=1E j) =

∞

∑j=1|µ|(E j)

which shows that |µ| is a measure as claimed.The following corollary is interesting. It concerns the case that µ is only finitely addi-

tive.

Corollary 20.2.4 Suppose (Ω,F ) is a set with a σ algebra of subsets F and supposeµ : F → C is only finitely additive. That is, µ

(∪n

i=1Ei)= ∑

ni=1 µ (Ei) whenever the Ei are

disjoint. Then |µ| , defined in the same way as above, is also finitely additive provided |µ|is finite.

Proof: Say E ∩F = /0 for E,F ∈F . Let π (E) ,π (F) suitable partitions for which thefollowing holds.

|µ|(E ∪F)≥ ∑A∈π(E)

|µ (A)|+ ∑B∈π(F)

|µ (B)| ≥ |µ|(E)+ |µ|(F)−2ε.

Similar considerations apply to any finite union.Now let E = ∪n

i=1Ei where the Ei are disjoint. Then letting π (E) be a partition of E,

|µ|(E)− ε ≤ ∑F∈π(E)

|µ (F)| ,

it follows that

|µ|(E) ≤ ε + ∑F∈π(E)

|µ (F)|= ε + ∑F∈π(E)

∣∣∣∣∣ n

∑i=1

µ (F ∩Ei)

∣∣∣∣∣≤ ε +

n

∑i=1

∑F∈π(E)

|µ (F ∩Ei)| ≤ ε +n

∑i=1|µ|(Ei)

which shows |µ| is finitely additive.In the case that µ is a complex measure, it is always the case that |µ|(Ω)< ∞.

Theorem 20.2.5 Suppose µ is a complex measure on (Ω,S ) where S is a σ algebra ofsubsets of Ω. That is, whenever, {Ei} is a sequence of disjoint sets of S ,

µ (∪∞i=1Ei) =

∞

∑i=1

µ (Ei) .

Then |µ|(Ω)< ∞.