13.2. VECTOR MEASURES 321

Now suppose |µ|(Ω)< ∞ and let E1 and E2 be sets of S such that E1∩E2 = /0 and let{Ai

1 · · ·Aini}= π(Ei), a partition of Ei which is chosen such that

|µ|(Ei)− ε <ni

∑j=1∥µ(Ai

j)∥ i = 1,2.

Such a partition exists because of the definition of the total variation. Consider the setswhich are contained in either of π (E1) or π (E2) , it follows this collection of sets is apartition of E1∪E2 denoted by π(E1∪E2). Then by the above inequality and the definitionof total variation,

|µ|(E1∪E2)≥ ∑F∈π(E1∪E2)

∥µ(F)∥> |µ|(E1)+ |µ|(E2)−2ε ,

which shows that since ε > 0 was arbitrary,

|µ|(E1∪E2)≥ |µ|(E1)+ |µ|(E2). (13.1)

Then 13.1 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. ■In the case that µ is a complex measure, it is always the case that |µ|(Ω) < ∞ this is

shown soon. However, first is an interesting corollary. It concerns the case that µ is onlyfinitely additive.

Corollary 13.2.4 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.

Since ε is arbitrary, |µ|(E ∩F) ≥ |µ|(E)+ |µ|(F) . Similar considerations apply to anyfinite union of disjoint sets. That is, if the Ei are disjoint, then |µ|

(∪n

i=1Ei)≥∑

ni=1 |µ|(Ei) .

Now let E = ∪ni=1Ei where the Ei are disjoint. Then letting π (E) be a suitable 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)

∣∣∣∣∣