622 CHAPTER 23. REPRESENTATION THEOREMS

Note that it makes sense to take finite sums because it is given that µ has values in avector space in which vectors can be summed. In the above, µ (Ei) is a vector. It might bea point in Rp or in any other vector space. In many of the most important applications, itis a vector in some sort of function space which may be infinite dimensional. The infinitesum has the usual meaning. That is ∑

∞i=1 µ(Ei) = limn→∞ ∑

ni=1 µ(Ei) where the limit takes

place relative to the norm on V .

Definition 23.2.2 Let (Ω,S ) be a measure space and let µ be a vector measuredefined on S . A subset, π(E), of S is called a partition of E if π(E) consists of finitelymany disjoint sets of S and ∪π(E) = E. Let

|µ|(E) = sup{ ∑F∈π(E)

∥µ(F)∥ : π(E) is a partition of E}.

|µ| is called the total variation of µ .

The next theorem may seem a little surprising. It states that, if finite, the total variationis a nonnegative measure.

Theorem 23.2.3 If |µ|(Ω) < ∞, then |µ| is a measure on S . Even if |µ|(Ω) =∞, |µ|(∪∞

i=1Ei) ≤ ∑∞i=1 |µ|(Ei) . That is |µ| is always subadditive and |µ|(A) ≤ |µ|(B)

whenever A,B ∈S with A⊆ B. In earlier terminology, |µ| is an outer measure.

Proof: Consider the last claim. Let a < |µ|(A) and let π (A) be a partition of A suchthat a < ∑F∈π(A) ∥µ (F)∥ . Then π (A)∪{B\A} is a partition of B and

|µ|(B)≥ ∑F∈π(A)

∥µ (F)∥+∥µ (B\A)∥> a.

Since this is true for all such a, it follows |µ|(B)≥ |µ|(A) as claimed.Let

{E j}∞

j=1 be a sequence of disjoint sets of S and let E∞ = ∪∞j=1E j. Then letting

a < |µ|(E∞) , it follows from the definition of total variation there exists a partition ofE∞, π(E∞) = {A1, · · · ,An} such that a < ∑

ni=1 ∥µ(Ai)∥. Also, Ai = ∪∞

j=1Ai∩E j and so bythe triangle inequality, ∥µ(Ai)∥ ≤ ∑

∞j=1∥∥µ(Ai∩E j)

∥∥. Therefore, by the above, and eitherFubini’s theorem or Lemma 2.5.4 on Page 65,

a <n

∑i=1

≥∥µ(Ai)∥︷ ︸︸ ︷∞

∑j=1

∥∥µ(Ai∩E j)∥∥= ∞

∑j=1

n

∑i=1

∥∥µ(Ai∩E j)∥∥≤ ∞

∑j=1|µ|(E j)

because{

Ai∩E j}n

i=1 is a partition of E j.Since a is arbitrary, this shows |µ|(∪∞

j=1E j) ≤ ∑∞j=1 |µ|(E j). If the sets, E j are not

disjoint, let F1 = E1 and if Fn has been chosen, let Fn+1 ≡ En+1 \∪ni=1Ei. Thus the sets, Fi

are disjoint and ∪∞i=1Fi = ∪∞

i=1Ei. Therefore,

|µ|(∪∞

j=1E j)= |µ|

(∪∞

j=1Fj)≤

∞

∑j=1|µ|(Fj)≤

∞

∑j=1|µ|(E j)

and proves |µ| is always subadditive as claimed, regardless of whether |µ|(Ω)< ∞.