604 CHAPTER 20. REPRESENTATION THEOREMS

Also,Ai = ∪∞

j=1Ai∩E j

and so by the triangle inequality, ||µ(Ai)|| ≤ ∑∞j=1 ||µ(Ai∩E j)||. Therefore, by the above,

and either Fubini’s theorem or Lemma 11.3.3 on Page 236

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 \∪n

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

{Ai1 · · ·Ai

ni}= 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). (20.2.5)