Measure, Integration & Real Analysis, 2021a

Section 9A Total Variation 265
If ε and n are as in the paragraph above, then
(
∣ ⋃ ∞
∣ν
k=1
E k
)
−
n−1
∑
k=1
ν(E k ) ∣ =
( ⋃ ∞ )
∣ lim ν j E k − lim
j→∞
k=1
∣
∣ ∞ ∣∣
= lim
j→∞
∑
k=n
≤ 2ε,
ν j (E k ) ∣
n−1
∑ j→∞ k=1
ν j (E k ) ∣
where the second line uses the countable additivity of the measure ν j and the third line
uses 9.21. The inequality above implies that ν ( ⋃ ∞k=1
E k
) = ∑
∞
k=1
ν(E k ), completing
the proof that ν ∈M F (S).
We still need to prove that lim k→∞ ‖ν − ν k ‖ = 0. To do this, suppose ε > 0. Let
m ∈ Z + be such that
9.22 ‖ν j − ν k ‖≤ε for all j, k ≥ m.
Suppose k ≥ m. Suppose also that E 1 ,...,E n ∈Sare disjoint subsets of X. Then
n
∑
l=1
|(ν − ν k )(E l )| = lim
j→∞
n
∑
l=1
|(ν j − ν k )(E l )|≤ε,
where the last inequality follows from 9.22 and the definition of the total variation
norm. The inequality above implies that ‖ν − ν k ‖≤ε, completing the proof.
EXERCISES 9A
1 Prove or give a counterexample: If ν is a real measure on a measurable
space (X, S) and A, B ∈Sare such that ν(A) ≥ 0 and ν(B) ≥ 0, then
ν(A ∪ B) ≥ 0.
2 Suppose ν is a real measure on (X, S). Define μ : S→[0, ∞) by
μ(E) =|ν(E)|.
Prove that μ is a (positive) measure on (X, S) if and only if the range of ν is
contained in [0, ∞) or the range of ν is contained in (−∞,0].
3 Suppose ν is a complex measure on a measurable space (X, S). Prove that
|ν|(X) =ν(X) if and only if ν is a (positive) measure.
4 Suppose ν is a complex measure on a measurable space (X, S). Prove that if
E ∈Sthen
{ ∞
|ν|(E) =sup ∑ |ν(E k )| : E 1 , E 2 ,... is a disjoint sequence in S
k=1
∞⋃
such that E = E k
}.
k=1
Measure, Integration & Real Analysis, by Sheldon Axler