Measure, Integration & Real Analysis, 2021a

Section 9A Total Variation 261
Now
n
∑
k=1
|ν(E k )| =
≥
=
=
∫
≥
≥
n
∑
k=1
n
∑
k=1
n
∑
k=1
∫
E
E
∫
E
∫
∣ h dμ∣
E k
∫
∣ g dμ∣ −
E k
|c k |μ(E k ) −
|g| dμ −
|g| dμ −
n
∑
∣
k=1
n ∫
∑
k=1
|h| dμ − 2ε.
n
∑
k=1
n ∣∫
∑
k=1
∫
∫
∣ (g − h) dμ∣
E k
∣
(g − h) dμ∣
E k
(g − h) dμ∣
E k
E k
|g − h| dμ
The inequality above implies that |ν|(E) ≥ ∫ E
|h| dμ − 2ε. Because ε is an arbitrary
positive number, this implies |ν|(E) ≥ ∫ E
|h| dμ, completing the proof.
Now we justify the terminology total variation measure.
9.11 total variation measure is a measure
Suppose ν is a complex measure on a measurable space (X, S). Then the total
variation function |ν| is a (positive) measure on (X, S).
Proof The definition of |ν| and 9.3(a) imply that |ν|(∅) =0.
To show that |ν| is countably additive, suppose A 1 , A 2 ,...are disjoint sets in S.
Fix m ∈ Z + . For each k ∈{1, . . . , m}, suppose E 1,k ,...,E nk ,k are disjoint sets in S
such that
9.12 E 1,k ∪ ...∪ E nk ,k ⊂ A k .
Then {E j,k :1≤ k ≤ m and 1 ≤ j ≤ n k } is a disjoint collection of sets in S that
are all contained in ⋃ ∞
k=1
A k . Hence
m
∑
k=1
n k
∑
j=1
m
∑
k=1
( ⋃ ∞
|ν(E j,k )|≤|ν|
k=1
A k
).
Taking the supremum of the left side of the inequality above over all choices of {E j,k }
satisfying 9.12 shows that
( ⋃ ∞
|ν|(A k ) ≤|ν| A k
).
k=1
Because the inequality above holds for all m ∈ Z + ,wehave
∞
( ⋃ ∞
|ν|(A k ) ≤|ν| A k
).
∑
k=1
k=1
Measure, Integration & Real Analysis, by Sheldon Axler