Measure, Integration & Real Analysis, 2021a

Section 2A Outer Measure on R 17
The union of the intervals (1, 4) and (3, 5) is the interval (1, 5). Thus
l ( (1, 4) ∪ (3, 5) ) 0. For each k ∈ Z + , let I 1,k , I 2,k ,... be a sequence of open intervals
whose union contains A k such that
Thus
2.9
∞
∑
k=1
∞
∑ l(I j,k ) ≤ ε
j=1
2 k + |A k|.
∞
∑ l(I j,k ) ≤ ε +
j=1
∞
∑
k=1
|A k |.
The doubly indexed collection of open intervals {I j,k : j, k ∈ Z + } can be rearranged
into a sequence of open intervals whose union contains ⋃ ∞
k=1
A k as follows, where
in step k (start with k = 2, then k = 3, 4, 5, . . . ) we adjoin the k − 1 intervals whose
indices add up to k:
I 1,1
, I 1,2 , I 2,1 , I 1,3 , I 2,2 , I 3,1 , I 1,4 , I 2,3 , I 3,2 , I 4,1 , I 1,5 , I 2,4 , I 3,3 , I 4,2 , I 5,1 ,... .
}{{} } {{ } } {{ } } {{ } } {{ }
2 3
4
5
sum of the two indices is 6
Inequality 2.9 shows that the sum of the lengths of the intervals listed above is less
than or equal to ε + ∑ ∞ k=1 |A k|. Thus ∣ ∣ ⋃ ∞
k=1
A k
∣ ∣ ≤ ε + ∑ ∞ k=1 |A k|. Because ε is an
arbitrary positive number, this implies that ∣ ∣ ⋃ ∞
k=1
A k
∣ ∣ ≤ ∑ ∞ k=1 |A k|.
Countable subadditivity implies finite subadditivity, meaning that
|A 1 ∪···∪A n |≤|A 1 | + ···+ |A n |
for all A 1 ,...,A n ⊂ R, because we can take A k = ∅ for k > n in 2.8.
The countable subadditivity of outer measure, as proved above, adds to our list of
nice properties enjoyed by outer measure.
Measure, Integration & Real Analysis, by Sheldon Axler