Measure, Integration & Real Analysis, 2021a

≈ 7π Chapter 2 Measures
Clearly a ∈ ã for each a ∈ [−1, 1]. Thus [−1, 1] =
Let V be a set that contains exactly one
element in each of the distinct sets in
{ã : a ∈ [−1, 1]}.
⋃
a∈[−1,1]
This step involves the Axiom of
Choice, as discussed after this proof.
The set V arises by choosing one
element from each equivalence
class.
In other words, for every a ∈ [−1, 1], the
set V ∩ ã has exactly one element.
Let r 1 , r 2 ,...be a sequence of distinct rational numbers such that
Then
[−2, 2] ∩ Q = {r 1 , r 2 ,...}.
[−1, 1] ⊂
∞⋃
(r k + V),
k=1
where the set inclusion above holds because if a ∈ [−1, 1], then letting v be the
unique element of V ∩ ã, wehavea − v ∈ Q, which implies that a = r k + v ∈
r k + V for some k ∈ Z + .
The set inclusion above, the order-preserving property of outer measure (2.5), and
the countable subadditivity of outer measure (2.8) imply
|[−1, 1]| ≤
∞
∑
k=1
|r k + V|.
We know that |[−1, 1]| = 2 (from 2.14). The translation invariance of outer measure
(2.7) thus allows us to rewrite the inequality above as
2 ≤
∞
∑
k=1
Thus |V| > 0.
Note that the sets r 1 + V, r 2 + V,...are disjoint. (Proof: Suppose there exists
t ∈ (r j + V) ∩ (r k + V). Then t = r j + v 1 = r k + v 2 for some v 1 , v 2 ∈ V, which
implies that v 1 − v 2 = r k − r j ∈ Q. Our construction of V now implies that v 1 = v 2 ,
which implies that r j = r k , which implies that j = k.)
Let n ∈ Z + . Clearly
n⋃
(r k + V) ⊂ [−3, 3]
k=1
|V|.
because V ⊂ [−1, 1] and each r k ∈ [−2, 2]. The set inclusion above implies that
2.19
However
2.20
n
∑
k=1
n⋃
∣ (r k + V) ∣ ≤ 6.
k=1
|r k + V| =
n
∑
k=1
|V| = n |V|.
Measure, Integration & Real Analysis, by Sheldon Axler
ã.