Measure, Integration & Real Analysis, 2021a

20 Chapter 2 Measures
Now we can prove that closed intervals have the expected outer measure.
2.14 outer measure of a closed interval
Suppose a, b ∈ R, with a < b. Then |[a, b]| = b − a.
Proof See the first paragraph of this subsection for the proof that |[a, b]| ≤b − a.
To prove the inequality in the other direction, suppose I 1 , I 2 ,...is a sequence of
open intervals such that [a, b] ⊂ ⋃ ∞
k=1
I k . By the Heine–Borel Theorem (2.12), there
exists n ∈ Z + such that
2.15 [a, b] ⊂ I 1 ∪···∪I n .
We will now prove by induction on n that the inclusion above implies that
2.16
n
∑ l(I k ) ≥ b − a.
k=1
This will then imply that ∑ ∞ k=1 l(I k) ≥ ∑ n k=1 l(I k) ≥ b − a, completing the proof
that |[a, b]| ≥b − a.
To get started with our induction, note that 2.15 clearly implies 2.16 if n = 1.
Now for the induction step: Suppose n > 1 and 2.15 implies 2.16 for all choices of
a, b ∈ R with a < b. Suppose I 1 ,...,I n , I n+1 are open intervals such that
[a, b] ⊂ I 1 ∪···∪I n ∪ I n+1 .
Thus b is in at least one of the intervals I 1 ,...,I n , I n+1 . By relabeling, we can
assume that b ∈ I n+1 . Suppose I n+1 =(c, d). Ifc ≤ a, then l(I n+1 ) ≥ b − a and
there is nothing further to prove; thus we can assume that a < c < b < d, as shown
in the figure below.
Hence
[a, c] ⊂ I 1 ∪···∪I n .
By our induction hypothesis, we have
∑ n k=1 l(I k) ≥ c − a. Thus
n+1
∑ l(I k ) ≥ (c − a)+l(I n+1 )
k=1
=(c − a)+(d − c)
= d − a
≥ b − a,
completing the proof.
Alice was beginning to get very tired
of sitting by her sister on the bank,
and of having nothing to do: once or
twice she had peeped into the book
her sister was reading, but it had no
pictures or conversations in it, “and
what is the use of a book,” thought
Alice, “without pictures or
conversation?”
– opening paragraph of Alice’s
Adventures in Wonderland, byLewis
Carroll
The result above easily implies that the outer measure of each open interval equals
its length (see Exercise 6).
Measure, Integration & Real Analysis, by Sheldon Axler