Measure, Integration & Real Analysis, 2021a

Section 2A Outer Measure on R 19
2.12 Heine–Borel Theorem
Every open cover of a closed bounded subset of R has a finite subcover.
Proof Suppose F is a closed bounded
subset of R and C is an open cover of F.
First consider the case where F =
[a, b] for some a, b ∈ R with a < b. Thus
C is an open cover of [a, b]. Let
To provide visual clues, we usually
denote closed sets by F and open
sets by G.
D = {d ∈ [a, b] : [a, d] has a finite subcover from C}.
Note that a ∈ D (because a ∈ G for some G ∈C). Thus D is not the empty set. Let
s = sup D.
Thus s ∈ [a, b]. Hence there exists an open set G ∈Csuch that s ∈ G. Let δ > 0
be such that (s − δ, s + δ) ⊂ G. Because s = sup D, there exist d ∈ (s − δ, s] and
n ∈ Z + and G 1 ,...,G n ∈Csuch that
Now
[a, d] ⊂ G 1 ∪···∪G n .
2.13 [a, d ′ ] ⊂ G ∪ G 1 ∪···∪G n
for all d ′ ∈ [s, s + δ). Thus d ′ ∈ D for all d ′ ∈ [s, s + δ) ∩ [a, b]. This implies that
s = b. Furthermore, 2.13 with d ′ = b shows that [a, b] has a finite subcover from C,
completing the proof in the case where F =[a, b].
Now suppose F is an arbitrary closed bounded subset of R and that C is an open
cover of F. Let a, b ∈ R be such that F ⊂ [a, b]. NowC∪{R \ F} is an open cover
of R and hence is an open cover of [a, b] (here R \ F denotes the set complement of
F in R). By our first case, there exist G 1 ,...,G n ∈Csuch that
Thus
completing the proof.
[a, b] ⊂ G 1 ∪···∪G n ∪ (R \ F).
F ⊂ G 1 ∪···∪G n ,
Measure, Integration & Real Analysis, by Sheldon Axler
Saint-Affrique, the small
town in southern France
where Émile Borel
(1871–1956) was born.
Borel first stated and
proved what we call the
Heine–Borel Theorem in
1895. Earlier, Eduard
Heine (1821–1881) and
others had used similar
results.
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