Measure, Integration & Real Analysis, 2021a

Section 2D Lebesgue Measure 53
Proof Let L denote the collection of sets A ⊂ R that satisfy (b). We have already
proved that every Borel set is in L (see 2.65). As a key part of that proof, which we
will freely use in this proof, we showed that L is a σ-algebra on R (see the proof
of 2.65). In addition to containing the Borel sets, L contains every set with outer
measure 0 [because if |A| = 0, we can take F = ∅ in (b)].
(b) =⇒ (c): Suppose (b) holds. Thus for each n ∈ Z + , there exists a closed set
F n ⊂ A such that |A \ F n | <
n 1 .Now
∞⋃
A \ F k ⊂ A \ F n
k=1
for each n ∈ Z + . Thus |A \ ⋃ ∞
k=1
F k |≤|A \ F n | <
n 1 for each n ∈ Z+ . Hence
|A \ ⋃ ∞
k=1
F k | = 0, completing the proof that (b) implies (c).
(c) =⇒ (d): Because every countable union of closed sets is a Borel set, we see
that (c) implies (d).
(d) =⇒ (b): Suppose (d) holds. Thus there exists a Borel set B ⊂ A such that
|A \ B| = 0. Now
A = B ∪ (A \ B).
We know that B ∈L(because B is a Borel set) and A \ B ∈L(because A \ B has
outer measure 0). Because L is a σ-algebra, the displayed equation above implies
that A ∈L. In other words, (b) holds, completing the proof that (d) implies (b).
At this stage of the proof, we now know that (b) ⇐⇒ (c) ⇐⇒ (d).
(b) =⇒ (e): Suppose (b) holds. Thus A ∈L. Let ε > 0. Then because
R \ A ∈L(which holds because L is closed under complementation), there exists a
closed set F ⊂ R \ A such that
|(R \ A) \ F| < ε.
Now R \ F is an open set with R \ F ⊃ A. Because (R \ F) \ A =(R \ A) \ F,
the inequality above implies that |(R \ F) \ A| < ε. Thus (e) holds, completing the
proof that (b) implies (e).
(e) =⇒ (f): Suppose (e) holds. Thus for each n ∈ Z + , there exists an open set
G n ⊃ A such that |G n \ A| <
n 1 .Now
( ⋂ ∞ )
G k \ A ⊂ G n \ A
k=1
for each n ∈ Z + . Thus | ( ⋂ ∞k=1
G k
) \ A| ≤|Gn \ A| ≤ 1 n for each n ∈ Z+ . Hence
| ( ⋂ ∞k=1
G k
) \ A| = 0, completing the proof that (e) implies (f).
(f) =⇒ (g): Because every countable intersection of open sets is a Borel set, we
see that (f) implies (g).
(g) =⇒ (b): Suppose (g) holds. Thus there exists a Borel set B ⊃ A such that
|B \ A| = 0. Now
A = B ∩ ( R \ (B \ A) ) .
We know that B ∈L(because B is a Borel set) and R \ (B \ A) ∈L(because this
set is the complement of a set with outer measure 0). Because L is a σ-algebra, the
displayed equation above implies that A ∈L. In other words, (b) holds, completing
the proof that (g) implies (b).
Our chain of implications now shows that (b) through (g) are all equivalent.
Measure, Integration & Real Analysis, by Sheldon Axler