Measure, Integration & Real Analysis, 2021a

98 Chapter 3 Integration
in other words,
|E k \ A k | + |A k \ E k |≤|G k \ A k | + |G k \ E k |
< ε
2|a k |n ;
∥ χAk − χ
Ek
∥ ∥1 <
ε
2|a k |n .
Now ∥∥∥ n ∥ ∥∥1 f − ∑ a k χ
Ek
≤
k=1
∥ f −
n ∥ ∥∥1
∑ a k χ
Ak
+
k=1
∥
n
∑
k=1
< ε 2 + n
∑
k=1|a k | ∥ ∥ χAk − χ
Ek
∥ ∥1
< ε.
a k χ
Ak
−
n
∑
k=1
a k χ
Ek
∥ ∥∥1
Each E k is a finite union of bounded intervals. Thus the inequality above completes
the proof because ∑ n k=1 a kχ
Ek
is a step function.
Luzin’s Theorem (2.91 and 2.93) gives a spectacular way to approximate a Borel
measurable function by a continuous function. However, the following approximation
theorem is usually more useful than Luzin’s Theorem. For example, the next result
plays a major role in the proof of the Lebesgue Differentiation Theorem (4.10).
3.48 approximation by continuous functions
Suppose f ∈L 1 (R). Then for every ε > 0, there exists a continuous function
g : R → R such that
‖ f − g‖ 1 < ε
and {x ∈ R : g(x) ̸= 0} is a bounded set.
For every a 1 ,...,a n , b 1 ,...,b n , c 1 ,...,c n ∈ R and g 1 ,...,g n ∈L 1 (R),
Proof
we have ∥∥∥ n ∥ ∥∥1 f − ∑ a k g k ≤
k=1
≤
∥ f −
∥ f −
n
∑
k=1
n
∑
k=1
a k χ
[bk , c k ]
∥ + ∥
1
a k χ
[bk , c k ]
∥ +
1
n
∑
k=1
n
∑
k=1
a k (χ
[bk , c k ] − g ∥
k)
∥
1
|a k |‖χ
[bk , c k ] − g k‖ 1 ,
where the inequalities above follow
from 3.43. By 3.47, we can choose
a 1 ,...,a n , b 1 ,...,b n , c 1 ,...,c n ∈ R to
make ‖ f − ∑ n k=1 a kχ
[bk , c k ] ‖ 1 as small
as we wish. The figure here then
shows that there exist continuous functions
g 1 ,...,g n ∈ L 1 (R) that make
∑ n k=1 |a k|‖χ
[bk , c k ] − g k‖ 1 as small as we
wish. Now take g = ∑ n k=1 a kg k .
The graph of a continuous function g k
such that ‖χ
[bk , c k ] − g k‖ 1 is small.
Measure, Integration & Real Analysis, by Sheldon Axler