Measure, Integration & Real Analysis, 2021a

64 Chapter 2 Measures
Proof Suppose ε > 0. Temporarily fix n ∈ Z + . The definition of pointwise
convergence implies that
2.86
∞⋃ ∞⋂
{x ∈ X : | f k (x) − f (x)| <
n 1 } = X.
m=1 k=m
For m ∈ Z + , let
A m,n =
∞⋂
{x ∈ X : | f k (x) − f (x)| <
n 1 }.
k=m
Then clearly A 1,n ⊂ A 2,n ⊂···is an increasing sequence of sets and 2.86 can be
rewritten as
∞⋃
A m,n = X.
m=1
The equation above implies (by 2.59) that lim m→∞ μ(A m,n )=μ(X). Thus there
exists m n ∈ Z + such that
2.87 μ(X) − μ(A mn , n) < ε
2 n .
Now let
∞⋂
E = A mn , n.
Then
n=1
(
μ(X \ E) =μ X \
( ⋃ ∞
= μ
≤
< ε,
n=1
∞⋂
A mn , n
n=1
)
)
(X \ A mn , n)
∞
∑ μ(X \ A mn , n)
n=1
where the last inequality follows from 2.87.
To complete the proof, we must verify that f 1 , f 2 ,...converges uniformly to f
on E. To do this, suppose ε ′ > 0. Let n ∈ Z + be such that 1 n < ε′ . Then E ⊂ A mn , n,
which implies that
| f k (x) − f (x)| < 1 n < ε′
for all k ≥ m n and all x ∈ E. Hence f 1 , f 2 ,...does indeed converge uniformly to f
on E.
Measure, Integration & Real Analysis, by Sheldon Axler