Measure, Integration & Real Analysis, 2021a

Section 7B L p (μ) 205
We now have a function f that is the pointwise limit (almost everywhere) of
f 1 , f 2 ,.... The definition of f shows that | f (x)| ≤g(x) for almost every x ∈ X.
Thus 7.22 shows that f ∈L p (μ).
To show that lim k→∞ ‖ f k − f ‖ p = 0, suppose ε > 0 and let n ∈ Z + be such that
‖ f j − f k ‖ p < ε for all j ≥ n and k ≥ n. Suppose k ≥ n. Then
(∫
‖ f k − f ‖ p =
(∫
≤ lim inf
j→∞
) 1/p
| f k − f | p dμ
= lim inf ‖ f k − f j ‖ p
j→∞
≤ ε,
) 1/p
| f k − f j | p dμ
where the second line above comes from Fatou’s Lemma (Exercise 17 in Section 3A).
Thus lim k→∞ ‖ f k − f ‖ p = 0, as desired.
The proof that we have just completed contains within it the proof of a useful
result that is worth stating separately. A sequence can converge in p-norm without
converging pointwise anywhere (see, for example, Exercise 12). However, the next
result guarantees that some subsequence converges pointwise almost everywhere.
7.23 convergent sequences in L p have pointwise convergent subsequences
Suppose (X, S, μ) is a measure space and 1 ≤ p ≤ ∞. Suppose f ∈L p (μ)
and f 1 , f 2 ,...is a sequence of functions in L p (μ) such that lim ‖ f k − f ‖ p = 0.
k→∞
Then there exists a subsequence f k1 , f k2 ,...such that
for almost every x ∈ X.
lim f k
m→∞ m
(x) = f (x)
Proof
Suppose f k1 , f k2 ,...is a subsequence such that
∞
∑ ‖ f km − f km−1 ‖ p < ∞.
m=2
An examination of the proof of 7.20 shows that lim
m→∞ f k m
(x) = f (x) for almost
every x ∈ X.
7.24 L p (μ) is a Banach space
Suppose μ is a measure and 1 ≤ p ≤ ∞. Then L p (μ) is a Banach space.
Proof
This result follows immediately from 7.20 and the appropriate definitions.
Measure, Integration & Real Analysis, by Sheldon Axler