Measure, Integration & Real Analysis, 2021a

78 Chapter 3 Integration
3.11 Monotone Convergence Theorem
Suppose (X, S, μ) is a measure space and 0 ≤ f 1 ≤ f 2 ≤···is an increasing
sequence of S-measurable functions. Define f : X → [0, ∞] by
f (x) = lim
k→∞
f k (x).
Then
∫
lim
k→∞
∫
f k dμ =
f dμ.
Proof The function f is S-measurable by 2.53.
Because f k (x) ≤ f (x) for every x ∈ X, wehave ∫ f k dμ ≤ ∫ f dμ for each
k ∈ Z + (by 3.8). Thus lim k→∞
∫
fk dμ ≤ ∫ f dμ.
To prove the inequality in the other direction, suppose A 1 ,...,A m are disjoint
sets in S and c 1 ,...,c m ∈ [0, ∞) are such that
3.12 f (x) ≥
m
∑ c j χ
Aj
(x) for every x ∈ X.
j=1
Let t ∈ (0, 1). Fork ∈ Z + , let
{
m }
E k = x ∈ X : f k (x) ≥ t ∑ c j χ
Aj
(x) .
j=1
Then E 1 ⊂ E 2 ⊂···is an increasing sequence of sets in S whose union equals X.
Thus lim k→∞ μ(A j ∩ E k )=μ(A j ) for each j ∈{1, . . . , m} (by 2.59).
If k ∈ Z + , then
m
f k (x) ≥ ∑ tc j χ
Aj
(x)
∩ E k
j=1
for every x ∈ X. Thus (by 3.9)
∫
f k dμ ≥ t
m
∑ c j μ(A j ∩ E k ).
j=1
Taking the limit as k → ∞ of both sides of the inequality above gives
∫
lim
k→∞
f k dμ ≥ t
Now taking the limit as t increases to 1 shows that
∫
lim
k→∞
f k dμ ≥
m
∑ c j μ(A j ).
j=1
m
∑ c j μ(A j ).
j=1
Taking the supremum of the inequality above over all S-partitions A 1 ,...,A m
of X and all c 1 ,...,c m ∈ [0, ∞) satisfying 3.12 shows (using 3.9) that we have
lim k→∞
∫
fk dμ ≥ ∫ f dμ, completing the proof.
Measure, Integration & Real Analysis, by Sheldon Axler