Measure, Integration & Real Analysis, 2021a

76 Chapter 3 Integration
3.6 Example integration with respect to counting measure is summation
Suppose μ is counting measure on Z + and b 1 , b 2 ,...is a sequence of nonnegative
numbers. Think of b as the function from Z + to [0, ∞) defined by b(k) =b k . Then
∫
∞
b dμ = ∑ b k ,
k=1
as you should verify.
Integration with respect to a measure can be called Lebesgue integration. The
next result shows that Lebesgue integration behaves as expected on simple functions
represented as linear combinations of characteristic functions of disjoint sets.
3.7 integral of a simple function
Suppose (X, S, μ) is a measure space, E 1 ,...,E n are disjoint sets in S, and
c 1 ,...,c n ∈ [0, ∞]. Then
Proof
∫ ( n ) n
∑ c k χ
Ek
dμ = ∑ c k μ(E k ).
k=1 k=1
Without loss of generality, we can assume that E 1 ,...,E n is an S-partition of
X [by replacing n by n + 1 and setting E n+1 = X \ (E 1 ∪ ...∪ E n ) and c n+1 = 0].
If P is the S-partition E 1 ,...,E n of X, then L ( ∑ n k=1 c kχ
Ek
, P ) = ∑ n k=1 c kμ(E k ).
Thus
∫ ( n ) n
∑ c k χ
Ek
dμ ≥ ∑ c k μ(E k ).
k=1
k=1
To prove the inequality in the other direction, suppose that P is an S-partition
A 1 ,...,A m of X. Then
( n )
L ∑ c k χ
Ek
, P =
k=1
=
≤
=
=
m
∑
j=1
μ(A j )
m n
∑ ∑
j=1 k=1
m n
∑ ∑
j=1 k=1
min
{i : A j ∩E i̸=∅} c i
μ(A j ∩ E k )
μ(A j ∩ E k )c k
n m
∑ c k ∑ μ(A j ∩ E k )
k=1 j=1
n
∑ c k μ(E k ).
k=1
min
{i : A j ∩E i̸=∅} c i
The inequality above implies that ∫ ( ∑ n k=1 c kχ
Ek
)
dμ ≤ ∑
n
k=1
c k μ(E k ), completing
the proof.
Measure, Integration & Real Analysis, by Sheldon Axler