Measure, Integration & Real Analysis, 2021a

Section 3A Integration with Respect to a Measure 79
The proof that the integral is additive will use the Monotone Convergence Theorem
and our next result. The representation of a simple function h : X → [0, ∞] in the
form ∑ n k=1 c kχ
Ek
is not unique. Requiring the numbers c 1 ,...,c n to be distinct and
E 1 ,...,E n to be nonempty and disjoint with E 1 ∪···∪E n = X produces what
is called the standard representation of a simple function [take E k = h −1 ({c k }),
where c 1 ,...,c n are the distinct values of h]. The following lemma shows that all
representations (including representations with sets that are not disjoint) of a simple
measurable function give the same sum that we expect from integration.
3.13 integral-type sums for simple functions
Suppose (X, S, μ) is a measure space. Suppose a 1 ,...,a m , b 1 ,...,b n ∈ [0, ∞]
and A 1 ,...,A m , B 1 ,...,B n ∈Sare such that ∑ m j=1 a jχ
Aj
= ∑ n k=1 b kχ
Bk
. Then
m
n
∑ a j μ(A j )= ∑ b k μ(B k ).
j=1
k=1
Proof We assume A 1 ∪···∪A m = X (otherwise add the term 0χ X \ (A1 ∪···∪A m ) ).
Suppose A 1 and A 2 are not disjoint. Then we can write
3.14 a 1 χ
A1
+ a 2 χ
A2
= a 1 χ
A1 \ A 2
+ a 2 χ
A2 \ A 1
+(a 1 + a 2 )χ
A1 ∩ A 2
,
where the three sets appearing on the right side of the equation above are disjoint.
Now A 1 =(A 1 \ A 2 ) ∪ (A 1 ∩ A 2 ) and A 2 =(A 2 \ A 1 ) ∪ (A 1 ∩ A 2 ); each
of these unions is a disjoint union. Thus μ(A 1 )=μ(A 1 \ A 2 )+μ(A 1 ∩ A 2 ) and
μ(A 2 )=μ(A 2 \ A 1 )+μ(A 1 ∩ A 2 ). Hence
a 1 μ(A 1 )+a 2 μ(A 2 )=a 1 μ(A 1 \ A 2 )+a 2 μ(A 2 \ A 1 )+(a 1 + a 2 )μ(A 1 ∩ A 2 ).
The equation above, in conjunction with 3.14, shows that if we replace the two
sets A 1 , A 2 by the three disjoint sets A 1 \ A 2 , A 2 \ A 1 , A 1 ∩ A 2 and make the
appropriate adjustments to the coefficients a 1 ,...,a m , then the value of the sum
∑ m j=1 a jμ(A j ) is unchanged (although m has increased by 1).
Repeating this process with all pairs of subsets among A 1 ,...,A m that are
not disjoint after each step, in a finite number of steps we can convert the initial
list A 1 ,...,A m into a disjoint list of subsets without changing the value of
∑ m j=1 a jμ(A j ).
The next step is to make the numbers a 1 ,...,a m distinct. This is done by replacing
the sets corresponding to each a j by the union of those sets, and using finite additivity
of the measure μ to show that the value of the sum ∑ m j=1 a jμ(A j ) does not change.
Finally, drop any terms for which A j = ∅, getting the standard representation
for a simple function. We have now shown that the original value of ∑ m j=1 a jμ(A j )
is equal to the value if we use the standard representation of the simple function
∑ m j=1 a jχ
Aj
. The same procedure can be used with the representation ∑ n k=1 b kχ
Bk
to
show that ∑ n k=1 b kμ(χ
Bk
) equals what we would get with the standard representation.
Thus the equality of the functions ∑ m j=1 a jχ
Aj
and ∑ n k=1 b kχ
Bk
implies the equality
∑ m j=1 a jμ(A j )=∑ n k=1 b kμ(B k ).
Measure, Integration & Real Analysis, by Sheldon Axler