6.2 The Lebesgue Integral For Nonnegative Simple Functions
To begin with, here is a useful lemma.
Lemma 6.2.1If f
(λ)
= 0 for all λ > a, where f is a decreasing nonnegative function, then
∫ ∞ ∫ a
f (λ)dλ = f (λ)dλ.
0 0
Proof:From the definition,
∫ ∞ ∫ R ∫ R
f (λ)dλ = lim f (λ)dλ = sup f (λ)dλ
0 R→ ∞ 0∫ R>1 0
= sup sup Rf (λ )∧M dλ
R>1 M 0
∫ R
= supsup f (λ )∧M dλ
M R>1 ∫0a
= supsup f (λ)∧ M dλ
M R>1 0
∫ a ∫ a
= suMp 0 f (λ)∧ M dλ ≡ 0 f (λ)dλ.■
Now the Lebesgue integral for a nonnegative function has been defined, what does it do to a
nonnegative simple function? Recall a nonnegative simple function is one which has finitely many
nonnegative real values which it assumes on measurable sets. Thus a simple function can be written in the
form
∑n
s (ω ) = ciXEi (ω)
i=1
where the ci are each nonnegative, the distinct values of s.
Lemma 6.2.2Let s
(ω)
= ∑i=1paiXEi
(ω)
be a nonnegative simple function where the Eiare distinct butthe aimight not be. Then
∫ ∑p
sdμ = aiμ (Ei). (6.1)
i=1
(6.1)
Proof:Without loss of generality, assume 0 ≡ a0< a1≤ a2≤
Lemma 6.2.3If a,b ≥ 0 and if s and t are nonnegative simple functions, then
∫ ∫ ∫
as +btdμ = a sdμ + b tdμ.
Proof: Let
∑n ∑m
s(ω ) = αiXAi(ω),t(ω) = βjXBj (ω)
i=1 i=1
where αi are the distinct values of s and the βj are the distinct values of t. Clearly as + bt
is a nonnegative simple function because it has finitely many values on measurable sets. In
fact,
∑m ∑n
(as+ bt)(ω) = (aαi + bβj)XAi∩Bj(ω )
j=1 i=1
where the sets Ai∩ Bj are disjoint and measurable. By Lemma 6.2.2,
∫ m n
∑ ∑
as+ btdμ = j=1 i=1(aαi + bβj)μ(Ai ∩ Bj)
n m m n
= ∑ a ∑ α μ(A ∩ B )+ b∑ ∑ β μ(A ∩B )
i=1 j=1 i i j j=1i=1 j i j
∑n m∑
= a αiμ(Ai )+ b βjμ(Bj)
i=1 j=1
∫ ∫
= a sdμ+ b tdμ. ■