336 CHAPTER 14. THE LEBESGUE INTEGRAL
Proof: From the definition,∫∞
0f (λ )dλ = lim
R→∞
∫ R
0f (λ )dλ = sup
R>1
∫ R
0f (λ )dλ = sup
R>1sup
M
∫ R
0f (λ )∧Mdλ
= supM
supR>1
∫ R
0f (λ )∧Mdλ = sup
MsupR>1
∫ a
0f (λ )∧Mdλ
= supM
∫ a
0f (λ )∧Mdλ ≡
∫ a
0f (λ )dλ .
Now the Lebesgue integral for a nonnegative function has been defined, what does itdo to a nonnegative simple function? Recall a nonnegative simple function is one whichhas finitely many nonnegative real values which it assumes on measurable sets. Thus asimple function can be written in the form s(ω) = ∑
ni=1 ciXEi (ω) where the ci are each
nonnegative, the distinct values of s.
Lemma 14.8.2 Let s(ω) = ∑pi=1 aiXEi (ω) be a nonnegative simple function where the
Ei are distinct but the ai might not be. Then∫sdµ =
p
∑i=1
aiµ (Ei) . (14.6)
Proof: Without loss of generality, assume 0≡ a0 < a1≤ a2≤ ·· · ≤ ap and that µ (Ei)<∞, i > 0. Here is why. If µ (Ei) = ∞, then the left side would be∫ ap
0µ ([s > λ ])dλ ≥
∫ ai
0µ ([s > λ ])dλ
= supM
∫ ai
0µ ([s > λ ])∧Mdλ ≥ sup
MMai = ∞
and so both sides are equal to ∞. Thus it can be assumed that for each i,µ (Ei)< ∞. Thenit follows from Lemma 14.8.1 and Lemma 14.6.2,∫
∞
0µ ([s > λ ])dλ =
∫ ap
0µ ([s > λ ])dλ =
p
∑k=1
∫ ak
ak−1
µ ([s > λ ])dλ
=p
∑k=1
(ak−ak−1)p
∑i=k
µ (Ei) =p
∑i=1
µ (Ei)i
∑k=1
(ak−ak−1) =p
∑i=1
aiµ (Ei)
Lemma 14.8.3 If a,b≥ 0 and if s and t are nonnegative simple functions, then∫(as+bt)dµ = a
∫sdµ +b
∫tdµ .
Proof: Let s(ω) = ∑ni=1 α iXAi(ω), t(ω) = ∑
mi=1 β jXB j(ω) where α i are the distinct
values of s and the β j are the distinct values of t. Clearly as+ bt is a nonnegative simplefunction because it has finitely many values on measurable sets. In fact,
(as+bt)(ω) =m
∑j=1
n
∑i=1
(aα i +bβ j)XAi∩B j(ω)