32.7. NONNEGATIVE SIMPLE FUNCTIONS 645

Proof: From the definition,∫∞

0f (λ )dλ = lim

R→∞

∫ R

0f (λ )dλ = sup

R>1

∫ R

0f (λ )dλ

= supR>1

supM

∫ R

0f (λ )∧Mdλ

= supM

supR>1

∫ R

0f (λ )∧Mdλ

= supM

supR>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 it doto a nonnegative simple function? Recall a nonnegative simple function is one which hasfinitely many nonnegative real values which it assumes on measurable sets. Thus a simplefunction can be written in the form

s(ω) =n

∑i=1

ciXEi (ω)

where the ci are each nonnegative, the distinct values of s.

Lemma 32.7.2 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) . (32.12)

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

≥ supM

Mai = ∞

and so both sides are equal to ∞. Thus it can be assumed that for each i,µ (Ei)< ∞. Thenit follows from Lemma 32.7.1 and Lemma 32.5.2,∫

∞

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 32.7.3 If a,b ≥ 0 and if s and t are nonnegative simple functions, then∫(as+bt)dµ = a

∫sdµ +b

∫tdµ .