14.8. NONNEGATIVE SIMPLE FUNCTIONS 335

[ f > λ ] which is what it means for f to be measurable. Also note that, in spite of the picture,in general we don’t know a good description of this set other than that it is measurable.

λ i

A B C

y = f (x)

Definition 14.7.1 Let (Ω,F , µ) be a measure space and suppose f : Ω→ [0,∞] ismeasurable. Then define∫

f dµ ≡∫

∞

0µ ([ f > λ ])dλ =

∫∞

0µ(

f−1 (λ ,∞))

dλ

which makes sense because λ → µ ([ f > λ ]) is nonnegative and decreasing. On the rightyou have an improper Riemann integral like what was discussed above.

Note that if f ≤ g, then∫

f dµ ≤∫

gdµ because µ ([ f > λ ])≤ µ ([g > λ ]) . Next I pointout that the integral is a limit of lower sums.

Lemma 14.7.2 In the situation of the above definition,∫f dµ = sup

h>0

∞

∑i=1

µ ([ f > hi])h

Proof: Let m(h,R) ∈N satisfy R−h < hm(h,R)≤ R. Then limR→∞ hm(h,R) = ∞ andso ∫

f dµ ≡∫

∞

0µ ([ f > λ ])dλ ≡ sup

Msup

R

∫ R

0µ ([ f > λ ])∧Mdλ

supM

supR

∫ hm(h,R)

0µ ([ f > λ ])∧Mdλ = sup

MsupR>0

suph>0

m(h,R)

∑k=1

(µ ([ f > kh])∧M)h

because the sum is just a lower sum for the integral∫ hm(h,R)

0 µ ([ f > λ ])∧Mdλ . Hence,switching the order of the sups, this equals

supR>0

suph>0

supM

m(h,R)

∑k=1

(µ ([ f > kh])∧M)h = supR>0

suph>0

limM→∞

m(h,R)

∑k=1

(µ ([ f > kh])∧M)h

= suph>0

supR

m(R,h)

∑k=1

µ ([ f > kh])h = suph>0

∞

∑k=1

µ ([ f > kh])h.

14.8 Nonnegative Simple FunctionsTo begin with, here is a useful lemma.

Lemma 14.8.1 If f (λ ) = 0 for all λ > a, where f is a decreasing nonnegative function,then

∫∞

0 f (λ )dλ =∫ a

0 f (λ )dλ .