202 CHAPTER 9. THE LEBESGUE INTEGRAL
Lemma 9.1.3 Let g be a decreasing nonnegative function defined on an interval [0,R] .Then ∫ R
0g∧Mdλ = sup
h>0
m(R,h)
∑i=1
(g(ih)∧M)h
where m(h,R) ∈ N satisfies R−h < hm(h,R)≤ R.
Proof: Since g∧M is a decreasing bounded function the lower sums converge to theintegral as h→ 0. Thus
∫ R
0g∧Mdλ = lim
h→0
(m(R,h)
∑i=1
(g(ih)∧M)h+(g(R)∧M)(R−hm(h,R))
)
Now the last term in the above is no more than Mh and so the above is
limh→0
(m(R,h)
∑i=1
(g(ih)∧M)h
)= sup
h>0
(m(R,h)
∑i=1
(g(ih)∧M)h
).■
9.1.2 The Lebesgue Integral for Nonnegative FunctionsHere is the definition of the Lebesgue integral of a function which is measurable and hasvalues in [0,∞].
Definition 9.1.4 Let (Ω,F , µ) be a measure space and suppose f : Ω→ [0,∞]is measurable. Then define
∫f dµ ≡
∫∞
0 µ ([ f > λ ])dλ which makes sense because λ →µ ([ f > λ ]) is nonnegative and decreasing.
Note that if f ≤ g, then∫
f dµ ≤∫
gdµ because µ ([ f > λ ])≤ µ ([g > λ ]) .For convenience ∑
0i=1 ai ≡ 0.
Lemma 9.1.5 In the above definition,∫
f dµ = suph>0 ∑∞i=1 µ ([ f > hi])h
Proof: Let m(h,R) ∈ N satisfy R−h < hm(h,R) ≤ R. Then limR→∞ m(h,R) = ∞ andso from Lemma 9.1.3,∫
f dµ ≡∫
∞
0µ ([ f > λ ])dλ = sup
Msup
R
∫ R
0µ ([ f > λ ])∧Mdλ
= supM
supR>0
suph>0
m(h,R)
∑k=1
(µ ([ f > kh])∧M)h
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. ■