160 CHAPTER 7. THE ABSTRACT LEBESGUE INTEGRAL

Lemma 7.1.3 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

).■

7.1.2 The Lebesgue Integral for Nonnegative FunctionsHere is the definition of the Lebesgue integral of a function which is measurable and hasvalues in [0,∞].

Definition 7.1.4 Let (Ω,F , µ) be a measure space and suppose f : Ω→ [0,∞] ismeasurable. Then define ∫

f dµ ≡∫

∞

0µ ([ 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 7.1.5 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→∞

m(h,R) = ∞

and so from Lemma 7.1.3,∫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