Kenneth Kuttler

Chapter 9

The Lebesgue IntegralThe presentation in terms of simple functions of the Lebesgue integral is presented in Prob-lems starting with 22 on Page 199. I will present it a different way here. The generalLebesgue integral requires a measure space, (Ω,F ,µ) and, to begin with, a nonnegativemeasurable function. I will use Lemma 2.5.3 about interchanging two supremums fre-quently. Also, I will use the observation that if {an} is an increasing sequence of points of[0,∞] , then supn an = limn→∞ an which is obvious from the definition of sup.

9.1 Nonnegative Measurable Functions9.1.1 Riemann Integrals for Decreasing FunctionsFirst of all, the notation [g < f ] means {ω ∈Ω : g(ω)< f (ω)} with other variants of thisnotation being similar. Also, the convention, 0 ·∞ = 0 will be used to simplify the presen-tation whenever it is convenient to do so. The notation a∧b means the minimum of a andb.

Definition 9.1.1 Let f : [a,b]→ [0,∞] be decreasing. Note that ∞ is a possiblevalue. Define ∫ b

af (λ )dλ ≡ lim

M→∞

∫ b

aM∧ f (λ )dλ = sup

∫ b

aM∧ f (λ )dλ

where a∧b means the minimum of a and b. Note that for f bounded,

supM

∫ b

aM∧ f (λ )dλ =

∫ b

af (λ )dλ

where the integral on the right is the usual Riemann integral because eventually M > f .For f a nonnegative decreasing function defined on [0,∞),∫

∞

0f dλ ≡ lim

R→∞

∫ R

0f dλ = sup

R>1

∫ R

0f dλ = sup

RsupM>0

∫ R

0f ∧Mdλ

Since decreasing bounded functions are Riemann integrable, the above definition iswell defined. For a discussion of this, see Calculus of One and Many Variables on the website or the single variable advanced calculus book. Now here is an obvious property.

Lemma 9.1.2 Let f be a decreasing nonnegative function defined on an interval [a,b] .Then if [a,b] =∪m

k=1Ik where Ik ≡ [ak,bk] and the intervals Ik are non overlapping, it follows∫ b

af dλ =

∑k=1

∫ bk

f dλ .

Proof: This follows from the computation,∫ b

af dλ ≡ lim

M→∞

∫ b

af ∧Mdλ = lim

M→∞

∑k=1

∫ bk

f ∧Mdλ =m

∑k=1

∫ bk

f dλ

Note both sides could equal +∞. ■In all considerations below, assume h is fairly small, certainly much smaller than R.

Thus R−h > 0.

201

Chapter 9The Lebesgue IntegralThe presentation in terms of simple functions of the Lebesgue integral is presented in Prob-lems starting with 22 on Page 199. I will present it a different way here. The generalLebesgue integral requires a measure space, (Q,.¥,) and, to begin with, a nonnegativemeasurable function. I will use Lemma 2.5.3 about interchanging two supremums fre-quently. Also, I will use the observation that if {a,} is an increasing sequence of points of(0, 09] , then sup, dn = limy—4.0d@, which is obvious from the definition of sup.9.1 Nonnegative Measurable Functions9.1.1 Riemann Integrals for Decreasing FunctionsFirst of all, the notation [g < f] means {@ € Q: g(@) < f(@)} with other variants of thisnotation being similar. Also, the convention, 0 - co = 0 will be used to simplify the presen-tation whenever it is convenient to do so. The notation a b means the minimum of a andb.Definition 9.1.1 Lez f : [a,b] > [0,] be decreasing. Note that ~ is a possiblevalue. Defineb b b[ fajar= lim MAS (A)dA =sup | MAS (A)dAa Mx Jaq M Jawhere a/b means the minimum of a and b. Note that for f bounded,b bsup [ Maf(ayaa = | f(A)dAwhere the integral on the right is the usual Riemann integral because eventually M > f.For f anonnegative decreasing function defined on [0,°),co -R R Ri fda = lim | fda =sup | fdA=supsup | fAMda0 Re JO R>1/0 R M>0/0Since decreasing bounded functions are Riemann integrable, the above definition iswell defined. For a discussion of this, see Calculus of One and Many Variables on the website or the single variable advanced calculus book. Now here is an obvious property.Lemma 9.1.2 Let f be a decreasing nonnegative function defined on an interval [a,b] .Then if [a,b] = UL. where I, = (ax, by) and the intervals l, are non overlapping, it followsb m by/ fdA=¥) | fda.Ja k=1° akProof: This follows from the computation,»b 4 b 4 m by 4 m by 4ddA = lim | A\MdaA = lim | A\MdiA = d[ f M2 Jaq f yim a f d ak fNote both sides could equal +-co.In all considerations below, assume h is fairly small, certainly much smaller than R.Thus R—h> 0.201