Chapter 7

The Abstract Lebesgue IntegralThe general Lebesgue integral requires a measure space, (Ω,F ,µ) and, to begin with, anonnegative measurable function. I will use Lemma 1.11.2 about interchanging two supre-mums frequently. Also, I will use the observation that if {an} is an increasing sequenceof points of [0,∞] , then supn an = limn→∞ an which is obvious from the definition of sup.Lebesgue integration is a theory which depends on absolute convergence. Thus we under-stand things in terms of nonnegative functions. For complex valued functions, we considerpositive and negative parts of real and imaginary parts. Thus one typically discusses non-negative functions in statements of the main theorems.

7.1 Nonnegative Measurable Functions7.1.1 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 7.1.1 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

M

∫ 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. See Theorem 5.1.11. Now here is an obvious property.

Lemma 7.1.2 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 =

m

∑k=1

∫ bk

ak

f dλ .

Proof: This follows from Theorems 5.1.7 and 5.1.11 along with the computation,∫ b

af dλ ≡ lim

M→∞

∫ b

af ∧Mdλ = lim

M→∞

m

∑k=1

∫ bk

ak

f ∧Mdλ =m

∑k=1

∫ bk

ak

f dλ

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

Thus R−h > 0.

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