Kenneth Kuttler

340 CHAPTER 14. THE LEBESGUE INTEGRAL

By Fatou’s lemma (Theorem 14.9.2),∫2gdµ ≤ lim inf

n→∞

∫2g−| f − fn|dµ =

∫2gdµ− lim sup

n→∞

∫| f − fn|dµ.

Subtracting∫

2gdµ , 0≤− limsupn→∞

∫| f − fn|dµ. Hence

0 ≥ lim supn→∞

(∫| f − fn|dµ

)≥ lim inf

n→∞

(∫| f − fn|dµ

)≥ lim inf

n→∞

∣∣∣∣∫ f dµ−∫

fndµ

∣∣∣∣≥ 0.

This proves the theorem by Theorem 4.10.10 because the limsup and liminf are equal.

Example 14.11.6 Let Ω≡N and let F be the set of all subsets of Ω. Let µ (E)≡ numberof entries in E. Then (N,F ,µ) is a measure space and the Lebesgue integral is summation.Thus all the convergence theorems mentioned above apply to sums.

First, why is µ a measure? If {Ei} are disjoint, then if infinitely many are nonempty,say En1 ,En2 , · · · . Then ∪iEi is infinite and so

µ (∪iEi) = ∞ =∞

∑i=1

µ (Ei) =∞

∑k=1

Enk ≥∞

∑k=1

1 = ∞

The alternative is that only finitely many Ei are nonempty and in this case, the assertionthat µ (∪iEi) = ∑

∞i=1 µ (Ei) is obvious. Hence µ is indeed a measure. Now let f : N→ R.

It is obviously measurable because the inverse image of anything is a subset of N. So iff (n)≥ 0 for all n, what is

∫f dµ?

f (i) =∞

∑k=1

f (k)X{k} (i) = limn→∞

∑k=1

f (k)X{k} (i)≡ fn (i)

Now fn is a nonnegative simple function and there is exactly one thing in {k}. Therefore,∫fndµ = ∑

nk=1 f (k) . Then, by the monotone convergence theorem,∫

f dµ = limn→∞

∫fndµ = lim

n→∞

∑k=1

f (k)≡∞

∑k=1

f (k)

When ∑k | f (k)|< ∞, one has∫

f dµ = ∑∞k=1 f (k).

This example illustrates how the Lebesgue integral pertains to absolute summabilityand absolute integrability. It is not a theory which can include conditional convergence.The generalized Riemann integral can do this. However, the Lebesgue integral is veryeasy to use because of this restriction. Of course one could make the same restrictionand consider those functions for which | f | is integrable in the context of the generalizedRiemann integral.

How does this new integral compare to the generalized Riemann Stieltjes integral? LetµF be the measure of Theorem 14.3.6 in what follows.

Theorem 14.11.7 Let F be an increasing integrator function and let f be continu-ous on [a,b] . Then ∫ b

af dF =

∫f dµF

Here the integral on the left is the Riemann Stieltjes integral.

340 CHAPTER 14. THE LEBESGUE INTEGRALBy Fatou’s lemma (Theorem 14.9.2),[sau <tim int, [2¢—|f—fuldu= [2¢du—tim sup | |f — fuldnoo n—ooSubtracting {2gdu,0<—limsup,,,.. f |f—fnldu. Hence0 > limsup ( / f hola) > lim inf ( / I~ foldn—yoo n—-roolim inf [ran | faan >0.VIVThis proves the theorem by Theorem 4.10.10 because the limsup and liminf are equal. JExample 14.11.6 Let Q = N and let F be the set of all subsets of Q. Let 1 (E) = numberof entries in E. Then (N, ¥ , 1) is a measure space and the Lebesgue integral is summation.Thus all the convergence theorems mentioned above apply to sums.First, why is a measure? If {£;} are disjoint, then if infinitely many are nonempty,say En,,En.,::: . Then U;£; is infinite and so1 (iE) == Ya (E) = Fn 2Yi==The alternative is that only finitely many £; are nonempty and in this case, the assertionthat p (U;E;) = Y=, uw (E;) is obvious. Hence pt is indeed a measure. Now let f: N—- R.It is obviously measurable because the inverse image of anything is a subset of N. So iff (n) > 0 for all n, what is f fdu?J=V) fk) gd = = lim ys ) Fg @) = fr)k=lNow f, is a nonnegative simple function and there is exactly one thing in {k}. Therefore,J frdut = LiL, f (k). Then, by the monotone convergence theorem,[fave = tim | frau = tim Yo Fb )= ErkWhen Y;|f (k)| < 0%, one has f fdu = Ye, f (k).This example illustrates how the Lebesgue integral pertains to absolute summabilityand absolute integrability. It is not a theory which can include conditional convergence.The generalized Riemann integral can do this. However, the Lebesgue integral is veryeasy to use because of this restriction. Of course one could make the same restrictionand consider those functions for which |f| is integrable in the context of the generalizedRiemann integral.How does this new integral compare to the generalized Riemann Stieltjes integral? LetLp be the measure of Theorem 14.3.6 in what follows.Theorem 14.11.7 Let F be an increasing integrator function and let f be continu-ous on [a,b]. Then‘b .[ far = | fangaHere the integral on the left is the Riemann Stieltjes integral.