Kenneth Kuttler

296 CHAPTER 10. THE ABSTRACT LEBESGUE INTEGRAL

Proof: Since F is an increasing function it can have only countably many disconti-nuities. The reason for this is that the only kind of discontinuity it can have is whereF (x+) > F (x−) . Now since F is increasing, the intervals (F (x−) ,F (x+)) for x a pointof discontinuity are disjoint and so since each must contain a rational number and the ra-tional numbers are countable, and therefore so are these intervals.

Let D denote this countable set of discontinuities of F . Then if l,r /∈D, [l,r]⊆ [a,b] , itfollows quickly from the definition of the Darboux Stieltjes integral that∫ b

aX[l,r)dF = F (r)−F (l) = F (r−)−F (l−) = µ ([l,r)) =

∫X[l,r)dµ.

Now let {sn} be the sequence of step functions of Lemma 10.10.3 such that these stepfunctions converge uniformly to f on [c,d] , say maxx | f (x)− sn (x)|< 1/n. Then∣∣∣∣∫ (X[c,d] f −X[c,d]sn

)dµ

∣∣∣∣≤ ∫ ∣∣X[c,d] ( f − sn)∣∣dµ ≤ 1

nµ ([c,d])

and∣∣∣∫ b

a(X[c,d] f −X[c,d]sn

)dF∣∣∣ ≤ ∫ b

a X[c,d] | f − sn|dF < 1n (F (b)−F (a)) .Also if sn is

given by the formula of Lemma 10.10.3,∫X[c,d]sndµ =

∫ mn

∑k=1

f(zn

k−1)X[zn

k−1,znk)

dµ =mn

∑k=1

∫f(zn

k−1)X[zn

k−1,znk)

dµ

=mn

∑k=1

f(zn

k−1)

µ([zn

k−1,znk))=

∑k=1

f(zn

k−1)(

F (znk−)−F

(zn

k−1−))

=mn

∑k=1

f(zn

k−1)(

F (znk)−F

(zn

k−1))

=mn

∑k=1

∫ b

af(zn

k−1)X[zn

k−1,znk)

dF =∫ b

asndF.

Therefore, ∣∣∣∣∫ X[c,d] f dµ−∫ b

aX[c,d] f dF

∣∣∣∣≤ ∣∣∣∣∫ X[c,d] f dµ−∫

X[c,d]sndµ

∣∣∣∣+

∣∣∣∣∫ X[c,d]sndµ−∫ b

asndF

∣∣∣∣+ ∣∣∣∣∫ b

asndF−

∫ b

aX[c,d] f dF

∣∣∣∣≤ 1

nµ ([c,d])+

1n(F (b)−F (a))

and since n is arbitrary, this shows∫

f dµ−∫ b

a f dF = 0. ■In particular, in the special case where F is continuous and f is continuous,

∫ ba f dF =∫

X[a,b] f dµ. Thus, if F (x) = x so the Darboux Stieltjes integral is the usual integral fromcalculus,

∫ ba f (t)dt =

∫X[a,b] f dµ where µ is the measure which comes from F (x) = x

as described above. This measure is often denoted by m. Thus when f is continuous∫ ba f (t)dt =

∫X[a,b] f dm and so there is no problem in writing

∫ ba f (t)dt for either the

Lebesgue or the Riemann integral. Furthermore, when f is continuous, you can computethe Lebesgue integral by using the fundamental theorem of calculus because in this case,the two integrals are equal. ■

296 CHAPTER 10. THE ABSTRACT LEBESGUE INTEGRALProof: Since F is an increasing function it can have only countably many disconti-nuities. The reason for this is that the only kind of discontinuity it can have is whereF (x+) > F (x—). Now since F is increasing, the intervals (F (x—) , F (x+)) for x a pointof discontinuity are disjoint and so since each must contain a rational number and the ra-tional numbers are countable, and therefore so are these intervals.Let D denote this countable set of discontinuities of F. Then if /,r ¢ D, [l,r] C [a,b], itfollows quickly from the definition of the Darboux Stieltjes integral that“b[ PitF =F OF O =F) -F(-) =H.) = | ZandNow let {s,} be the sequence of step functions of Lemma 10.10.3 such that these stepfunctions converge uniformly to f on [c,d], say max, |f (x) — 5p (x)| < 1/n. Then| (ica Ficayn) an < | \Fica (f—sn)|du < “1 (c,d)and(Beat cain) aF| < f° Boa |f—Sn|dF < 4 (F (b)—F (a)) Also if sp isgiven by the formula of Lemma 10.10.3,| %easau= [YF (CL) %ep pend =¥ [FE1) Fig oqk=1 k=1=P) ([ch_4.2%)) = Yr) (F (t-)—F (1)mn mn pb b=f (di) (Fld —F (41)) = y | f(a) Zig and? = | snd F.k=1 k=174 aTherefore,b[cata [ %eatar|<|[ %eatdn— | Acasnan|+|f Seat | “sya +< “a (c d)) +—(F(b) —F(a))b b| 5,dF — / Kel fa|and since n is arbitrary, this shows [ fdu — i? fdF =0.08In particular, in the special case where F is continuous and f is continuous, [ iM fdF =J Zianfdu. Thus, if F (x) =x so the Darboux Stieltjes integral is the usual integral fromcalculus, [? f (t) dt = J 2iay|fdu where wu is the measure which comes from F (x) = xas described above. This measure is often denoted by m. Thus when f is continuous{of (at = J 2\ay)fdm and so there is no problem in writing {of @adt for either theLebesgue or the Riemann integral. Furthermore, when f is continuous, you can computethe Lebesgue integral by using the fundamental theorem of calculus because in this case,the two integrals are equal.