Kenneth Kuttler

7.3. THE RIEMANN DARBOUX INTEGRAL∗ 181

Proof: By assumption, there is a partition P1 of [a,b] and one for [b,c] P2 such thatU ( f ,P1)−L( f ,P1) <

2 , U ( f ,P2)−L( f ,P2) <ε

2 . Thus if P = P1 ∪P2, then U ( f ,P)−L( f ,P) < ε and so there is no space between Ī and I . Thus the function is integrable on[a,c] and also, using the partitions just described,

−ε < L( f ,P)−U ( f ,P) = L( f ,P1)+L( f ,P2)−U ( f ,P)

≤∫ b

af dx+

∫ c

bf dx−

∫ c

af dx ≤U ( f ,P1)+U ( f ,P2)−L( f ,P) =U ( f ,P)−L( f ,P)< ε

Thus∫ b

a f dx+∫ c

b f dx−∫ c

a f dx ∈ [−ε,ε] and ε is arbitrary so∫ b

a f dx+∫ c

b f dx−∫ c

a f dx = 0.

Definition 7.3.11 Define∫ a

b f dx ≡−∫ b

a f dx.

Theorem 7.3.12 Let f be integrable on [min(p,q,r) ,max(p,q,r)]. Then∫ q

p f dx+∫ rq f dx =

∫ rp f dx.

Proof: The case where a < b < c was just done. Suppose a < c < b. Then from whatwas just done, ∫ c

af dx+

∫ b

cf dx =

∫ b

af dx

and so ∫ c

af dx =

∫ b

af dx−

∫ b

cf dx =

∫ b

af dx+

∫ c

bf dx

Other cases are similar.

Theorem 7.3.13 If f is continuous on [a,b] , then f is integrable on [a,b]. Also, iff is integrable on [a,b] and is changed at finitely many points, the resulting function is alsointegrable and has the same integral as f .

Proof: I will show there is no gap between the upper and lower integrals. Let ε > 0 begiven. Let n be so large that if |x− y| < 2δ , then | f (x)− f (y)| < ε/((b−a)+1). Sucha δ exists because f is uniformly continuous due to the fact that [a,b] is compact. SeeTheorem 4.7.2. Now let n be so large that b−a

n < δ . Then let P = {x0,x1, ...,xn} be auniform partition, each xk − xk−1 =

b−an . Then

U ( f ,P)−L( f ,P) =n

∑k=1

f (zk)(xk − xk−1)−n

∑k=1

f (wk)(xk − xk−1)

where f (zk) is the maximum value of f on [xk−1,xk] and f (wk) the minimum value of fon [xk−1,xk] . Thus |wk − zk|< 2δ and so the above is no larger than

∑k=1

(b−a)+1

(b−a

)= (b−a)

(b−a)+1< ε

Thus, from Proposition 7.3.7, f is integrable.

7.3. THE RIEMANN DARBOUX INTEGRAL* 181Proof: By assumption, there is a partition P; of [a,b] and one for [b,c] P) such thatU(f,Pi)-L(f,P1) < §,U(f,P) -L(f,P2) < §. Thus if P= P, UP), then U(f,P) —L(f,P) < € and so there is no space between J and J . Thus the function is integrable on[a,c] and also, using the partitions just described,—€<L(f,P)—U(F,P) =LF, PI +L (FP) —U (FP)b C c< [fact [ fax [ fax <U(f,P)+U(f.P) LRP) =U (FP) -LUP) <€a b aThus |? fdx-+ ff fdx— < fdx € [—€, €] and € is arbitrary so [? fdx+ ff fdx—f< fdx =0.|Definition 7.3.11 Define [f fax =— J? fax.Theorem 7.3.12 Let f be integrable on [min(p,q,r) ,max(p,q,r)]. Then Jp fax+J fdx = fi fax.Proof: The case where a < b < c was just done. Suppose a < c < b. Then from whatwas just done,Cc b b[ tare | fax= | faxa c a[tac= [Ptae— [pac= [pace [pasOther cases are similar. Jjand soTheorem 7.3.13 If f is continuous on |a,b|, then f is integrable on [a,b]. Also, iff is integrable on |a, b| and is changed at finitely many points, the resulting function is alsointegrable and has the same integral as f.Proof: I will show there is no gap between the upper and lower integrals. Let € > 0 begiven. Let n be so large that if |x—y| < 26, then |f (x) — f(y)| < €/((b—a) +1). Sucha 6 exists because f is uniformly continuous due to the fact that [a,b] is compact. SeeTheorem 4.7.2. Now let n be so large that pa < 6. Then let P = {x0,x1,...,Xn} be auniform partition, each x, —x,_) = pa Thenn nU(f,P)—LF,P) = YF (ze) (xe —xn-1) — YF (we) (Xe — e-1)k=1k=1where f (z;) is the maximum value of f on [xg_1,xx] and f (w,) the minimum value of fon [xx_1,xg]. Thus |wx — zz] < 26 and so the above is no larger thanb-—a €a €XL basi ( n )=0-4 (b-a)+1 ~*Thus, from Proposition 7.3.7, f is integrable.