9.3. FUNDAMENTAL DEFINITIONS AND PROPERTIES 199

Then letting r (ε)≡ ε

6(g(b)−g(a)+1)n ,

U ( f ,P)−L( f ,P)<n

∑i=1

( f (tk)+ r (ε)− ( f (sk)− r (ε)))(g(xk)−g(xk−1))

<n

∑i=1

2r (ε)(g(xk)−g(xk−1))+

∣∣∣∣∣∑P f (tk)(g(xk)−g(xk−1))− I

∣∣∣∣∣+

∣∣∣∣∣∑P f (sk)(g(xk)−g(xk−1))− I

∣∣∣∣∣< ε

3+

ε

3+

ε

3= ε

Example 9.3.11 Let g(x) = x be increasing and continuous and let f be decreasing. Thenfor P = a = x0 < · · ·< xn = b a partition of equally spaced points,

U ( f ,P)−L( f ,P) =n

∑k=1

f (xk−1)(xk− xk−1)−n

∑k=1

f (xk)(xk− xk−1)

=b−a

n

n

∑k=1

( f (xk−1)− f (xk)) =b−a

n( f (a)− f (b))

Thus for n large enough, U ( f ,P)−L( f ,P) < ε . It follows∫ b

a f dx exists. A similar argu-ment shows that if f is increasing, then the integral exists.

From this important result, one can obtain fairly easily the fact that various functionsof Riemann integrable functions are Riemann integrable.

Definition 9.3.12 Let k : D×D→ R satisfy∣∣k (a,b)− k(â, b̂)∣∣≤ K

(|a− â|+

∣∣b− b̂∣∣) .

Such a function is called Lipschitz and K is called the Lipschitz constant.

Theorem 9.3.13 Let f ,h ∈ R([a,b] ,g) for g an increasing function and supposef ([a,b]) ,g([a,b]) are both contained in D. Let k : D×D→ R be Lipschitz with constantK. Then k ( f ,h) ∈ R([a,b] ,g).

Proof: By assumption and Theorem 9.3.13, along with the observation that if P ⊆ Q,then U ( f ,P)≥U ( f ,Q) and L( f ,P)≤ L( f ,Q) , there exists a partition P such that

U ( f ,P)−L( f ,P)< ε, U (h,P)−L(h,P)< ε.

Now let

Mki ≡ sup{k ( f (x) ,h(x)) : x ∈ [xi−1,xi]} ,

mki ≡ inf{k ( f (x) ,h(x)) : x ∈ [xi−1,xi]}

Mhi ≡ sup{h(x) : x ∈ [xi−1,xi]} ,mh

i ≡ inf{h(x) : x ∈ [xi−1,xi]} ,

a similar convention holding for M fi ,m

fi . Here P = {x0,x1, · · · ,xn}. Then

U (k ( f ,h))−L(k ( f ,h)) =n

∑k=1

(Mk

i −mki

)(g(xi)−g(xi−1))

9.3. FUNDAMENTAL DEFINITIONS AND PROPERTIES 199Then letting r(e) = COE OESU(f,P)—L(F.P) <¥ (F(t) +r (€) —(F (se) —r(€))) (@ (xe) — ¢ (%e-1))i=|< oe (¢ (xz) —g (v1) +E—€ E€ €EN<y+y+ z=e |+ ILA Sk) — 8 (Xk-1)) — 37343Example 9.3.11 Let g(x) = x be increasing and continuous and let f be decreasing. Thenfor P=a=xo < +++ <%X, =ba partition of equally spaced points,U(f,P)—-L(f,P) = Sf (%K-1) Oe = Xe-1) — Df (Xk) (KR — K-1)five=1= hu Sf (xe-1) — f (xk)) =(f(a) —f (2)Thus for n large enough, U (f,P) —L(f,P) < €. It follows p? fdx exists. A similar argu-ment shows that if f is increasing, then the integral exists.From this important result, one can obtain fairly easily the fact that various functionsof Riemann integrable functions are Riemann integrable.Definition 9.3.12 Let k: Dx D— R satisfy( b|).Such a function is called Lipschitz and K is called the Lipschitz constant.|k(a,b) —k (4,6) | <Theorem 9.3.13 Lez f,h © R((a,b],g) for g an increasing function and supposef ({a,5]) ,g ([a,b]) are both contained in D. Let k : D x D + R be Lipschitz with constantK. Then k(f,h) € R(|a,5],g).Proof: By assumption and Theorem 9.3.13, along with the observation that if P C Q,then U(f,P) >U(f,Q) and L(f,P) < L(f,Q), there exists a partition P such thatU(f,P)-L(f,P) <€, U (h,P) —L(h,P) <€Now letMy = sup{k(f(x),h(x)) sx € [x-t2i]f,me = inf {k(f (x) ,h(x)) 2x € [xi-1,2i]}M? = sup {h(x) : x € [x_-1,xi]} ,m” = inf {h(x) : x € [x_-1,2i]},a similar convention holding for Ved ym! . Here P = {x0,%1,°+- ,Xn}. ThenU (RFI) — ERI) = Ye (ME mb) (slo) 26-0)