1549901369-Elements_of_Real_Analysis__Denlinger_

682 Appendix C • Answers & Hints for Selected Exercises

Let c > 0. Then 3 'P1 3 S..(f, 'P1) ?: f: f - ~ and 3 'P2 3 S_(g, 'P2) ?: f:g - ~.
Taking Q ='Pi U 'P2, - -

S..(f, Q) + S..(g, Q)?: f:f + f:g -c. (4)

By (3) and ( 4) together, f: (f + g) ?: f: f + f: g -c. Apply the forcing principle.

(b) Use f(x) = 1 if xis rationa l, -1 if xis irrational, and g = -f.

(c) Modify the proof of (a), using Mi(!), Mi(g), Mi(!+ g), upper sums,

and other appropriate changes to prove that f: (f + g) :::; f: f + f: g.

(d) If f,g are integrable on [a,b], then from (a), (c), and Thm. 7.2.7,

f: 1+ f: g = f:1 + f:g:::; f:u+g):::; f:u+g):::; f:1+ f:g = f: 1 + f: g.

3. Suppose f is integrable on [-a, a]. Let {'P~} be a sequence of tagged parti-
   tions of [O, a] 3 ll'Pn 11 -t 0. By Thm. 7 .3.6, R(f, 'P~) --+ f 0 a f.
   For each 'Pn = {O,xi ,x2,· ··Xmn} with tags {x:: i = 1, 2,··· ,mn}, let
   Qn = {-Xmn> · · · , -X2, -xi, 0, X1, X2, · · · Xmn} with tags
   {-x:: i = 1, 2, · · · , mn} U {x: : i = 1, 2, · · · , mn}· Then {Q~} is a sequence of
   tagged partitions of [O, a] 3 II Qn II --+ 0. By Thm. 7.3.6, R(f, Q~) --+ f~a f.
   fin mn fin
   But R(f, Q~) = I: f (-x:)6i + I: f (x:)6i = I:[! (-x:) + f (xnJ6i. If f
   i=l i=l i=i
   is even, then f (- x:) + f (x:) = 2j (x:), while if f is odd, f (-x:) + f (x:) = 0.
   b n


4. (a) In this case, fa f ?: S..(f, 'P) = L mi6i ?: 0.
   
   
   • i=l
     (b) For 'P = {O,x1,X2, .. · Xn,}, m:::; m i:::; M i:::; M, so f:f?: S..(f,'P) =
     n n b _ n n
     L mi6i ?: L m6i = m(b - a) and fa f :::; S(j, 'P) = L Mi6i :::; L M 6i =
     i=l i=i i=i i=l
     M(b - a).
     (c) By (b), -M(b - a):::; f: f :::; M(b - a); i.e., If: ti :::; M(b - a).
     (d) Vx E [a, b], g(x)-f(x) ?: 0, so by (a), f:(g-f) ?: O; i.e., f: g-f: f?: 0.
   
   
   
   


5. T is regulated and g is piecewise continuous, so T, g are integrable on [O, l].
   However, go Tis the Dirichlet function, which is not integrable on [O, 1].


6. Let f(x) = x if 0 < x:::; 1, 1 if x = 0. Note that 1/ f is unbounded on [O, l].


7. Let f be the constant f(x) = 1, and g denote the Dirichlet function.


8. By Ex. 7.2.13, J_g_f:::; rb j_g_h:::; rb J_g_9 rb and Jaf:::; rb Jah:::; rb Ja9· rb Smee. f,g are

integrable on [a, b], this means f: f :::; f:h:::; f: g and f: f:::; f:h:::; f: g. But

f: f = f: g, so his integrable on [a, b] and f: h = f: f = f : g.