Bartle - Introduction To Real Analysis - Chapter 7 Solutions [PDF]

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Bartle - Introduction to Real Analysis - Chapter 7 Solutions Section 7.1 Problem 7.1-3. Show that f : [a, b] → R is Riemann integrable on [a, b] if and only if there exists L ∈ R such that for every  > 0 there exists δ > 0 such that if P˙ is any tagged partition with the norm ||P˙ || ≤ δ , then |S(f ; P˙ ) − L| ≤ . Solution: Suppose there is an L ∈ R where, for /2 > 0, there is a δ 0 > 0 such that if ||P˙ || ≤ δ 0 , then |S(f ; P˙ ) − L| ≤ /2. Let δ = δ 0 . Clearly, then, if ||P˙ || < δ , then |S(f ; P˙ ) − L| < /2 < . As a result, f ∈ R[a, b]. Now suppose that f ∈ R[a, b]. Then for  > 0, there is a δ 0 > 0 such that if ||P˙ || < δ 0 , then |S(f ; P˙ ) − L| < . Let δ = δ 0 /2. Therefore, if ||P˙ || ≤ δ , then |S(f ; P˙ ) − L| < , from which it follows that |S(f ; P˙ ) − L| ≤ . Problem 7.1-4. Let P˙ be a tagged partition of [0, 3]. (a) Show that the union U1 of all subintervals in P˙ with tags in [0, 1] satisfies [0, 1 − ||P˙ ||] ⊆ U1 ⊆ [0, 1 + ||P˙ ||]. (b) Show that the union U2 of all subintervals in P˙ with tags in [1, 2] satisfies [1 + ||P˙ ||, 2 − ||P˙ ||] ⊆ U1 ⊆ [1 − ||P˙ ||, 2 + ||P˙ ||]. S Solution: Part (a) Suppose P˙ = {(Ii = [xi−1 , xi ], ti ). Let T1 = {tk : tk ∈ [0, 1]} and U1 = tk ∈T1 Ik . Because P˙ is a partition of [0, 1], the maximum length of a subinterval must be less than or equal to 1. Therefore, there is at least one tag in [0, 1], so T1 is not empty. First we will show [0, 1 − ||P˙ ||] ⊆ U1 . Let tm be the greatest tag in T1 . By definition, tm is the tag of subinterval [xm−1 , xm ]; further, tm ≤ 1. If we assume that xm ≤ 1 − ||P˙ ||, there is a subinterval [xm , xm+1 ] that does not have a tag less than or equal to one. Because xm+1 − xm ≤ ||P˙ ||, it follows that xm+1 ≤ xm + ||P˙ || ≤ 1, which means that tm+1 ≤ 1 and, consequently, tm+1 ∈ T1 . But this contradicts that tm is the greatest tag in T1 . Therefore, xm > 1 − ||P˙ ||. In addition, because xm−1 ≤ tm ≤ 1, it follows that xm−1 ≤ 1. We further know that x0 = 0 and t0 ≥ 0 because no subinterval of P˙ contains points less than zero. Because tk ≤ tk+1 , we infer that tk ∈ [0, 1] for 1 ≤ k ≤ m. Therefore, Ik ⊆ U1 for 1 ≤ k ≤ m. Let v ∈ [0, 1 − ||P˙ ||]. Because x0 = 0 and xm > 1 − ||P˙ ||, it follows that x0 ≤ v ≤ xm . Consequently, v ∈ Ik for some 1 ≤ k ≤ m. Because tk ∈ [0, 1], we have Ik ⊆ U1 , from which it follows that v ∈ U1 . Accordingly, [0, 1 − ||P˙ ||] ⊆ U1 . Now we will show U1 ⊆ [0, 1 + ||P˙ ||]. Let w ∈ U1 . Clearly, w ∈ Ik for some 1 ≤ k ≤ m. Therefore, 0 ≤ w ≤ xk ≤ xm . Because xm ≤ xm−1 + ||P˙ || and xm−1 ≤ 1, it follows that w ≤ 1 + ||P˙ ||. It follows that w ∈ [0, 1 + ||P˙ ||]. Accordingly, U1 ⊆ [0, 1 + ||P˙ ||. S Part (b) Suppose T2 = {tk : tk ∈ [1, 2]} and U2 = tk ∈T2 Ik . As shown part (a), T2 is not empty. There is a tm that is the greatest tag in T2 , which is also the tag of [xm−1 , xm ]. Using a similar argument to that in part (a), xm > 2 − ||P˙ || and xm−1 < 2. Because xm − xm−1 ≤ ||P˙ ||, we infer that xm ≤ xm−1 + ||P˙ || ≤ 2 + ||P˙ ||. Let tp be the smallest tag in T2 , which is the tag of [xp−1 , xp ]. Because tp ≤ xp , we infer that xp ≥ 1. Similar to the argument above, 1 − ||P˙ || ≤ xp−1 ≤ 1. Since xn < tn < xn+1 , it must be that tk ∈ T2 for p ≤ k ≤ m, and therefore Ik ⊆ U2 . Let v ∈ [1 + ||P˙ ||, 2 − ||P˙ ||]. Because xp−1 ≥ 1 and xm ≤ 2 − ||P˙ ||, it follows that w ∈ Ik for some k such that p ≤ k ≤ m. Since Ik ⊆ U2 , we infer that w ∈ U2 , from which it follows that [1 + ||P˙ ||, 2 − ||P˙ ||] ⊆ U2 . Now let w ∈ U2 . As shown above, 1 − ||P˙ || ≤ xp−1 ≤ w ≤ xm ≤ 2 + ||P˙ ||. Accordingly, w ∈ [1 − ||P˙ ||, 2 + ||P˙ ||], from which it follows that U2 ⊆ [1 − ||P˙ ||, 2 + ||P˙ ||]. Problem 7.1-9. If f ∈ R[a, b] and if (P˙n ) is any sequence of tagged partitions of [a, b] such that ||P˙n || → 0, prove that Rb f = limn S(f ; P˙n ). a Solution: Since f is Riemann integrable on [a, b] by hypothesis, there is an L ∈ R such that for any  > 0, there is a δ > 0 such that if ||P˙ || < δ , then |S(f ; P˙ ) − L| < . If lim ||P˙n || = 0, there is a K(δ ) > 0 such that for n ≥ K(δ ), we have | ||P˙n || − 0 | = ||P˙n || < δ . As a result, for n ≥ K(δ ), the Riemann integrability of f requires that |S(f ; P˙n ) − L| < , Rb from which it follows that lim S(f ; P˙n ) = L = a f . Problem 7.1-10. Let g(x) := 0 if x ∈ [0, 1] is rational and g(x) := 1/x if x ∈ [0, 1] is irrational. Explain why g ∈ / R[0, 1]. However, show that there exists a sequence (P˙n ) of tagged partitions of [a, b] such that ||P˙n || → 0 and limn S(g; P˙n ) exists.



Solution: Assume g is bounded on [0, 1] by M ≥ 1. Let x0 = q/M where q ∈ (0, 1) and is irrational. Clearly, x0 ∈ (0, 1) and is irrational. We therefore have g(x0 ) = M/q > M , which contradicts that g is bounded. Because g is unbounded on [0, 1], under Theorem 7.1.6, g is not Riemann integrable on [0, 1]. Now let (P˙n ) be a sequence where P˙n is a tagged partition of a given [a, b] ⊆ [0, 1] with n equally-spaced subintervals, each tagged by a rational number (which is possible because of the density of rational numbers under Theorem Pn 2.4.8). For each subinterval, ||P˙n || = (b − a)/n, from which it follows that lim ||P˙n || = 0. In addition, S(g; P˙n ) = i=1 f (ti )(xi − Pn xi−1 ) = i=1 0 · (xi − xi−1 ) = 0. Accordingly, lim S(g; P˙n ) = 0, as we sought to show. Notwithstanding this result, of course, g ∈ / R[0, 1] because partitions of [a, b] with irrational tags will not meet the definition of Riemann integrability. Problem 7.1-11. Suppose f is bounded on [a, b] and that there exists two sequences of tagged partitions of [a, b] such that ||P˙n || → 0 and ||Q˙ n || → 0, but such that limn S(f ; P˙n ) 6= limn S(f ; Q˙ n ). Show that f is not in R[a, b]. Rb Solution: If f ∈ R[a, b], then there is an L ∈ R such that L = a f and L is unique. From Exercise 7.1-9, L = lim S(f ; P˙n ) and L = lim S(f ; P˙n ). But this is impossible because, by hypothesis, lim S(f ; P˙n ) 6= lim S(f ; Q˙ n ). Therefore, f ∈ / R[a, b]. Problem 7.1-13. Suppose c ≤ d are points in [a, b]. If φ : [a, b] → R satisfies φ(x) = α > 0 for x ∈ [c, d] and φ(x) = 0 Rb elsewhere in [a, b], prove that φ ∈ R]a, b] and that a φ = α(d − c). Solution: Suppose P˙ is any tagged partition of [a, b]. Let P˙1 be the subset of P˙ with tags in [a, c), let P˙2 be the subset of P˙ with tags in [c, d], and let P˙3 be the subset of P˙ with tags in (d, b]. It follows that S(φ; P˙ ) = S(φ; P˙1 )+S(φ; P˙2 )+S(φ; P˙3 ). Since φ(ti ) = 0 for all tags in P˙1 and P˙3 , the Riemann sums of those partitions are zero. Therefore, S(φ; P˙ ) = S(φ; P˙2 ). To prove that the Riemann integral of φ on [a, b] exists and is equal to α(d − c), we must show that for some δ > 0, if ||P˙ || < δ , then |S(φ; P˙2 ) − α(d − c)| < . We will now find an upper and lower bound of S(φ; P˙2 ), from which we can find an upper bound on δ . Let δ > 0 be a value to be determined for a given  > 0. We may assume without loss of generality that δ < d − c. This restriction will ensure that there is at least one subinterval in P˙2 . If ||P˙ || ≥ δ , the condition of the Riemann integral is trivially true. We will now assume that ||P˙ || < δ . Let U2 be the union of the subintervals in P˙2 . Suppose tm is the smallest tag in P˙2 . It must be that xm−1 ≥ c − ||P˙ || > c − δ . If it were to the contrary (that xm−1 ≤ c − δ ), then xm < xm−1 + δ < c ≤ c, from which it would follow that tm < c. This contradicts that tm is the smallest tag of P˙2 . It also must be that xm−1 ≤ c + ||P˙ || < c + δ . If the contrary were true (that xm−1 ≥ c + δ ), then if m − 1 > 1, then xm−2 > xm−1 − δ ≥ c, so tm−1 ≥ c, which contradicts that tm is the smallest tag in P˙1 . On the other hand, if m − 1 = 1, then xm−1 = a, in which case it must be that xm−1 ≤ c < c+ . Now suppose tp is the largest tag in P˙2 . Following similar arguments to those above, we find that xp < d + δ and xp > d − δ  . Based on these results, we infer that: [c + δ , d − δ ] ⊆ U2 ⊆ [c − δ , d + δ ]. Since each tag ti in P˙2 is in [d, c], we know that φ(ti ) = α. Because the total length of the subintervals spanned by P˙2 are between d − c − 2δ and d − c + 2δ , the Riemann sum of P˙2 is bounded by: α(d − c − 2δ ) ≤ S(φ, P˙2 ) ≤ α(d − c + 2δ ). Simple manipulation of the expression results in: |S(φ, P˙2 ) − α(d − c)| = |S(φ, P˙ ) − α(d − c)| ≤ 2αδ . By our assumption above, δ must be less than d − c. If we let δ < inf{d − c, /(4α)}, then S(φ, P˙ ) − α(d − c)| < Rb 2α/(4α) < . Because  and P˙n are arbitrary, it follows that a φ = α(d − c). Problem 7.1-14. Let 0 ≤ a ≤ b, let Q(c) := x2 for x ∈ [a, b] and let P := {[xi−1 , xi ]} be a partition of [a, b]. For each i, let qi be the positive square root of 1/3(x2i + xi xi−1 + x2i−1 ). (a) Show that qi satisfies 0 ≤ xi−1 ≤ qi ≤ xi . (b) Show that Q(qi )(xi − xi−1 = 1/3(x3i − x3i−1 ).



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˙ = 1/3(b3 − a3 ). (c) If Q˙ is the tagged partition with the same subintervals as P and the tags qi , show that S(Q; Q) (d) Use the argument in Example 7.1.4(c) to show that Q ∈ R[a, b] and Z



b



b



Z



x2 dx =



Q= a



a



1 3 (b − a3 ). 3



Solution: Part (a) We infer from the hypothesis that xi ≥ 0 for 1 ≤ i ≤ n. Because xi > xi−1 , we have: r r 1 2 1 2 q1 = (x + xi xi−1 + x2i−1 ) < (x + x2i + x2i ) = xi . 3 i 3 i Similarly, we have: r 1 2 q1 > + x2i−1 + xi−1 i2 ) = xi−1 ≥ 0. (x 3 i−1 As a result, 0 ≤ xi−1 ≤ qi ≤ xi . Because qi ∈ [xi−1 , xi ], we may use qi as tags of the subintervals in P . Part(c) We have: 1 2 (x + xi xi−1 + x2i−1 )(xi − xi−1 ) = x3i − x3i−1 . 3 i Part(c) Based on the result in part (b), the Riemann sum of Q˙ is: Q(qi )(xi − xi−1 ) =



˙ = S(Q, Q)



n X



n



Q(qi )(xi − xi−1 ) =



i=1



1 1X 1 Q(qi )(x3i − x3i−1 ) = (xn − x0 ) = (b3 − a3 ), 3 i=1 3 3



where we have relied upon the fact that the terms of the sum telescope. Part(d) Suppose P˙ be an arbitrary tagged partition of [a, b] with tags ti and ||P˙ || < δ for a δ > 0 to be determined. Let Q˙ be a tagged partition of [a, b] with the same subintervals as P˙ but with tags qi as specified in part (a). Clearly, ˙ < δ and |ti − qi | < δ. ||Q|| The difference of the two Riemann sums of the partitions is: n n X X ˙ = Q(qi )(xi − xi−1 ) Q(ti )(xi − xi−1 ) − |S(Q; P˙ ) − S(Q; Q)| i=1 i=1 n n X X = (Q(ti ) − Q(qi ))(xi − xi−1 ) ≤ |(Q(ti ) − Q(qi )|(xi − xi−1 ) i=1



=



i=1



n X



|t2i − qi2 |(xi − xi−1 ) =



n X



i=1




0 be given. Suppose P˙ = {([xi−1 , xi ], ti )}ni=1 is an arbitrary tagged partition of [a, b] where ||P˙ || < δ . Rb Because f is Riemann integrable, we have |S(f ; P˙ ) − L| <  where L = a f . We note that: S(f ; P˙ ) =



n X



f (ti )(xi − xi−1 ).



i=1



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The lengths of [a, b] and [a + c, b + c] are equal. Let Q˙ = {([xi−1 + c, xi + c], ti + c)}ni=1 be tagged partition of ˙ = ||P˙ ||. Its Riemann sum is: [a + c, b + c]. Clearly, ||Q||



˙ = S(g; Q)



n X



g(ti + c)[(xi + c) − (xi−1 + c)] =



i=1



n X



f (ti + c − c)(xi − xi−1 ) =



i=1



n X



f (ti )(xi − xi−1 ) = S(f ; P˙ ).



i=1



˙ − L| = |S(f ; P˙ ) − L| < . It follows that g ∈ R[a + c, b + c] and As a result, |S(g; Q)



R b+c a+c



g=



Rb a



f.



Section 7.2 Problem 7.2-5. If J is any subinterval of [a, b] and if φJ (x) := 1 for x ∈ J and φJ (x) := 0 elsewhere on [a, b], we say that φJ is an elementary step function on [a, b]. Show that every step function is a linear combination of elementary step functions. Solution: Let f : [a, b] → C be a step function where C ⊆ R. By the definition of a step function, f has n ∈ N non-overlapping subintervals where f is constant on each subinterval. Let the subintervals be I1 , I2 , . . . , In , and let φk be an elementary step function where φk (x) = 1 on Ik and φk (x) = 0 elsewhere on [a, b]. We will prove by mathematical induction that f is equal to a linear combination of n elementary step functions. For f with n = 1 subintervals, f (x) = α1 for x ∈ I1 = [a, b]. Therefore, f (x) = α1 φ1 (x), which is a linear combination of a single step function. Assume that any step function Snwith n subintervals may be written as a linear combination of step functions. For f with n + 1 subintervals, let Ia = k=1 In and let f 0 : Ia → C be a restriction of f where f 0 (x) = f (x). By the inductive hypothesis, we have: f 0 (x) =



n X



αk φk (x).



k=1



Let f 00 : [a, b] → C where f 00 (x) = 0 for x ∈ In+1 and f 00 (x) = f 0 (x) for x ∈ Ia . We then have f (x) = αn+1 on In+1 . Because αn+1 φn+1 (x) = αn+1 on In+1 and is zero on Ia , we may express f as: n+1 X



f (x) = f 00 (x) + αn+1 φn+1 (x) =



αk φk (x),



k=1



which is a linear combination of n + 1 elementary step functions. This result holds for any step function with n ∈ N subintervals, which includes all possible step functions. Accordingly, every step function is a linear combination of elementary step functions. Problem 7.2-8. Suppose that f is continuous on [a, b], that f (x) ≥ 0 for all x ∈ [a, b] and that for all x ∈ [a, b].



Rb a



f = 0. Prove that f (x) = 0



Solution: Suppose f (c) > 0 for some c ∈ [a, b]. Because f is continuous on [a, b], we have limx→c f (x) = f (c) > 0. Consequently, thereR is a δ > 0 such that for d = c − δ/2 and e = c + δ/2, if x ∈ V = [d, e] ∩ [a, b], then f (x) > 0. By e Theorem 7.1.5(c), d f > 0, from which it follows from the Additivity Theorem that: Z



b



d



Z f=



a



e



Z



a



f+ e



Because f (x) ≥ 0 for x ∈ [a, d] ∪ [e, b], it must be that b



which contradicts that



Rb a



Rd a



b



Z f+



a



f ≥ 0 and



b



Z f>



a



Z f>



d



Z



b



Z



f+



b



f. e



Rb e



f ≥ 0, in which case:



b



Z



f ≥ 0,



f+ e



a



f = 0. Therefore, f (x) = 0 for all x ∈ [a, b].



Problem 7.2-10. If f and g are continuous on [a, b] and



Rb a



f=



Rb a



Page 4



g, prove that there exists c ∈ [a, b] such that f (c) = g(c).



Solution: Because



Rb a



f=



Rb



g, we have:



a



b



Z



b



Z f−



b



Z



(f − g) = 0.



g= a



a



a



Assume f (x) 6= g(x) for all x ∈ [a, b]. We may assume without loss of generality that there is a α ∈ [a, b] where f (α) > g(α). By Bolzano’s Intermediate Value Theorem, it then cannot be that there is a β ∈ [a, b] where f (β)−g(β) < 0. If it were otherwise, then because f (β) − g(β) < 0 < f (α) − g(α), there would be an h ∈ [a, b] between α and β such that f (h) − g(h) = 0, which is not possible by our assumption that f and g never have the same value. Therefore, f (x) > g(x) for all x ∈ [a, b]. Rb As a result, for all x ∈ [a, b], it follows that f (x) − g(x) > 0. Therefore, by Theorem 7.1.5(c), a (f − g) > 0, which Rb contradicts the hypothesis that a (f − g) = 0. As a result, there must be an x ∈ [a, b] where f (x) = g(x). Problem 7.2-11. If f is bounded by M on [a, b] and if the restriction of f to every interval [c, b] where c ∈ [a, b] is Riemann Rb Rb integrable, show that f ∈ R[a, b] and the a f → a f as c → a+ . Solution: By hypothesis, −M ≤ f (x) ≤ M for all x ∈ [a, b]. For c ∈ (a, b), let αc (x) = −M and ωc (x) = M for x ∈ [a, c) and αc (x) = ωc (x) = f (x) for x ∈ [c, b]. On [a, c], the functions αc and ωc are Riemann integrable because f if Riemann integrable. On [a, c), the functions αc and ωc are scaled step functions and are therefore Riemann integrable by Lemma 7.2.5. We can then apply the Squeeze Theorem. If x ∈ [a, b], then αc (x) ≤ f (x) ≤ ωc (x). We also have: Z



b



b



Z (ωc − αc ) =



Z ωc −



a



Z



Rb a



a



b



b



a



f = limc→a+



Rb c



a



c→a



b



Z



f, c



b



Z



Z αc = M (a − c) +



a



Taking the limit as c → a+ , we have: " Z b # Z lim+ M (a − c) + f = lim+ Rb



αc = 2M (c − a).



a



a



a



It follows that



ωc −



ωc = M (c − a) +



f≥



c



c



Z



b



Z



a



Z



c



(ωc − αc ) = 2M (/4M ) = /2 < . It follows from the Squeeze Theorem



f≤ and:



Z αc =



a



For a given  > 0, if c = a + /4M , then that f ∈ R[a, b]. We can write the integral of f as:



c→a



b



c



b



b



f. c



Z f≤ a



"



b



f ≤ lim+ M (c − a) + c→a



Z c



b



#



Z



f = lim+ c→a



b



f. c



f.



Problem 7.2-15. If f is bounded and there is a finite set E such that f is continuous on every point of [a, b]\E, show that f ∈ R[a, b]. Solution: We can establish this result by the Squeeze Theorem. Suppose that there are n ∈ N points of discontinuity. Let  > 0 be given. Let P˙ be a tagged partition of [a, b] with N subintervals where kP˙ k < δ for some δ > 0 to be determined. Because f is bounded on [a, b], on any subinterval Ik in P˙ has a minimum at f (uk ) and a maximum f (vk ) at some uk , vk ∈ Ik . Let α (x) = uk and ω (x) = vk for x ∈ Ik (in other words, for each subinterval in P˙ , the functions α and ω are each step functions with the minimum and maximum values of f on those subintervals, respectively). Clearly, α (x) ≤ f (x) ≤ ω (x) for all x ∈ [a, b]. In addition, ω − α ≥ 0 for all x ∈ [a, b]. Now let P˙1 be a tagged partition containing the subintervals and tags of P˙ where f is continuous on every point of each subinterval. Let P˙2 be the tagged partition containing the subintervals and tags of P˙ where f is discontinuous on at least one point of each subinterval. We see that for any function g defined on [a, b], it must be that S(g; P˙ ) = S(g; P˙1 ) + S(g; P˙3 ).



Page 5



Let C be the set of the indices of the tags in P˙1 and D be the set of the indices of the tags in P˙2 . Because α and ω are step functions, we have: Z



b



0≤



(ω − α ) = a



=



X



N X



(ω (ti ) − α (ti ))(xi − xi−1 )



i=1



(ω (ti ) − α (ti ))(xi − xi−1 ) +



k∈C



X



(ω (ti ) − α (ti ))(xi − xi−1 )



k∈D



Note that the summation on the left is over the subintervals where f is always continuous and the summation on the right is over the subintervals where f is somewhere discontinuous. We will now establish a value for δ. Because f is uniformly continuous on the subintervals of P˙1 , there is a δ 0 > 0 where if |vi − ui | < δ 0 , then |f (vi ) − f (ui )| < /2(b − a). Now observe that there are at most 2N subintervals in P˙2 (because any discontinuity point may result in two subintervals in P˙2 if it is an endpoint in both). Accordingly, if we make δ = inf{δ 0 , /4M }, the maximum length of all the subintervals of P˙2 is /2. We will now address each summation above in turn. For the summation over P˙1 : X X  (xk − xk−1 ) ≤ /2. (ω (ti ) − α (ti ))(xi − xi−1 ) < 2(b − a) k∈C



k∈C



For the summation over P˙2 : X



(ω (ti ) − α (ti ))(xi − xi−1 ) ≤ 2M



k∈D



X



(xi − xi−1 ) ≤ /2.



k∈D



Accordingly: Z 0≤



b



(ω − α ) < /2 + /2 = . a



By the Squeeze Theorem, f ∈ R[a, b]. Problem 7.2-16. If f is continuous on [a, b], a < b, show that there exists c ∈ [a, b] such that we have This result is sometimes called the Mean Value Theorem for Integrals.



Rb a



f = f (c)(b − a).



Solution: Since f is continuous on [a, b], it is Riemann integrable on that interval. It also has an absolute minimum M∗ = f (x1 ) and an absolute maximum M ∗ = f (x2 ) for some x1 , x2 ∈ [a, b] and M∗ , M ∗ ∈ R. It follows that M∗ ≤ f (x) ≤ M ∗ for x ∈ [a, b]. By Theorem 7.1.5(c): b



Z



f ≤ M ∗ (b − a).



M∗ (b − a) ≤ a



We can define two non-negative quantities α = ( follows that:



Rb a



f )/(b − a) − M∗ ) and β = (M ∗ − Z



(M∗ + α)(b − a) =



Rb a



f )/(b − a), from which it



b



f = (M ∗ − β)(b − a).



a



Because α ≥ 0 and β ≥ 0, we have M∗ ≤ M∗ + α ≤ M ∗ . By Bolzano’s Intermediate Value Theorem, there is a Rb c ∈ [a, b] between x0 and x1 where f (c) = M∗ + α. Accordingly, a f = f (c)(b − a). Problem 7.2-19. Suppose that a > 0 and that f ∈ R[−a, a]. Ra Ra (a) If f is even (that is, if f (−x) = f (x) for all x ∈ [0, a]), show that −a f = 2 0 f . Ra (b) If f is odd (that is, if f (−x) = −f (x) for all x ∈ [0, a]), show that −a f = 0. Ra R0 Rf Solution: Part (a) Since f ∈ R[−a, a], by the Additive Theorem −a f = −a f + 0 (and f is Riemann integrable on the subintervals on the right-hand side). Suppose  > 0 is given and P˙ = {[xi−1 , xi ], ti }ni=1 is an arbitrary tagged partition of [−a, 0]. Let P˙ 0 = {[x0i−1 , x0i ], t0i }ni=1 be a tagged partition where x0n−i = −xi and t0n−i+1 = −ti . Because xi−1 ≤ ti ≤ xi , it follows that x0j−1 ≤ t0j ≤ x0j . Moreover, x00 = −xn = 0 and x0n = −x0 = a. The partition P˙ 0 is therefore on [0, a].



Page 6



Given that f is an even function, we can show: S(f ; P˙ ) =



n X



f (ti )(xi − xi−1 ) =



i=1



=



n X



n X



f (−t0n−i+1 )[(−x0n−i ) − (−x0n−i+1 )]



i=1



f (−t0j )(x0j



0



− xj−1 ) =



j=1



n X



f (t0j )(x0j − x0j−1 ) = S(f ; P˙ 0 ),



j=1



where we replaced the summation index i with j = n − i + 1. Ra Because 0 f is Riemann integrable, there is a δ > 0 such that if kP˙ 0 k < δ , then: Z a Z a = S(f ; P˙ ) − S(f ; P˙ 0 ) − f < . f 0



0



R0 Ra Ra R0 Ra Ra Because kP˙ k = kP˙ 0 k and P˙ is arbitrary, −a f = 0 f . Therefore, −a f = −a f + 0 f = 2 0 f . Part (b) The argument is pretty much identical to that in part (a) but relies on f (x) = −f (−x) for x ∈ [−a, a] to R0 Ra Ra conclude that S(f ; P˙ ) = −S(f ; P˙ 0 ). It then follows that −a f = − 0 f and, consequently, −a f = 0 by the Additivity Theorem.



Section 7.3 Problem 7.3-1. Extend the proof of the Fundamental Theorem 7.3.1 to the case of an arbitrary finite set E. Solution: Suppose E is an arbitrary finite set where F 0 (x) 6= f (x) if x ∈ E. Let  > 0 be given. Let n be the number of elements of E. We may index the elements of E as ei in increasing order starting at 1 and let e0 = a and en+1 = b. This results in a list of points in [a, b] where e0 = a < e1 < e2 < · · · < en < en+1 = b. Let Ji = [ei−1 , ei ] for 1 ≤ i ≤ n + 1. Because E is finite, it cannot cannot contain any subintervals of [a, b] (otherwise, Sn+1 any such subinterval would give rise to an infinite number of points contained in E). As a result, i=1 Ji = [a, b]. Therefore: b



Z



f= a



n+1 X Z ei i=1



f.



ei−1



From Theorem 7.3.1, for Ji where 1 ≤ i ≤ n + 1, it follows that F (ei ) − F (ei−1 ) = intervals, we have: Z



b



f= a



n+1 X



R ei ei−1



f . Summing over all the



(F (ei ) − F (ei−1 )) = F (b) − F (a),



i=1



thus establishing the Fundamental Theorem for an arbitrary finite E. Problem 7.3-6. If f ∈ R[a, b] and if c ∈ [a, b], the function defined by Fc (z) := integral of f with basepoint c. Find a relation between Fa and Fc . Solution: Based on the definition above, Fa (z) = Rc Fa (z) = Fc (z) + a f .



Rz a



Rz c



f for z ∈ [a, b] is called the indefinite



f . By the Additivity Theorem,



Rc a



f+



Rz c



f =



Rz a



f . Therefore,



Problem 7.3-14. Show there does not exist a continuously differentiable function f on [0, 2] such that f (0) = −1, (2) = 4, and f 0 (x) ≤ 2 for 0 ≤ x ≤ 2 (Apply the Fundamental Theorem). Solution: Assume to the contrary that such a function f exists. Because f is continuously differentiable on [0, 2], its derivative f 0 is continuous on that interval and therefore f 0 ∈ R[0, 2]. By the Fundamental Theorem of Calculus, f (2) − R2 R2 f (0) = 0 f 0 . However, since f 0 (x) ≤ 2 for x ∈ [0, 2], it follows from Theorem 7.1.5(c) that 0 f 0 ≤ 2(2 − 0) = 4. Consequently:



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Z



2



f (2) − f (0) = 5 =



f 0 ≤ 4,



0



which is a contradiction. Therefore, such a function f cannot exist. Problem 7.3-15. If f : R → R is continuous and c > 0, defined g : R → R by g(x) := differentiable on R and find g 0 (x).



R x+c x−c



f (t)dt. Show that f is



Rx Solution: Because f is continuous on R, the indefinite integral defined by h(x) = 0 f for x ∈ R is differentiable on R and h0 (x) = f (x). By the Additivity Theorem: Z x+c Z x+c Z x−c g(x) = f= f− f = h(x + c) − h(x − c). x−c



0



0



Since h is differentiable on R, it follows that g is differentiable on R, and g 0 (x) = h0 (x+c)−h0 (x−c) = f (x+c)−f (x−c). Problem 7.3-16. If f :∈ [0, 1] → R is continuous and



Rx



f=



0



R1 x



f for all x ∈ [0, 1], show that f (x) = 0 for all x ∈ [0, 1].



Solution: R xLet x ∈ [0, 1] be given. Since f is continuous on [0, 1], by the Fundamental Theorem of Calculus there is an F (x) = 0 f where F 0 (x) = f (x) for x ∈ [0, 1]. By the Additivity Theorem: Z



x



Z



1



f+ 0



As a result, F (x) = 1/2



R1 0



Z f = 2F (x) =



x



1



f. 0



f = k where k ∈ R is a constant. It follows that F 0 (x) = f (x) = 0 for x ∈ [0, 1].



Problem 7.3-20. (a) If Z1 and Z2 are null sets, show that Z1 ∪ Z S2∞is a null set. (b) More generally, if Zn is a null set for each n ∈ N, show that n=1 Zn is a null set. Solution: Part(a) We will first solve part (b). The solution there will establish part (a) as special case. S∞ Part(b) For n ∈ N, there is a countable setSCn = {Jnk = (ank , bnk )}∞ k=1 such that Zn ⊆ k=1 Jnk P∞ Let  > 0 be given. ∞ and k=1 (bnk − S ank ) < /2n . For convenience, we define J = J , which is the union of all intervals of C nk n. n k=1 S∞ ∞ Suppose Z = Z . Now let C = C , which is itself a countable set by Theorem 1.3.12. We can then define n n n=1 n=1 S∞ J = i=1 Jn , which the union of all intervals in C. If z ∈ Z, then z ∈ Zn for some n ∈ N. Clearly, then, z ∈ Jn ⊆ J. It follows that Z ⊆ J. The sum of the lengths of all the intervals in J is: ∞ X ∞ X i=1 j=1



(bij − aij ) ≤



  ∞ X 1  =  − 1 = . 2i 1 − 1/2 i=1



We conclude that Z is a null set. S∞ Part (a) Redux Suppose Z1 and Z2 are null sets. Observe that Z = Z1 ∪ Z2 = Z1 ∪ Z2 ∪ i=3 ∅. In addition, ∅ is obviously a null set because it contains no intervals. If we let Zn = ∅ for n > 2 and apply part (a), we see that Z is a null set.



Section 7.4 Problem 7.4-3. Let f and g be bounded functions on I := [a, b]. If f (x) ≤ g(x) for all x ∈ I, show that L(f ) ≤ L(g) and (U (f ) ≤ U (g). Solution: Since f and g are bounded on [a, b], there is a supremum and infimum of the image of any subinterval of [a, b] under f and g. Suppose P is an arbitrary partition of [a, b] with n ∈ N subintervals. Let mf k = inf{f (x) : x ∈ [xk−1 , xk ]} and Mf k = sup{f (x) : x ∈ [xk−1 , xk ]}, with similar definitions for mgx and Mgx . Because f (x) ≤ g(x) for all x ∈ [a, b] by hypothesis, mf k ≤ mgk and Mf k ≤ Mgk for all 1 ≤ k ≤ n.



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As a result, the lower and upper sums of f and g on P are: L(f ; P ) =



n X



mf k (xk − xk−1 ) ≤



k=1



and U (f ; P ) =



n X



n X



mgk (xk − xk−1 ) = L(g; P ),



k=1



Mf k (xk − xk−1 ≤



k=1



n X



Mgk (xk − xk−1 ) = U (g; P ).



k=1



Assume L(f ) > L(g). If  = L(f ) − L(g), then there is a P ∈ P(I) such that L(f ; P ) > L(f ) −  = L(g), so L(f ; P ) is an upper bound to {L(g; P 0 ) : P 0 ∈ P(I)}. But this contradicts that L(f : P ) ≤ L(g; P ). Therefore, L(f ) ≤ L(g). By the same argument, U (f ) ≤ U (g). Problem 7.4-5. Let f , g, h be bounded functions on I := [a, b] such that f (x) ≤ g(x) ≤ h(x) for all x ∈ I. Show that if f Rb Rb Rb Rb and h are Darboux integrable and if a f = a h, then g is also Darboux integrable with a g = a f . Solution: Because f , g, and h are bounded, there are supremum and infimum of the image of any subinterval of I under any f , g, and h. Let P be a partition of I with n ∈ N subintervals. Clearly, mf k ≤ mgk and Mgk ≤ Mhk for all subintervals of P . Therefore, L(f ; P ) ≤ L(g; P ) and U (g; P ) ≤ U (h; P ). Since this is true for every partition of Rb Rb I, it follows that L(f ) ≤ L(g) and U (g) ≤ U (h). If a f = a h, then L(f ) = U (f ) = L(h) = U (h). Therefore, L(f ) ≤ L(g) ≤ U (g) ≤ U (h), from which it follows that L(g) = U (g) = L(f ). Therefore, g is Darboux integrable with Rb Rb Rb g = a f = a h. a Problem 7.4-8. Let f be continuous on I := [a, b] and assume f (x) ≥ 0 for all x ∈ I. Prove if L(f ) = 0, then f (x) = 0 for all x ∈ I. Solution: Because f is continuous on [a, b], it is bounded on [a, b]. For any partition P of I, it must be that mk ≥ 0 because f is always non-negative on I. Therefore, L(f ; P ) ≥ 0. If L(f ) = 0, then 0 ≤ L(f ; p) ≤ L(f ) = 0, from which it follows that L(f ; P ) = 0. Now assume that f (c) > 0 for some c ∈ I. Because f is continuous on I, it follows that limx→c f (x) > 0. By Theorem 4.2.9, there is a δ > 0 such that if x ∈ [c − δ, c + δ] ∩ I, then f (x) > 0. Let P 0 be a partition of I where for some subinterval k, we have [xk−1 , xk ] ⊆ [c − δ, c + δ]. Therefore, mk > 0. Because mi ≥ 0 for all other subintervals i of P 0 , we have L(f ; P 0 ) > 0, which contradicts that L(f ; P 0 ) = 0 for all partitions of I. Therefore, f (x) = 0 for all x ∈ I. Problem 7.4-9. Let f1 and f2 be bounded functions on [a, b]. Show that L(f1 ) + L(f2 ) ≤ L(f1 + f2 ). Solution: For any partition P of [a, b] with n ∈ N subintervals, it follows that L(f1 ; P ) ≤ L(f1 ) and L(f2 ; P ) ≤ L(f2 ). Let g = f1 + f2 . Now let m1k , m2k , and mgk be the infima of f1 , f2 , and g on subinterval k of P , respectively, with similar definitions for M1k , M2k , and Mgk for the applicable suprema on subinterval k. Clearly, mgk = m1k + m2k and Mgk = M1k + M2k . As a result: L(g; P ) =



n X k=1



mgk (xk − xk−1 ) =



n X



(m1k + m2k )(xk − xk−1 ) = L(f1 ; P ) + L(f2 ; P ) ≤ L(g),



k=1



since L(g) is the supremum of the lower sums of all P on [a, b]. Let S = {L(f1 ; P ) + L(f2 ; P ) : P ∈ P[a, b]}. By definition, L(g) = L(f + g) = sup S. By Exercise 2.4.8, L(f + g) ≥ sup{L(f1 ; P ) : P ∈ P[a, b]} + sup{L(f2 ; P ) : P ∈ P[a, b]} = L(f1 ) + L(f2 ). Problem 7.2-11. If f is a bounded function on [a, b] such that f (x) = 0 except for x is {c1 , c2 , . . . , cn }, show that U (f ) = L(f ) = 0.



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Solution: Let  > 0 be given and C = {c1 , c2 , . . . , cn }. Let mck = mk and Mck = Mk where subinterval k of a given partition contains ck . Suppose β = inf{f (x) : x ∈ [a, b, ]}, suppose γ = sup{f (x) : x ∈ [a, b, ]}, and suppose τ = sup{|β|, |γ|}. Clearly τ > 0 (because f is non-zero at some point on [a, b]) and for any partition, if a subinterval k includes some c ∈ C, then −τ ≤ mk ≤ Mk ≤ τ . We will solve this problem two ways: One using the definition of the Darboux integral, and the other using the Integrability Criterion. Let P be a partition of [a, b] where P = {a, c1 − /(4nτ ), c1 + /(4nτ ), . . . , cn − /(4nτ ), cn − /(4nτ ), b}. The lower sum is : L(f ; P ) =



n X



mk (xk − xk−1 ) = mc1



k=1



       =− . + . . . + mcn ≥ −τ + ... + 2nτ 2nτ 2nτ 2nτ 2



The upper sum is: U (f ; P ) =



n X



Mk (xk − xk−1 ) = Mc1



k=1



       = . + . . . + Mcn ≤τ + ... + 2nτ 2nτ 2nτ 2nτ 2



From Theorem 7.4.5, − < L(f ; P ) ≤ L(f ) ≤ U (f ) ≤ U (f ; P ) < . It then follows that L(f ) = U (f ) = 0. This is easily seen because for any a ∈ R and  > 0, if − < a < , then 0 ≤ |a| < , from which it follows from Theorem 2.1.9 Rb that |a| = 0, implying that a = 0. Therefore, f is Darboux integrable and a f = 0. We will now prove the statement using the Integrability Criterion. Let P be a partition of [a, b] where the sum of the length of the subintervals containing some c ∈ C is less than /(2τ ). Let K = {i ∈ N : ci ∈ [xk−1 , xk ], ci ∈ C}. For 1 ≤ j ≤ n, if j ∈ / K, then mj = Mj = 0, and if j ∈ K, then −τ ≤ mj ≤ 0 and 0 ≤ Mj ≤ τ . Therefore, the lower sum is bounded by: X  0 ≥ L(f ; P ) ≥ −τ (xk − xk−1 ) < − , 2 k∈K



and the upper sum is bounded by: 0 ≤ U (f ; P ) ≤ τ



X



(xk − xk−1 )
0 such that for n ≥ K(), we have |an − L| < . Accordingly, L −  = L − L + M = M < an . But an ∈ S, which contradicts that M is the supremum of S. Therefore, sup S ≥ L. Now let m = inf S and assume m > L. Then if 0 = m − L, there is a K 0 (0 ) > 0 such that if n ≥ K 0 (0 ), then |an − L| < 0 . It follows that an < L + 0 = L + m − L = m. But an ∈ S, which contradicts that m is the infimum of S. Therefore, inf S ≤ L. Problem 7.4-13. Let P be the partition whose existence is asserted in the Integrability Criterion 7.4.8. Show that if P is any refinement of P , then U (f ; P ) − L(f ; P ) < . Solution: By hypothesis, there is a P for  > 0 such that U (f ; P ) − L(f ; P ) < . If P is a refinement of P , then Theorem 7.4.2 requires that L(f ; P ) ≥ L(f ; P ) and U (f ; P ) ≤ L(f ; P ). By Lemma 7.4.3, U (f ; P ) ≥ L(f ; P ). Therefore, 0 ≤ U (f ; P ) − L(f ; P ) ≤ U (f ; P ) − L(f ; P ) < .



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