solution 104 9

7/23/2019 Solution 104 9

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MATH 104-06: INTRODUCTION TO ANALYSIS

SOLUTION 9

(1) P279, 32.2

Proof. (a) For partition P = {0 = t0 < t1 < t2 < · · · < tn = b} of [0, b], wecompute the upper and lower Darboux sum of P . Since M (f, [tk−1, tk]) = tk andm(f, [tk−1, tk]) = 0, we have

U (f, P ) =n

k=1

M (f, [tk−1, tk]) · (tk − tk−1) =n

k=1

tk · (tk − tk−1)

and

L(f, P ) =n

k=1

m(f, [tk−1, tk]) · (tk − tk−1) =n

k=1

0 · (tk − tk−1) = 0.

Since U (f, P ) >

n

k=1tk+tk−1

2 · (tk − tk−1) =

n

k=1

t2k−t

2

k−1

2 = t2n−t

2

0

2 = b2

2 for any

partition P , so we have that the upper Darboux integral U (f ) ≥ b2

2 . Actually, for

the decomposition given by tk = kb

n , we have U (f, P ) =

n

k=1 tk · (tk − tk−1) =n

k=1kb

n · b

n = b2

n2

n

k=1 k = b2(n+1)2n , which goes to b2

2 when n goes to infinity, so the

Darboux integral U (f ) equals b2

2 .

For the lower Darboux integral, since each partition has lower Darboux integralequals 0, we have the lower Darboux integral L(f ) equals 0.

(b) Since U (f ) = b2

2 and L(f ) = 0 which are not equal to each other, so f is notintegrable on [0, b].

(2) P280, 32.7

Proof. We can first suppose that f and g only differ by one point x0 ∈ [a, b], thenthe proof for the general case follows by induction.

Since f is integrable on [a, b], f is bounded on [a, b], so there exists M > 0 suchthat |f (x)| < M for any x ∈ [a, b]. Since g only differs from f at one point, we canalso enlarge M such that |g(x)| < M also holds for any x ∈ [a, b].

For any ǫ > 0, since f is integrable on [a, b], we can find a partition P of [a, b]such that U (f, P ) − L(f, P ) < ǫ. By adding x0 to the partition P , we get a finerpartition P ′ ⊇ P . Let P ′ = {a = t0 < · · · < tk−1 < tk = x0 < tk+1 < · · · < tn = b},and let δ = ǫ

4M . If tk+1 − x0 (or x0 − tk−1) is greater than δ , then we add a new

point in (x0 − δ, x0) (or a point in (x0, x0 + δ )) to P ′ to get a new partition P ′′ ⊇ P ,with U (f, P ′′) − L(f, P ′′) ≤ U (f, Q) − L(f, Q) < ǫ.

Now we estimate the upper and lower Darboux sums for g about P ′′. We stilldenote the partition P ′′ by {a = t0 < · · · < tk−1 < tk = x0 < tk+1 < · · · < tn = b}

1

7/23/2019 Solution 104 9


with tk+1 − x0, x0 − tk−1 < δ . Then

|U (g, P ′′) − U (f, P ′′)| =

nm=1

M (g, [tm−1, tm]) − M (f, [tm−1, tm])

· (tm − tm−1)

≤M (g, [tk−1, tk]) − M (f, [tk−1, tk])

· (tk − tk−1) +

M (g, [tk, tk+1]) − M (f, [tk, tk+1])

· (tk+1 − tk)

<2M · δ + 2M · δ = ǫ.

So we have that |U (g, P ′′) − U (f, P ′′)| < ǫ and the same estimation implies that|L(g, P ′′) − L(f, P ′′)| < ǫ. So we have that U (g, P ′′) − L(g, P ′′) < U (f, P ′′) −L(f, P ′′) + 2ǫ < 3ǫ, which implies that g is integrable on [a, b]. Moreover, we

have that b

a f −

b

a g ≤ U (f, P ′′) − L(g, P ′′) < U (f, P ′′) − L(f, P ′′) + ǫ < 2ǫ and

b

a f −

b

a g ≥ L(f, P ′′)−U (g, P ′′) ≥ L(f, P ′′)−U (f, P ′′)−ǫ > −2ǫ. So |

b

a f −

b

a g| <

2ǫ holds for any ǫ > 0, so b

a f =

b

a g holds.

(3) P280, 32.8

Proof. We need only to show that for any ǫ > 0, there is a partition P of [c, d] suchthat U (f, P ) − L(f, P ) < ǫ.

Since f is integrable on [a, b], there is a partition Q of [a, b] such that U (f, Q) −L(f, Q) < ǫ. By adding c and d to the partition Q, we get a finer partition Q′ suchthat Q ⊆ Q′, and U (f, Q′) − L(f, Q′) ≤ U (f, Q) − L(f, Q) < ǫ.

Let Q′ = {a = t0 < · · · < tm = c < · · · < tn = d < · · · < tk = b}, thenP = {tm = c < · · · < tn = d} is a partition of [c, d]. Then we have

ǫ > U (f, Q′) − L(f, Q′) =kl=1

M (f, [tl−1, tl]) − m(f, [tl−1, tl])

· (tl − tl−1)

≥n

l=m+1

M (f, [tl−1, tl]) − m(f, [tl−1, tl])

· (tl − tl−1) = U (f, P ) − L(f, P ).

Then U (f, P ) − L(f, P ) < ǫ implies that f is integrable on [c, d].

(4) P289, 33.4

Proof. Let f : [0, 1] → R be defined by

f (x)

1 x ∈ [0, 1] ∩Q−1 x ∈ [0, 1] \Q

Then |f |(x) = 1 for any x ∈ [0, 1], which is a continuous function. So |f | isintegrable on [0, 1].

However, for any partition P = {0 = t0 < · · · < tn = 1}, we have M (f, [tk−1, tk]) =1 and m(f, [tk−1, tk]) = −1. So we can compute the upper and lower Darboux sums:U (f, P ) =

n

k=1 M (f, [tk−1, tk]) · (tk − tk−1) =

n

k=1(tk − tk−1) = tn − t0 = 1 andL(f, P ) =

n

k=1 m(f, [tk−1, tk]) · (tk − tk−1) =

n

k=1(−1) · (tk − tk−1) = −(tn − t0) =−1. Then the upper and lower Darboux integrals are U (f ) = 1 and L(f ) = −1,which are not equal. So f is not integrable on [0, 1].

(5) P289, 33.7

7/23/2019 Solution 104 9


Proof. (a) We first show that for any interval [c, d] ⊆ [a, b], M (f 2, [c, d])−m(f 2, [c, d]) ≤

2B

M (f, [c, d]) − m(f, [c, d])

holds.

For any ǫ > 0, there exists x, y ∈ [c, d], such that f 2(x) > M (f 2, [c, d]) − ǫ andf 2(y) < m(f 2, [c, d]) + ǫ. So we have M (f 2, [c, d]) < f 2(x) + ǫ and −m(f 2, [c, d]) <

−f 2

(y) + ǫ, which impliesM (f 2, [c, d]) − m(f 2, [c, d]) < f 2(x) − f 2(y) + 2ǫ

≤|f (x) + f (y)| · |f (x) − f (y)| + 2ǫ ≤ 2B

M (f, [c, d]) − m(f, [c, d])

+ 2ǫ.

Since the inequality holds for any ǫ > 0, we have M (f 2, [c, d]) − m(f 2, [c, d]) ≤

2B

M (f, [c, d]) − m(f, [c, d])

.

So for P = {a = t0 < t1 < · · · < tn = b}, we have

U (f 2, P ) − L(f 2, P ) =

nk=1

M (f 2, [tk−1, tk]) − m(f 2, [tk−1, tk])

· (tk − tk−1)

≤

nk=1

2B

M (f, [tk−1, tk]) − m(f, [tk−1, tk])

· (tk − tk−1) = 2B

U (f, P ) − L(f, P )

.

(b) For any ǫ > 0, since f is integrable on [a, b], there exists a partition P of [a, b] such that U (f, P ) − L(f, P ) < ǫ. Then the previous inequality implies thatU (f 2, P ) − L(f 2, P ) < 2Bǫ, so f 2 is also integrable on [a, b].

(6) P290, 33.9

Proof. For any ǫ > 0, since (f n) uniformly converges to f on [a, b], there exists N

such that for any n > N , we have |f n(x) − f (x)| < ǫ for any x ∈ [a, b]. We fix ann > N , since f n is integrable on [a, b], there is a partition P = {t0 = a < t1 < · · · <

tk = b} such that U (f n, P ) − L(f n, P ) < ǫ.Now we estimate |U (f n, P ) − U (f, P )| and |L(f n, P ) − L(f, P )|. Since |f n(x) −

f (x)| < ǫ for any x ∈ [a, b], we have thatM (f n, [tk−1, tk]) − M (f, [tk−1, tk])

< ǫ

andm(f n, [tk−1, tk]) − m(f, [tk−1, tk])

< ǫ for any k ∈ {1, · · · , n}. So

|U (f n, P ) − U (f, P )| ≤n

k=1

M (f n, [tk−1, tk]) − M (f, [tk−1, tk]) · (tk − tk−1)

≤n

k=1

ǫ(tk − tk−1) = ǫ(b − a).

The same estimation gives that |L(f n, P ) − L(f, P )| ≤ ǫ(b − a).So we have that U (f, P ) − L(f, P ) < U (f n, P ) − L(f n, P ) + ǫ(b − a) + ǫ(b − a) =

ǫ · (1 + 2(b − a)), so f is integrable on [a, b].Since f and f n are both integrable on [a, b], |f n−f | is also integrable on [a, b]. Since

|f n(x)−f (x)| < ǫ for any x ∈ [a, b], we have that b

a(f n−f )

≤ b

a |f n−f | ≤ ǫ(b−a).

So limn→∞

b

a f n =

b

a f holds.

solution 104 9

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