Exercises in Classical Real Analysis
Themis Mitsis

Contents
Chapter 1.

Numbers

5

Chapter 2.

Sequences, Series and Limits

11

Chapter 3.

Topology

23

Chapter 4.

Measure and Integration

29

3

CHAPTER 1

Numbers
E 1.1. Let a, b, c, d be rational numbers and x an irrational number such that cx + d 0. Prove that (ax + b)/(cx + d) is irrational if and only if ad bc.
S. Suppose that (ax + b)/(cx + d) = p/q, where p, q ∈ Z. Then
(aq − cp) x = d p − bq, and so we must have d p − bq = aq − cp = 0, since x is irrational. It follows that ad = bc. Conversely, if ad = bc then (ax + b)/(cx + d) = b/d ∈ Q.
E 1.2. Let a1 ≤ a2 ≤ · · · ≤ an and b1 ≤ b2 ≤ · · · ≤ bn be real numbers. Prove that 







n i=1 



ai  




n j= 1





 b j ≤ n



n

ak bk k=1 and that equality obtains if and only if either a1 = an or b1 = bn .
S. Since {ai }n=1 and {bi }n=1 are both increasing, we have i i
 n  n 





0≤
(ai − a j )(bi − b j ) = 2n ak bk − 2  ai  





1≤i, j≤n

k =1

i= 1

n j=1 



 b j .



If we have equality then the above implies (ai − a j )(bi − b j ) = 0 for all i, j. In particular
(a1 − an )(b1 − bn ) = 0, and so either a1 = an or b1 = bn .
E 1.3. (a) If a1 , a2 , . . . , an are all positive, then
 n  n


1




 a 
 ≥ n2





i 

 ai  i=1 i=1 and equality obtains if and only if a1 = a2 = · · · = an .
(b) If a, b, c are positive and a + b + c = 1, then
(1/a − 1)(1/b − 1)(1/c − 1) ≥ 8 and equality obtains if and only if a = b = c = 1/3.
S. (a) By the Cauchy-Schwarz inequality we have
 n 1/2  n
1/2
n
1/2

1



1/2 1

 a 
.


 n= ≤ ai  i 

 ai ai  i=1 i=1 i=1 (b) Since a + b + c = 1, (a) implies 1/a + 1/b + 1/c ≥ 9 and therefore
(1/a − 1)(1/b − 1)(1/c − 1) = 1/a + 1/b + 1/c − 1 ≥ 8.

5

CHAPTER 1. NUMBERS
E 1.4. Prove that for all n ∈ N we have
135
2n − 1
1
· · ···
≤√
246
2n
3n + 1 end equality obtains if and only if n = 1.
S. Note that

√
2k − 1
3k − 2
≤√
2k
3k + 1 and therefore the product telescopes.
E 1.5. (a) For all n ∈ N we have
√
√
√
√
1
n + 1 − n < √ < n − n − 1. n (b) If n ∈ N ad n > 1 then
√
2 n+1−2<

n k=1 S. (a) We have
√
√ n+1− n= √
√
√ n− n−1= √

√
1
√ < 2 n − 1. k 1
√ < √, n+1+ n 2 n
1

1
1
> √.
√
n+ n−1 2 n

(b) Sum inequalities (a) for k = 2, 3, . . . , n.
E 1.6. Let n ∈ N and x ∈ R. Then
(a) −1 < x < 0 implies (1 + x)n ≤ 1 + nx + (n(n − 1)/2) x2 .
(b) x > 0 implies (1 + x)n ≥ 1 + nx + (n(n − 1)/2) x2 .
S. Induction on n.
E 1.7. If n ∈ N, then n! ≤ ((n + 1)/2)n .
S. In the Geometric-Arithmetic Means Inequality, take ak = k.
E 1.8. If b1 , b2 , . . . , bn are positive real numbers, then n ≤ (b1 b2 · · · bn )1/n .
1
1
1
+ b2 + · · · bn b1 S. In the Geometric-Arithmetic Means Inequality, take ak = 1/bk .
E 1.9. If x, y ∈ R and n ∈ N, then
(a) [ x + y] ≥ [ x] + [y],
(b) [[ x]/n] = [ x/n], n−1 (c)

[ x + k/n] = [nx].

k=0

6

CHAPTER 1. NUMBERS
S. (a) [ x] + [y] is an integer and satisfies [ x] + [y] ≤ x + y, therefore [ x] + [y] ≤
[ x + y].
(b) We claim that [ x/n] ≤ [ x]/n. Indeed, if this were not the case we would have
[ x]/n < [ x/n] ≤ ([ x] + )/n, for some 0 ≤ < 1. Therefore [ x] < n[ x/n] ≤ [ x] + , a contradiction since n[ x/n] is an integer. It follows that [ x/n] ≤ [[ x]/n]. The converse inequality is obvious.
(c) Let n− 1

f ( x) =

[ x + k/n] − [nx]. k =0

Then f is periodic with period 1/n and vanishes on the interval [0, 1/n]. So, f = 0 identically.
E 1.10. (a) If a, b, c are positive real numbers then
1
1
1
a+ b+ c
2
3
6

2

12 12 12 a+ b+ c
2
3
6

≤

with equality if and only if a = b = c.
(b) If a1 , . . . , an and w1 , . . . , wn are positive real numbers with








2



ai wi  ≤



n

n i=1 wi = 1 then

n

a2 wi i i=1

i=1

with equality if and only if a1 = a2 = · · · = an .
S. (a),(b) Cauchy-Schwarz inequality.
E 1.11. If n ∈ N, then n (a) k =1
2n

(b)

n2 k =

(−1)k

k =1

2n n 2n 2 k .
2n
n

= (−1)n

.

S. (a) By the Binomial Theorem we have
2n

(1 + x)2n = k =0

But
2n

(1 + x)

2n k
x.
k





= (1 + x) (1 + x) = 

 n n

n

i=0

= i, j

n n i+ j x= ij


 n

 n j n i 



 x x 



j i j=0

2n

xk k =0

i+ j=k

nn
.
ij

Equating the coefficients of x we get n nn
2n
=
=
ij n i+ j=n

n i=0 7

nn
=
i n−i

n i=0 2

n
.
i

CHAPTER 1. NUMBERS
(b) As in (a) we have
2n

2n
(−1)k x2k k (1 − x2 )2n = k=0 and
(1 − x2 )2n = (1 − x)2n (1 + x)2n






=



i= 0

2n 2n
(−1)i xi+ j = i j

= i, j

  2n




2n
2n j 




(−1)i xi   x 



i j j=0

2n

4n

xk

(−1)i

k =0

i+ j=k

2n 2n
.
i j Equating the coefficients of x2n we get
(−1)n

2n
2n 2n
=
(−1)i
=
n i j i+ j=2n

E 1.12. If m, n ∈ N, then 1 +

m

n+k k k=1

=

2n

2

(−1)i i= 0

n+m+1 m 2n
.
i

.

S. m 1+ k =1

n+k
=1+
k

m

n+k+1 n+k − k k−1

k=1

=

n+m+1
.
m

E 1.13. Prove Lagrange’s inequality for real numbers

 n
2  n
n










 a b  =  a2   b2  −




(ak b j − a j bk )2 .



k k





k k 
1≤k< j≤n

k=1

k =1

k =1

S. We have
(ak b j − a j bk )2 =
1≤k< j≤n

(a2 b2 + a2 b2 − 2ak b j a j bk ). jk kj
1≤k< j≤n

But a2 b2 kj a2 b2 jk +

1≤k< j≤n





=



k=1

1≤k< j≤n

and





2ak b j a j bk = 


1≤k< j≤n

n

n k =1






a2  

k 




n



b2 

k

k=1

2



ak bk  −



n

a2 b2 kk − k =1

n

a2 b2 . kk k =1

The result follows.
E 1.14. Given a real x and an integer N > 1, prove that there exist integers p and q with 0 < q ≤ N such that |qx − p| < 1/N .
N
S. For k = 0, 1, . . . , N let ak = kx − [kx]. Then {ak }k=0 ⊂ [0, 1), and therefore there exist 0 ≤ k1 , k2 ≤ N such that |ak1 − ak2 | < 1/N .

E 1.15. If x is irrational prove that there are infinitely many rational numbers p/q with q > 0 and such that | x − p/q| < 1/q2 .
8

CHAPTER 1. NUMBERS
S. Assume there are finitely many, say, p1 /q1 , . . . , pn /qn . Then, by the preceding exercise, there exists p/q such that | x − p/q| < 1/(qN ) with q ≤ N and 1/N < min{| x − pi /q1 | : 1 ≤ i ≤ n}. (The minimum is positive because x is irrational.)

9

CHAPTER 2

Sequences, Series and Limits
E 2.1. Evaluate lim

n

n→∞ k=0

S. If a

k

(1 + a2 ) where a ∈ C.

1, then for all n ∈ N we have n+ 1

n k (1 + a2 ) = k=0 1 − a2
.
1−a

Therefore the sequence converges to 1/(1 − a) for |a| < 1. It diverges for |a| > 1 or a = 1.
The limit does not exist if |a| = 1 and a 1.
E 2.2. Evaluate lim

n

n→∞ k=1

√1 . n2 +k

S. Note that n √ n2 + n

n

≤ k =1

1
≤ 1.
√
2+k n Therefore the sum converges to 1.
√
√
E 2.3. Let x = 2 + 2 and y = 2 − 2. Then n ∈ N implies
(a) xn + yn ∈ N and xn + yn = [ xn ] + 1.
(b) lim ( xn − [ xn ]) = 1. n→∞ S. (a) By the Binomial Theorem, we have n xn + yn = k=0 n k+ n−k
2 2 (1 + (−1)n−k ) = k 0≤k≤n n−k even

n k+1+ n−k
2 ∈ N.
2
k

Since xn + yn − 1 < xn < xn + yn , we conclude that [ xn ] = xn + yn − 1.
(b) By (a), xn − [ xn ] = 1 − yn → 0 as n → ∞.
E 2.4. If { xn }∞ 1 ⊂ R, {yn }∞ 1 ⊂ (0, ∞) and { xn /yn }∞ 1 is monotone, then the n= n= n= sequence {zn }∞ 1 defined by n= x1 + · · · + xn zn = y1 + · · · + yn is also monotone.
S. Assume that { xn /yn }∞ 1 is increasing and prove inductively that zn ≤ zn+1 ≤ n= xn+1 /yn+1 using the fact a a+c c ac
≤⇒≤
≤. bd b b+d d
11

CHAPTER 2. SEQUENCES, SERIES AND LIMITS
E 2.5. Let 0 < a < b < ∞. Define
√
x1 = a, x2 = b, x2n+1 = x2n x2n−1 ,

x2n+2 =

x2n + x2n−1
.
2

Then the sequence { xn }∞ 1 converges. n= S. Note that [ x2n+1 , x2n+2 ] ⊂ [ x2n−1 , x2n ] and x2n − x2n−1 x2 − x1
≤ · · · ≤ n−1 → 0. x2n+2 − x2n+1 ≤
2
2
Therefore the sequence converges and
∞

lim xn =

n→∞

[ x2n−1 , x2n ]. n=1 E 2.6. Let 0 < a < b < ∞. Define xn + xn+1
.
2 converges and determine its limit.

x1 = a,
Prove that the sequence { xn }∞ 1 n= x2 = b,

xn+2 =

S. Note that xn+1 − xn = (−1/2)n−1 ( x2 − x1 ). Therefore n−2 xn = x1 + ( x2 − x1 )

1
2 a + 2b
(− )k → a + (b − a) =
.
2
3
3 k =0

E 2.7. Let { xn }∞ 1 ⊂ R satisfy 0 < xn < 1 and 4 xn+1 (1 − xn ) ≥ 1 for all n ∈ N. n= Show that lim xn = 1/2. n→∞ S. Note that

1
≥ xn .
4(1 − xn )
Therefore the sequence is increasing. Since it is bounded, it converges to a limit l which must satisfy 4l(1 − l) ≥ 1. We conclude that l = 1/2. xn+1 ≥

E 2.8. Let 1 < a < ∞, x = 1, and xn+1 = a(1 + xn )/(a + xn ). Show that
√
xn → a. by S
√ . Prove inductively that the sequence is decreasing and bounded from below
a.

E 2.9. Define x0 = 0, x1 = 1, and
1
n xn+1 = xn−1 + xn . n+1 n+1
Prove that { xn }∞ 1 converges and determine its limit. n= S. Note that xn+1 − xn = (−1)n /(n + 1)!, and so n−1 xn = k =0

1
(−1)k
→.
(k + 1)! e E 2.10. Let a ∈ R, a {0, 1, 2} and define x1 = a, xn+1 = 2 − 2/ xn for n ∈ N.
Find the limit points of the sequence { xn }∞ 1 . n= 12

CHAPTER 2. SEQUENCES, SERIES AND LIMITS
S. Note that xn+4 = xn for all n ∈ N. Therefore the sequence takes on the values { x1 , x2 , x3 , x4 } only.
E 2.11. For n ∈ N, write n = 2 j−1 (2k − 1) where j, k ∈ N and write
11
Sn = + . jk Find all limit points of the sequence {S n }∞ 1 . Evaluate lim S n and lim S n . n= S. Let A be the set of limit points of {S n }∞ 1 . We claim that A = {0} ∪ {1/n : n= n ∈ N}. Indeed, let nk = 2k−1 (2k − 1) and m p,k = 2 p−1 (2k − 1).Then
2
1
11
→ 0, S m p,k = + → as k → ∞. k pk p Hence A ⊃ {0} ∪ {1/n : n ∈ N}. Now take l ∈ A, l 0. Then there exists a subsequence
{S nm }∞=1 such that S nm → l. Write nm = 2 jm −1 (2km − 1). Note that at least one of the m sets { jm : m ∈ N}, {km : m ∈ N} is unbounded, and so we may assume, without loss of generality, that there exists { jmi }∞ 1 with jmi → ∞. Then, since S nmi → l, we have kmi → i= 1/l. Therefore {kmi }∞ 1 is eventually constant and l ∈ {1/kmi : i ∈ N}. lim S n = inf A = 0, i= lim S n = sup A = 1.
S nk =

E 2.12. Prove that (n/e)n < n! for all n ∈ N.
S. Induction on n. It is clearly true for n = 1. Assuming (n/e)n < n! we have n+1 e

n+1

=

n+1
1
1+ e n

n

n e n

<

n+1 e n! = (n + 1)!. e E 2.13. Evaluate
(a) lim ((2n)!/(n!)2 )1/n , n→∞ (b) lim (1/n)[(n + 1)(n + 2) · · · (n + n)]1/n , n→∞ (c) lim [(2/1)(3/2)2 (4/3)3 · · · ((n + 1)/n)n ]1/n . n→∞ S. Let an =

(2n)!
,
(n!)2 cn =

bn =
2
1

(n + 1)(n + 2) · · · (n + n)
,
nn
2

3
2

4
3

3

···

n

n+1 n .

Then an+1 (2n + 1)(2n + 2)
=
→ 4, an (n + 1)2

n bn+1 = bn n+1

cn+1
1
= 1+ cn n+1
Therefore

√ n an → 4,

n

bn →

(2n + 1)(2n + 2)
4
→, e (n + 1)2

n+1

4
,
e

√
E 2.14. Evaluate lim n → ∞( n n − 1)n .
13

n

→ e.
√
n cn → e.

CHAPTER 2. SEQUENCES, SERIES AND LIMITS
√
√
S. Since n n → 1, there exists n0 ∈ N such that 0 < n n − 1 < 1/2 for all
√
√
√
n ≥ n0 , and so 0 < ( n n − 1)n < (1/2)n . Therefore 0 ≤ lim( n n − 1)n ≤ lim( n n)n ≤ 0. We
√
conclude that lim ( n n − 1)n = 0. n→∞ E 2.15. If { xn }∞ 1 ⊂ (0, ∞) and xn → x, then ( x1 · · · xn )1/n → x. n= S. By the Harmonic-Geometric-Arithmetic Means Inequality we have n x1 + · · · + xn
≤ ( x1 · · · xn )1/n ≤
.
1 n + · · · + x1 x 1

Therefore ( x1 · · · xn )

1/n

n

→ x.

E 2.16. (a) Let S n =

n k=1 1/k for n ∈ N. Then lim |S n+ p − S n | = 0 for all p ∈ N, n→∞ but {S n }∞ 1 diverges to ∞. n= (b) Find a divergent sequence { xn }∞ 1 in R such that lim | xn2 − xn | = 0. n= n→∞

S. (a) |S n+ p − S n | = 1/(n + 1) + · · · + 1/(n + p) ≤ p/(n + 1) → 0 k ( n) k(n)+1 (b) For n ≥ 4 let k(n) be the unique integer such that 22 ≤ n < 22 and define xn =

k(n) j= 1

1/ j. Note that k(n) → ∞ and k(n2 ) = k(n) + 1. Therefore xn → ∞ and | xn2 − xn | =

1/(k(n) + 1) → 0.
E 2.17. There exist two divergent series an and bn of positive terms with a1 ≥ a2 ≥ · · · and b1 ≥ b2 ≥ · · · such that if cn = min{an , bn }, then cn converges.
S. Let ak = 1/2k ,

bk = 1/2n

if 2n ≤ k < 2n+1 , n even

ak = 1/2n ,

bk = 1/2k

if 2n ≤ k < 2n+1 , n odd.

and

E 2.18. Evaluate the sums
∞
(a)
1/(n(n + 1)(n + 2)), n=1 ∞

(b)

(n − 1)!/(n + p)!, where p ∈ N is fixed.

n=1

S. (a) Note that
1
1
1
1
=
−
.
n(n + 1)(n + 2) 2 n(n + 1) (n + 1)(n + 2)
Consequently
n
11
1
1
1
=
−
→.
k(k + 1)(k + 2) 2 2 (n + 1)(n + 2)
4
k=1
(b) We have
(n − 1)!
1
=
=
(n + p)! n · · · (n + p)
Therefore
n
1
(k − 1)!
=
(k + p)! p k=1

1
1
1
−
. p n · · · (n + p − 1) (n + 1) · · · (n + p)
1
1
1
−
→
. p! (n + 1) · · · (n + p) p p!
14

CHAPTER 2. SEQUENCES, SERIES AND LIMITS

E 2.19. Let an be a convergent series of nonnegative terms. Then
(a) lim nan = 0,
(b) possibly lim nan > 0,
(c) if an ≥ an+1 for all n > n0 , then lim nan = 0.
S. (a) Suppose that lim nan > c > 0 for some c. Then there exists n0 ∈ N such that nan > c for n ≥ n0 . Consequently,
N

N

1
→∞
n

an > c n=n0 a contradiction.
(b) Let ak = 1/2k if k

n=n0

2n and ak = 1/2n if k = 2n . Then

N

∞

N

ak = k =1

as n → ∞,

ak + k=2n k 2n

1
1
1
+
n.

n=1

∞

bn . n=1 On the other hand
∞

∞

∞

am,n = m=1 n=1

m=1

E 2.21. (a) Prove that
(b) Prove that

∞

mcm

∞ n=1 ∞









1/n2 < 2.
∞

m=1 n=1

∞

1
=
cm . n(n + 1) m=1 n=m 

1

 = ∞.


2
(m + n)
15

CHAPTER 2. SEQUENCES, SERIES AND LIMITS
S. (a) We have
∞
n=1

∞

1
=1+
n2

n=2

1 1 and let d be a digit (0 ≤ d < b). Let A denote the set of all k ∈ N such that the b-adic expansion of k fails to contain the digit d.
(a) If ak = 1/k for k ∈ A and ak = 0 otherwise, then

∞

k=1

ak < ∞.

(b) For n ∈ N let A(n) denote the number of elements of A that are ≤ n. Then lim (A(n)/n) = 0.

n→∞

S. Let
An = {k : k is an n-digit number and does not contain the digit d}
= {k : bn−1 ≤ k < bn } ∩ A.
Note that |An | = (b − 2)(b − 1)n−1 .
(a) We have
∞

∞

∞

ak = k =1

(b) If b

ak ≤ n=1 k∈An

n=1

∞

|An | b−1 = (b − 2) n−1 b b n=1

n−1

< ∞.

2 then
A(n) ≤

(b − 1)k ≤ n1/ logb−1 b − 1.

|Ak | = (b − 2) k:bk−1 ≤n

k:bk ≤n

If b = 2 then A(n) = |{k : 2k ≤ n}| ≤ log2 n. Therefore lim (A(n)/n) = 0. n→∞ E 2.23. Let 0 < x < 1. Then x has a terminating decimal expansion if and only if there exist nonnegative integers m and n such that 2m 5n x is an integer.
S. If x has a terminating decimal expansion, then x = p/10k = p/(2k 5k ).
Conversely, if 2m 5n x = N ∈ N for some, say, m ≤ n, then x = 2n−m N /10n .
E 2.24. Evaluate lim (n!e − [n!e]). n→∞ S. Let S n =

n

1/k!. Then, using the error estimate for the “tail”, we have

k=0

0 < n!e − n!S n < 1/n. We conclude that [n!e] = n!S n and therefore n!e − [n!e] → 0.
E 2.25. Show that lim n sin(2πen!) = 2π. n→∞ 16

CHAPTER 2. SEQUENCES, SERIES AND LIMITS
S. Since lim (en! − [en!]) = 0 we have n→∞ sin(2πen! − 2π[en!]) sin(2πen!) = 1 ⇒ lim
= 2π. n→∞ en! − [en!]
2πen! − 2π[en!]
Note that the error estimate for the Maclaurin series expansion of e implies
1/(n + 1) < en! − [en!] < 1/n, and so lim n(en! − [en!]) = 1. It follows that lim n→∞

n→∞

n sin(2πen!) = n(en! − [en!])

sin(2πen!)
→ 2π. en! − [en!]

E 2.26. Find the sum of the series
∞
n=1

S.
∞
n=1

1
.
√
√
(n + 1) n + n n + 1

√
√

n
n + 1



 = 1.


=
−
√


√ n+1 (n + 1) n + n n + 1 n=1 n
∞

1

E 2.27. Let an > 0 for each n ∈ N. Then
∞
∞√
(a)
an < ∞ implies an an+1 < ∞, n=1 n=1

(b) the converse of (a) is false,
∞

(c)

n=1

an < ∞ implies

∞

1
(a−1 + a−+1 )−1 < ∞, n n

n=1

(d) the converse of (c) is false.
S. By the Harmonic-Geometric-Arithmetic Means Inequality, we have
√
1
1
2(a−1 + a−+1 )−1 ≤ an an+1 ≤ (an + an+1 ), n n
2
proving (a) and (c). For (b) and (d), let an = 1/n if n is even and an = 1/n3 if n is odd.
E 2.28. Suppose that dn > 0 for all n ∈ N and

∞ n=1 = ∞. What can be said of

the following series?
∞

(a)
(b)
(c)

n=1
∞
n=1
∞
n=1

dn /(1 + dn ), dn /(1 + ndn ),
2
dn /(1 + dn ).

S. (a) If {dn }∞ 1 is bounded then 1/(1 + dn ) is bounded from below, therefore n= ∞ n=1 dn
≥C
1 + dn

∞

dn = ∞. n=1 If {dn }∞ 1 is unbounded then there exists a subsequence {dkn }∞ 1 with dkn → ∞. Therefore n= n= there exists n0 such that dkn /(1+dkn ) > 1/2 for all n ≥ n0 . Consequently
17

∞

n=1

dn /(1+dn ) = ∞.

CHAPTER 2. SEQUENCES, SERIES AND LIMITS
(b) Let dn = 1 for all n ∈ N. Then
∞

∞

dn = n=1 Let dk = 1/2k if k

n=1

dn
= ∞.
1 + ndn

2n and dk = 2n if k = 2n . Then

1

dk
 2k
=  k+n
2
1 + kdk  1+4n

∞

Therefore

n=1

∞ n=1 dn = ∞ and

if k 2n , if k = 2n .

dn /(1 + ndn ) < ∞.
∞

(c) Let dn = 1 for all n. Then

n=1

∞.

2 dn /(1 + dn ) = ∞. Let dn = n2 . Then

∞ n=1 2 dn /(1 + dn ) <

E 2.29. Let 0 < a < b < ∞ and define x1 = a, x2 = b, and xn+2 = n ∈ N. Find lim xn .

√

xn xn+1 for

n→∞

S. Let yn = log xn and use Exercise 2.6.
−
E 2.30. Let 0 < a < b < ∞ and define x1 = a, y1 = b, xn+1 = 2( xn 1 + y−1 )−1 , n √
∞
∞ and yn+1 = xn yn . Then { xn }n=1 and {yn }n=1 both converge and have the same limit.

S. Prove inductively, using the Harmonic-Geometric Means Inequality, that
1
a < xn ≤ xn+1 ≤ yn+1 ≤ yn < b and yn+1 − xn+1 ≤ (yn − xn ).
2
∞

E 2.31. Show that if x ∈ (0, 1) there is a subseries

∞ k=1 k=1

ak = 1 and 0 < an ≤

∞

, n = 1, 2, . . . , then for every

k=n+1

ank whose sum is x.

S. Note that, since the sum of the series is 1 and x ∈ (0, 1), there exists n1 ∈ N such that
∞

∞

ak > x k=n1 implying

ak ≤ x

and k=n1 +1

∞

ak > x − an1

and an1 ≤ x.

k=n1 +1

Therefore there exists n2 > n1 such that
∞

ak > x − an1

≤ x − an1 .

and k=n2 +1

k=n2

Continuing this way, we can find a sequence of integers n1 < n2 < · · · such that
∞

m

0≤ x−

ank < k =1

Letting m → ∞, we conclude that

∞ k =1

ak . k=nm +1

ank = x.
18

CHAPTER 2. SEQUENCES, SERIES AND LIMITS
E 2.32. Show that if an , bn ∈ R, (an + bn )bn
∞

and

∞

(an /bn )2 converge, then

n=1

n=1

0, n = 1, 2, . . . , and both

∞ n=1 an /bn

an /(an + bn ) converges.

S. Choose k0 ∈ N such that |1 + ak /bk | ≥ 1/2 for all k ≥ k0 . Then
1
2
.
≤
|ak bk + b2 | |bk |2 k Note that

n k=k0 n n a2 ak ak k =
−
ak + bk k=k bk k=k ak bk + b2 k 0
0

and

n

We conclude that

n=1

ak 2
.
bk

≤2

ak bk + b2 k k=k0
∞

n

a2 k k=k0

an /(an + bn ) converges.
∞

E 2.33. Show that if bn

0 and

R such that an /bn → 1 as n → ∞ and
S. Let

∞ n=1 n=1

bn = ∞, then there is a sequence {an }∞ 1 ⊂ n= (−1)n an diverges.

n

Sn =

bk ,

an = bn + (−1)n

k =1

bn
.
Sn

Note that an > 0 for large n and m m

m

(−1)n an = n=1 (−1)n bn + n=1 n=1

bn
.
Sn

The first series in the above sum converges, being alternating, while the second diverges by
∞

Abel’s Theorem. Therefore as n → ∞.

n=1

an diverges. On the other hand, an /bn = 1 + (−1)n /S n → 1

E 2.34. Show that if n ≥ 2, then

∞

(1 − (1 − 2−k )n )

log n.

k =1

S. Note that
1
= m+1 ∞

1
0

xm dx ≤ k=0 and similarly

∞ k=1 ∞

1−1/2k+1

xm dx =
1−1/2k

1
1
1− k
2k
2

k=1 m ≤

1
1
1− k
2k
2

m

2
.
m+1

Therefore
∞

1− 1− k =1

1
2k

∞ n−1

n

=

1
1
1− k
2k
2 k=1 m=0 n 1 m m=1

log n.
19

n−1 ∞

m

= m=0 k=1

1
1
1− k
2k
2

m

CHAPTER 2. SEQUENCES, SERIES AND LIMITS

∞ −x e (sin( x n→∞ 0

E 2.35. Show that if rn ∈ R, then lim

+ rn ))n dx = 0.

S.
∞

∞

|e− x (sin( x + rn ))n |dx =

0

e− x | sin( x + rn mod 2π)|n dx

0

∞

= ern mod 2π

e− x | sin( x)|n dx

rn mod 2π
∞

≤ e2π

e− x | sin( x)|n dx.

0

Note that | sin( x)|n → 0 almost everywhere, and so, by the Dominated Convergence Theo∞ rem, 0 e− x | sin( x)|n dx → 0.
E 2.36. Let f : [0, 1] → R be defined by
 x log x

 x−1 if 0 < x < 1,



 f ( x ) = 0 if x = 0,



1

if x = 1.
Show that

∞

1

f ( x)dx = 1 −
0

n=2

S. Note that x log x
=
x−1

∞ n=0 1
.
− 1)

n2 (n

x(1 − x)n n+1 and the convergence is uniform on [0, 1] by Weierstrass M-test. Therefore
∞

1

f ( x)dx =
0

n=0

∞

∞

1

1 n+1 x(1 − x)n dx =
0

n=0

1
(n + 1)2 (n + 2)

1
.
n2 (n − 1)

=1− n=2 E 2.37. Show that kn lim

n→∞

Conclude that

∞ j=1 S. Note that

j+1 j j=n

1
= log k. j (−1) j+1
= log 2. j dx 1
≤≤
x j j+1 j dx
.
x−1

Therefore kn+1 n

dx
≤
x

kn j= n

1
≤
j
20

k n+ 1 n dx
,
x−1

CHAPTER 2. SEQUENCES, SERIES AND LIMITS and consequently
1
≤ n log k +

kn j=n 1 k ≤ log k +
.
j n−1 Taking the limit as n → ∞, we obtain the first assertion. To prove the second assertion, note that
2n
j=1

Since

(−1) j+1
=
j

∞ (−1) j+1 j=1 j

2n j=1 1
−2
j

n j=1 1
=
2j

2n

1
=
j j=n+1 2n j=n 11
− → log 2, as n → ∞. jn ∞ (−1) j+1 j=1 j

converges by Leibniz, we conclude that

E 2.38. Show that e x that its limit as x → ∞ is 0.

2

/2

∞ −t2 /2 e dt x = log 2.

is a decreasing function of x on [0, ∞) and

S. By L’Hospital’s Rule we have
∞ −t2 /2 e dt x 2 /2 e− x

lim

x→∞

Now let g( x) = e x

2

∞

/2

2

−e− x
1
= lim
= lim = 0. x→∞ − xe− x2 /2 x→∞ x

2

/2

dt

and h( x) =

x

Then

∞

2

e−t
∞

e− x /2
−
x

e−t

2

/2

dt.

x
2

e− x /2
< 0. x2 x
Hence h is strictly decreasing. Note that lim h( x) = 0. therefore h( x) > 0 and consequently x→∞ g ( x) < 0. g ( x) = xe x

2

/2

e−t

2

/2

dt − 1 and

21

h ( x) = −

CHAPTER 3

Topology
E 3.1. Let X be a 2nd countable space. Show that if {Gi }i∈I is an arbitrary family of open sets in X then there exists a countable subset J ⊂ I such that i∈I Gi = i∈ J G i .
S. Suppose {Uk }k∈N is a basis for the topology of X. Let
K = {k ∈ N : ∃ i(k) ∈ I such that Uk ⊂ Gi(k) } and put J = {i(k) : k ∈ K }.
E 3.2. Let X be a 2nd countable space, and let A ⊂ X be an uncountable set.
Prove that A has at least one condensation point.
S. Suppose that for each x ∈ A there is an open set U x ⊂ X with x ∈ U x and
|A ∩ U x | ≤ ℵ0 . Since X is 2nd countable there exists { xn }∞ 1 ⊂ A such that x∈A U x = n= ∞
∞
n=1 (U xn ∩ A) and therefore U xn0 ∩ A must be uncountable for some n=1 U xn . Hence A = n0 , a contradiction.
E 3.3. If X is a 2nd countable space and A is a closed subset of X , then there exist a perfect set P and a countable set N , such that A = P ∪ N . Conclude that any subset of a 2nd countable space can have only countably many isolated points.
S. Let P = { x ∈ X : for each nbd U x of x, U x ∩ A is uncountable}. Using the preceding exercise, P is perfect and A \ P is countable.
E 3.4. Prove the following assertions.
(a) If A is nonempty perfect subset of a complete metric space then A is uncountable.
(b) Any countable closed subset of a complete metric space has infinitely many isolated points.
(c) There exists a countable closed subset of R having infinitely many limit.points.
S. Suppose X is a complete metric space.
(a) Note that since A is a closed subset of X , it is complete as a metric space. If A is countable then by the Baire category theorem, at least one of its points must be isolated.
(b) Assume that there exists a countable closed subset of X with finitely many isolated points. Removing these points results in a countable perfect set, contradicting (a).
(c) Take infinite copies of a convergent sequence together with its limit.
E 3.5. It is impossible to express [0, 1] as a union of disjoint closed nondegenerate intervals of length < 1.
S. Suppose [0, 1] = i∈I [ xi , yi ], where {[ xi , yi ]}i∈I is disjoint. Note that I must be countable. Then the set of endpoints ({ xi : i ∈ I } ∪ {yi : i ∈ I }) \ {0, 1} is a countable perfect set, a contradiction by the preceding exercise.
23

CHAPTER 3. TOPOLOGY
E 3.6. It is impossible to express [0, 1] as a countable union of disjoint closed sets. S. Suppose [0, 1] = ∞ 1 Fn with the Fn ’s closed and pairwise disjoint. Since n= F1 ∩ F2 = ∅, we can find a closed interval I1 such that I1 ∩ F1 = ∅, I1 ∩ F2 ∅, I1 \ F2 ∅.
We repeat the same procedure inside I1 with I1 ∩ F2 playing the role of F1 and I1 ∩ Fk playing the role of F2 , where Fk is the first set in the sequence {Fn }∞ 3 intersecting I1 . We n= thereby construct a decreasing sequence of closed intervals {In }∞ 1 such that In ∩ Fn = ∅, a n= contradiction.
E 3.7. Let A be a bounded subset of R which is not closed. Construct explicitly an open cover of A that has no finite subcover.
S. Let x ∈ R \ A be a point such that ( x − , x + ) ∩ A ∅ for all > 0. For each n choose xn ∈ ( x − 1/n, x + 1/n) ∩ A. Without loss of generality we may assume that
{ xn }∞ 1 is monotone. If x1 < · · · < xn < · · · x, consider the cover {(−∞, xn )}∞ 1 ∪ {( x, ∞)}. n= n=
If x1 > · · · > xn > · · · x, then take the covering {( xn , ∞)}∞ 1 ∪ {(−∞, x)}. n= E 3.8. Let (X, ρ) be a metric space and A, B ⊂ X disjoint closed sets. Show that there exists a continuous function f : X → R such that f | A = 0 and f | B = 1.
S. Let

ρ( x, A)
.
ρ( x, A) + ρ( x, B)
Then f is well-defined and has the required properties. f ( x) =

E 3.9. If X is a connected metric space with at least two points, then X is uncountable. S. Let x, y ∈ X be two distinct points. By the preceding exercise, there exists a continuous function f : X → R with f ( x) = 0 and f (y) = 1. Since X is connected, f has the intermediate value property. Therefore [0, 1] ⊂ f (X ). We conclude that X is uncountable. E 3.10. Let S be a nonempty closed subset of R and let f : S → R be continuous. Then there exists a continuous g : R → R such that f ( x) = g( x) for all x ∈ S and sup x∈R |g( x)| = sup x∈S | f ( x)|. This is false for every nonclosed S ⊂ R.
S. Write R \ S = ∞ 1 In , where the In ’s are disjoint open intervals and extend n= f on each In linearly (if (−∞, a) or (a, ∞) appear among the In ’s take f to be constant on these intervals). If S is not closed we can find a point x S and, say, an increasing sequence x1 < · · · < xn < · · · < x of points in S such that limn xn = x. Any continuous function f on R \ { x}, and therefore on S , with f ( xn ) = n cannot be extended to the whole line. E 3.11. Let X be a topological space, Y a metric space, f : X → Y an arbitrary function and define A f = { x ∈ X : f is continuous at x}.
(a) Prove that A f is a Gδ set.
(b) Assume that there exists a set D ⊂ X such that D and X \ D are both dense in
X . Prove that for any Gδ set G ⊂ X there exists a function f : X → R such that
A f = G.
(c) Show that there is no function f : R → R which is continuous at each rational and discontinuous at each irrational.
24

CHAPTER 3. TOPOLOGY
(d) Construct explicitly a function f : R → R which is continuous at each irrational and discontinuous at each rational.
P. For any point x ∈ X define the oscilation of f at x by w f ( x) = inf {diam( f (U )) : U is a nbd of x}.
(a) Note that x ∈ A f if and only if w f ( x) = 0. Therefore
∞

Af =

{ x ∈ X : w f ( x) < 1/n}. n=1 The sets in the intersection are open, hence A f is Gδ .
(b) Write G = ∞ 1 Gn where each Gn is open and X = G1 ⊃ G2 ⊃ · · · . Define n= f : X → R by

0

if x ∈ G,



 f ( x) = 1/n
.
if x ∈ D ∩ (Gn \ Gn+1 ),



−1/n if x ∈ (X \ D) ∩ (G \ G )

n n+1 (c) If such a function existed, Q would be Gδ by (a).
(d)

1/n if x = m/n, (m, n) = 1,


.
f ( x) = 
0
 if x is irrational
E 3.12. Construct a strictly increasing function that is continuous at each irrational and discontinuous at each rational.
S. Let {rn : n ∈ N} be an enumeration of the rationals and define f : R → R by f ( x) = rn < x 1/2n . Note that f (rn −) = f (rn ) = f (rn +) − 1/2n and f ( x−) = f ( x) = f ( x+) for all x + inR \ Q.
E 3.13. Let X be a topological space and (Y, ρ) a metric space. Suppose that
{ fn }∞ 1 is a sequence of continuous functions from X into Y and that f : X → Y is some n= function such that limn fn ( x) = f ( x) for all x ∈ X .
(a) Show that there exists a set E ⊂ X that is of 1st category in X such that f is continuous at each point of X \ E . In particular, if X is a complete metric space, then f is continuous at every point of a dense subset of X .
(b) f −1 (V ) is an Fσ set in X for every open V ⊂ Y .
(c) There is no sequence { fn }∞ 1 of continuous real functions on R such that fn ( x) → n= 1 for x ∈ Q and fn ( x) → 0 for x ∈ R \ Q.
(d) Show that χQ , the characteristic function of Q, is the pointwise limit of a sequence of functions, so that each of them is the pointwise limit of a sequence of continuous functions.
S. (a) Let Ak,m = { x ∈ X : ρ( fm ( x), fn ( x)) ≤ 1/k for all n ≥ m}. Then each Ak,m is closed, and so Ak,m \ A◦,m is nowhere dense. Now let k ∞

∞

∞

∞

A◦,m , E = k G= k=1 m=1

(Ak,m \ A◦,m ). k k=1 m=1

Then E is of 1st category, X \ G ⊂ E (since X = ∞=1 Ak,m for all k), and each x ∈ G is a m point of continuity of f .
(b) f −1 (V ) = ∞ 1 ∞=1 { x ∈ X : ρ( fn ( x), Y \ V ) ≥ 1/k for all n ≥ m}. m k=
25

CHAPTER 3. TOPOLOGY
(c) If such a sequence existed then the characteristic function of Q would be continuous at some point by (a).
(d) Let φ : R → R be defined by φ( x) = |2 x − 2k − 1| for x ∈ [k, k + 1], k ∈ Z. Then lim lim φ(m! x)n = χQ ( x) for all x ∈ R.

m→∞ n→∞

E 3.14. Every compact metric space X is the continuous image of the Cantor space {0, 1}N .
S. Construct inductively a family of nonempty closed sets { Bs } s∈{0,1} min f is taken on at least three times.
E 3.16. Suppose that f : [a, b] → R satisfies f −1 ({y}) is closed for all y ∈ R and f ([c, d]) is connected for all [c, d] ⊂ [a, b]. Prove that f is continuous.
S. Let x ∈ [a, b] and take { xn }∞ 1 ⊂ [a, b] such that xn ↑ x. Then I = n= f ([ xn , x]) is an interval containing f ( x). We claim that I = { f ( x)} and therefore f ( xn ) → f ( x). Indeed, take f (y) ∈ I . Then there exist tn ∈ [ xn , x] such that f (tn ) = f (y). Hence tn → x and tn ∈ f −1 ({ f (y)}). Since f −1 ({ f (y)}) is closed, it follows that x ∈ f −1 ({ f (y)}), and so f ( x) = f (y).
∞
n=1

E 3.17. Let (X, ρ) be a metric space. Then there exists a continuous f : X → R that is not uniformly continuous on X if and only if there exist two nonempty disjoint closed sets A and B such that dist(A, B) = 0.
26

CHAPTER 3. TOPOLOGY
S. Suppose that A and B are disjoint closed sets with dist(A, B) = 0. Define f : X → R by ρ( x, A)
.
f ( x) = ρ( x, A) + ρ( x, B)
Then f is continuous but not uniformly continuous. Now if f is a real continuous function on X which is not uniformly continuous, then we can inductively choose points xn , yn ∈ X such that ρ( xn , yn ) < 1/n, | f ( xn ) − f (yn )| ≥ 0 , for a certain 0 , and { xn }∞ 1 ∩ {yn }∞ 1 = ∅. n= n=
The sets { xn : n ∈ N} and {yn : n ∈ N} have the required properties.
E 3.18. Let f : R → R be continuous and satisfy | f ( x) − f (y)| ≥ c| x − y| for all x, y ∈ R, where c does not depend on x and y. Then f (R) = R.
S. Note that f is one-to-one and that
| lim f ( x)| = | lim f ( x)| = ∞. x→∞ x→−∞

E 3.19. Let f : R → R be arbitrary. Show that the set E of x ∈ R such that f has a simple discontinuity at x is at most countable.
S. Suppose that E is uncountable. Then at least one of the sets A = { x : f ( x +) f ( x−)} and B = { x : f ( x+) = f ( x−), f ( x) f ( x+)} must be uncountable.
Without loss of generality, we may assume that A is uncountable, and so there exists a number 0 such that the set { x : | f ( x+) − f ( x−)| > 0 } is uncountable and therefore has a point of accumulation a. Then we can find two sequences { xn }∞ 1 and {yn }∞ 1 such that n= n= xn ↑ a, yn ↑ a and | f ( xn ) − f (yn )| ≥ 0 /2, contradicting the fact that lim x↑a f ( x) exists.
E 3.20. If f : R → R has a local maximum at each x ∈ R, then f (R) is countable. S. For every a ∈ f (R) choose xa ∈ R with f ( xa ) = a and an open interval Ia with rational endpoints such that xa ∈ Ia and for each x ∈ Ia , f ( x) ≥ f ( xa ) = a. Then the function f (R) a → Ia ∈ {( p, q) : p, q ∈ Q} is one-to-one.
E 3.21. Let α ∈ R \ Q. Then the set A = {mα + n : m, n ∈ Z} is dense in R.
S. Note that all the elements of A are distinct since α is irrational. So, the set {mα − [mα] : m ∈ N} is an infinite subset of [0, 1] and therefore has a limit point.
Consequently, there exists {rn } ⊂ A with 0 < rn ↓ 0. Now let x > 0, > 0. Choose n ∈ N with rn < and let m be the smallest integer such that mrn > x. Then (m − 1)rn ≤ x and so,
0 < mrn − x ≤ rn < .

27

CHAPTER 4

Measure and Integration
E 4.1. Let {φn }∞ 1 be an approximate identity in L1 (R) (that is, φn ≥ 0, φn = n= 1, limn→∞ |t|≥δ φn (t)dt = 0 for all δ > 0). Show that limn→∞ φn p = ∞ for all p > 1.
S. Let M > 0. Then there exists n0 ∈ N such that for all n ≥ n0
3/4 ≤

φn (t)dt ≤
|t|≤1/(8 M )

φn (t)dt +
{t:|t|≤1/(8 M )}∩[φn ≤ M ]

≤ 1/4 +

φn (t)dt

[φn ≥ M ]

φn (t)dt.

[φn ≥ M ]

It follows that φn (t)dt ≥ 1/2
[φn ≥ M ]

and therefore p φn −1 (t)φn (t)dt ≥ M p−1

p φn (t)dt ≥
[φn ≥ M ]

We conclude that φn

p

φn (t)dt ≥ 1/2 M p−1 .

[φn ≥ M ]

→ ∞.

E 4.2. Let A ⊂ R be a measurable set with |A| > 0. Then for any n ∈ N, A contains arithmetic progressions of length n.
S. Let x0 be a point of density of A. Choose 0 > 0 such that n 0 < 1/8. Then there exists l > 0 such that |( x0 − l , x0 + l ) ∩ A| ≥ 2(1 − 0 )l for all 0 < l ≤ l. Now choose
> 0 such that n2 < 1/8l. Then for k = 0, 1, . . . , n − 1 we have
|( x0 − l, x0 + l) ∩ k + A| = | k + ( x0 − l − k, x0 + l − k) ∩ A|
≥ | k + ( x0 − l + n, x0 + l − n) ∩ A|
= |( x0 − l + n, x0 + l − n) ∩ A|
≥ 2(1 − 0 )(l − n).
29

CHAPTER 4. MEASURE AND INTEGRATION
Hence
n−1

n−1

( x0 − l, x0 + l) \

( x0 − l, x0 + l) \ k + A

k+A = k =0

k=0 n−1 ≤

|( x0 − l, x0 + l) \ k + A| k=0 n−1

=

(2l − |( x0 − l, x0 + l) ∩ k + A|) k=0 n−1

≤

(2l − 2(1 −

0 )(l

− n))

k=0

< 2 n2 + 2n 0 l < l/4 + l/4 = l/2.
−1
Therefore | n=0 k + A| > 0. In particular, there exists x ∈ k , . . . , x − (n − 1) ∈ A.

n−1 k =0

k + A, and so, x, x −

E 4.3. Let A be a measurable set of reals with arbitrarily small periods (there exist positive numbers pn with pn → 0 so that pn + A = A for all n). Then either A or its complement has measure zero.
S. Suppose that |A| > 0 and |A | > 0. Let x1 be a point of density of A and x2 a point of density of A with x1 < x2 . Then there exists δ > 0 such that
|( x1 − δ, x1 + δ) ∩ A| ≥ 3δ/2,

|( x2 − δ, x2 + δ) ∩ A | > 3δ/2.

It follows that
|( x2 − δ, x2 + δ) ∩ x2 − x1 + A| = |( x2 − x1 ) + ( x1 − δ, x1 + δ) ∩ A|
= |( x1 − δ, x1 + δ) ∩ A| ≥ 3δ/2.
Consider the function φ : [0, ∞) → R defined by φ( x) = |( x2 − δ, x2 + δ) ∩ x + A|.
Then φ is continuous and therefore constant, since it is constant on the dense set {mpn : m, n ∈ N}. Therefore
|( x2 − δ, x2 + δ) ∩ A| = |( x2 − δ, x2 + δ) ∩ x2 − x1 + A| ≥ 3δ/2.
But this is impossible since |( x2 − δ, x2 + δ) ∩ A | ≥ 3δ/2.
E 4.4. Let f : R → R be a measurable function with periods s and t whose quotient is irrational. Prove that f is constant a.e.
S. Note that since s/t is irrational, the set {ns + mt : m, n ∈ Z} is dense in R.
Therefore the set f −1 ([a, b]) has arbitralily small periods and hence has either full or zero measure for all a < b. If it has zero measure for all a < b then f = +∞ or f = −∞ almost everywhere. Suppose that f −1 (I1 ) has full measure for some interval I1 . Divide I1 into two subintervals of equal length. Then the inverse image of one of these subintervals must have full measure. Call this interval I2 . Continuing this way we obtain a decreasing sequence
I1 ⊃ I2 ⊃ · · · of closed intervals whose length tends to zero. Let {r} = ∞ 1 In . Then the n= set f −1 ({r}) = ∞ 1 f −1 (In ) has full measure and therefore f = r almost everywhere. n= 30

CHAPTER 4. MEASURE AND INTEGRATION
E 4.5. Let A, B ⊂ R be measurable sets of positive measure. Show that A − B contains an interval.
S. Let x1 be a point of density of A and x2 a point of density of B. Then there exists δ > 0 such that
|( x1 − δ, x1 + δ) ∩ A| ≥ 3δ/2,

|( x2 − δ, x2 + δ) ∩ B| ≥ 3δ/2.

It follows that
|( x2 − δ, x2 + δ) ∩ x2 − x1 + A| = |( x2 − x1 ) + ( x1 − δ, x1 + δ) ∩ A|
= |( x1 − δ, x1 + δ) ∩ A| ≥ 3δ/2.
Therefore |( x2 − δ, x2 + δ) ∩ ( x2 − x1 + A) ∩ B| > 0. Now consider the function φ( x) = |( x2 − δ, x2 + δ) ∩ ( x + A) ∩ B|.
Then φ is continuous and φ( x2 − x1 ) > 0. Hence there is an interval I such that φ( x) > 0 for all x ∈ I . It follows that ( x + A) ∩ B ∅ for all x ∈ I and so, I ⊂ B − A.
E 4.6. Suppose (X, µ) is a σ-finite measure space and let f : X → C be a measurable function such that | f g| < ∞ for all g ∈ L p (X ). Show that f ∈ Lq (X ) where q is the exponent conjugate to p.
S. Write X = ∞ 1 Ak with Ak disjoint and µ(Ak ) < ∞. Suppose that f ≥ 0, k= f q = ∞ and let Bn = [2n ≤ f < 2n+1 ], n ∈ Z. Then
∞

∞

fq =

∞=

fq
Bn ∩Ak

Bn

n=−∞
∞

∞

fq = n=−∞ k=1
∞
qN (i)

∞

2qn µ( Bn ∩ Ak ) = n=−∞ k=1

2

µ( BN (i) ∩ A M(i) ).

i=1

Let

n

2qN (k) µ( BN (k) ∩ A M(k) )

Sn = k =1

and

∞

g= i=1 Then

2qN (i)/ p χ B N (i ) ∩ A M (i ) .
S N (i)

∞

2qN (i) p µ( BN (i) ∩ A M (i) ) < ∞
S N (i) i=1 by Abel’s Theorem. On the other hand gp =

∞

fg = i= 1
∞

= i= 1

2qN (i)/ p
S N (i)

∞

f≥
BN (i) ∩A M(i)

i=1

qN (i)

2 µ( BN (i) ∩ A M(i) ) = ∞
S N (i)

By Abel’s Theorem again.

31

2qN (i)/ p 2N (i) µ( BN (i) ∩ A M(i) )
S N (i)