solutions MANUAL
Principles of Analysis
Measure, Integration, Functional
Analysis, and Applications
by
Hugo D. Junghenn
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solutionS MANUAL
Principles of Analysis
Measure, Integration, Functional
Analysis, and Applications
by
Hugo D. Junghenn
Boca Raton London New York
CRC Press is an imprint of the
Taylor & Francis Group, an informa business
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CRC Press
Taylor & Francis Group
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Boca Raton, FL 33487-2742
© 2018 by Taylor & Francis Group, LLC
CRC Press is an imprint of Taylor & Francis Group, an Informa business
No claim to original U.S. Government works
Printed on acid-free paper
Version Date: 20180324
International Standard Book Number-13: 978-1-4987-7328-7 (Hardback)
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Library of Congress Cataloging-in-Publication Data
Names: Junghenn, Hugo D. (Hugo Dietrich), 1939- author.
Title: Principles of real analysis : measure, integration, functional
analysis, and applications / Hugo D. Junghenn.
Description: Boca Raton : CRC Press, Taylor & Francis Group, 2018. | Includes
bibliographical references and index.
Identifiers: LCCN 2017061660 | ISBN 9781498773287
Subjects: LCSH: Functions of real variables--Textbooks. | Mathematical
analysis--Textbooks.
Classification: LCC QA331.5 .J86 2018 | DDC 515/.8--dc23
LC record available at https://lccn.loc.gov/2017061660
Visit the e-resources at: https://www.crcpress.com/9781498773287
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Contents
1 Measurable Sets
1
2 Measurable Functions
15
3 Integration
21
4 Lp Spaces
37
5 Differentiation
43
6 Fourier Analysis on Rd
53
7 Measures on Locally Compact Spaces
59
8 Banach Spaces
65
9 Locally Convex Spaces
79
10 Weak Topologies on Normed Spaces
83
11 Hilbert Spaces
89
12 Operator Theory
95
13 Banach Algebras
101
3
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Chapter 1
Measurable Sets
1.1 (a) Consider cases.
(b) 1(A B)c = 1 − 1A B = 1 − |1A − 1B | = |1Ac − 1B )|, etc.
(c) 1Ac B c = |1Ac − 1B c | = |1 − 1A − (1 − 1B )| = |1A − 1B )|.
(d) 1(A∩C) (B∩C) = |1A∩C − 1B∩C | = |1C (1A − 1B )| = 1C∩(A B) .
(e) By De Morgan’s law, the left side is
∞
∞
∞
∞
An ∩
Bnc ∪
Bn ∩
Acn ,
n=1
n=1
n=1
n=1
which is contained in the right side.
1.2 Parts (a) and (b) follow directly from the definitions; (c) follows from (a) and (b); (d)
and (e) follow from De Morgan’s laws; and (g) and (h) follow from set distributive laws.
Parts (f) and (i) may be proved using (a) and (b).
For strict inclusion in (c), let {rn } be an enumeration of the rationals in [0, 1] and
An = [0, rn ). For strict inclusion in (f) define
A2n = [−1/2n, 0),
A2n+1 = (−1/(2n + 1), 0],
B2n = [0, 1/2n)
B2n+1 = (0, 1/(2n + 1)).
For strict inclusion in (i) take complements.
1.3 Note that Cn → C iff C ⊆ limn Cn and limn Cn ⊆ C, which is equivalent to
(i) x ∈ C ⇒ x ∈ Cn eventually and (ii) x ∈ Cn infinitely often ⇒ x ∈ C.
Parts (a) and (b) are readily verified from this. Part (c) follows from Ex. 1.2(d,e)and
part (d) follows from (a)–(c).
1.4 For (a), 1B (x) = 1 iff x ∈ Aj eventually iff inf j≥n 1Aj (x) = 1 for all large n iff
limn 1An (x) = 1.
For (b), 1C (x) = 1 iff x ∈ Aj for infinitely many j iff supj≥n 1Aj (x) = 1 for all n iff
limn 1An (x) = 1.
1.5 (a) x ∈ lim An ⇒ x < an for infinitely many n ⇒ x ≤ limn an .
(b) x < limn an ⇒ x < supk≥n ak for all n. By the approximation property of suprema,
x < an for infinitely many n, that is, x ∈ lim An .
(c), (d), and (e) are similar to (a) and (b).
1.6 (a), (b) P(X).
(c) All unions of {1}, {2}, {3, 4}, {5}, and {6}.
1
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2
Principles of Analysis
1.7 Let G denote the collection of all such sets. Clearly, G ⊆ σ F ∪ {E} . Taking A = X
and B = ∅ shows that E ∈ G. Similarly E c ∈ G. Since A = (A ∩ E) ∪ (A ∩ E c ), F ⊆ G.
It therefore remains to show that G is a σ-field. Closure under countable unions follows
from
c
c
(An ∩ E) ∪ (Bn ∩ E ) =
An ∩ E ∪
Bn ∩ E .
n
n
n
Closure under complements follows from
c
[(A ∩ E) ∪ (B ∩ E c )] = (Ac ∪ E c ) ∩ (B c ∪ E) = (Ac ∩ B c ) ∪ (Ac ∩ E) ∪ (B c ∩ E c ).
1.8 F is closed under complements, so A, B ∈ F ⇒ A ∪ B = (Ac ∩ B c )c = (Ac \ B)c ∈ F.
1.9 Let {xn : n ∈ N} be a set of distinct points in X and consider {x2n : n ∈ N}.
1.10 A, B ∈ F ⇒ A, Ac ∈ Fm and B ∈ Fn for some m, n ⇒ Ac , A ∪ B ∈ Fk , where
k = max{m, n}. Therefore, F is a field. For the example, let Fn denote the field of all
finite unions of members of the partition of [0, 1) consisting of the intervals
[j/2n , (j + 1)/2n ), j = 0, 1, . . . , 2n − 1.
∞
Then for all n, Fn ⊆ Fn+1 and [0, 1/2n ) ∈ Fn but n=1 [0, 1/2n ) ∈ F.
1.11 Let F = {∅, X, {1}, {2, 3}}, G = {∅, X, {2}, {1, 3}}. Then F ∪ G is not closed under
intersections.
1.12 F is σ-field of all subsets of (0, 1) that are either countable or cocountable. Since all
singletons are closed they are Borel sets, hence F ⊆ B(0, 1). No proper open subinterval
of (0, 1) is countable or cocountable.
1.13 F is clearly closed underfinite intersections and complements, hence also under finite
unions. Since [0, 1/2] = n [0, 1/2 + 1/n), F is not a σ-field .
1.14 A ⊆ ϕ(A) ⇒ σ(A) ⊆ σ ϕ(A) , and A ⊆ σ(A) ⇒ ϕ(A) ⊆ σ(A) ⇒ σ ϕ(A) ⊆ σ(A).
1.15 Every countable set is a countable union of finite sets hence is a member of σ(Ff ).
Therefore, Fc ⊆ σ(Ff ). Since Ff ⊆ Fc , σ(Ff ) ⊆ Fc .
1.16 (a) Every open set is a countable union of closed balls, hence O ⊆ σ(K) ⊆ B(Rd ).
(b) Let F = σ(Ir ). Taking unions using xn ↓ a1 ∈ R, xn ∈ Q, we have (a1 , ∞)×R · · ·×R ∈
F. Taking complements and then intersections, (a1 , b1 ] × R · · · × R ∈ F. Similarly for
the other coordinates. Taking intersections, (a1 , b1 ] × · · · × (ad , bd ] ∈ F. These generate
B(R).
∞
∞
1.17 (a) ⇒ (b) is clear. For (b) ⇒ (c), n=1 Bn = n=1 An ∈ F, whereA1 = B1 and
n
A
Bn−1 . For (c) ⇒ (a) and a sequence (Ak ) in F set Bn = k=1 Ak to get
n∞= Bn \ ∞
k=1 Ak =
n=1 Bn ∈ F.
1.18 Let F be the collection of all A ∈ σ(A) such that A ∩ E ∈ σ(A ∩ E). Then F is a σ-field
containing A, hence by minimality F = σ(A), that is, σ(A) ∩ E ⊆ σ(A ∩ E). For the
reverse inclusion, note that σ(A) ∩ E is a σ-field containing A ∩ E and use minimality.
1.19 This follows from Ex. 1.18, since the relative topology of E consists of all sets U ∩ E,
where U is open in X.
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Measurable Sets
3
1.20 Let F denoted the collection of all the sets displayed in the statement of the exercise.
Check that F is a σ-field . By Ex. 1.19, B(a, b) = B[a, b] ∩ (a, b) ⊆ B[a, b]. Therefore,
F ⊆ B[a, b]. Let U ⊆ [a, b] be open. Then W := U ∩ (a, b) is open in (a, b) and hence
is in B(a, b). Since U must be one of the sets W , W ∪ {a}, W ∪ {b}, or W ∪ {a, b},
U ∈ F. By minimality, F = B[a, b].
1.21 By 1.2.4, the left side is σ (A1 ∩ E1 ) × · · · × (Ad ∩ Ed ) = σ (A1 × · · · × Ad ) ∩ E .
Now apply Ex. 1.18.
1.22 Let F denote the collection of all Borel sets B such that B + x is a Borel set. Then F is
a σ-field containing all intervals, hence F = B(Rd ). A similar argument works for rB.
1.23 F is clearly a field and A ⊆ F ⊆ σ(A). Also, F is closed under countable unions, hence
is a σ-field . Indeed, ifAn ∈ F, then there exists a countable
family Cn ⊆ A such that
An ∈ σ(Cn ). Let C = n Cn . Then C is countable and n An ∈ σ(C) ⊆ F.
εm
, where
1.24 For each m-tuple ε = (ε1 , . . . εm ) with εj = ±1, let B ε = B1ε1 ∩ · · · ∩ Bm
Bj if εj = 1,
εj
Bj =
Bjc if εj = −1.
The sets B ε are disjoint since two distinct ε’s will differ in some coordinate j and the
intersection of the corresponding B ε ’s will then be contained in Bj ∩ Bjc . Also, every Bj
is a union of those B ε for which εj = 1. Denoting the nonempty B ε ’s by A1 , . . . , An , we
see that F is generated by the partition {A1 , . . . , An }.
1.25 Let B1 , B2 , . . . be an infinite sequence of distinct members
∞ ofε F not in {∅, X}. For each
sequence ε = (ε1 , ε2 , . . .) with εj = ±1, define B ε = j=1 Bj j , where
ε
Bj j
=
Bj
Bjc
if εj = 1,
if εj = −1.
The sets B ε are disjoint since two distinct ε’s will differ in some coordinate j and the
intersection of the corresponding B ε ’s will then be contained in Bj ∩ Bjc . Denote the
nonempty B ε ’s by A1 , A2 , . . . Since the Bj ’s are unions of the An ’s, the sequence (An )
must be infinite. Now let x ∈ (0, 1) have binary expansion x = .d1 d2 . . . and define T (x)
to be the union of those An for which dn = 1. Removing superfluous expansions, T is
one-to-one hence T (0, 1) has cardinality of the continuum.
For the example, consider the field F consisting of subsets of N that are finite or cofinite.
Each such set may be identified with a rational number, hence F is countable.
n
∞
1.26 (a) If Bn ∈ M, then An := j=1 Bj ∈ M and An ↑ j=1 Bj . An analogous argument
holds for intersections.
(b) Straightforward.
(c) De Morgan’s laws and the monotonicity of m(F) imply that A is monotone. Since F
is closed under complements, F ⊆ A, hence by minimality m(F) = A.
(d) That B is monotone is straightforward. Since F is is closed under finite unions,
F ⊆ B. By minimality, m(F) ⊆ B.
(e) That C is monotone is straightforward. By (d), F ⊆ C. Therefore, by minimality,
m(F) ⊆ C, which is the assertion that m(F) is closed under finite unions.
(f) Follows from (e) and (a).
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4
Principles of Analysis
1.27 Apply countable additivity to the constant sequence An = ∅.
1.28 For 1.3.3(c), let {An } be a disjoint sequence in F and A = n An . If An is countable
for every n, then A is countable and µ(An ) = µ(A) = 0. If Am is cocountable for some
m then A is cocountable and An , as a subset of Acm , is countable
m, hence
for n =
µ(Am ) = µ(A) = 1 and µ(An ) = 0, n = m. In either case, µ(A) = n µ(An ).
For 1.3.3(d), let {An } and A be as above. If x ∈ Am for some n, then µ(Am ) = µ(A) = 1
and µ(An ) = 0 for all n = m. If x ∈ An for every n, then µ(A) = µ(An ) = 0. In either
case, µ(A) = n µ(An ).
1.29 Let µ be counting measure on N and An = {n, n + 1, . . .}.
1.30 (a) Finite additivity is easily established. If X = N then µ({1, 2, . . . , n}) = 0 does not
converge to µ(N) = 1.
(b) Suppose X is uncountable. Let An , A ∈ F and An ↑ A. If some Am is cofinite then
A and An , n ≥ m, are cofinite hence µ(An ) ↑ µ(A). Now suppose every An is finite.
Then Ac cannot be finite so A must be finite, hence again µ(An ) ↑ µ(A).
1.31 From A ⊆ (A B) ∪ B we have µ(A) ≤ µ(A B) + µ(B).
1.32 Since A ∪ B is the union of the pairwise disjoint sets A ∩ B c , A ∩ B, and B ∩ Ac , by
additivity
µ(A ∪ B) = µ(A ∩ B c ) + µ(A ∩ B) + µ(B ∩ Ac ).
Similarly,
µ(A) + µ(B) = µ(A ∩ B c ) + 2µ(B ∩ A) + µ(B ∩ Ac ).
It follows that µ(A ∪ B) = ∞ iff µ(A) + µ(B) = ∞, which proves the assertion in the
infinite case. For the finite case, subtract the above equations.
1.33 By monotonicity, µ(A ∩ B) = 0, hence, by inclusion-exclusion and monotonicity,
µ(A∪B) = µ(A)+µ(B)−µ(A∩B) = µ(A) and µ(A) = µ(A\B)+µ(A∩B) = µ(A\B).
1.34 (By induction on n). The case n = 2 is given in Ex. 1.32. Assume the assertion holds for
n ≥ 2 and set B = A1 ∪ · · · ∪ An . Then the following hold:
µ(A1 ∪ · · · ∪ An+1 ) = µ(B ∪ An+1 ) = µ(B) + µ(An+1 ) − µ(B ∩ An+1 ),
µ(B) + µ(An+1 ) =
n+1
i=1
µ(B ∩ An+1 ) =
n
i=1
µ(Ai ) −
n
µ(Ai ∩ Aj ) + · · · + (−1)n−1 µ(A1 ∩ · · · ∩ An ),
1≤i<j≤n
n
µ(Ai ∩ An+1 ) −
1≤i<j≤n
µ(Ai ∩ Aj ∩ An+1 )
+ · · · + (−1)n−1 µ(A1 ∩ · · · ∩ An ∩ An+1 ).
The assertion for n + 1 follows.
1.35 Let α = µ(X). From 1.34,
n
µ(Aci ) −
α − µ A1 ∩ · · · ∩ An = µ Ac1 ∪ · · · ∪ Acn =
i=1
+
n
n
µ(Aci ∩ Acj )
1≤i<j≤n
µ(Aci ∩ Acj ∩ Ack ) − · · · + (−1)n−1 µ(Ac1 ∩ · · · ∩ Acn ).
1≤i<j<k≤n
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Measurable Sets
5
The general term on the right is
n
(−1)p−1
µ(Aci1 ∩ · · · ∩ Acip ) = (−1)p−1
1≤i1 <···<ip ≤n
= (−1)
p−1
n
α − (−1)p−1
p
n
n
[α − µ(Ai1 ∪ · · · ∪ Aip )]
1≤i1 <···<ip ≤n
µ(Ai ∪ · · · ∪ Aip ).
1≤i1 <···<ip ≤n
Summing,
n
n
n
α−
(−1)p−1
(−1)p−1
α − µ A1 ∩ · · · ∩ An =
p
p=1
p=1
n
µ(Ai ∪ · · · ∪ Aip ).
1≤i1 <···<ip ≤n
Since
n
(−1)p−1
p=1
n
n
n
n
n
=−
=1−
= 1 − (1 − 1)n = 1,
(−1)p
(−1)p
p
p
p
p=1
p=0
the result follows.
1.36 Set B1 := A1 , C1 := ∅, and for n ≥ 2
c
Bn := An ∩ An−1 ∪ · · · ∪ A1 , Cn := An ∩ An−1 ∪ · · · ∪ A1 .
The sets Bn are disjoint, n An = n Bn , µ(An ) = µ(Bn ) + µ(Cn ), and µ(Cn ) = 0,
hence
∞
∞
∞
∞
µ
An = µ
Bn =
µ(Bn ) =
µ(An ).
n=1
∞
n=1
n=1
n=1
∞
1.37 (a) Set Bk := j=k Aj , B := k=1 Bk = limk Ak . Then µ(Bk ) ≤ µ(Ak ) and Bk ↑ B, so
µ(B) ≤ limk µ(Ak ).
∞
∞
(b) Set Ck := j=k Aj , C := k=1 Ck = limk Ak . Then µ(Ck ) ≥ µ(Ak ) and Ck ↓ C, so
µ(C) ≥ limk µ(Ak ).
∞
(c) By countable subadditivity, µ(Ck ) ≤ j=k µ(Aj ) → 0 as k → ∞.
1.38 Let An ↑ A. If {x1 , x2 } ⊆ An for some n, then µ(Am ) = 1 = µ(A) for all m ≥ n. If
{x1 , x2 } ⊆ An for every n then µ(An ) = 0 = µ(A) for all n. But µ is not finitely additive:
µ{x1 , x2 } = µ{x1 } + µ{x2 }.
∞
n
1.39 Let A = n=1 An be a disjoint union of members of F and set Bn = k=1 Ak . Then
Bnc ↓ Ac , hence, by finite additivity and continuity from above,
n
k=1
µ(Ak ) = µ(Bn ) = µ(X) − µ(Bnc ) → µ(X) − µ(Ac ) = µ(A).
1.40 Since µ is clearly finitely additive, it suffices to show continuity from below. Let Aj ↑ A
in F and choose any r < µ(A). Choose n so that r < µn (A), and then choose jn so
that r < µn (Aj ) for all j ≥ jn . For such j, r < µn (Aj ) ≤ µ(Aj ) ≤ µ(A). Therefore,
µ(Aj ) ↑ µ(A).
1.41 Apply Ex. 1.40 to partial sums.
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6
Principles of Analysis
1.42 Let {Ai : i ∈ I} be any family of pairwise disjoint sets of positive measure. Since µ is
finite, {i ∈ I : µ(Ai ) ≥ 1/n} is finite for every n ∈ N. Its union I is therefore countable.
1.43 Let Xn ↑ X with µ(Xn ) < ∞. Suppose first that µ(A) < ∞. Set Em = {E ∈ E :
µ(A ∩ E) ≥ 1/m}. If E1 , E2 , . . . ∈ Em then
1
,
µ(A ∩ Ek ) ≥
µ(A) ≥
m
k
k
hence Em must be finite. Since E = m Em , E is countable. In the general case, for each
n, µ(A ∩ Xn ∩ E) > 0 for at most countably many E ∈ E.
1.44 By monotonicity, µ0 (A) ≤ µ(A). Moreover, if µ(A) is finite, then A is one of the sets in
the definition of µ0 (A), hence µ(A) = µ0 (A)
To show finite additivity, let A1 , A2 ∈ F with A1 ∩A2 = ∅ and let B ∈ F with B ∈ A1 ∪A2
and µ(B) < ∞. Set Bi = Ai ∩ B. Then µ(Bi ) < ∞, hence
µ(B) = µ(B1 ) + µ(B2 ) ≤ µ0 (A1 ) + µ0 (A2 ).
Since B was arbitrary, µ0 (A1 ∪ A2 ) ≤ µ0 (A1 ) + µ0 (A2 ). For the reverse inequality,
let Bi ∈ F, Bi ⊆ Ai , and µ(Bi ) < ∞. Then B1 ∪ B2 ⊆ A1 ∪ A2 and µ(B1 ∪ B2 ) =
µ(B1 ) + µ(B2 ) < ∞, so
µ(B1 ) + µ(B2 ) ≤ µ0 (A1 ∪ A2 ).
Taking the supremum over all B1 and then over all B2 yields µ0 (A1 ) + µ0 (A2 ) ≤
µ0 (A1 ∪ A2 ). Therefore µ0 is finitely additive.
Let An ∈ F and An ↑ A ∈ F. Since µ0 is monotone, α := limn µ0 (An ) exists in [0, ∞]
and α ≤ µ0 (A). For the reverse inequality, let B ∈ F with B ⊆ A and µ(B) < ∞, and
set Bn = B ∩ An . Then Bn ↑ B and µ(Bn ) ≤ µ0 (An ). Taking limits yields µ(B) ≤ α.
Since B was arbitrary, µ0 (A) ≤ α. By 1.3.2, µ0 is a measure.
Now assume the stated condition holds and let A ∈ F. To show that µ(A) = µ0 (A) it
suffices to show that µ(A) ≤ µ0 (A), and for this we may assume µ0 (A) < ∞. Choose
Bn ∈ F with Bn ⊆ A, µ(Bn ) < ∞, and µ(Bn ) → µ0 (A). Replacing Bn by B1 ∪ · · · ∪ Bn ,
we may assume that Bn is increasing, say Bn ↑ B. Then µ(B) = µ0 (A) < ∞. If
µ(A) = ∞, then µ(A \ B) = ∞ and we may apply the condition to get a C ∈ F
with C ⊆ A \ B and 0 < µ(C) < ∞. But then B ∪ C ⊆ A, µ(B ∪ C) < ∞, and
µ(B ∪ C) > µ(B) = µ0 (A), contradicting the definition of µ0 (A). Therefore, µ(A) must
be finite, so µ(A) = µ0 (A).
Conversely, if µ(A) = ∞ and µ(B) = 0 for all B ⊆ A with finite measure, then µ0 (A) = 0,
so µ0 (A) = µ(A).
1.45 For finite subsets S of N define
Ej ∩
Ejc and E =
ES
ES =
j∈S
j∈S c
S
Thus x ∈ ES iff x ∈ Ek for precisely the
∞indices k ∈ S. There are countably many such
sets, they are pairwise disjoint, and k=1 Ek ⊇ E. Moreover, k ∈ S c ⇒ Ek ∩ ES = ∅,
and k ∈ S ⇒ Ek ⊇ ES . Therefore, for D ⊆ E,
s(D) =
∞
k=1
µ(Ek ∩ D) =
=
∞ k=1 S
S k∈S
K29551_SM_Cover.indd 12
µ(Ek ∩ ES ∩ D) =
µ(ES ∩ D) =
S
∞
S k=1
µ(ES ∩ D ∩ Ek )
|S|µ(ES ∩ D),
(†)
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Measurable Sets
7
where |S| denotes the cardinality of S. Now take D to be the union of all ES with
|S| = m. Then D = A and (a) follows from (†). Similarly for (b) and (c).
1.46 The σ-field generated by F and all subsets of X \ E.
1.47 A member of Gν is of the form B = A ∪ N , where A ∈ G, N ⊆ M and M ∈ G with
ν(M ) = 0. Then B ∈ Fµ and ν(B) = ν(A) = µ(A) = µ(B).
1.48 For A, M ∈ F, N ⊆ M , and µ0 (M ) = 0 (= µ(M )), we have
µ0 (A ∪ N ) = µ0 (A) = sup{µ(B) : B ∈ F, B ⊆ A, µ(B) < ∞},
and
µ 0 (A ∪ N ) = sup{µ(C) : C ∈ Fµ , C ⊆ A ∪ N, µ(C) < ∞}.
Let B ∈ F, B ⊆ A and µ(B) < ∞. Then B ⊆ A ∪ N , hence µ(B) ≤ µ 0 (A ∪ N ).
Since B was arbitrary, µ0 (A ∪ N ) = µ0 (A) ≤ µ 0 (A ∪ N ). For the reverse inequality,
let C ∈ Fµ , C ⊆ A ∪ N, and µ(C) < ∞. Then µ(C) = µ(D) for some D ∈ F, D ⊆ C.
Set B = A ∩ D. Since D = B ∪ (Ac ∩ D) and Ac ∩ D ⊆ Ac ∩ C ⊆ Ac ∩ (A ∪ N ) ⊆ M ,
µ(B) = µ(D). Since B is a set in the definition of µ0 (A ∪ N ), µ(B) ≤ µ0 (A ∪ N ).
Therefore, µ(C) ≤ µ0 (A ∪ N ). Since C was arbitrary, µ 0 (A ∪ N ) ≤ µ0 (A ∪ N ).
1.49 For fixed i, if A, M ∈ Fi , N ⊆ M , and µ(M ) = 0, then clearly, A ∪ N ∈ Gµ . Therefore,
H ⊆ Gµ , hence Hµ ⊆ Gµ . Conversely, Fi ⊆ Hµ for every i, hence G ⊆ Hµ and so
Gµ ⊆ Hµ .
1.50 Let B ∈ Fµ , say B = A ∪ N , where N ⊆ M , A, M ∈ F, µ(M ) = 0. Then ν(M ) =
η(M ) = 0, hence B ∈ Fν ∩ Fη and
ν(B) + η(B) = ν(A) + η(A) = µ(A) = µ(B).
1.51 Let B ∈ FµE , say
B = A ∪ N, N ⊆ M, A, M ∈ F and µE (M ) = µ(M ∩ E) = 0.
Then B ∩ E = (A ∩ E) ∪ (N ∩ E), N ∩ E ⊆ M ∩ E, hence B ∩ E ∈ Fµ and
µE (B) = µ(B ∩ E) = µ(A ∩ E) = µE (A) = µE (B).
Therefore, FµE ∩ E ⊆ Fµ ∩ E, and µE = µE on FµE . The reverse inclusion is similar.
1.52 Note first that µ∗ ≤ µ∗ , both set functions are monotone, and µ∗ = µ = µ∗ on F. Let
G = {E ⊆ X : µ∗ (E) = µ∗ (E)}, A, M ∈ F, N ⊆ M , µ(M ) = 0. Then
µ∗ (A ∪ N ) ≤ µ∗ (A ∪ M ) = µ(A ∪ M ) = µ(A) = µ∗ (A) ≤ µ∗ (A ∪ N ),
hence A ∪ N ∈ G. Therefore, Fµ ⊆ G.
Conversely, if E ∈ G, then there exist sequences {Bn } and {An } in F such that
An ⊆ E ⊆ Bn with µ∗ (E) ≤ µ(Bn ) ≤ µ∗ (E) +
1/n and µ∗ (E) − 1/n ≤ µ(An ) ≤
nµ∗ (E).
n
The inequalities still hold if Bn is replaced by j=1 Bj and An is replaced by j=1 Aj ,
hence we may assume that Bn ↓ B ∈ F and An ↑ A ∈ F. Then A ⊆ E ⊆ B,
µ∗ (E) = µ(B), and µ∗ (E) = µ(A). Set N = E \ A and M = B \ A. Then E = A ∪ N ,
N ⊆ M and µ(M ) = 0, so E ∈ Fµ .
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1.53 E measurable iff ∀ C = ∅: µ(E ∩ C) + µ(E c ∩ C) = 1; iff µ(E c ∩ C) = 0 and µ(E ∩ C) = 1
or µ(E c ∩ C) = 1 and µ(E ∩ C) = 0; iff C ⊆ E or C ⊆ E c . This is not possible if
∅ E X. Therefore, M = {∅, X}.
1.54 Let E ⊆ X. A typical sum in the definition of µ∗ (E) is
s :=
δ0 (In ), In ∈ OI , E ⊆
In .
n
n
Suppose 0 ∈ E. Then 0 ∈ In for some n, hence s ≥ 1 and so µ∗ (E) ≥ 1. Since the
∗
intervals (−∞, −1), (−2, 2), (1, ∞) cover E, µ
that 0 ∈ E. Take
(E) = 1. Now suppose
I2n+1 = (−n, 0) and I2n = (0, n). Then E ⊆ n In and s = 0, so µ∗ (E) = 0.
For any C ⊆ X, µ∗ (C) = µ∗ (C ∩ E) + µ∗ (C ∩ E c ) by considering the cases 0 ∈ C and
0 ∈ C. Therefore, M(µ∗ ) = P(R).
1.55 Let E ⊆ X. Then µ∗ (E) ≤ µ(X) = 1. A typical sum in the definition of µ∗ (E) is
µ(An ), An ∈ F, E ⊆
An .
s :=
n
n
If E is countable, then we may take as a cover the single set E hence µ∗ (E) = 0. If
µ∗ (E) < 1, then there exists an s < 1, which implies that µ(An ) < 1 for every n and
hence that E is countable. Therefore, if E uncountable, µ∗ (E) = 1.
If µ∗ (C) = µ∗ (C ∩E)+µ∗ (C ∩E c ) for every C ⊆ X, then, in particular, µ∗ (E)+µ∗ (E c ) =
1, so E or E c is countable. Conversely, if E is cocountable, then for C countable both
sides of µ∗ (C) = µ∗ (C ∩ E) + µ∗ (C ∩ E c ) are zero, and for C uncountable both sides
are one.
1.56 By definition, for A ⊆ X,
∗
(µE ) (A) = inf
∗
n
∗
µ(E ∩ An ) : An ∈ F, A ⊆
(µ )E (A) = µ (E ∩ A) = inf
n
n
An
µ(Bn ) : Bn ∈ F, E ∩ A ⊆
and
n
Bn
.
A typical member
of the first set may be written s := n µ(Bn ), where Bn∗:= E ∩ An .
Since E ∩ A ⊆ n Bn , s is also a member of the second set. Therefore, (µ )E (A) ≤ s,
and since s was arbitrary, (µ∗ )E (A) ≤ (µE )∗ (A).
Now consider a typical member t := n µ(Bn ) of the second set. Then the sets E c and
Bn cover A, hence
µ(E ∩ Bn ) ≤
µ(Bn ) ≤ t.
(µE )∗ (A) ≤ µ(E ∩ E c ) +
n
n
Therefore, (µE )∗ (A) ≤ (µ∗ )E (A).
1.57 Let E ⊆ X. A typical sum in the definition of µ∗b (E) is
µ(Bn ), where Bn ∈ B and E ⊆
Bn .
s :=
n
n
We show that µ∗a (E) ≤ s, and for
assume that s < ∞. Let ε > 0 and for each
this we may
n choose Aj,n such that Bn ⊆ j Aj,n and j µ(Aj,n ) < µ∗a (Bn ) + ε/2n = µ(Bn ) + ε/2n .
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Then E ⊆ j,n Aj,n hence µ∗a (E) ≤ j,n µ(Aj,n ) ≤ s + ε. Therefore, µ∗a (E) ≤ s, hence
µ∗a (E) ≤ µ∗b (E). Similarly, µ∗a (E) ≥ µ∗b (E).
Counterexample: Take X = {1, 2}, A := P(X), B := {X, ∅}, and µ = counting measure.
Then µ∗a {1} = 1 and µ∗b {1} = 2.
1.58 By measurability of A,
µ∗ (E ∪ A) = µ∗ [(E ∪ A) ∩ A] + µ∗ [(E ∪ A) ∩ Ac ] = µ(A) + µ∗ (E).
1.59 By measurability of A,
µ∗ E ∩ (A ∪ B) = µ∗ [E ∩ (A ∪ B)] ∩ A + µ∗ [E ∩ (A ∪ B)] ∩ Ac
= µ∗ (E ∩ A) + µ∗ (E ∩ B).
Now let (An ) be a disjoint sequence in M(µ∗ ) with union A. By countable subadditivity,
monotonicity, and induction
∞
n
n
∗
∗
∗
µ (E ∩ Ak ) ≥ µ (E ∩ A) ≥ µ E ∩
Ak =
µ∗ (E ∩ Ak ).
k=1
k=1
k=1
1.60 Let C ⊆ X, and An ∈ A with C ⊆ n An . By monotonicity, countable subadditivity,
and the hypothesis,
µ∗ (An ∩ E) +
µ∗ (An ∩ E c ) =
µ∗ (An ).
µ∗ (C ∩ E) + µ∗ (C ∩ E c ) ≤
n
n
n
Therefore, µ∗ (C ∩ E) + µ∗ (C ∩ E c ) ≤ µ∗ (C), hence E is measurable.
1.61 A1 × A2 is clearly a π system. Let Ai , Bi ∈ Ai withB1 × B2 ⊆ A1 × A2 , and let
ni
Ci,j ∈ Ai , j = 1, . . . , ni be disjoint such that Ai \ Bi = j=1
Ci,j , i = 1, 2. Then
(A1 × A2 ) \ (B1 × B2 ) = [(A1 \ B1 ) × B2 ] ∪ [A1 × (A2 \ B2 )] ∪ [(A1 \ B1 ) × (A2 \ B2 )],
and (A1 \ B1 ) × B2 =
ni
j=1
C1,j × B2 , a disjoint union of members of A1 × A2 , etc.
1.62 (a) Clearly A ∩ E is a π-system. Let A, B, Cj ∈ A such that A \ B =
Then
n
(A ∩ E) \ (B ∩ E) = (A \ B) ∩ E =
Cj ∩ E.
n
j=1
Cj (disjoint).
j=1
(b) Let F ⊆ E. A typical sum in the definition of µ∗ (F ) is of the form
s :=
µ(An ), where An ∈ A and F ⊆
An .
n
n
∗
Then F ⊆ n (An ∩ E), hence s ≥ n µ(An ∩ E) ≥ ν (F ). Therefore, µ∗ (F ) ≥ ν ∗ (F ).
For the reverse inequality, note that a typical sum in the definition of ν ∗ (F ) is of the
form
µ(Bn ), where Bn ∈ A ∩ E and F ⊆
Bn .
t :=
n
∗
n
∗
∗
Since Bn ∈ A, t ≥ µ (F ). Therefore, ν (F ) ≥ µ (F ).
(c) follows from (b).
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Principles of Analysis
1.63 (a) Let E ⊆ n An , An ∈ A. Then ν(E) ≤ n ν(An ) = n µ(An ), hence ν(E) ≤
µ∗ (E) = µ(E).
(b) We
may assume A is a ring. Choose disjoint An ∈ A such that E ⊆ A := n An
and n µ(An ) < µ(E) + ε. Then
µ(E) ≤ µ(A) = ν(A) = ν(E) + ν(A \ E) ≤ ν(E) + µ(A \ E) ≤ ν(E) + ε.
take A = X. Now let µ∗ (E) < ∞ and for each
1.64 If µ∗ (E) = ∞, then µ∗ (X) = ∞, hence
∞
k ∈ N choose Aj,k ∈ A such that E ⊆ j=1 Aj,k and j µ(Aj,k ) < µ∗ (E) + 1/k. Define
n ∞
An = k=1 j=1 Aj,k . Then E ⊆ An ,
µ(Aj,n ) < µ∗ (E) + 1/n,
µ∗ (E) ≤ µ(An ) ≤
j
∗
and An ↓ A ∈ σ(A), hence µ (E) = µ(A).
1.65 Choose A, B ∈ σ(A) such that E ⊆ A, F ⊆ B, µ∗ (E) = µ(A), and µ∗ (F ) = µ(B)
(Ex. 1.64). Then, by monotonicity and inclusion-exclusion,
µ∗ (E ∪ F ) + µ∗ (E ∩ F ) ≤ µ(A ∪ B) + µ(A ∩ B) = µ(A) + µ(B) = µ∗ (E) + µ∗ (F ).
1.66 Let E ⊆ n An , An ∈ A. Then
µ(An ) +
ν(An ) =
(µ + ν)(An ).
µ∗ (E) + ν ∗ (E) ≤
n
n
∗
∗
n
∗
Since the An were arbitrary, µ (E) + ν (E) ≤ (µ + ν) (E). Now let A, B ∈ σ(A) such
that E ⊆ A ∩ B, µ∗ (E) = µ(A), and ν ∗ (E) = ν(B) (Ex. 1.64). Then
µ∗ (E) + ν ∗ (E) = µ(A ∩ B) + ν(A ∩ B) = (µ + ν)(A ∩ B) ≥ (µ + ν)∗ (E).
Therefore, µ∗ (E) + ν ∗ (E) = (µ + ν)∗ (E).
For the inclusion, let C ⊆ X. Since
µ∗ (C) = µ∗ (C ∩ E) + µ∗ (C ∩ E c ) and ν ∗ (C) = ν ∗ (C ∩ E) + ν ∗ (C ∩ E c ),
we have (µ∗ + ν ∗ )(C) = (µ∗ + ν ∗ )(C ∩ E) + (µ∗ + ν ∗ )(C ∩ E c ).
The inclusion may be strict. For example, let (X, F, µ) be as in 1.3.3(c) and let ν be the
measure on F that assigns ∞ to every nonempty set. Then M(µ∗ ) ∩ M(ν ∗ ) = F and
M(µ∗ + ν ∗ ) = P(X).
α ≤ µ∗ (E). Now let
1.67 Let E ⊆ X. By monotonicity, α := limn µ∗n (E) exists in [0, ∞] and ∗
An ∈ σ(A) with E ⊆ An and µn (An ) = µn (E) (Ex. 1.64). Set A = n An . Then E ⊆ A
and µn (A) = µ∗n (E) for every n, so α = limn µn (A) = µ(A) ≥ µ∗ (E).
1.68 Follows from Ex. 1.66 and 1.67 applied to the partial sums.
1.69 Let En ⊆ X and En ↑ E. By Ex. 1.6.4, there exist An , A ∈ σ(A) such that
En ⊆ An , µ∗ (En ) = µ(An ), E ⊆ A, and µ∗ (E) = µ(A).
Since En ⊆ Ej ⊆ A ∩ Aj for j ≥ n,
En ⊆ Bn := A ∩
j≥n
Aj ⊆ A ∩ An .
Then Bn ↑ B ∈ σ(A), E ⊆ B, µ∗ (En ) = µ(Bn ) and µ(A) = µ∗ (E) ≤ µ(B) ≤ µ(A), so
lim µ∗ (En ) = lim µ(Bn ) = µ(B) = µ∗ (E).
n
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1.70 By 1.64, choose A, B ∈ σ(A) such that
E ⊆ A, µ∗ (E) = µ(A), and E c ⊆ B, µ∗ (E c ) = µ(B).
Then A ∪ B = X, hence, by inclusion-exclusion,
µ(A ∩ B) = µ(A) + µ(B) − µ(A ∪ B) = µ∗ (E) + µ∗ (E c ) − µ(X) = 0.
Since E ∩ B ⊆ A ∩ B we have µ∗ (E ∩ B) = 0 and so E ∩ B ∈ M(µ∗ ). Therefore,
E = B c ∪ (E ∩ B) ∈ M(µ∗ ).
1.71 If I = (a, b] := (a1 , b1 ] × · · · ×(ad , bd ], then int(I) = n In , In := j (aj , bj − 1/n] hence
λ(int I) = limn λ(In ) = limn j (bj − 1/n − aj ) = |I|. Similarly, λ(cl I) = |I|. It follows
that
∞
∞
∗
(†)
λ (E) = inf
λ(An ) : An ∈ A and E ⊆
An
n=1
n=1
for A = OI or CI . Since OI ⊆ O and each open set is a union of open intervals, (†)
holds for A = O. Since CI ⊆ K and K ∈ K is contained in Un = {x : d(x, K) < 1/n},
which is open and ↓ K, (†) holds for A = K.
1.72 N can contain no open d-dimensional interval, hence every such interval must contain a
point of N c .
of length, λ(x+I) = λ(I) for all I ∈ H
1.73 (a) By properties I . A typical sum in the definition
of λ∗ (E) is s := n λ
(In ), where In ∈ HI and E ⊆ n In . Then Jn := In + x ∈ HI ,
E + x ⊆ n Jn , and n λ(Jn ) = s, hence s ≥ λ∗ (E + x). Therefore λ∗ (E + x) ≥ λ∗ (E).
Replacing x by −x and E by E + x yields the reverse inequality.
(b) Let C ⊆ Rd and D = C − x. Then
C ∩ (x + E) = x + D ∩ E and C ∩ (x + E)c = x + D ∩ E c ,
hence, by (a) and measurability of E,
λ∗ C ∩ (x + E) + λ∗ C ∩ (x + E)c = λ∗ (D ∩ E) + λ∗ (D ∩ E c ) = λ∗ (D) = λ∗ (C).
Therefore, x + E ∈ M(λ).
1.74 Both assertions are clear if r = 0. Consider next r = −1. By properties of length,
∗
λ(−I) = λ(I) for all d-dimensional intervals I. A
typical sum in the definition of λ (E)
is s =:
(In ), where In ∈ OI andE ⊆ n In (using Ex. 1.71 with A = OI ).
n λ
Then −E ⊆ n (−In ), −In ∈ OI , and n λ(−In ) = s, hence s ≥ λ∗ (−E). Therefore,
λ∗ (E) ≥ λ∗ (−E). Replacing E by −E yields the reverse inequality.
Let C ⊆ Rd , D := −C, and F := −E. Then
C ∩ F = −[D ∩ E] and C ∩ F c = −[D ∩ E c ],
hence, by measurability of E and the first part,
λ∗ C ∩ F + λ∗ C ∩ F c = λ∗ (D ∩ E) + λ∗ (D ∩ E c ) = λ∗ (D) = λ∗ (C).
Therefore, F ∈ M(λ). This proves (a) and (b) for r = −1.
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Principles of Analysis
It remains to treat the case r > 0. If I = (a1 , b1 ] × · · · × (ad , bd ], then rI = (ra1 , rb1 ] ×
· · · × (rad , rbd ] hence
λ(rI) =
d
j=1
(rbj − raj ) = rd
d
(bj − aj ) = rd λ(I).
j=1
Now let E ⊆ Rd and Ij ∈ HI such that E ⊆ j Ij . Then rE ⊆
r−d λ∗ (rE) ≤ r−d
λ(rIj ) =
λ(Ij ).
j
j
rIj hence
j
Since the Ij ’s were arbitrary, r−d λ∗ (rE) ≤ λ∗ (E) or λ∗ (rE) ≤ rd λ∗ (E). Write E =
r−1 (rE) to get the reverse inequality. Therefore, (a) holds for all r.
Now let C ⊆ Rd and set D := r−1 C. Then
r−1 [C ∩ (rE)] = D ∩ E and r−1 [C ∩ (rE)c ] = D ∩ E c ,
hence
r−d λ∗ C ∩ (rE) + r−d λ∗ C ∩ (rE)c = λ∗ (D ∩ E) + λ∗ (D ∩ E c ) = λ∗ (D) = r−d λ∗ (C).
Therefore, rE ∈ M(λ), so (b) holds for all r.
∞
1.75 For d = 1, take U = n=1 (rn − ε/2n+1 , rn + ε/2n+1 ), where {r1 , r2 , . . . , } is an enumeration of the rationals. A similar argument proves the case d > 1.
1.76 λ∗ (A ∩ B c ) ≤ λ([0, 1] ∩ B c ) = λ([0, 1]) − λ(B) = 0, hence
λ∗ (A) ≤ λ∗ (A ∩ B) + λ∗ (A ∩ B c ) = λ∗ (A ∩ B) ≤ λ∗ (A).
1.77 Since λ(E)/r > λ(E), there exists a cover of E by be closed bounded intervals In such
that
λ(In ) < λ∗ (E)/r ≤
λ∗ (In ∩ E)/r.
n
n
Therefore, λ(In ) ≤ λ (E ∩ In )/r for some n.
∗
1.78 Let g(x, y) = f (x) − y. Then g is continuous and G = g −1 ({0}), hence G is a Borel set.
For the second assertion, let a < b and set
Ga,b := {(x, y) : y = f (x), a ≤ x ≤ b}
By uniform continuity, given ε > 0, there exists δ > 0 such that u, v ∈ [a, b] and
|u − v| < δ imply |f (u) − f (v)| < ε/2(b − a). Set aj = a + j(b − a)/k, j = 0, 1, . . . , k,
where k > (b − a)/δ. Then the intervals
[aj−1 , aj ] × [f (aj−1 ) − ε/2(b − a), f (aj−1 ) + ε/2(b − a)]
cover Ga,b and the measure of their union is ε. Since ε was arbitrary, λ(Ga,b ) = 0. Since
G is a countable union of such sets, λ(G) = 0.
1.79 (a) µ(B) = ν(B ∩ Z), where ν is counting measure on Z. (b) µ(B) = λ(B ∩ [0, 1]).
1.80 The sum and product of right continuous functions is right continuous. If F, G are
nonnegative distribution functions and a < b, then
F G(b) − F G(a) = [F (b) − F (a)]G(b) + [G(b) − G(a)]F (a) ≥ 0.
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1.81 Let a ∈ R and ε > 0. Choose N so that n>N pn < ε, then choose δ > 0 so that
(a, a + δ) contains none of the numbers c1 , c2 , . . . , cN . If a < x < a + δ, then
pn ≤
pn < ε.
0 ≤ F (x) − F (a) =
n:a<cn ≤x
n>N
Therefore, F is right continuous at a.
If a = cn for every n, we may choose δ so that (a − δ, a] contains none of the numbers
c1 , c2 , . . . , cN . If a − δ < x < a, then, as before,
0 ≤ F (a) − F (x) =
pn < ε.
n:x<cn ≤a
Therefore, F is left continuous at a. On the other hand, if x < ck then
F (ck ) − F (x) =
pn ≥ pk ,
n:x<cn ≤ck
so F is not left continuous at ck .
1.82 For example, for (a) use µ(a, b) = limn µ(a, b − 1/n] (continuity from below), and for (c)
use µ[a, b) = limn µ(a − 1/n, b] (continuity from above).
1.83 F is continuous and the limits limx→±∞ F (x) are finite.
1.84 For each t at which f is discontinuous, choose a rational number rt in (at , bt ). Since the
correspondence t → rt is one-to-one, the set of discontinuities of f is countable.
1.85 Choose n such that µ A ∩ [−n, n] > b. Define G(x) = µ A ∩ [−n, x] . Then G is
continuous, G(−n) = 0 and G(n) > b, so by the intermediate value theorem, G(x) = b
for some x ∈ [−n, n]. Take B = A ∩ [−n, x].
1.86 The Cantor set is Borel measurable, has measure 0, and contains a set that is not Borel
measurable.
1.87 (a) By 1.8.1, there exists a compact set K ⊆ A such that λ(A \ K) = λ(A) − λ(K) <
λ(A)/2, hence λ(K) > λ(A)/2 > 0. Thus if the result holds for K it holds for A.
(b) By 1.8.1, there exists an open set U ⊇ A such that λ(U ) − λ(A) < λ(A). For any
a, b ∈ U and y ∈ U c , |a − b| + |b − y| ≥ |a − y| ≥ d(a). Taking the infimum over y yields
|a − b| + d(b) ≥ d(a) or |a − b| ≥ d(a) − d(b). Therefore, |a − b| ≥ |d(a) − d(b)|, hence d
is continuous. That d is positive on U follows from the fact that U c is closed.
(c) If x + a ∈ U c for some a ∈ A, then |x| = |(x + a) − a| ≥ r. Therefore, |x| < r ⇒
x + A ⊆ U . Moreover, A cannot be disjoint from A + x, otherwise, by translation
invariance,
2λ(A) = λ(A) + λ(A + x) = λ A ∪ (A + x) ≤ λ(U ),
contradicting the choice of U in part (b). Thus (x + A) ∩ A =
∅ for all x with |x| < r,
hence (−r, r) ∈ A − A.
1.88 If A is such a subgroup, then (−r, r) ⊆ A − A ⊆ A for some r > 0, hence (−nr, nr) ∈ A
for all n.
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1.89 (a) By induction, λ(En ) =
n
j=1 bj .
(b) If I is an open interval contained in En then, because I is connected, I must be
contained in one of the intervals comprising En , hence λ(I) < 1/2n . Therefore, E can
contain no open interval.
(c) Then bn < 1 − r < 1, hence λ(En ) < (1 − r)n → 0.
n
n
(d) Use the hint with an = 1 − a1/2 , so that bn = a1/2 and so
ln
∞
n=1
∞
bn =
ln bn = ln a.
n=1
1.90 Set A0 = [0, 1]. Since d1 = 3, A must be contained in the set A1 obtained by removing
from A0 the interval [.3, .4). Similarly, since d2 = 3, A is contained in the set A2 obtained
by removing from A1 the nine intervals of the form [.d1 3, .d1 4), d1 = 3. Having obtained
An−1 in this way, we see that A must be contained in the set An gotten by removing
from An−1 the 9n−1 intervals of the form [.d1 d2 . . . dn−1 3, .d1 d2 . . . dn−1 4), dj = 3. Since
each of these intervals has length 10−n , λ(An ) = λ(An−1 ) − 9n−1 10−n , hence
λ(Am ) = 1 +
m
n=1
λ(An ) − λ(An−1 ) = 1 − (.1)
m
n=1
(.9)n−1 → 0 as m → 0.
By construction A ⊆ n An . But since every member of the intersection has a decimal
expansion of the required type, equality holds. Therefore, A is a Borel set and λ(A) =
limn λ(An ) = 0.
To show that A is uncountable, note that A ⊇ B, where B consists of all x ∈ [0, 1] having
a decimal expansion .d1 d2 . . . with no digit dj odd. The mapping which takes such an x
to .e1 e2 . . .(base 5), where en = dn /2 may be used to show that B is uncountable.
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