REAL ANALYSIS - UFPRpeople.ufpr.br/~jrrb/RoydenFitz.pdf · real analysis fourth edition (2010), first printing royden and fitzpatrick partial scrutiny, solutions of selected problems,

REAL ANALYSISFOURTH EDITION (2010), FIRST PRINTING

Royden and Fitzpatrick

PARTIAL SCRUTINY,SOLUTIONS OF SELECTED PROBLEMS,

COMMENTS, SUGGESTIONS AND ERRATA

Jose Renato Ramos Barbosa

2013

Departamento de Matematica

Universidade Federal do Parana

Curitiba - Parana - Brasil

[email protected]

=============================================================================Errata by Fitzpatrick1

• On the dedication page, To John Slavins, H. L. Royden was inadvertently omitted;

• Contents:

– In the title of 15.1, change Helley to Helly;

– In the title of 19.2, change p ≤ ∞ to p < ∞;

– In the title of 19.4, change p < 1 to p < ∞;

• Preface:

– On p. x, Replace 1998 by 1990;

– On p. xi, replace Helley by Helly, twice.

==========================================================================================================================================================PART ONE==========================================================================================================================================================1==========================================================================================================================================================Errata, p. 4, ll. -10 and -9

Delete either define or is defined !=============================================================================Errata, p. 9Between The Completeness Axiom and The triangle inequality, S should be E.=============================================================================PROBLEMS 4-5, p. 10Cf. [Fit06]2, EXERCISES 17,20, pp. 11-12. Here comes the solution of EX. 17, p. 11:

• b. x ∈ S, 0 < r < min{

b2−c2b , b

}⇒ (b− r)2 > b2 − 2br > c > x2 ⇒ x < b− r!

• c. 0 < r < min{

c−b2

2b+1 , 1}⇒ (b+ r)2 < b2 + 2br+ r < c ⇒ b+ r ∈ S!

=============================================================================PROB. 6, p. 11There is a b ∈ R such that b ≤ x for all x ∈ E. Hence −x ≤ −b for all x ∈ E. Thus, if s = sup {−x | x ∈ E},

−s ≤ x for all x ∈ E and b ≤ −s for each such b.=============================================================================Errata by Fitzpatrick

• (iii), PROB. 15, p. 13: Replace rn−1 by rn+1.

• Lines -9 and -8, p. 14: The image of g is contained in the integers, not the natural numbers. So redefineg as follows: define g(x) = 2((p+ q)2 + q) for x = p/q > 0, g(x) = g(−x) + 1 for x = p/q < 0, andg(0) = 1.

=============================================================================PROB. 21, p. 16Let f be an invertible mapping from {1, . . . , (n+ 1) +m} onto {1, . . . , n+ 1}. If f (n + 1 + m) = n + 1,

define g(i) = f (i) for each i ∈ {1, . . . , n+m}. Otherwise, define g(i) = f (i) for each i ∈ {1, . . . , n+m} ∼{f−1(n+ 1)

}and g( f−1(n+ 1)) = f (n+ 1+m). In any case, g is an invertible mapping from {1, . . . , n+m}

onto {1, . . . , n}. This contradicts the induction hypothesis.=============================================================================Comment, p. 17, Proof, Prop. 9

1Patrick Fitzpatrick’s errata placed on http://www2.math.umd.edu/∼pmf/RealAnalysis/index.html.2Patrick Fitzpatrick’s Advanced Calculus, Thomson Brooks/Cole, 2006.

• “By the definition of bx, there is a number y > w such that (x, y) ⊆ O, ...” : Suppose (x, y) 6⊆ O for each

y > w. Thus, since y ≤ w for each (x, y) ⊆ O, w = bx!

• “... {Ix}x∈O is disjoint.” : If Ix1 ∩ Ix2 6= ∅, either one of ax1 , bx1 , ax2 and bx2 belongs to O, which is not

possible, or Ix1 = Ix2 = Ix for some x ∈ O.

=============================================================================Errata, p. 17, l. -7

“Therefore every open interval that contains x ...” .=============================================================================Errata, p. 18, Proof, Prop. 11

X ∼ E should be R ∼ E .=============================================================================Comment, The Heine-Borel Theorem, p. 18Concerning its Proof, if c < b then there is an ǫ′ < ǫ such that {O1, . . . ,Ok,O} covers [a, c+ ǫ′]!=============================================================================Errata/Comment, p. 19, Proof, The Nested Set Theorem

• 4 should be 11 and F should be F1;

• The hypothesis that F1 is bounded works with The Heine-Borel Theorem.

=============================================================================Errata, p. 20, l. 6

Shouldn’t “... Proposition 4 ...” be “... Proposition 11 ...” ?=============================================================================PROB. 27, p. 20Q is open (thus R ∼ Q is closed) since its points are isolated;Q is not closed (thus R ∼ Q is not open) since R ∼ Q ⊂ Q.=============================================================================Errata by Fitzpatrick, p. 23, last lineReplace the lim inf by lim sup.=============================================================================PROB. 39, p. 24For (i), let En = {ak | k ≥ n} and sn = sup En. Thus limn→∞ sn = l iff for every ǫ > 0, there is an index N

such that sn ∈ (l − ǫ, l + ǫ) for every index n ≥ N. Hence, concerning the ⇒ part of (i), for each index n ≥ N,since l − ǫ is not an upper bound for En, there is a kn ≥ n such that akn > l − ǫ, and, since l + ǫ is an upperbound for En, ak ≤ l + ǫ for all k ≥ n. Also note that, since E1 = {a1, . . . , aN , . . .} , E2 = {a2, . . . , aN , . . .} , . . .,we can have at most N − 1 indices n with an > l + ǫ. Now, for the ⇐ part of (i), since there are only finitelymany indices n with an > l + ǫ, there is an index N such that an ≤ l + ǫ for all n ≥ N. Therefore sn ≤ l + ǫprovided n ≥ N. (The problem with the ≤ is the equals sign!) Also note that there cannot be an index N′ forwhich sN′ ≤ l − ǫ. Otherwise an ≤ l − ǫ for all n ≥ N′. (Thus there are only finitely many indices n withan > l − ǫ!) Hence the decreasing sequence sn is at least bounded. Now use Theo. 15, p. 21;

For (ii), (iii) (via PROB. 6, p. 11) and (v), the proof is trivial;For (iv), if a = ∞, use (ii) and (iii). Otherwise, the ⇒ part of (iv) follows from

{an} → a ∈ R

(i)︸︷︷︸⇒ lim sup {an} = a

{−an} → −a︸︷︷︸=⇒ lim inf {an}

(iii)︸︷︷︸= − lim sup {−an} = −(−a).

Now, for the ⇐ part of (iv), suppose {an} 6→ a ∈ R. Hence there is some ǫ > 0 so that for all N there is ann ≥ N with either an ≥ a+ ǫ or an ≤ a− ǫ. Thus there are infinitely many n where this happens.

=============================================================================Errata by Fitzpatrick, p. 24

• PROB. 42: Replace > by ≥.

• PROB. 44: Replace 0 < x < 1 by 0 ≤ x ≤ 1 and replace q/pn by q/pn , 0 < p < q.

=============================================================================Errata, last paragraph before PROBLEMS, p. 27Replace “... the Monotone Convergence Theorem for Sequence for Real Sequences that if {xn} is a sequence

...” by “... the Monotone Convergence Criterion for Real Sequences that if {xn} is a decreasing sequence ...”.=============================================================================PROB. 58, p. 28Let O be an open set. Thus, by Prop. 22, p. 25, there is an open set U such that f−1(O) = R ∩ U , which is

open. Then, if F is a closed set, f−1(F) is closed since R ∼ f−1(F) = f−1(R) ∼ f−1(F) = f−1(R ∼ F) is open.Now, f−1(B) is a Borel set if B is a Borel set. In fact, see PROB. 46, p. 53.

==========================================================================================================================================================2==========================================================================================================================================================PROB. 1, p. 31Since B = A ∪ (B ∼ A) ∪ ∅ ∪ ∅ ∪ . . ., m(B)−m(A) = m(B ∼ A) +m(∅) +m(∅) + · · · ≥ 0.=============================================================================PROB. 2, p. 31Since A ∪ ∅ ∪ ∅ ∪ . . . = A and m(A) < ∞, m(∅) +m(∅) + · · · = m(A)−m(A) = 0.=============================================================================m∗(A), p. 31{

∑∞k=1 l(Ik) | A ⊆ ∪∞

k=1 Ik}6= ∅ since A ⊆ R = I1 ∪

(∪∞j=1(I2j ∪ I2j+1)

)for I1 = (−r, r), I2j = (j− r, j+ r),

I2j+1 = (−j− r,−j+ r) and r > 1.=============================================================================Ex., p. 31

• ∑∞k=1

12k

=12

1− 12

= 1;

• Errata by Fitzpatrick: Replace E by C.

=============================================================================Comment, p. 33, Proof, Prop. 2

For the conclusion: Jk = (ck, dk), ck − y = ak and dk − y = bk ⇒ Jk = Ik + y with Ik = (ak, bk) .=============================================================================PROB. 5, p. 34Suppose that [0, 1] is countable. Then m∗([0, 1]) = 0 (Ex., p. 31). But m∗([0, 1]) = 1 (Prop. 1, p. 31).=============================================================================PROB. 6, p. 34Let B be the set of rational numbers in [0, 1]. Then, since m∗(B) = 0 (Ex., p. 31), m∗(A) = m∗(A ∪ B) =

m∗([0, 1]) = 1 by PROB. 9, p. 34.=============================================================================PROB. 9, p. 34Since m∗ is finitely subadditive (p. 34), m∗(A ∪ B) ≤ m∗(A) + m∗(B) = m∗(B). Since m∗ is monotone (p.

31) and B ⊆ A ∪ B, m∗(B) ≤ m∗(A ∪ B).=============================================================================Errata by Fitzpatrick, p. 35, l. 5Replace Constantine by Constantin.=============================================================================Errata, p. 35, l. 11‘))’ should be ‘)’.=============================================================================Errata by Fitzpatrick, p. 39, l. 2Change the second A1 to A2.=============================================================================Comment, p. 39, Proof, Prop. 10

m∗(A ∩ [E+ y])m∗ is translation invariant︸︷︷︸

= m∗({A ∩ [E+ y]} − y) = m∗([A− y] ∩ E) since

x ∈ [A− y] ∩ E ⇔ x+ y ∈ A and x+ y ∈ E+ y ⇔ x+ y ∈ A ∩ [E+ y] ⇔ x ∈ {A ∩ [E+ y]} − y.

=============================================================================PROB. 11, p. 39A σ-algebra (see definition on p. 38), besides containing R = (−∞,∞), that contains intervals of the form

(a,∞), also contains intervals of the form:

• (−∞, a] = R ∼ (a,∞);

• (a, b] = (a,∞) ∩ (−∞, b] with a < b;

• (a, b) = ∪∞k=1(a, xk] with a < b, x1 = a+b

2 and xk+1 = xk+b2 for each index k ≥ 1. In fact, since {xk} is

bounded (a = a+a2 < x1 <

b+b2 = b and, if a < xk < b, a = a+a

2 < xk+1 <b+b2 = b) and monotone

(xk+1 >xk+xk

2 = xk), there exists limk→∞ xk = β. Then β = β+b2 , that is, β = b. It follows that (a, b) ⊃

∪∞k=1(a, xk] (easy!) and (a, b) ⊂ ∪∞

k=1(a, xk] (if x ∈ (a, b), then there is an index k such that a < x < xk.Otherwise, limk→∞ xk ≤ x < b!);

• [b, b] = {b} = (a, b] ∼ (a, b) if a < b;

• (−∞, a) = (−∞, a] ∼ {a};• [a,∞) = (a,∞) ∪ {a};• [a, b) = [a,∞) ∩ (−∞, b) with a < b.

=============================================================================PROB. 12, p. 39

• A way to solve the problem: The Borel σ-algebra contains intervals of the form (a,∞). Now use the lastproblem.

• Another way to solve the problem: For all {a, b} ⊂ R with a < b, [a,∞) = R ∼ (−∞, a), (−∞, a] =R ∼ (a,∞), (a, b] = (a,∞) ∩ (−∞, b], [a, b) = [a,∞) ∩ (−∞, b) and [a, b] = R ∼ [(−∞, a) ∪ (b,∞)]are Borel sets since open intervals are Borel sets. Note that the degenerate interval [a, a] = {a} =⋂∞

k=1

[a− 1

k , a+1k

]is a Borel set.

=============================================================================Comment/Errata, p. 41, Proof, Theo. 10

• ll. 6 and 7: E ⊆ R = A1 ∪(∪∞j=1(A2j ∪ A2j+1)

)for A1 = (−1, 1), A2j = (−2j,−(2j− 1)] ∪ [2j − 1, 2j)

and A2j+1 = (−(2j+ 1),−2j]∪ [2j, 2j+ 1). Then each Ek = Ak ∩ E is measurable, E = ∪∞k=1Ek, m

∗(Ek) ≤m∗(Ak) ≤ 2 and Ek ∩ Ek′ = ∅ for k 6= k′;

• l. 12: ‘))’ should be ‘)’;

• l. 15: ‘O’ should be ‘Ok’.

=============================================================================PROB. 16, p. 43For measurability of E ⇒ (iii), let E be a measurable set and ǫ > 0. Then EC = R ∼ E is measurable and,

since measurability ⇒ (i), there is an open set O ⊃ EC with m∗(O ∼ EC) < ǫ. Thus F = R ∼ O is a closed setsuch that F ⊂ E and m∗(E ∼ F) = m∗((R ∼ EC) ∼ (R ∼ O)) = m∗(O ∼ EC) < ǫ. For (iii) ⇒ (iv), assume(iii) holds for E. Choose a closed set Fk ⊂ E with m∗(E ∼ Fk) < 1/k. Define F = ∪∞

k=1Fk. Then F is an Fσ setcontained in E. Moreover, since for each k, E ∼ F ⊆ E ∼ Fk, by the monotonicity of outer measure,

m∗(E ∼ F) ≤ m∗(E ∼ Fk) < 1/k.

Therefore m∗(E ∼ F) = 0 and so (iv) holds. For (iv) ⇒ measurability of E, since a set of measure zero ismeasurable, as is a Fσ set, and the measurable sets are an algebra, the set

E = F ∪ [E ∼ F]

is measurable.=============================================================================Comment, p. 46, “Therefore almost all x ∈ R fail to belong to ∩∞

n=1

[∪∞k=nEk

]... , l. 7

If E = R and E0 = ∩∞n=1

[∪∞k=nEk

], m(E0) = 0 and the property of failing to belong to E0 holds for all

x ∈ E ∼ E0 (see the 3 last lines of p. 45).=============================================================================PROB. 24, p. 47That this equality holds is trivial if E1 ∩ E2 = ∅ (by using m∗(E1 ∩ E2) = 0 and Finite Additivity, p. 46)

or E2 ⊂ E1 (by using E1 ∪ E2 = E1 and E1 ∩ E2 = E2). Thus suppose E1 ∩ E2 6= ∅ and neither E2 ⊂ E1nor E1 ⊂ E2. Then, since E1 ∪ E2 = (E1 ∼ E2) ∪ (E1 ∩ E2) ∪ (E2 ∼ E1), E1 = (E1 ∼ E2) ∪ (E1 ∩ E2) andE2 = (E2 ∼ E1) ∪ (E1 ∩ E2) are disjoint unions,

m(E1 ∪ E2) +m(E1 ∩ E2) = m(E1 ∼ E2) +m(E1 ∩ E2) +m(E2 ∼ E1) +m(E1 ∩ E2) = m(E1) +m(E2)

by Finite Additivity.=============================================================================Comment/Errata, p. 47, Proof, Lemma 16

• U = ∪λ∈Λ (λ + E) is bounded and therefore has finite measure. In fact, since E ⊂ [−b, b] and Λ ⊂[−β, β], U ⊂ [−(b+ β), b+ β]. Then, by monotonicity, m (U) ≤ 2(b+ β);

• Line -2: Shouldn’t > be ≥ ?

• m (E) = 0 since the RHS of (15) equals m (E)∑λ∈Λ λ0< ∞ 3.

=============================================================================Comment, p. 48, Proof, Theo. 17Another contradiction: By the Countable Monotonicity (p. 46), m(E) = 0!=============================================================================Comment, p. 49, Proof, Theo. 18

m∗(A) = m∗ ([A ∩ E] ∪[A ∩ EC

]) disjoint union︸︷︷︸= m∗ (A ∩ E) +m∗ (A ∩ EC

).

=============================================================================PROB. 29, p. 49(i) and (ii) are trivial. For (iii), if R is the relation, since πR0 and 0Rπ but π − π = 0, R is not transitive on

R, whereas, since the difference between two rational numbers is not an irrational number, R is not a relationon Q.

=============================================================================Comment, p. 49, Question 2Before going any further, a countable set C = ∪c∈C {c} is also Borel since each {c} is closed (see p. 20)!=============================================================================Comment, p. 52, Proof, Prop. 20

m(O) = 1 since 1 = m([0, 1])

disjoint union︸︷︷︸= m(C)︸︷︷︸

0

+ m(O).

=============================================================================Errata/Comment, p. 52, Prop. 21 and its Proof

• C should be C;

• l. -14: See PROB. 45, p. 53;

• l. -13: ψ(O) is open since {0, 2} ∩ ψ(O) = ∅ 4 and, by Prop. 22, p. 25, there is an open set U such that

ψ(O) =(

ψ−1)−1

(O) = {[0, 2] ∩ U} ∼ {0, 2} = (0, 2) ∩ U .

Thus ψ(C) = [0, 2] ∼ ψ(O) = [0, 2] ∩ [R ∼ ψ(O)] is closed.

3∑λ∈Λ λ0 = 1+ 1+ . . ..4{0, 1} ⊂ C ⇒ {0, 2} ⊂ ψ(C).

=============================================================================PROB. 45, p. 53Cf. [Fit06], Theo. 3.29, p. 78.=============================================================================PROB. 46, p. 53For f defined on an interval, which is a Borel set by PROB. 12, p. 39, if O is an open set, then f−1(O) is a

Borel set by Prop. 22, p. 25. Therefore f−1(∅) is a Borel set, and, if f−1(E) is a Borel set, then f−1(R ∼ E) =f−1(R) ∼ f−1(E) = f−1(R) ∩

[R ∼ f−1(E)

]is a Borel set. Now, f−1

(∪∞k=1Ek

)= ∪∞

k=1 f−1(Ek) is a Borel set if

each f−1(Ek) is a Borel set.=============================================================================PROB. 47, p. 53Let B be a Borel set contained in the interval where a continuous strictly increasing function f is defined

on. Thus, since f−1 is continuous, f (B) =(f−1

)−1(B) is a Borel set.

==========================================================================================================================================================3==========================================================================================================================================================Comment, p. 56, Proof, (ii), Prop. 5Since {x ∈ E | f (x) > c} = {x ∈ D | f (x) > c} ∪ {x ∈ E ∼ D | f (x) > c}, f is measurable if f |D and f |E∼D

are measurable. Since D and E ∼ D are measurable sets, {x ∈ D | f (x) > c} = D ∩ {x ∈ E | f (x) > c} and{x ∈ E ∼ D | f (x) > c} = (E ∼ D) ∩ {x ∈ E | f (x) > c}, it follows that f |D and f |E∼D are measurable if f ismeasurable.

=============================================================================Comment on remarks above Theo. 6, p. 56If f and g are finite a.e. on E, since E0 = {x ∈ E | f (x), g(x) 6= ±∞}, then m (E ∼ E0) = 0. Thus, since

{x ∈ E ∼ E0 | ( f + g)(x) > c} ⊂ E ∼ E0 for each real number c,(( f + g)|E∼E0

)−1(c,∞) is measurable5. There-

fore ( f + g)|E∼E0 is measurable. Hence, if ( f + g)|E0 is measurable, f + g is measurable on E.=============================================================================Errata/Comment, pp. 57-58, Ex.

• l. -5, p. 57: “... Lemma ...” should be “... Proposition ...” ;

• Remark on χA, ll. -2 and -1, p. 57: See p. 61;

• Conclusion, p. 58: x ∈ f−1(I) ⇔ f (x) ∈ I ⇔ χA

(ψ−1(x)

)∈ I ⇔ ψ−1(x) ∈ A ⇔ x ∈ ψ(A).

=============================================================================Comment, p. 58, Proof, Prop. 7x ∈ ( f ◦ g)−1(O) ⇔ f (g(x)) ∈ O ⇔ g(x) ∈ f−1(O) ⇔ x ∈ g−1

(f−1(O)

).

=============================================================================Comment on consequences of Prop. 7 and Prop. 8, pp. 58-9

• | f |p = | − |p ◦ f with | − |p = h ◦ g, g(t) = |t| and h(u) = up for each (t, u) ∈ R× [0,∞).

• Let f be measurable on E. On the one hand, by Prop. 8, f+ = max { f , 0} is measurable on E. On theother hand, since {x ∈ E | − f (x) > c} = {x ∈ E | f (x) < −c} is measurable for each real number c, − fis measurable on E. Thus f− = max {− f , 0} is measurable by Prop. 8.

=============================================================================PROB. 1, p. 59If E0 ⊂ [a, b], m(E0) = 0 and f (x) = g(x) for all x ∈ [a, b] ∼ E0, then E0 = ∅ or f = g on E0. Otherwise,

since f (c) 6= g(c) for some c ∈ E0, limx→c f (x) 6= limx→c g(x). Hence there is an interval I ⊂ [a, b] containingc such that f (x) 6= g(x) for all x ∈ I, which is absurd since m(I) > 0!

Now let f and g be continuous on E = [a, b] ∪ {c} with c 6∈ [a, b] 6 and f (c) 6= g(c). Therefore f (x) = g(x)for all x ∈ E ∼ {c}.

5m∗((

( f + g)|E∼E0

)−1(c,∞)

)≤ m∗(E ∼ E0) = 0 (see p. 31) and any set of outer measure zero is measurable (see p. 35).

6limx→c f (x) = f (c), that is, for each ǫ > 0 there is a δ > 0 such that f (x) ∈ ( f (c)− ǫ, f (c) + ǫ) whenever x ∈ (c− δ, c+ δ) ∩ E . In

fact, if one takes δ > 0 with [a, b] ∩ (c− δ, c+ δ) = ∅, then x = c. Hence f (x) = f (c).

=============================================================================PROB. 2, p. 59

Consider D = (−∞, 0), E = [0,∞) and f : R → R defined by f (x) =

{0 if x < 0,1 if x ≥ 0.

=============================================================================PROB. 3, p. 59Consider E is a measurable set and f : E → R is continuous except on a finite subset D of E. Since

{x ∈ D | f (x) > c} is empty or finite for each real number c and f is continuous on E ∼ D, f |D and f |E∼D aremeasurable. Therefore f is measurable. 7

=============================================================================PROB. 7, p. 59Let M be the σ-algebra8 of measurable sets and A =

{A | f−1(A) ∈ M

}, which is a σ-algebra since:

• ∅ ∈ A since f−1(∅) = ∅ ∈ M;

• If A ∈ A, since f−1(R ∼ A) = f−1(R) ∼ f−1(A) = E ∼ f−1(A) ∈ M, then R ∼ A ∈ A;

• f−1(∪∞k=1Ak

)= ∪∞

k=1 f−1(Ak) ∈ M if each Ak ∈ A.

Now let B be the σ-algebra of Borel sets. On the one hand, since (c,∞) ∈ B for each real number c, if f−1(A) ∈M for all A ∈ B, then f is measurable. On the other hand, if f is measurable and A ∈ B, since A contains allopen sets 9, B ⊂ A 10 and hence f−1(A) ∈ M.

=============================================================================Comment/Errata, p. 60, Remarks above Prop. 9

• Define fn and f by

fn(x) = xn for all x ∈ [0, 1] and f (x) =

{0 if 0 ≤ x < 1,1 if x = 1.

Hence the pointwise limit f is discontinuous at x = 1. For a pointwise limit of Riemann integrablefunctions which is not Riemann integrable, see Example on p. 70;

• l. -13: measureable should be measurable .

=============================================================================Proof, Prop. 9, pp. 60-1

• “Since m(E0) = 0, ... f is measurable iff its restriction to E ∼ E0 is measurable.” , p. 60:

f |E0 is measurable since {x ∈ E0 | f (x) > c} is measurable for each real number c;11

• Errata, l. 3, p. 61: measureable should be measurable .

=============================================================================Comment, p. 61, “... χA is measurable iff A is measurable.” , l. 11

On the one hand, if χA is measurable, χ−1A (c,∞) = A is measurable for 0 ≤ c < 1.12 On the other hand,

if A is measurable, then χA|A and χA|R∼A are measurable since they are constant functions. Hence χA ismeasurable13.

=============================================================================Errata/Comment, p. 62, Proof, The Simple Approximation Lemma

7The argument remains valid if one replaces finite by countable. As a matter of fact, the argument still works if m(D) = 0. Thus

f is continuous a.e. on E ⇒ f is measurable .

8See Definition on p. 19, Theo. 9 on p. 39.9See Prop. 2, p. 55.10See p. 39.11m∗

((f |E0

)−1(c,∞)

)≤ m∗(E0) = 0 (see p. 31) and any set of outer measure zero is measurable (see p. 35).

12χ−1A (c,∞) = ∅ for c ≥ 1, whereas χ−1

A (c,∞) = R for c < 0.13See Prop. 5, p. 56.

• l. 4: “Defne ...” should be “Define ...” ;

• l. 6: See PROB. 7, p. 59, PROB. 12, p. 39.

=============================================================================Errata, p. 63, ll. 2 and 6Change ψn to ϕn.=============================================================================PROB. 14, p. 63Consider E0 ⊂ E, m(E0) = 0 and f (x) 6= ±∞ for all x ∈ E ∼ E0. Hence, by Lusin’s Theorem, p. 66, for

each ǫ > 0, there is a continuous function g on R and a closed set F ⊂ E ∼ E0 for which f = g on F andm((E ∼ E0) ∼ F) <

ǫ2 . Thus F ⊂ E is closed, which implies that F is measurable14, m(E ∼ F) <

ǫ2 (since

E ∼ F = [(E ∼ E0) ∼ F]∪ E0 implies m(E ∼ F) = m((E ∼ E0) ∼ F) + 0 by Finite Additivity, p. 46), m(F) < ∞

(by Monotonicity, p. 46, since m(E) < ∞) and f = g is finite on F.15 Now consider En = E ∩ [−n, n] for eachnatural number n. Hence, since ∪∞

n=1En = E and {En}∞n=1 is an ascending collection of measurable sets, it

follows that, by the Continuity of Measure, p. 44,

limn→∞

m(En) = m (∪∞n=1En) = m(E) < ∞.

Therefore there is an index n0 such that, for all n ≥ n0,

m (E ∼ [−n, n]) = m ((E ∼ E) ∪ (E ∼ [−n, n])) = m (E ∼ (E ∩ [−n, n])) = m(E ∼ En) = m(E)−m(En) <ǫ

2.

Thus F0 = F ∩ [−n0, n0] is bounded and closed, which implies that F0 is measurable, f = g is finite andbounded on F0 by The Extreme Value Theorem, p. 26, m(F0) < ∞ since m(F) < ∞, and

m(E ∼ F0) = m (E ∼ (F ∩ [−n0, n0])) = m ((E ∼ F) ∪ (E ∼ [−n0, n0])) <16 ǫ

2+

ǫ

2= ǫ.

=============================================================================Comment/Errata, p. 65, Proof, Lemma 10

• Since f is real-valued, there are no points at which f and fk take infinite values of opposite signs;

• f is measurable by Prop. 9, p. 60, and its Proof;

• Since f and fk are measurable, so is | f − fk| by Theo. 6 (and its previous remarks), p. 56, and the imme-diate consequence of Prop. 7, p. 58;

• En = ∩∞k=n {x ∈ E | | f (x)− fk(x)| < η};

• In the last two lines, change ǫ to δ, twice, and replace En by EN .

=============================================================================Comment, p. 65, “... m(E ∼ F) < ǫ ...” , l. -1

F ⊂ A ⊂ E and E ∼ F = (E ∼ A) ∪ (A ∼ F).=============================================================================Comment, p. 66, Proof, Prop. 11E ∼ F = ∪n

k=1 [Ek ∼ Fk]. In fact:

• ⊂ part: If x ∈ E ∼ F, then x ∈ Ek ∼ F for some k. Thus x ∈ Ek ∼ Fk;

• ⊃ part: If x ∈ Ek ∼ Fk for some k, then x ∈ E ∼(∪i 6=kEi

)(which implies that x 6∈ ∪i 6=kFi) and x 6∈ Fk.

=============================================================================Errata, l. 1, p. 67≤ should be <.=============================================================================

14See Theo. 9, p. 39.15If F is bounded, by The Extreme Value Theorem, p. 26, then f is bounded on F. What if F is not bounded? The rest of the proof shows

that, concerning Lusin’s Theorem, “... a closed set F ...” can be replaced by “... a bounded closed set F ...” if m(E) < ∞.16Outer measure is countably subadditive, Prop. 3, p. 33.

PROB. 27, p. 67See the second half of the Proof of Lemma 10, p. 65.=============================================================================PROB. 28, p. 67See the first three items concerning the Proof of Lemma 10 above. In particular, if f is finite on E ∼ E0 with

m(E0) = 0, | f − fk| is (properly defined and) measurable on E ∼ E0. Now use Prop. 5, p. 56.==========================================================================================================================================================4==========================================================================================================================================================PROBLEMS 2-6, pp. 70-71Cf. [Fit06], Lemmas 6.2-6.4, pp. 139-140, Theo. 6.8, p. 143, and Theo. 6.18, p. 156.=============================================================================Comment, p. 72, Lemma 1 (and its Proof)Besides some of the ai’s may be equal, ∪n

i=1Ei may not be equal to E (see (1), p. 71).=============================================================================Comment, p. 72, Proof, Prop. 2

Let j(i) and k(i) be indices such that Ei = ϕ−1(aj(i)

)∩ ψ−1

(bk(i)

)6= ∅. Then Ei1 ∩ Ei2 = ∅ for i1 6= i2

17

and ∪ni=1Ei = E.18

=============================================================================Errata, lines -13,-11, p. 73Shouldn’t the comma and the period be placed after the right braces?=============================================================================Comment, p. 74, Proof, Theo. 3On the one hand, if Rl =

{(R)

∫I ϕ | ϕ a step function, ϕ ≤ f

}and Ll =

{∫I ϕ | ϕ simple, ϕ ≤ f

}, since

Rl ⊂ Ll , then sup Rl ≤ sup Ll . On the other hand, if Ru ={(R)

∫I ψ |ψ a step function, f ≤ ψ

}and Lu ={∫

I ψ |ψ simple, f ≤ ψ}, since Ru ⊂ Lu, then inf Lu ≤ infRu. Therefore, since supRl ≤ sup Ll ≤ inf Lu ≤

infRu and sup Rl = infRu, it follows that sup Rl = sup Ll = inf Lu = infRu.=============================================================================Comment on the comment before Theo. 5, p. 75See Theo. 17, p. 48, and the comment on characteristic functions, p. 61.=============================================================================Comment, pp. 75-76, Theo. 5Let A be a nonempty set of real numbers that is bounded below and α > 0. Hence αA is a nonempty set

of real numbers that is bounded below and inf αA = α inf A. In fact, on the one hand, since inf A ≤ x for allx ∈ A, α inf A ≤ αx for all x ∈ A. Thus αA is bounded below and α inf A ≤ inf αA. On the other hand, for allx ∈ A, since inf αA ≤ αx, that is, 1

α inf αA ≤ x, it follows that 1α inf αA ≤ inf A, that is, inf αA ≤ α inf A.

Now, if A is a nonempty set of real numbers that is bounded above and α < 0, by similar reasoning,

inf αA = 19 − sup {−αx | x ∈ A} = − sup (−αA) = −(−α) sup A = α sup A.

Now, suppose A and B are nonempty bounded sets of real numbers. Then A+ B is a nonempty bounded setof real numbers such that inf (A + B) = inf A + inf B and sup (A + B) = sup A + sup B. In fact, let x ∈ Aand y ∈ B. On the one hand, since inf A+ inf B ≤ x + y ≤ sup A+ sup B, it follows that A+ B is bounded,inf (A+ B) ≥ inf A+ inf B and sup (A+ B) ≤ sup A+ sup B. On the other hand, since inf (A+ B) ≤ x+ y ≤sup (A + B), it follows that inf (A + B) − x ≤ y ≤ sup (A + B) − x. Therefore inf (A + B) − x ≤ inf B andsup B ≤ sup (A+ B)− x, that is, inf (A+ B)− inf B ≤ x and x ≤ sup (A+ B)− sup B. Hence inf (A+ B)−inf B ≤ inf A and sup A ≤ sup (A + B) − sup B, that is, inf (A + B) ≤ inf A + inf B and sup A + sup B ≤sup (A+ B).

=============================================================================Errata, l. 10, p. 76Delete the extra bound.

17If Ei1 ∩ Ei2 6= ∅, then ϕ−1(aj(i1)

)∩ ϕ−1

(aj(i2)

)6= ∅ and ψ−1

(bk(i1)

)∩ψ−1

(bk(i2)

)6= ∅. Hence, since aj(i1) = aj(i2) and bk(i1) = bk(i2),

it follows that Ei1 = Ei2 , that is, i1 = i2.18If x ∈ E, there is an index i such that ϕ(x) = aj(i) and ψ(x) = bk(i).19See PROB. 6, p. 11.

=============================================================================Comment, p. 77, Proof, Prop. 8

• f is bounded. In fact, if r > 0 and f is not bounded above, then f (xr) > r for some xr ∈ E. Thus, for each0 < ǫ < r, there is an index N such that

fn(xr) = f (xr)− ( f (xr)− fn(xr)) > r− | f (xr)− fn(xr)| > r− ǫ

for all n ≥ N. Hence fn is not bounded above for all n ≥ N. Now, if f is not bounded below, since − f isnot bounded above, there is an index N such that − fn is not bounded above for all n ≥ N. Therefore fnis not bounded below for all n ≥ N.

• Since it is clear that uniform convergence implies pointwise convergence, f is measurable by Prop. 9, p.60.

=============================================================================Ex., p. 78Cf. [Fit06], Ex. 9.25, p. 243.=============================================================================Proof, The Bounded Convergence Theorem, p. 78| f | ≤ M on E since, for all ǫ > 0, there is an index N such that, for all n ≥ N, | f | = | f − fn + fn| ≤

| f − fn|+ | fn| < ǫ + M.=============================================================================PROB. 9, p. 79Since outer measure is monotone (see p. 31) and any set of outer measure zero is measurable (see p. 35), f

is measurable byDefinition (see p. 55). Therefore, since f is bounded, f is integrable (see Theo. 4, p. 74). Now,since the integral of each simple function over E is zero byDefinition (pp. 71-72),

∫E f = 0.

=============================================================================PROB. 10, p. 79On the one hand, if ϕ is a simple function such that ϕ ≤ f on A, since ϕ · χA is simple and ϕ · χA ≤ f · χA

on E, then∫A ϕ =

∫E ϕ · χA ≤

∫E f · χA, which implies that

∫A f ≤

∫E f · χA. On the other hand, if ψ is a simple

function such that f ≤ ψ on A, since ψ ·χA is simple and f ·χA ≤ ψ ·χA on E, then∫A ψ =

∫E ψ ·χA ≥

∫A f ·χA,

which implies that∫A f ≥

∫A f · χA.

=============================================================================Remark before Def., p. 79 (PROB. 18, p. 84)If f ≡ 0 on E, then

∫E0

f = 0 for each E0 ⊂ E such that m(E0) < ∞. If f (x) 6= 0 for some x ∈ E, then x ∈ E0for each E0 ⊂ E such that f ≡ 0 on E ∼ E0 and m(E0) < ∞. Let E′

0 and E′′0 be two of the E0’s and I = E′

0 ∩ E′′0 .

Therefore, since f ≡ 0 on E′0 ∼ I and f ≡ 0 on E′′

0 ∼ I,

∫

E′0f =

∫

E′0∼If +

∫

If =

∫

If =

∫

E′′0∼If +

∫

If =

∫

E′′0f

by Cor. 6, p. 76.=============================================================================Proof, Prop. 9, p. 80

• Errata: X should be E;

•∫E ϕ = 0. In fact, if E0 ⊂ E with ϕ ≡ 0 on E ∼ E0 and m(E0) = 0, then

∫E0

ϕ = 0 by PROB. 9, p. 79.

Hence, on the one hand, if m(E) < ∞, use that∫E ϕ =

∫E∼E0

ϕ +∫E0

ϕ (see Cor. 6, p. 76), and, on the

other hand, if m(E) = ∞, use that∫E ϕ =

∫E0

ϕ (see remark before Def., p. 79).

=============================================================================Proof, Theo. 11, p. 82∫E0

f = 0 since f = 0 a.e. on E0. In fact, if E1 = {x ∈ E0 | f (x) > 0}, then m(E1) = 0 since m(E0) = 0.=============================================================================Proof, Fatou’s Lemma, pp. 82-3

• E0 is the finite support of h (see p. 79) and h vanishes outside a set of finite measure which contains E0.Thus, since E0 = {x ∈ E | h(x) > 0} = ∪∞

k=1 {x ∈ E | h(x) ≥ 1/k} is measurable, m(E0) < ∞;

• hn is measurable by Prop. 8, p. 58;

• Concerning the first and the last equal sign, l. -1, p. 82, use Additivity Over Domains of Integration andthe vanishing of each hn on E ∼ E0;

•∫E hn ≤

∫E fn follows from hn ≤ fn viaMonotonicity of Integration, p. 80;

•∫E h = lim inf

∫E hn (see Prop. 19, (iv), p. 23).

=============================================================================Errata, l. -5, p. 83∫

fn should be∫E fn.

=============================================================================PROB. 17, p. 84Since f is a nonnegative20 measurable21 function on E and f = 0 a.e. on E 22,

∫E f = 0 by Prop. 9, p. 80.

=============================================================================Errata, PROB. 24.(ii), p. 85Remove the last comma.=============================================================================l. -11, p. 85On the one hand, if f is measurable, then f+ and f− are measurable (see p. 59). On the other hand, let f+

and f− be measurable. Hence, since there is no point at which f+ and − f− take infinite values of oppositesign23, f = f+ − f− is a properly defined measurable function.

=============================================================================Proof, Prop. 15, p. 86

• Errata: Remove the first comma, l. 7;

• f = f+ − f− is finite a.e. on E since f+ and f− are finite a.e. on E.24

=============================================================================Proof, Cor. 18, p. 88Since there is no point at which the measurable functions f , χA and χB

25 take infinite values of oppositesign26, f · χA and f · χB are properly defined measurable functions on E.

=============================================================================Errata, (22), p. 89f ′(x) should be f ′(x) < ∞.=============================================================================PROB. 28, p. 89

For C measurable,∫C f =

∫E f · χC for each nonnegative extended real-valued measurable function f on E .27

Hence, if f is integrable over E, it follows that∫C f =

∫C f+ −

∫C f− =

∫E f+ · χC −

∫E f− · χC =

∫E f · χC.

28

20 f ≡ ∞ on E.21For each real number c, {x ∈ E | f (x) > c} = E is measurable (see Def., p. 55).22 f ≡ 0 on E ∼ E and m(E) = 0.23

• The sum of two measurable functions on E is measurable on E if the functions are finite a.e. on E in order to avoid points at whichthe functions take infinite values of opposite sign (see p. 56);

• f+(x) = f (x) and − f−(x) = 0 if f (x) = +∞. f+(x) = 0 and − f−(x) = f (x) if f (x) = −∞.

24| f | is finite a.e. on E, 0 ≤ f+ ≤ | f | and 0 ≤ f− ≤ | f |.25See p. 61.26The product of two measurable functions on E is measurable on E if the functions are finite a.e. on E in order to avoid points at which

the functions take infinite values of opposite sign (see p. 56).27See PROB. 10 and Def., p. 79.28The second equal sign follows from the previous box. For the last equal sign, f · χC is integrable over E by the Integral Comparison

Test since | f · χC | ≤ | f | on E. For the first equal sign, f is integrable over C since | f | is integrable over C. In fact,∫

C| f | =

∫

E| f | · χC ≤

∫

E| f | < ∞,

where = follows from the previous box, ≤ follows from Theo. 10, p. 80, and < follows fromDef., p. 84.

=============================================================================l. 9, Proof, Theo. 20, p. 91

∑nk=1

∫Ek

f =∫∪nk=1Ek

f = 29∫E f · χn.

=============================================================================PROB. 37, p. 91

m(∩∞n=1En) = 0 since ∩∞

n=1En ⊂ {−∞,∞}30. Therefore 31∣∣∣∫∩∞n=1En

f∣∣∣ ≤

∫∩∞n=1En

| f | = 0. Now use Theo. 21,

(ii), p. 91.=============================================================================Proof, Prop. 23, pp. 92-3

• Errata, l. -6, p. 92: Shouldn’t E0 be E?

• Errata, l. 1, p. 93: Shouldn’t δ be δ0?

• Are the last two lines really necessary?32

=============================================================================Proof, The Vitali Convergence Theorem, p. 94

• Errata, first line: “Propositions ...” should be “Proposition ...” ;

• (29): For the first equal sign, since { fn} is uniformly integrable over E and m(E) < ∞, fn is integrableover E 33 for each index n. Now use the Linearity of Integration, p. 87.

• Errata, l. -8: Shouldn’t “... for any measurable subset of A ...” be “... for any measurable subset A of E ...” ?

• Errata, l. -3: The last fn should be f .

=============================================================================Proof, Theo. 26, p. 95

• For (30), use∫A hn +

∫E∼A hn =

∫E hn;

• Errata: On the line after (30), replace “... Propositions 23 and 24, ...” by “... Proposition 24, ...” .

=============================================================================PROB. 40, p. 95Since | f | is integrable over R and, by Additivity Over Domains of Integration34,

∫I | f | ≤

∫R | f | for each

interval I, f is integrable over I. Hence, by Prop. 15, p. 86, F(x) is properly defined. Now let x ∈ R and ǫ > 0.Thus, by Prop. 23, p. 92, there is a δ > 0 such that if |x− y| < δ, then35:

• |F(x)− F(y)| = |∫ yx f | ≤

∫ yx | f | < ǫ for x ≤ y;

• |F(x)− F(y)| = |∫ xy f | ≤

∫ xy | f | < ǫ for x > y.

Then f is continuous at x.==========================================================================================================================================================5==========================================================================================================================================================“... fn = χ[n,n+1] ... { fn} is uniformly integrable over R ...” , l. -5, p. 97

29PROB. 28, p. 89.30If there is an index n0 such that E ⊂ (−n0, n0), then ∩∞

n=1En = ∅ since En0 = ∅.31Use Prop. 9, p. 80, and the Integral Comparison Test, p. 86.32By Theo. 11, p. 82,

∫E f < N. Now use Def., p. 84.

33Prop. 23, p. 92.34See p. 82.35Use Additivity Over Domains of Integration and the Integral Comparison Test, pp. 86-7.

Let A ⊂ R be measurable. Since 36∫A fn ≤

∫R fn =

∫ n+1n fn = 1, which implies that

∫A fn < ǫ for ǫ > 1,

and 37∫A fn = 0 if A ∩ [n, n + 1] = ∅, it suffices to consider 0 < ǫ ≤ 1 and A ∩ [n, n + 1] 6= ∅. Hence, if

0 < δ ≤ ǫ and m(A) < δ, then

∫

Afn =

∫

A∼[n,n+1]fn +

∫

A∩[n,n+1]fn = 0+m(A ∩ [n, n+ 1]) ≤ m(A) < ǫ

by Additivity Over Domains of Integration, p. 82, and Def., pp. 71-2.=============================================================================Errata,Def., p. 98Delete an a.=============================================================================Comment preceding The Vitali Convergence Theorem, p. 98Let E0 ⊂ E with m(E0) < ∞. Since

{f |E0 | f ∈ F

}is uniformly integrable over E0, each one of its elements

is integrable over E0 by Prop. 23, p. 92. Then∫E0| f | < ∞. Now take E0 with

∫E∼E0

| f | < ∞ for all f ∈ F .

Hence, by Theo. 11, p. 82, ∫

E| f | =

∫

E∼E0| f |+

∫

E0| f | < ∞

for all f ∈ F .=============================================================================Proof, The Vitali Convergence Theorem, p. 99Since 38

∫E∼E0

| f | < ∞ and∫E0| f | < ∞, f is integrable over E.39

=============================================================================PROB. 1, p. 99On the one side, use The (General) Vitali Convergence Theorem (p. 98) as The Vitali Convergence Theo-

rem (p. 94) is used in the Proof of Theo. 26, p. 95. On the other side, use PROB. 2 (p. 99) as Prop. 24 (p. 93) isused in the Proof of Theo. 26, p. 95.

=============================================================================PROB. 2, p. 99By Prop. 24, p. 93, it suffices to prove that { fk}nk=1 is tight over E. In fact, for ǫ > 0 and k ∈ {1, . . . , n}, let

Ek be a set of finite measure for which∫E∼Ek

| fk| < ǫ and E0 = ∪nk=1Ek. Hence, since

∫

E∼Ek

| fk| =∫

E∼E0| fk|+

∫

E0∼Ek

| fk|

by Additivity Over Domains of Integration, p. 82,

∫

E∼E0| fk| ≤

∫

E∼Ek

| fk| < ǫ.

=============================================================================Def., p. 99fn − f is measurable.40 Then | fn − f | is measurable.41 Hence

{x ∈ E

∣∣ | fn(x)− f (x)| > η}is measurable.42

=============================================================================Last line of the Proof, p. 100Use Monotonicity, p. 46.=============================================================================Errata by Fitzpatrick, p. 101

• Concerning l. 5, replace fnk by fnk(x);

• In displayed equation (6), replace 0 by f ≡ 0.

36Use Additivity Over Domains of Integration, p. 82, the comments preceding theDefinition, p. 79, and Theo. 3, p. 73.37Use Prop. 9, p. 80.38See lines -7 and -3, p. 98.39By Theo. 11, p. 82,

∫E | f | =

∫E∼E0

| f |+∫E0

| f | < ∞.40See Theo. 6, p. 56.41See p. 58.42See Def., p. 55.

=============================================================================PROB. 9, p. 102Take f ≡ 0 on E = R and fn = χ[n,∞).

43

=============================================================================Proof, Lemma 6, p. 103

• Errata, l. 4: “... are ϕ∗ and ψ∗ properly defined ...” should be “... ϕ∗ and ψ∗ are properly defined ...” .

• Errata by Fitzpatrick, lines 4-5: Replace “... real-valued ...” by “... real ...” .

• Concerning “... ϕ∗ and ψ∗ ... are measurable since each is the pointwise limit of a sequence of measurablefunctions.”, lines 4-6, and “Therefore f is measurable.”, l. ±15, use Prop. 9, p. 60.

• “Observe that since 0 ≤ f − ϕ1 ≤ ψ1 − ϕ1 on E and ψ1 and ϕ1 are integrable over E, we infer from theintegral comparison test that f is integrable over E.”, lines -15 and -14. In fact, since f − ϕ1 (from theintegral comparison test) and ϕ1 are integrable over E, it follows that f = f − ϕ1 + ϕ1 is integrable overE by Linearity of Integration, p. 87.

=============================================================================Errata, p. 104, l. -15

... U( f , Pn) and L( f , P′n) upper ... ” should be ... U( f , Pn) and L( f , P′

n) are upper ... ” .=============================================================================PROB. 16, p. 106Assume boundedness. f is Riemann integrable over [a, b] by Theo. 8, p. 104. Therefore f is measurable by

the comments preceding Theo. 8, p. 104.==========================================================================================================================================================7==========================================================================================================================================================First paragraph, p. 136Suppose f (x), g(x) 6= ±∞ for all x ∈ E ∼ E1, f0(x) = f (x) for all x ∈ E ∼ E2, g0(x) = g(x) for all x ∈ E ∼

E3 and m(E1) = m(E2) = m(E3) = 0. Hence, since f0(x), g0(x) 6= ±∞ and α f0(x) + βg0(x) = α f (x) + βg(x)for all x ∈ E ∼ (E1 ∪ E2 ∪ E3) with m (E1 ∪ E2 ∪ E3) = 0, it follows that [α f0 + βg0] = [α f + βg].

=============================================================================“... if f ∼= g, then

∫E | f |p =

∫E |g|p.” , l. 15, p. 136

See Prop. 15, p. 86.=============================================================================(1), p. 136If max {|a|, |b|} = |a|, then |a+ b|p ≤ 2p|a|p ≤ 2p|a|p + 2p|b|p.=============================================================================Comments at the very end of p. 136If f = g ∈ F on E ∼ A, where m(A) = 0, { fn} → f pointwise on E ∼ B, where m(B) = 0, and,

for each index n, fn = gn ∈ F on E ∼ En, where m(En) = 0, then {gn} = { fn} → f = g pointwise onE ∼

[A ∪ B ∪

(∪∞n=1En

)], where m

(A ∪ B ∪

(∪∞n=1En

))= 0.

=============================================================================Nonnegativity, First Example, p. 137See Prop. 9, p. 80.=============================================================================The statement after (2), p. 138

For each positive integer n, there is some Mn ≥ 0 such that || f ||∞ + 1n > Mn and | f (x)| ≤ Mn for all

x ∈ E ∼ En with m(En) = 0. Hence

| f | < || f ||∞ +1

non E ∼ En.

=============================================================================Errata by Fitzpatrick, l. 1, p. 138

43Note that m{x ∈ E

∣∣ | fn(x)− f (x)| > 12

}= ∞ for all n.

Replace positivity by nonnegativity.=============================================================================Errata by Fitzpatrick, l. -3, p. 138Insert f before ∈.=============================================================================PROB. 1, p. 139By PROB. 6 on p. 71, The Extreme Value Theorem on p. 26, Theo. 3 on p. 73 and, concerning [Fit06],

The Mean Value Theorem for (Riemann) Integrals on p. 166, || − ||1 : C[a, b] → R is the restriction of the

norm 44 || − ||1 : L1[a, b] → R to C[a, b] and∫ ba | f | = | f (x0)|(b − a) for some x0 ∈ [a, b], which implies

|| − ||1 ≤ (b− a)|| − ||max.=============================================================================Errata by Fitzpatrick, PROB. 2, p. 139Insert “be” after “to”.=============================================================================Errata by Fitzpatrick, l. -4, p. 139L1 should be Lp.=============================================================================Errata by Fitzpatrick, p. 140On the line after (3), Lq(X, µ) should be Lq(E).=============================================================================Errata by Fitzpatrick, second line of the Proof, p. 140Replace f by g, twice.=============================================================================Last line of the Proof, pp. 140-1f ∗ ∈ Lq(E) and || f ∗||q = 1 since

∫

E| f ∗|q =

∫

E|| f ||(1−p)q

p | f |(p−1)q = || f ||−pp

∫

E| f |p = 1.

=============================================================================Proof, Cor. 3, p. 143

• “... g = χE belongs to Lq(E) since m(E) < ∞.” follows from the line after the definition of characteristic

function, p. 61, the consequence of Prop. 7, p. 58, and Theo. 7, p. 103;

•∫E |g|q = m(E) follows from the Def. on pp. 71-2.


• In the first Example, α < should be α ≤;

• In the second Example, (1,∞) should be (0,∞), ln x should be | ln x|, and for x > 1 should be deleted.

=============================================================================PROB. 6, p. 143

∫

E

∣∣∣∣f

|| f ||p· g

||g||q

∣∣∣∣ ≤∣∣∣∣∣∣∣∣

f

|| f ||p

∣∣∣∣∣∣∣∣p

·∣∣∣∣∣∣∣∣

g

||g||q

∣∣∣∣∣∣∣∣q

⇔ 1

|| f ||p · ||g||q

∫

E| f · g| ≤ || f ||p

|| f ||p· ||g||q||g||q

.

=============================================================================PROB. 7, p. 143f ∈ Lp1(E) ∼ Lp2(E) since

∫

E

∣∣xα∣∣p1dx =

∫

E

(x−α

)−p1 dx

<

∫

E

(x1/p1

)−p1dx =

∫

E

1

xdx =

∫

E

(x1/p2

)−p2dx

≤∫

E

(x−α

)−p2 dx =∫

E

∣∣xα∣∣p2dx

44See first Example on p. 137.

and∫E

1x dx = ∞. In fact, since E ∋ x 7→ g(x) = 1

x is a nonnegative measurable45 function and,46 for eachx ∈ E, if

hx(t) =

{0 for t ∈ (0, x),1t for t ∈ [x, 1],

then∫E hx ∈

{∫E h | h bounded, measurable, of finite support and 0 ≤ h ≤ g on E

}, it follows that

∫E g = ∞ by

the Def. on p. 79.=============================================================================PROB. 8, p. 143Since

∣∣ ∫E f · g

∣∣ ≤∫E | f · g| < ∞ (by Theo. 1 on p. 140 (p = q = 2) and the Integral Comparison Test47),∫

E f 2 < ∞,∫E g2 < ∞ and

∫E(λ f + g)2 < ∞, it follows that

0 ≤ λ2∫

Ef 2 + 2λ

∫

Ef · g+

∫

Eg2 ≤ λ2

∫

Ef 2 + 2λ

∫

E| f · g|+

∫

Eg2 < ∞.

Hence 4(∫

E | f · g|)2 − 4

(∫E f 2

) (∫E g2

)≤ 0.

=============================================================================Errata by Fitzpatrick, PROBLEMS, pp. 143-4

• In 9, Replace a = b = 1 by ap = bq;

• In 12 and 20, replace p < ∞ by p ≤ ∞;

• In 22, replace 1 < p by 1 ≤ p.

=============================================================================Errata, Proof, Prop. 4, p. 145Shouldn’t || − ||p be || − ||?=============================================================================Errata, l. -3, p. 145

Shouldn’t “... lemma ...” be “... proposition ...” ?=============================================================================Proof, Prop. 5, p. 146

• Errata, line preceding (7): Shouldn’t “... nonnegative ...” be “... positive ...” ?

• Since ∑∞k=1 ǫk < ∞, limk→∞ ǫk = 0. Hence, for k large enough, 0 < ǫ2k < ǫk < 1. Thus ∑

∞k=1 ǫ2k < ∞.

=============================================================================Proof, Theo. 6, p. 147, sentence after (12)If gk = | fn+k − fn|p for each index k, then {gk} → | f − fn|p pointwise a.e. on E. Hence 48

∫

E| f − fn|p ≤ lim inf

∫

Egk.

=============================================================================CommentRichard Hevener observed that theGeneral Lebesgue Dominated Convergence Theorem, p. 89, provides

a short, direct proof of Theo. 7, p. 148.=============================================================================Errata by Fitzpatrick

• Line -7, p. 148: Replace the second and fourth fn by f ;

• Change f to fn in the last line of the statement of Theo. 8, p. 149.

45See Prop. 3, p. 55.46ln x =

∫ x1

1t dt for x > 0, limx→0 ln x = −∞ and

∫ 1x

1t dt = − ln x for x ∈ E = (0, 1].

47See p. 86.48See p. 82.

=============================================================================PROB. 25, p. 149Use Cor. 3, p. 142.=============================================================================PROB. 30, p. 150

• Errata by Fitzpatrick: In the sentence “Prove that ...” , replace fk by fn , twice.

• Use the definition of || − ||max49 for the first ≤. Use (6) on p. 146 for the second ≤.

• And now, the conclusion: A sequence { fn} satisfying the previous hypotheses is 50 uniformly Cauchy on[a, b] and therefore converges uniformly to a function f ∈ C[a, b].

=============================================================================PROB. 31, p. 150Let { fn} be Cauchy in C[a, b]. Then we may inductively choose a strictly increasing sequence of natural

numbers {nk} for which || fnk+1− fnk ||max ≤ (1/

√2)k for all k. The subsequence

{fnk

}satisfies the hypotheses

of the PROB. 30 since the geometric series with ratio 1/√2 converges. Then 51

{fnk

}converges to a function f

in C[a, b] since{fnk

}→ f uniformly on [a, b]. Now use Prop. 4, p. 145.

=============================================================================PROB. 32, p. 150

• Errata by Fitzpatrick: In the sentence “Prove that ...” , replace fk by fn , twice.

• Use the definition of || − ||∞ 52 for the first ≤. Use (6) on p. 146 for the second ≤;

• And now, the conclusion: A sequence { fn} satisfying the previous hypotheses is 53 uniformly Cauchy onE ∼ E0 and therefore converges uniformly to a function f ∈ L∞(E) on E ∼ E0.

=============================================================================PROB. 33, p. 150Let { fn} be Cauchy in L∞(E). Then we may inductively choose a strictly increasing sequence of natural

numbers {nk} for which || fnk+1− fnk ||∞ ≤ (1/

√2)k for all k. The subsequence

{fnk

}satisfies the hypotheses

of PROB. 32 since the geometric series with ratio 1/√2 converges. Then 54

{fnk

}converges to a function f in

L∞(E) since, for some E0 ⊂ E with m(E0) = 0,{fnk

}→ f uniformly on E ∼ E0. Now use Prop. 4, p. 145.

=============================================================================Proof, Prop. 9, p. 151

• First paragraph: Let n be a positive integer. Then 55 there are simple functions ϕn and ψn defined onE ∼ E0 such that

ϕn ≤ g ≤ ψn and 0 ≤ ψn − ϕn <1

non E ∼ E0,

which implies that 0 ≤ g− ϕn ≤ ψn − ϕn <1n on E ∼ E0. Hence {ϕn} → g uniformly on E ∼ E0, which

implies that 56 {ϕn} → g in L∞(E).

• Last sentence: Since {|ϕn − g|p} → 0 pointwise on E, it follows that 57 limn→∞ ||ϕn − g||p = 0.

=============================================================================Proof, Prop. 10, pp. 151-2

• Errata by Fitzpatrick, l. -2, p. 151: “..., and ...” should be “. We ...” ;

49See the last Ex., p. 138.50Concerning [Fit06], see theDef. on p. 247, Theo. 9.29, p. 247, and Theo. 9.31, p. 249.51See the first two lines on p. 145.52See the last Ex., p. 137.53Concerning [Fit06], see theDef. on p. 247, Theo. 9.29, p. 247, and EX. 7, p. 249.54See the first paragraph on p. 145.55See p. 61.56See the first paragraph on p. 145.57See p. 88.

• Errata, l. 1, p. 152: “... [A ∼ U ] ∪ [U ∩ A]” should be “... [A ∼ U ] ∪ [U ∼ A]” ;

• (14), p 152: |χA − χU |p = |χA − χU | = χ[A∼U ]∪[U∼A].

=============================================================================Comments between the Def. and Theo. 11, p. 152Use (13) and the comments preceding Prop. 9, p. 151, and PROB. 38, p. 153.=============================================================================Ex., pp. 152-3Suppose η(x1) = η(x2). Hence

∣∣∣∣∣∣χ[a,x1]

− χ[a,x2]

∣∣∣∣∣∣∞=

∣∣∣∣∣∣χ[a,x1]

− fη(x1) + fη(x2) − χ[a,x2]

∣∣∣∣∣∣∞≤

∣∣∣∣∣∣χ[a,x1]

− fη(x1)

∣∣∣∣∣∣∞+

∣∣∣∣∣∣χ[a,x2]

− fη(x2)

∣∣∣∣∣∣∞< 1.

Therefore, since∣∣∣∣∣∣χ[a,x1]

− χ[a,x2]

∣∣∣∣∣∣∞= 1 for x1 < x2, we have x1 = x2.

=============================================================================PROB. 36, p. 153

Let g ∈ X. On the one hand, if S is dense in X, then there is a function fn ∈ S satisfying || fn − g|| < 1n

for each positive integer n. Hence { fn} → g in X. On the other hand, if ǫ > 0 and limn→∞ || fn − g|| = 0 withfn ∈ S for each positive integer n, then there is an index n such that || fn − g|| < ǫ. Thus S is dense in X.

=============================================================================PROB. 37, p. 153If h ∈ H and ǫ > 0, then there is a function g ∈ G (which implies the existence of a function f ∈ F such

that || f − g|| < ǫ2 ) for which ||g− h|| < ǫ

2 . Hence || f − h|| < ǫ.=============================================================================PROB. 38, p. 153Since Q is countable, the cartesian product of finitely many countable sets is countable and the union of a

countable collection of countable sets is countable,

⋃

n∈N∪{0}

{n

∑k=0

akxk

∣∣∣∣ (a0, a1, . . . , an) ∈ Qn

}

is countable.=============================================================================Errata by Fitzpatrick, PROBLEMS, pp. 153-4

• 39: Replace p < ∞ by p ≤ ∞;

• 48: Replace ||Φ( f )− Φ(g)||p by ||Φ( f )− Φ(g)||pp.

==========================================================================================================================================================8==========================================================================================================================================================Errata, l. -10, p. 155Replace “... to said ... by “... is said ...”.=============================================================================Errata by Fitzpatrick, l. -7, p. 155Replace “... Helley ... by “... Helly ...”.=============================================================================“... (5) holds for M = ||T||∗.”, l. -4, p. 156Since 0 = |T(0)| ≤ ||T||∗ · ||0|| = 0, suppose there is some f0 ∈ X ∼ {0} such that |T( f0)| > ||T||∗ · || f0||,

that is,|T( f0)||| f0|| > ||T||∗. Then |T( f0)|

|| f0|| is not a lower bound forM = {M |M ≥ 0, |T( f )| ≤ M · || f || for all f ∈ X}.Hence there is some M0 ∈ M such that

|T( f0)||| f0|| > M0, that is, |T( f0)| > M0 · || f0||, which is absurd!

=============================================================================Errata, (8), p. 157

Shouldn’t T( f ) be |T( f )|?58=============================================================================Proof, Prop. 2, p. 157

• Errata by Fitzpatrick: Replace 59 q− 1 by 1− q;

• Comment by Fitzpatrick: The Proof 60 does not cover the case p = 1. For p = 1, argue by contradiction.If ||g||∞ > ||T||∗,61 there is a set E0 of finite positive measure on which |g| > ||T||∗62 and one gets acontradiction by choosing f to be 1/m(E0) · sgn(g) · χE0 .

63

=============================================================================Proof, Lemma 4, p. 158

• “The function ϕn is integrable over E since it is dominated on E by the integrable function g.”, l. 14: UseThe Integral Comparison Test, p. 86.

• “Therefore, since ϕn is simple, it has finite support, and hence fn belongs to Lp(E).”, l. 15: If λ > 0is less than the smallest positive value taken by ϕn, by Chebychev’s Inequality, p. 80, and since ϕn is

integrable, it follows that m{x ∈ E

∣∣ ϕn(x) 6= 0}= m

{x ∈ E

∣∣ ϕn(x) ≥ λ}≤ 1

λ

∫E ϕn < ∞. Hence, even

if m(E) = ∞, since | fn|p = ϕqn is simple but has finite support, we have

∫E | fn|p =

∫E0| fn|p < ∞ with

m(E0) < ∞ and fn ≡ 0 on E ∼ E0.64

• Errata, l. -8: |ϕ|qn should be ϕqn.

• Errata by Fitzpatrick, l. -3: Remove the second comma.

=============================================================================Errata, Theo. 5, p. 159∫I should be

∫ ba .

=============================================================================Proof, The Riesz Representation Theo. for the Dual of Lp(E), p. 161

• || f ||p = || f ||p, l. 7. In fact,∫R | f |p = 65

∫R∼[−n,n] | f |p +

∫[−n,n] | f |p =

∫ n−n | f |p.

• ||Tn||∗ ≤ ||T||∗, l. 9. In fact, ||Tn||∗ = inf{M |M ≥ 0, |Tn( f )| ≤ M · || f ||p for all f ∈ Lp[−n, n]

}.

• Suggestion, l. 9: Replace “The preceding theorem tells ...” by “The preceding theorem and Prop. 2 on p.157 tell ...”.

• First paragraph after (16):

– Let f be the extension of f ∈ Lp[−n, n] to [−n− 1, n+ 1] that vanishes outside [−n, n]. It follows

that R(gn+1|[−n,n]) =∫ n−n (gn+1|[−n,n]) · f =

∫ n+1−n−1 gn+1 · f = Tn+1( f ) = T( f ) = T( f ) = Tn( f ) =∫ n

−n gn · f = Rgn .

– For f ∈ Lp(R) that vanishes outside a bounded set, there is some natural number n for which fvanishes outside [−n, n]. Then T( f ) = Tn( f |[−n,n]) =

∫ n−n gn · ( f |[−n,n]) =

∫R g · f .

• Last sentence before the last paragraph: Use PROB. 5, p. 162.

58See the solution of PROB. 1, p. 162.59Insert the first here.60Insert (from “On the other hand, ...” on) here.

61Insert since ||T||∗ is not an essential upper bound for g (see p. 136), here.62As a matter of fact, the Comment by Fitzpatrick replaces |g| > ||T||∗ by ||g||∞ > ||T||∗.63First note that f ∈ L1(E) and || f ||1 ≤ 1. And now, the contradiction: Integrating both sides of |g| > 1

m(E0)·∫E0

|g|, we have

∫

E0

|g| >(

1

m(E0)·∫

E0

|g|)·m(E0)!

64See pp. 71 and 79.65Use Additivity Over Domains of Integration, p. 82.

• Last paragraph:

– Errata: “Then T is a bounded ...” should be “Then T is a bounded ...”.

– Let f be the extension of f ∈ Lp(E) to all of R that vanishes outside E. Thus T( f ) = T( f ) =∫R g · f = 66

∫E g · f = Rg.

=============================================================================Errata by Fitzpatrick, l. -8, p. 161Replace Lebesgue-Stieltjes by Riemann-Stieltjes=============================================================================PROB. 1, p. 162Since |T( f )| ≤ ||T||∗ · || f || for all f ∈ X (“... (5) holds for M = ||T||∗.”, l. -4, p. 156), it follows that

|T( f )| ≤ ||T||∗ for all f ∈ X with || f || ≤ 1. Hence ||T||∗ is an upper bound for T ={|T( f )|

∣∣ f ∈ X, || f || ≤ 1}.

As a matter of fact, ||T||∗ = sup T . Otherwise, there is some M0 ≥ 0 such that |T( f )| ≤ M0 < ||T||∗ for allf ∈ X with || f || ≤ 1. Hence, for all f ∈ X ∼ {0}, we have |T( f/|| f ||)| ≤ M0, that is, |T( f )| ≤ M0 · || f ||, whichis absurd since ||T||∗ = inf {M |M ≥ 0, |T( f )| ≤ M · || f || for all f ∈ X}.

=============================================================================PROB. 2, p. 162The Nonnegativity is trivial. Since

sup{|α| · |T( f )|

∣∣ f ∈ X, || f || ≤ 1}= |α| · sup

{|T( f )|

∣∣ f ∈ X, || f || ≤ 1},

the Positive Homegeneity holds. Since

sup{|(T1 + T2)( f )|

∣∣ f ∈ X, || f || ≤ 1}≤ sup

{|T1( f )|+ |T2( f )|

∣∣ f ∈ X, || f || ≤ 1}=

sup{|T1( f )|

∣∣ f ∈ X, || f || ≤ 1}+ sup

{|T2( f )|

∣∣ f ∈ X, || f || ≤ 1},

The Triangle Inequality holds.=============================================================================PROB. 5, p. 162Let f ∈ Lp(E) and x0 ∈ E. Fix a natural number n. Define fn ∈ Lp(E) by

fn(x) =

{f (x) for x ∈ [x0 − n, x0 + n] ∩ E;0 for x ∈ E ∼ [x0 − n, x0 + n].

Since {| fn − f |} → 0 pointwise a.e. on E, {| fn − f |p} → 0 pointwise a.e. on E. Hence limn→∞

∫E | fn − f |p = 0

by The Lebesgue Dominated Convergence Theorem, p. 88. Then 67 { fn} → f in Lp(E). Now use PROB. 36,p. 153.

=============================================================================Errata by Fitzpatrick, PROB. 10, p. 162Replace TV( f ) by TV(g).=============================================================================Errata by Fitzpatrick, l. 3, p. 163Replace Radamacher by Rademacher.=============================================================================Ex., p. 163

• The definition does not say what happens to fn on [(2n − 1)/2n, 1].

• For 1 ≤ p < ∞, || fn − fm||p ≥ 2(p−1)/p holds since

∫ 1

0| fn(x)− fm(x)|p ≥ 2p · 1

2.

=============================================================================Comment following the Proof of Prop. 6, pp. 163-4

66Use Additivity Over Domains of Integration, p. 87.67See the last sentence of the first paragraph, p. 145.

• In order not to confuse the limits f1 and f2 with the first two terms of { fn}, concerning the limits, fshould be replaced by another letter.

• For the first equals sign on p. 164, use Theo. 1, p. 140.

=============================================================================Lines -6, -5 and -4, p. 166Since g belongs to the linear span of F , there are n αk’s in R and gk’s in F for which

limn→∞

∫

Efn · g = lim

n→∞

∫

Efn ·

n

∑k=1

αkgk = limn→∞

n

∑k=1

αk

∫

Efn · gk =

n

∑k=1

αk limn→∞

∫

Efn · gk =

n

∑k=1

αk

∫

Ef · gk =

∫

Ef ·

n

∑k=1

αkgk =∫

Ef · g.

=============================================================================Proof, Theo. 10, p. 167First note that the set of simple functions in Lq(E) is the linear span of itself. Therefore, on the one hand, if

{ fn} ⇀ f in Lp(E) and A ⊂ E is measurable, then

limn→∞

∫

Afn

PROB. 10, p. 79︸︷︷︸= lim

n→∞

∫

Efn · χA

Prop. 9, p. 166︸︷︷︸=

∫

Ef · χA

PROB. 10, p. 79︸︷︷︸=

∫

Af .

On the other hand, if ψ = ∑ni=1 ai · χEi on E, where each Ei = ψ−1(ai) (see p. 71), then

limn→∞

∫

Efn · ψ =

n

∑i=1

ai limn→∞

∫

Efn · χEi

PROB. 10, p. 79︸︷︷︸=

n

∑i=1

ai limn→∞

∫

Eifn,

which equals, provided that limn→∞

∫A fn =

∫A f for every measurable subset A of E,

n

∑i=1

ai

∫

Eif

PROB. 10, p. 79︸︷︷︸=

n

∑i=1

ai

∫

Ef · χEi =

∫

Ef · ψ.

Hence, by Prop. 9 (p. 166), { fn} ⇀ f in Lp(E)=============================================================================Errata/Comment by Fitzpatrick, p. 167

• Change the font on the last line of Theo. 11.

• In the Riemann-Lebesgue Lemma, replace “... corollary ...” by “... theorem ...”. Also, extend the lemmato p = 1 by using Theo. 10, the density of the simple functins in L∞, and Theo. 12 of Chapter 2.

• l. -3: Replace the second 1 by 0.

=============================================================================Ex., pp. 167-8

• For 1 < p < ∞, { fn} ⇀ f in Lp(R) also follows from Theo. 12, p. 168.

• The last sentence means

limn→∞

∫

R1 · fn = 1 and

∫

R1 · f = 0

Prop. 6, p. 163︸︷︷︸=⇒ { fn} 6⇀ f in L1(R).

=============================================================================Errata by Fitzpatrick, p. 168, l. 14Replace 11 by 10.=============================================================================Comment, pp. 168-9, Proof, The Radon-Riesz Theo.At the very end, use Prop. 6, p. 163.

=============================================================================PROB. 12, p. 169Since { fn} is a subsequence of itself, use “Therefore no subsequence of { fn} is Cauchy in Lp(I) and hence

no subsequence can converge in Lp(I).”, Ex., p. 163.=============================================================================PROB. 15, p. 170

• Errata: “For each natural number n, ...” should be “For each natural number n, ...”.

• For “Is this true for p = 1?” , see the conclusion of the Ex., pp. 167-8.


• PROB. 17 (iii): Replace the second fn by f .

• PROB. 19: Replace 1 ≤ p < ∞ by 1 < p < ∞.

=============================================================================PROB. 25, p. 171

• Errata: “... there is bounded linear functional ...” should be “... there is a bounded linear functional ...”.

• For (i), if { fn} ⇀ f1, f2 in X, then limn→∞ T( fn) = T( f1), T( f2) in R. Hence T( f1) = T( f2), which impliesthat T( f1 − f2) = 0. Therefore, since T( f1 − f2) = || f1 − f2||, f1 = f2.

68

=============================================================================Errata by Fitzpatrick, p. 171Replace Helley by Helly, twice.=============================================================================Errata by Fitzpatrick, p. 172

• l. 10: Replace the second69 “... is ...” by “... in ...”.

• l. -9: Replace 6 by 7 and Helley by Helly.

=============================================================================Errata/Comment by Fitzpatrick, p. 173

• l. 11: Replace 5 by 6.

• In the Remark, remove the comma in the first line. Also, add the assumption that m(E) < ∞, which isnecessary in this version of the Dunford-Pettis Theorem. In the case m(E) = ∞, one needs to assumetightness and uniform integrability.

=============================================================================PROB. 32, p. 174“Proposition 2, with p and q interchanged and the observation that p is the conjugate of q, tells us that each

Tn is a bounded linear functional on X and ||Tn||∗ = || fn||p.” is where the failure takes place. (In fact, if p = 1,then q = ∞ 6∈ [1,∞).) Another point of failure is “Moreover, according to Theorem 11 of Chapter 7, since1 < q < ∞, X = Lq(E) is separable.” since q = ∞.


• PROB. 33: Replace 1 ≤ p by 1 < p.

• PROB. 36: Replace Helley by Helly.

=============================================================================

68In order not to confuse the limits f1 and f2 with the first two terms of { fn}, concerning the limits, f should be replaced by anotherletter.

69As a matter of fact, the third!

==========================================================================================================================================================PART TWO==========================================================================================================================================================13==========================================================================================================================================================Comment on the Definition, p. 254Is the anomalous case c1 · c2 = 0 really necessary?=============================================================================PROB. 1, p. 255Let ∅ 6= S ⊂ X. For the sufficient condition, suppose S is a subspace of X. On the one hand, clearly

S+ S ={x+ y

∣∣ x, y ∈ S}⊂ S and λ · S =

{λ · x

∣∣ x ∈ S}⊂ S with λ ∈ R. On the other hand, if x ∈ S, then

x = λ(λ−1 · x

)∈ λ · S for λ 6= 0, and, since x − x = 0 ∈ S, x = x + 0 ∈ S + S. Therefore S ⊂ λ · S and

S ⊂ S+ S. Now, for the necessary condition, suppose S+ S ⊂ S and λ · S ⊂ S, λ 6= 0. Thus x+ y,λ · x ∈ S forevery x, y ∈ S and λ ∈ R, λ 6= 0. (Notice that 0 · x = 0 ∈ S since 0 = x+ (−1) · x ∈ S.) Hence S is a subspace.

=============================================================================PROB. 2, p. 255

• Y+ Z is a subspace of X.In fact, let yi ∈ Y and zi ∈ Z, i = 1, . . . , n. Hence, since Y and Z are subspaces of X, it follows that

∑ni=1 αiyi ∈ Y and ∑

ni=1 βizi ∈ Z for each αi, βi ∈ R, i = 1, . . . , n. Therefore, if λi ∈ R, i = 1, . . . , n,

∑ni=1 λi (yi + zi) = ∑

ni=1 λiyi + ∑

ni=1 λizi ∈ Y+ Z.

• Y+ Z ⊂ span [Y ∪ Z].In fact, if y+ z ∈ Y+ Z with y ∈ Y ⊂ Y ∪ Z and z ∈ Z ⊂ Y ∪ Z, then y+ z = 1 · y+ 1 · z ∈ span [Y ∪ Z].

• span [Y ∪ Z] ⊂ Y+ Z.Due to Y ⊂ Y+ Z and Z ⊂ Y+ Z,70 if λi ∈ R, i = 1, . . . , n, and x = ∑

ni=1 λixi ∈ span [Y ∪ Z] with xi ∈ Y

or xi ∈ Z, i = 1, . . . , n, then each xi ∈ Y+ Z, which implies that x ∈ Y+ Z.

=============================================================================PROB. 3, p. 255(i) Consider Yi is a subspace of X for each index i belonging to a set I and x1, . . . , xn ∈ ∩i∈IYi. Hence

x1, . . . , xn ∈ Yi for each i ∈ I. Therefore, if λ1, . . . ,λn ∈ R, then x = ∑nk=1 λkxk ∈ Yi for each i ∈ I. Then

x ∈ ∩i∈IYi.(ii) Let ∩i∈IYi be the intersection of all the linear subspaces of X that contains S. Hence, since S ⊂ ∩i∈IYi,

span [S] ⊂ ∩i∈IYi by (i). Concerning ∩i∈IYi ⊂ span [S], it suffices to show that span [S] is a subspace. In fact,

consider x = ∑nk=1 λkxk and x′ = ∑

n′k=1 λ′

kx′k are two linear combinations in span [S] with n ≤ n′. Therefore, if

λ and λ′ are real numbers, then

λx+ λ′x′ =n

∑k=1

(λλkxk + λ′λ′

kx′k

)+ λ′λ′

n+1x′n+1 + · · ·+ λ′λ′

n′x′n′ ∈ span [S]

where λ′λ′n+1x

′n+1 + · · ·+ λ′λ′

n′x′n′ occurs provided that n+ 1 ≤ n′.

(iii) Now let ∩i∈IYi be the intersection of all the closed linear subspaces of X that contains S, argue as in (ii),consider PROB. 23, p. 190, and use that Yi ⊂ Yi for each i ∈ I. Therefore, on the one hand,

span [S] ⊂ ∩i∈IYi ⇒ span [S] ⊂ ∩i∈IYi

⊂ ∩i∈IYi

⊂ ∩i∈IYi.

On the other hand, since span [S] is closed and contains S, span [S] = Yi for some i ∈ I.71 Thus ∩i∈IYi ⊂span [S].

=============================================================================Errata/Comment, p. 256

70In fact, y = y+ 0 and z = 0+ z for each y ∈ Y and z ∈ Z.71“... the closure of a linear subspace of X is a linear subspace.”, p. 254, at the very end.

• l. -11: ‘Lp(E)’ should be ‘Lp(E)’.

• l. -2: “... (1) holds for M = ||T||.”On the one hand, clearly, it holds if u = 0. On the other hand, if u 6= 0, then

||T(u)||||u|| is a lower bound for

{M

∣∣ (1) holds}. Thus, since

||T(u)||||u|| ≤ ||T||, (1) holds for M = ||T||.

=============================================================================Suggestion/Errata, p. 257Notice the underlined sign/letter:

• Proof, Theo. 1: “..., set λ≤δ/||u|| ...”;

• Prop. 2: “Let X and Y be normed linear spaces.”

=============================================================================PROB. 10, p. 258It follows from the following two facts:

1. ||T|| = g.l.b.{M

∣∣ (1) holds};72

2. (2) holds.73

=============================================================================PROB. 11, p. 258Since ||T(u)|| ≤ ||T|| · ||u|| for all u ∈ X,74 it follows that ||T(u)|| ≤ ||T|| for all u ∈ X with ||u|| ≤ 1. Hence

||T|| is an upper bound for T ={||T(u)||

∣∣ u ∈ X, ||u|| ≤ 1}. As a matter of fact, ||T|| = l.u.b. T . Otherwise,

there is some M0 ≥ 0 such that ||T(u)|| ≤ M0 < ||T|| for all u ∈ X with ||u|| ≤ 1. Hence, for all u ∈ X ∼ {0},we have ||T(u/||u||)|| ≤ M0, that is, ||T(u)|| ≤ M0||u||, which is absurd since ||T|| = g.l.b.

{M

∣∣ (1) holds}.

=============================================================================Errata/(Errata by Fitzpatrick)/Comment, p. 260

• 1 line after formula (5): ‘ei’ should be ‘e1’.

• 3 lines after formula (5):

– ‘= M√n||x||∗’ should be ‘≤ M

√n||x||∗’;

– Due to the Cauchy-Schwarz inequality on Rn, |x1| + · · · + |xn| ≤ √n√

x21 + · · ·+ x2n. In fact, if

x, y ∈ Rn, then∣∣〈x, y〉

∣∣ ≤ ||x||2||y||2. Now take x = (|x1|, . . . , |xn|) and y = (1, . . . , 1).

• Notice that c1 > 0!

• ||x|| ≥ m · ||x||∗ holds for x = 0; if x 6= 0, since x/||x||∗ ∈ S,75 it holds due to

∣∣∣∣x/||x||∗∣∣∣∣ = f (x1/||x||∗, . . . , xn/||x||∗) ≥ m.

=============================================================================Comment on “..., Y is a closed subset of the metric space X, ...” , p. 261, Proof, Cor. 6

Suppose {xn} is a sequence in Y that converges to the point x ∈ X. Hence {xn} is Cauchy (in Y). Since Y iscomplete, {xn} converges to x′ ∈ Y. As the limit of a convergent sequence is unique, x = x′. Thus x ∈ Y.


• line before Proof of Riesz’s Theorem‘||x0 − y′||’ should be ‘||x− y′||’.

• PROB. 28Change ‘compact’ to ‘countable’.

72See p. 256.73Ibid.74Ibid.75∣∣∣∣x/||x||∗

∣∣∣∣∗ = 1.

=============================================================================PROB. 24, p. 262

Without loss of generality, consider 0 < ǫ < 1. Thus, since 0 < 1− ǫ < 1, it follows that 11−ǫ > 1. Now

argue from ||x− y1|| < d1−ǫ , the new formula (8).

=============================================================================

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REAL ANALYSIS - UFPRpeople.ufpr.br/~jrrb/RoydenFitz.pdf · real analysis fourth edition (2010), first printing royden and fitzpatrick partial scrutiny, solutions of selected problems,