Roy Den Fitz

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REALANALYSISFOURTHE DITION(2010), FIRSTPRINTING

Royden and Fitzpatrick

PARTIALS CRUTINY,SOLUTIONS OFSELECTEDPROBLEMS,

COMMENTS, SUGGESTIONS ANDE RRATAJose Renato Ramos Barbosa

2013

Departamento de Matematica

Universidade Federal do Parana

Curitiba - Parana - Brasil

[email protected]


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=============================================================================ErratabyFitzpatrick1

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 top < ; 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. 9BetweenThe Completeness AxiomandThe triangle inequality,S should beE.=============================================================================PROBLEMS4-5, p.10Cf. [Fit06]2,EXERCISES 17,20,pp. 11-12. Here comes the solution ofEX. 17, p. 11:

b. xS, 0 < r < min b2c2b , b(b r)2 > b2 2br > c > x2 x < b r! c.0 < r < min

cb22b+1 , 1

(b+r)2 < b2 +2br+r < cb+rS!

=============================================================================PROB.6, p.11There is abR such thatbx for allxE. Hence x bfor allxE. Thus, ifs = sup{x|xE},

sx for allxE andb sfor each suchb.=============================================================================ErratabyFitzpatrick

(iii),PROB.15, p.13: Replacern1 byrn+1.

Lines -9 and -8, p. 14: The image ofg is contained in the integers, not the natural numbers. So redefinegas 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 fbe 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}f1(n+1)

and g(f1(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 Fitzpatricks errata placed on http://www2.math.umd.edu/pmf/RealAnalysis/index.html.2Patrick FitzpatricksAdvanced Calculus, Thomson Brooks/Cole, 2006.


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By the definition ofbx, there is a numbery > wsuch that(x,y) O, ... : Suppose(x,y) O for eachy > w. Thus, sinceyw for each(x,y) O,w = bx!

... {Ix}xO is disjoint. : IfIx1Ix2= , either one ofa x1 , bx1 , a x2 andbx2 belongs toO, which is notpossible, orIx1 = Ix2 = Ixfor somex O.

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

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

XE should be RE .=============================================================================Comment,The Heine-Borel Theorem, p. 18Concerning itsProof, ifc < bthen there is an < such that {O1, . . . , Ok, O} covers[a, c+ ]!=============================================================================Errata/Comment, p. 19,Proof,The Nested Set Theorem

4 should be 11 andF should beF1; The hypothesis thatF1is bounded works withThe Heine-Borel Theorem.

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

Shouldnt ... Proposition 4 ... be ... Proposition 11 ... ?=============================================================================PROB.27, p.20Qis open (thus RQ is closed) since its points are isolated;Qis not closed (thus RQ is not open) since RQQ.=============================================================================

ErratabyFitzpatrick, p. 23, last lineReplace the lim inf by limsup.=============================================================================PROB.39, p.24For(i), letEn ={ak| kn}and sn = supEn. Thus limn sn = liff for every >0, there is an index N

such thatsn(l , l+ )for every index nN. Hence, concerning the part of(i), for each indexnN,sincel is not an upper bound for En, there is a kn nsuch that akn > l , and, since l + is an upperbound for En,ak l+ for all k n. Also note that, since E1 ={a1, . . . , aN, . . .} , E2 ={a2, . . . , aN, . . .} , . . . ,we can have at most N 1 indicesn withan > l+. Now, for thepart of(i), since there are only finitelymany indicesn with an > l+, there is an index Nsuch thatan l+ for all n N. Thereforesn l+providedn N. (The problem with theis the equals sign!) Also note that there cannot be an indexNforwhichsN l . Otherwisean l for all n N. (Thus there are only finitely many indices n withan > l !) Hence the decreasing sequencesnis at least bounded. Now useTheo. 15, p. 21;

For(ii),(iii)(viaPROB.6, p.11) and(v), the proof is trivial;For(iv), ifa = , use(ii)and(iii). Otherwise, the part of(iv)follows from

{an}aR(i) limsup {an}= a {an} a = liminf{an} (iii)= limsup {an}=(a).

Now, for thepart of(iv), suppose{an} a R. Hence there is some > 0 so that for all Nthere is annNwith eitherana + or ana . Thus there are infinitely many n where this happens.

=============================================================================ErratabyFitzpatrick, p. 24

PROB.42: Replace >by

.

PROB.44: Replace 0 < x < 1 by 0x1 and replace q/pn byq/pn , 0 < p < q.


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=============================================================================Errata, last paragraph beforePROBLEMS, 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.28

LetObe an open set. Thus, by Prop. 22, p.25, there is an open setUsuch that f1(O) = R U, which isopen. Then, ifF is a closed set, f1(F)is closed sinceR f1(F) = f1(R) f1(F) = f1(RF)is open.Now, f1(B)is a Borel set ifB is a Borel set. In fact, see PROB.46, p.53.

==========================================================================================================================================================2==========================================================================================================================================================PROB.1, p.31SinceB = A (BA) . . . ,m(B) m(A) =m(BA) +m() +m() + 0.=============================================================================PROB.2, p.31

SinceA . . .= Aandm(A) < ,m() +m() + = m(A) m(A) =0.=============================================================================m(A), p. 31k=1l(Ik) |A k=1 Ik

= since AR = I1 j=1(I2j I2j+1)for I1 = (r, r), I2j = (j r,j+r),I2j+1= (j r, j+r)andr > 1.

=============================================================================Ex., p. 31

k=1 12k =12

1 12=1;

ErratabyFitzpatrick: ReplaceE byC.

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

For the conclusion: Jk= (ck, dk),cky= akand dky= bkJk= Ik+ywithIk= (ak, bk) .=============================================================================PROB.5, p.34Suppose that[0, 1]is countable. Thenm([0, 1]) =0 (Ex., p. 31). Butm([0, 1]) =1 (Prop. 1, p. 31).=============================================================================PROB.6, p.34Let Bbe 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 byPROB.9, p.34.=============================================================================

PROB.9, p.34Sincem is finitely subadditive (p.34),m(A B) m(A) +m(B) = m(B). Sincem is monotone (p.

31) andBA B,m(B)m(A B).=============================================================================ErratabyFitzpatrick, p. 35, l. 5Replace Constantine by Constantin.=============================================================================Errata, p. 35, l. 11)) should be ).=============================================================================ErratabyFitzpatrick, p. 39, l. 2Change the secondA1to A2.



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m(A [E+y])m is translation invariant

= m({A [E+y]} y) =m([A y] E)sincex[A y] Ex +yA and x+yE+yx +yA [E+y]x{A [E+y]} y.

=============================================================================PROB.11, p.39

A-algebra (see definition on p. 38), besides containingR = (,), that contains intervals of the form(a,), also contains intervals of the form:

(, a] =R(a,); (a, b] = (a,) (, b]witha < b; (a, b) =k=1(a, xk]with a < b, x1 = a+b2 andx k+1 = xk+b2 for each indexk 1. In fact, since{xk}is

bounded (a = a+a2 < x1 < b+b

2 = b and, ifa < xk < b, a = a+a

2 < xk+1 < b+b

2 = b) and monotone

(xk+1 > xk+xk

2 = xk), there exists limkxk = . Then = +b

2 , that is, = b. It follows that (a, b)k=1(a, xk] (easy!) and(a, b) k=1(a, xk](ifx (a, b), then there is an index ksuch that a < x < xk.Otherwise, limkxk x < b!);

[b, b] ={b}= (a, b](a, b)ifa 0. Then EC = R E is measurable and,

sincemeasurability(i), there is an open setO EC withm(O EC) < . ThusF = R Ois a closed setsuch that F Eand m(E F) = m((R EC) (R O)) = m(O EC) < . For(iii)(iv), assume(iii)holds for E. Choose a closed setFk Ewithm(E Fk) < 1/k. DefineF =k=1Fk. ThenF is an Fsetcontained inE. Moreover, since for eachk,EFEFk, by the monotonicity of outer measure,

m(EF)m(EFk) < 1/k.Thereforem(E F) = 0 and so (iv) holds. For (iv) measurability of E, since a set of measure zero ismeasurable, as is aFset, and the measurable sets are an algebra, the set

E= F [EF ]


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is measurable.=============================================================================Comment, p. 46, Therefore almost allxR fail to belong to n=1

k=nEk... , l. 7If E = R and E0 =n=1

k=n Ek, m(E0) = 0 and the property of failing to belong to E0 holds for allxEE0(see the 3 last lines of p. 45).

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

PROB.24, p.47That this equality holds is trivial ifE1 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 supposeE1 E2= and neither E2 E1nor E1 E2. Then, sinceE1 E2 = (E1 E2) (E1 E2) (E2 E1), E1 = (E1 E2) (E1 E2) andE2 = (E2E1) (E1 E2)are disjoint unions,

m(E1 E2) +m(E1 E2) =m(E1E2) +m(E1 E2) +m(E2E1) +m(E1 E2) =m(E1) +m(E2)by Finite Additivity.

=============================================================================Comment/Errata, p. 47,Proof,Lemma 16

U =

( +E) is bounded and therefore has finite measure. In fact, since E

[

b, b] and

[,],U[(b+), b+]. Then, by monotonicity, m (U)2(b+); Line -2: Shouldnt > be ? m (E)= 0 since the RHS of (15) equalsm (E)0 < 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 ECdisjoint union

= m(A E)+mA EC.=============================================================================PROB.29, p.49(i) and (ii) are trivial. For (iii), ifR is the relation, since R0 and 0Rbut = 0, R is not transitive on

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

=============================================================================Comment, p. 49,Question 2Before going any further, a countable set C =cC {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. 21and itsProof

Cshould beC; l. -14: SeePROB.45, p.53; l. -13: (O)is open since {0, 2} (O) = 4 and, byProp. 22, p. 25, there is an open setUsuch that

(O) =1

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

Thus(C) = [0, 2](O) = [0, 2] [R(O)]is closed.30 =1+1+. . ..4{0, 1}C{0, 2}(C).


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=============================================================================PROB.45, p.53Cf. [Fit06],Theo. 3.29,p. 78.=============================================================================PROB.46, p.53For f defined on an interval, which is a Borel set byPROB.12, p.39, ifOis an open set, then f1(O)is a

Borel set byProp.22, p.25. Therefore f1

()is a Borel set, and, if f1

(E)is a Borel set, then f1

(R E) =f1(R) f1(E) = f1(R) R f1(E)is a Borel set. Now, f1k=1Ek=k=1f1(Ek)is a Borel set ifeach f1(Ek)is a Borel set.

=============================================================================PROB.47, p.53Let Bbe a Borel set contained in the interval where a continuous strictly increasing function fis defined

on. Thus, since f1 is continuous, f(B) =

f11

(B)is a Borel set.==========================================================================================================================================================3==========================================================================================================================================================

Comment, p. 56,Proof,(ii),Prop. 5Since {xE | f(x) > c}={xD| f(x) > c} {xED| f(x) > c}, fis measurable if f|Dand f|EDare measurable. SinceD andE D are measurable sets,{xD| f(x) > c} = D {xE | f(x) > c}and{xED| f(x) > c} = (E D) {xE | f(x) > c}, it follows that f|D and f|ED are measurable if f ismeasurable.

=============================================================================Commenton remarks aboveTheo. 6, p. 56If f and g are finite a.e. on E, since E0 ={xE | f(x),g(x)=}, then m (EE0) = 0. Thus, since

{xEE0| (f+ g)(x) > c}EE0for each real numberc,

(f+g)|EE01

(c,)is measurable5. There-fore(f+g)|EE0 is measurable. Hence, if(f+ g)|E0 is measurable, f+ gis measurable on E.

=============================================================================Errata/Comment, pp.57-58,Ex.

l. -5, p.57: ... Lemma ... should be ... Proposition ... ; Remark onA, ll. -2 and -1, p.57: See p.61; Conclusion, p.58: x f1(I) f(x)IA

1(x)

I1(x)Ax(A).=============================================================================Comment, p. 58,Proof,Prop. 7x(fg)1(O) f(g(x)) O g(x) f1(O)xg1f1(O).=============================================================================Commenton consequences ofProp. 7andProp. 8, pp.58-9

|f|p =| |p fwith | |p =h g,g(t) =|t| andh(u) =up for each(t, u)R [0,).

Let fbe measurable on E. On the one hand, by Prop. 8, f+ = max {f, 0} is measurable on E. On theother hand, since{xE | f(x) > c}={xE |f(x) < c} is measurable for each real numberc,fis measurable onE. Thus f = max {f, 0} is measurable byProp. 8.

=============================================================================PROB.1, p.59IfE0 [a, b],m(E0) = 0 and f(x) = g(x)for all x [a, b] E0, thenE0 = or f = gon E0. Otherwise,

since f(c)= g(c)for somecE0, limxc f(x)= limxcg(x). Hence there is an intervalI [a, b]containingcsuch that f(x)= g(x)for allxI, which is absurd sincem(I) > 0!

Now let fandg be continuous on E = [a, b] {c} withc[a, b] 6 and f(c)= g(c). Therefore f(x) = g(x)for allxE{c}.

5m

(f+ g)|EE0

1(c,)

m(EE0) =0 (see p.31) and any set of outer measure zero is measurable (see p. 35).

6limxc f(x) = f(c), that is, for each > 0 there is a > 0 such that f(x)(f(c) , f(c) + )wheneverx(c , c+ ) E . Infact, if one takes > 0 with[a, b] (c , c+ ) = , thenx = c. Hence f(x) = f(c).


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

ConsiderD = (, 0),E = [0,)and f :RR defined by f(x) =

0 ifx < 0,1 ifx0.

=============================================================================PROB.3, p.59Consider E is a measurable set and f : E

R is continuous except on a finite subset D ofE. Since

{xD| f(x) > c} is empty or finite for each real number c and fis continuous onED, f|Dand f|EDaremeasurable. Therefore fis measurable. 7

=============================================================================PROB.7, p.59Let Mbe the -algebra8 of measurable sets and A= A | f1(A) M, which is a-algebra since: A since f1() = M; IfA A, since f1(RA) = f1(R) f1(A) =E f1(A) M, thenRA A; f1k=1Ak=k=1f1(Ak) M if each Ak A.

Now let Bbe the-algebra of Borel sets. On the one hand, since(c,) Bfor each real numberc, if f1(A)Mfor all A B, then f is measurable. On the other hand, if fis measurable and A B, sinceAcontains allopen sets 9 , B A 10 and hence f1(A) M.

=============================================================================Comment/Errata, p. 60, Remarks aboveProp. 9

Define fnand fby

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

0 if 0x < 1,1 ifx = 1.

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

l. -13: measureable should be measurable .=============================================================================Proof,Prop. 9, pp.60-1

Sincem(E0) =0, ... fis measurable iff its restriction toEE0is measurable. , p.60:f|E0 is measurable since {xE0 |f(x) > c} is measurable for each real number c;11

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

=============================================================================Comment, p. 61, ...Ais measurable iffA is measurable. , l. 11

On the one hand, ifA is measurable, 1A (c,) = Ais measurable for 0 c < 1.12 On the other hand,ifA is measurable, then A|A and A|RA are measurable since they are constant functions. HenceA 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 ifm(D) =0. Thus

fis continuous a.e. onE f is measurable .

8SeeDefinitionon p.19,Theo. 9on p.39.9SeeProp. 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).121A (c,) = forc1, whereas1A (c,) =R for c < 0.

13SeeProp. 5, p. 56.


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l. 4: Defne ... should be Define ... ; l. 6: SeePROB.7, p.59,PROB.12, p.39.

=============================================================================Errata, p. 63, ll. 2 and 6Changento n.=============================================================================PROB.14, p.63ConsiderE0 E,m(E0) = 0 and f(x)=for all x E E0. Hence, by Lusins Theorem, p.66, for

each > 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) < 2 . Thus F Eis closed, which implies that F is measurable14, m(E F) < 2 (sinceEF =[(EE0)F ] E0impliesm(EF) =m((EE0)F) + 0 by Finite Additivity, p.46),m(F) < (by Monotonicity, p.46, sincem(E) < ) and f = gis finite on F.15 Now considerEn = E [n, n]for eachnatural number n. Hence, sincen=1En = E and{En}n=1 is an ascending collection of measurable sets, itfollows that, bythe Continuity of Measure, p. 44,

limn m(En) =m (

n=1En) =m(E) < .

Therefore there is an indexn0such that, for allnn0,m (E[n, n]) =m ((EE) (E[n, n])) =m (E(E [n, n])) =m(EEn) =m(E) m(En) <

2.

Thus F0 = F [n0, n0] is bounded and closed, which implies that F0 is measurable, f = g is finite andbounded on F0by The Extreme Value Theorem, p. 26,m(F0) < sincem(F) < , and

m(EF0) =m (E(F [n0, n0]))= m ((EF) (E[n0, n0])) < 16 2

+

2=.

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

Since fis real-valued, there are no points at which fand f

ktake infinite values of opposite signs;

fis measurable byProp. 9, p. 60, and itsProof; Since fand fkare measurable, so is|f fk|by Theo. 6(and its previous remarks), p.56, and the imme-

diate consequence ofProp. 7, p. 58;

En =k=n {xE | |f(x) fk(x)| < }; In the last two lines, change to , twice, and replaceEnby EN.

=============================================================================Comment, p. 65, ...m(EF) < ... , l. -1FAE and EF = (EA) (AF).=============================================================================Comment, p. 66,Proof,Prop. 11EF =nk=1[Ek Fk]. In fact: part: IfxEF , thenxEk F for somek. ThusxEkFk; part: IfxEk Fkfor somek, thenxE

i=kEi(which implies that x i=kFi) andx Fk.=============================================================================Errata, l. 1, p.67 should be


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PROB.27, p.67See the second half of theProofofLemma 10, p. 65.=============================================================================PROB.28, p.67See the first three items concerning the ProofofLemma 10above. In particular, if fis finite onEE0with

m(E0) =0, |f fk| is (properly defined and) measurable on EE0. Now useProp. 5, p. 56.==========================================================================================================================================================4==========================================================================================================================================================PROBLEMS2-6, pp.70-71Cf. [Fit06],Lemmas 6.2-6.4,pp. 139-140,Theo. 6.8,p. 143, andTheo. 6.18,p. 156.=============================================================================Comment, p. 72,Lemma 1(and itsProof)Besides some of theais may be equal, ni=1Ei may not be equal toE (see (1), p.71).=============================================================================Comment, p. 72,Proof,Prop. 2

Let j(i)and k(i)be indices such that Ei = 1 aj(i) 1 bk(i)= . ThenEi1 Ei2 = fori1= i2 17

and ni=1Ei = E.18=============================================================================Errata, lines -13,-11, p.73Shouldnt the comma and the period be placed after the right braces?=============================================================================Comment, p. 74,Proof,Theo. 3On the one hand, ifRl =

(R)

I | a step function, f

and Ll =

I | simple, f

, since

Rl Ll , then supRl supLl . On the other hand, ifRu =

(R)

I | a step function, f

and Lu =I | simple, f

, since Ru Lu, then infLu infRu. Therefore, since sup Rl supLl infLu

infRuand supRl =infRu, it follows that supRl =supLl =infLu = infRu.=============================================================================

Comment on the comment beforeTheo. 5, p. 75SeeTheo. 17, p. 48, and the comment on characteristic functions, p. 61.=============================================================================Comment, pp.75-76,Theo. 5Let Abe a nonempty set of real numbers that is bounded below and > 0. Hence Ais a nonempty set

of real numbers that is bounded below and infA = infA. In fact, on the one hand, since infA xfor allx A, infAxfor allx A. ThusAis bounded below and infAinfA. On the other hand, for allxA , since infAx, that is, 1infAx , it follows that 1infAinfA, that is, infA infA.

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

infA= 19 sup{x | xA}=sup (A) =() supA= supA.Now, suppose Aand B are nonempty bounded sets of real numbers. Then A +B is a nonempty bounded setof real numbers such that inf(A+B) = infA+infB and sup (A+B) = supA+supB. In fact, letx Aandy B. On the one hand, since infA+infB x+y supA+sup B, it follows that A+B is bounded,inf(A+B)infA+infBand sup (A+B)supA+sup B. On the other hand, since inf(A+B)x +ysup (A+B), it follows that inf(A+B) x y sup (A+B) x. Therefore inf(A+B) x infBandsupB sup (A+B) x, that is, inf(A+B) infB xand x sup (A+B) supB. Hence inf(A+B) infB infA and supA sup (A+ B) sup B, that is, inf(A+ B) infA+infB and supA+sup Bsup (A+B).

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

17IfEi1 Ei2= , then1

aj(i1)

1

aj(i2 )

= and1

bk(i1 )

1

bk(i2)

= . Hence, sinceaj(i1) = aj(i2 )andbk(i1) = bk(i2 ),

it follows thatEi1

=Ei2

, that is,i1

= i2

.18IfxE , there is an indexisuch that(x) = aj(i)and(x) =bk(i).19SeePROB.6, p.11.


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fis bounded. In fact, ifr > 0 and fis not bounded above, then f(xr) > rfor somexrE. Thus, for each0 < < r, there is an index Nsuch that

fn(xr) = f(xr)

(f(xr)

fn(xr)) > r |

f(xr)

fn(xr)|> r

for allnN. Hence fnis not bounded above for all nN. Now, if fis not bounded below, since f isnot bounded above, there is an index Nsuch that fnis not bounded above for allnN. Therefore fnis not bounded below for allnN.

Since it is clear that uniform convergence implies pointwise convergence, fis 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 > 0, there is an index Nsuch that, for all n N,|f| =|f fn+ fn| |f fn| + |fn| < +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 fis bounded, fis integrable (seeTheo. 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 is a simple function such that fonA, since Ais simple and A fA

onE, then

A =

E A

E fA, which implies that

A f

E fA. On the other hand, ifis a simplefunction such that f onA, since Ais simple andfA AonE, then A

=

E A A fA,

which implies that A f A fA.=============================================================================Remark beforeDef., p. 79(PROB.18, p.84)If f 0 onE, then E0 f =0 for eachE0E such thatm(E0) < . If f(x)=0 for somexE, thenxE0

for eachE0 E such that f 0 onEE0and m(E0) < . LetE0and E 0 be two of theE0s andI= E 0 E0 .Therefore, since f 0 onE0Iand f 0 onE0 I,

E0f =

E0I

f+

If =

I

f =

E0 If+

I

f =

E0f

byCor. 6, p. 76.=============================================================================Proof,Prop. 9, p. 80

Errata: Xshould beE; E = 0. In fact, ifE0 Ewith 0 onE E0 andm(E0) = 0, then E0 = 0 byPROB.9, p.79.

Hence, on the one hand, ifm(E) < , use that

E =

EE0 +

E0 (see Cor. 6, p.76), and, on the

other hand, ifm(E) = , use that

E =

E0(see remark beforeDef., p. 79).

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

E0f =0 since f =0 a.e. onE0. In fact, ifE1 ={xE0 | f(x) > 0}, thenm(E1) =0 sincem(E0) =0.

=============================================================================Proof,Fatous Lemma, pp.82-3

E0is the finite support ofh (see p.79) andh vanishes outside a set of finite measure which contains E0.Thus, sinceE0 ={xE | h(x) > 0}=k=1 {xE | h(x)1/k} is measurable,m(E0) < ;


12/26

hnis measurable byProp. 8, p. 58; Concerning the first and the last equal sign, l. -1, p. 82, useAdditivity Over Domains of Integrationand

the vanishing of eachhnon EE0;

Ehn

E fnfollows fromhn fnvia Monotonicity of Integration, p. 80;

Eh = lim infEhn(seeProp. 19,(iv), p. 23).=============================================================================Errata, l. -5, p.83

fnshould be

E fn.=============================================================================PROB.17, p.84Since fis a nonnegative20 measurable21 function onEand f =0 a.e. onE 22 ,

E f =0 byProp. 9, p.80.

=============================================================================Errata,PROB.24.(ii), p.85Remove the last comma.=============================================================================l. -11, p.85

On the one hand, if fis measurable, then f+ and fare measurable (see p.59). On the other hand, let f+and f be measurable. Hence, since there is no point at which f+ andf take infinite values of oppositesign23, f = f+ fis a properly defined measurable function.

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

Errata:Remove the first comma, l. 7; f = f+ fis finite a.e. on Esince f+ and fare finite a.e. onE.24

=============================================================================Proof,Cor. 18, p. 88Since there is no point at which the measurable functions f,A andB

25 take infinite values of opposite

sign26, fAand fBare properly defined measurable functions on E.=============================================================================Errata, (22), p.89f(x)should be f(x) < .=============================================================================PROB.28, p.89

ForCmeasurable,

C f =

E fCfor each nonnegative extended real-valued measurable function fonE .27Hence, if fis integrable over E, it follows that

C f =

C f+ C f = E f+ C E f C= E fC.28

20f onE.21For each real numberc,

{x

E

|f(x) > c

}= E is measurable (seeDef., p. 55).

22f 0 on EE andm(E) =0.23

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

the functions take infinite values of opposite sign (see p.56).27SeePROB.10 andDef., p. 79.28The second equal sign follows from the previous box. For the last equal sign, fC is integrable over E bythe Integral Comparison

Testsince |fC | |f| onE. For the first equal sign, fis integrable overC since |f| is integrable overC. In fact,

C |f|= E |f| C E |f| < ,where=follows from the previous box, follows fromTheo. 10, p. 80, and < follows fromDef., p. 84.


13/26

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

nk=1

Ek

f =nk=1 Ek f =

29

E fn.=============================================================================PROB.37, p.91

m(

n=1En) =0 since

n=1En

{,

}30. Therefore 31 n=1 En f n=1 En |f|= 0. Now useTheo. 21,(ii), p. 91.

=============================================================================Proof,Prop. 23, pp.92-3

Errata, l. -6, p.92: ShouldntE0be E? Errata, l. 1, p.93: Shouldntbe 0? 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 integrableoverE 33 for each indexn. Now use theLinearity of Integration, p. 87.

Errata, l. -8: Shouldnt ... for any measurable subset ofA ... be ... for any measurable subsetA ofE ... ?

Errata, l. -3: The last fnshould be f.

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

For (30), use Ahn+ EAhn = Ehn; Errata:On the line after (30), replace ... Propositions 23 and 24, ... by ... Proposition 24, ... .

=============================================================================PROB.40, p.95Since|f|is integrable over R and, byAdditivity Over Domains of Integration34, I |f| R |f|for each

intervalI, fis integrable over I. Hence, byProp. 15, p. 86,F(x)is properly defined. Now letxR and > 0.Thus, byProp. 23, p. 92, there is a > 0 such that if|x y| < , then35:

|F(x) F(y)|=| yx f| yx|f| < forxy;|F(x) F(y)|=|

x

y f| x

y|f| < forx > y.

Then fis continuous atx.==========================================================================================================================================================5==========================================================================================================================================================... fn = [n,n+1]... {fn} is uniformly integrable overR ... , l. -5, p.97

29PROB.28, p.89.30If there is an indexn0such thatE(n0, n0), then n=1En = sinceEn0 = .31UseProp. 9, p. 80, andthe Integral Comparison Test, p. 86.32ByTheo. 11, p. 82,

Ef < N. Now useDef., p. 84.

33

Prop. 23, p. 92.34See p.82.35UseAdditivity Over Domains of Integrationandthe Integral Comparison Test, pp.86-7.


14/26

Let A Rbe measurable. Since 36A fn R fn = n+1n fn = 1, which implies that A fn < for > 1,and 37

A fn = 0 ifA [n, n+1] = , it suffices to consider 0 < 1 and A [n, n+1]= . Hence, if

0 < andm(A) < , thenA

fn =

A[n,n+1]fn+

A[n,n+1]

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

byAdditivity Over Domains of Integration, p. 82, andDef., pp.71-2.=============================================================================Errata,Def., p. 98Delete ana.=============================================================================Comment precedingThe Vitali Convergence Theorem, p. 98LetE0 Ewithm(E0) < . Since

f|E0 | f F

is uniformly integrable overE0, each one of its elements

is integrable over E0 byProp. 23, p.92. Then

E0|f| < . Now take E0 with

EE0 |f| < for all f F.

Hence, byTheo. 11, p. 82, E|f|=

EE0

|f| +

E0|f| <

for all f F.=============================================================================Proof,The Vitali Convergence Theorem, p. 99Since38

EE0 |f| < and

E0

|f| < , fis integrable overE.39=============================================================================PROB.1, p.99On the one side, useThe(General)Vitali Convergence Theorem(p.98) as The Vitali Convergence Theo-

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

=============================================================================PROB.2, p.99ByProp. 24, p.93, it suffices to prove that{fk}nk=1 is tight overE. In fact, for >0 andk{1 , . . . , n}, let

Ekbe a set of finite measure for which EEk |fk| < andE0 =nk=1Ek. Hence, since

EEk|fk|=

EE0

|fk| +

E0Ek|fk|

byAdditivity Over Domains of Integration, p. 82,EE0

|fk|

EEk|fk| < .

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

xE

|fn(x) f(x)| >

is measurable.42

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

Last line of theProof, p. 100Use Monotonicity, p.46.=============================================================================ErratabyFitzpatrick, p. 101

Concerning l. 5, replace fnkby fnk(x); In displayed equation (6), replace 0 by f 0.

36UseAdditivity Over Domains of Integration, p. 82, the comments preceding theDefinition, p. 79, andTheo. 3, p. 73.37UseProp. 9, p. 80.38See lines -7 and -3, p. 98.39ByTheo. 11, p. 82,

E |f|=

EE0 |f| +

E0|f| < .

40

SeeTheo. 6, p. 56.41See p.58.42SeeDef., p. 55.


15/26

=============================================================================PROB.9, p.102Take f 0 onE = R and fn = [n,).43=============================================================================Proof,Lemma 6, p. 103

Errata, l. 4: ... areand properly defined ... should be ...and are properly defined ... . ErratabyFitzpatrick, lines 4-5: Replace ... real-valued ... by ... real ... . Concerning ... and ... are measurable since each is the pointwise limit of a sequence of measurable

functions., lines 4-6, and Therefore fis measurable., l. 15, useProp. 9, p. 60. Observe that since 0 f1 1 1onE and 1 and 1 are integrable overE, we infer from the

integral comparison test that f is integrable over E., lines -15 and -14. In fact, since f 1 (from theintegral comparison test) and 1are integrable overE, it follows that f = f 1+ 1is integrable overEbyLinearity of Integration, p. 87.

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

Errata, p. 104, l. -15...U(f, Pn)andL(f, Pn)upper ... should be ...U(f, Pn)andL(f, Pn)are upper ... .

=============================================================================PROB.16, p.106Assume boundedness. fis Riemann integrable over[a, b]by Theo. 8, p. 104. Therefore fis measurable by

the comments precedingTheo. 8, p. 104.==========================================================================================================================================================7==========================================================================================================================================================First paragraph, p.136

Suppose f(x),g(x)= for allxEE1, f0(x) = f(x)for allxEE2,g0(x) = g(x)for allxEE3and m(E1) = m(E2) = m(E3) = 0. Hence, since f0(x),g0(x)= andf0(x) +g0(x) = f(x) +g(x)for allxE(E1 E2 E3)withm (E1 E2 E3)= 0, it follows that[f0+g0] = [f+ g].

=============================================================================... if f= g, then E |f|p = E |g|p. , l. 15, p.136

SeeProp. 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, wherem A B n=1En = 0.=============================================================================Nonnegativity, FirstExample, p. 137SeeProp. 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 allxEEnwithm(En) =0. Hence

|f| < ||f||+ 1n

onEEn.=============================================================================ErratabyFitzpatrick, l. 1, p.138

43Note thatm

xE |fn (x) f(x)| > 12= for alln.


16/26

Replace positivity by nonnegativity.=============================================================================ErratabyFitzpatrick, l. -3, p.138Insert fbefore .=============================================================================PROB.1, p.139

By 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) Integralson p. 166,|| ||1 : C[a, b] R is the restriction of thenorm 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.

=============================================================================ErratabyFitzpatrick,PROB.2, p.139Insert be after to.=============================================================================ErratabyFitzpatrick, l. -4, p.139L1 should beLp.=============================================================================ErratabyFitzpatrick, p. 140On the line after (3), Lq(X,)should beLq(E).=============================================================================ErratabyFitzpatrick, second line of theProof, p. 140Replace fbyg, twice.=============================================================================Last line of theProof, pp.140-1fLq(E)and ||f||q= 1 since

E|f|q =

E||f||(1p)qp |f|(p1)q =||f||pp

E|f|p =1.

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

...g = Ebelongs toLq(E)sincem(E) < . follows from the line after the definition ofcharacteristicfunction, p. 61, the consequence ofProp. 7, p. 58, andTheo. 7, p. 103;

E |g|q =m(E)follows from theDef.on pp.71-2.=============================================================================ErratabyFitzpatrick, p. 143

In the firstExample, < should be; In the secondExample,(1,)should be(0,), ln xshould be | ln x|, and forx > 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|fg| ||f||p||f||p

||g||q||g||q .

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

E

xp1 dx = E

x

p1 dx 0, limx0ln x= and 1x 1t dt = ln xfor xE = (0, 1].47See p.86.48See p.82.


18/26

=============================================================================PROB.25, p.149UseCor. 3, p. 142.=============================================================================PROB.30, p.150

ErratabyFitzpatrick: In the sentence Prove that ... , replace fkby fn, twice. Use the definition of| || |max49 for the first . Use (6) on p.146for the second . And now, the conclusion: A sequence{fn}satisfying the previous hypotheses is 50 uniformly Cauchyon

[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 allk. The subsequence

fnk

satisfies the hypotheses

of thePROB.30 since the geometric series with ratio 1/

2 converges. Then51

fnk

converges to a function finC[a, b]since fnk funiformly on[a, b]. Now useProp. 4, p. 145.=============================================================================

PROB.32, p.150

ErratabyFitzpatrick: In the sentence Prove that ... , replace fkby fn, twice. Use the definition of| || | 52 for the first . Use (6) on p.146for the second ; And now, the conclusion: A sequence{fn}satisfying the previous hypotheses is 53 uniformly Cauchyon

EE0and therefore converges uniformly to a function f L(E)onEE0.=============================================================================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 allk. The subsequence fnksatisfies the hypothesesofPROB.32 since the geometric series with ratio 1/

2 converges. Then 54

fnk

converges to a function f inL(E)since, for some E0E withm(E0) =0,

fnk funiformly onEE0. Now useProp. 4, p. 145.

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

First paragraph: Let n be a positive integer. Then 55 there are simple functions n and n defined onEE0such that

n gnand 0n n < 1n

onEE0,

which implies that 0g nn n < 1n onEE0. Hence {n}g uniformly onEE0, whichimplies that56 {n}g in L(E).

Last sentence: Since {|n g|p}0 pointwise onE, it follows that 57 limn ||n g||p = 0.=============================================================================Proof,Prop. 10, pp.151-2

ErratabyFitzpatrick, l. -2, p.151: ..., and ... should be . We ... ;49See the lastEx., p. 138.50Concerning [Fit06], see theDef.on p.247,Theo. 9.29, p. 247, andTheo. 9.31, p. 249.51See the first two lines on p. 145.52See the lastEx., p. 137.53Concerning [Fit06], see theDef.on p.247,Theo. 9.29, p. 247, andEX. 7, p. 249.54See the first paragraph on p.145.55

See p.61.56See the first paragraph on p.145.57See p.88.


19/26

Errata, l. 1, p.152: ...[A U] [U A] should be ...[A U] [U A] ;

(14), p152:|A U|p =|A U|= [AU][UA].

=============================================================================Comments between theDef.andTheo. 11, p. 152

Use (13) and the comments preceding Prop. 9, p. 151, andPROB.38, p.153.=============================================================================Ex., pp.152-3Suppose(x1) =(x2). Hence[a,x1] [a,x2] =

[a,x1 ] f(x1)+ f(x2) [a,x2] [a,x1] f(x1) +

[a,x2] f(x2)< 1.

Therefore, since[a,x1] [a,x2] =1 forx1 < x2, we havex1 = x2.=============================================================================

PROB.36, p.153

Let g X. On the one hand, ifSis dense in X, then there is a function fn Ssatisfying||fng|| < 1nfor each positive integer n. Hence

{fn

}gin X. On the other hand, if > 0 and limn

||fn

g

||= 0 with

fn Sfor each positive integern, then there is an indexn such that ||fn g|| < . Thus Sis dense inX.=============================================================================PROB.37, p.153Ifh H and > 0, then there is a function g G (which implies the existence of a function f Fsuch

that ||fg|| < 2 ) for which ||g h|| < 2 . Hence ||f h|| < .=============================================================================PROB.38, p.153SinceQ is countable, the cartesian product of finitely many countable sets is countable and the union of a

countable collection of countable sets is countable,

nN{0}

n

k=0

akxk

(a0, a1, . . . , an)Qn

is countable.=============================================================================ErratabyFitzpatrick,PROBLEMS, pp.153-4

39: Replace p < byp; 48: Replace ||(f) (g)||pby ||(f) (g)||pp.

==========================================================================================================================================================8

==========================================================================================================================================================Errata, l. -10, p.155Replace ... to said ... by ... is said ....=============================================================================ErratabyFitzpatrick, 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 for M={M |M0, |T(f)| M ||f|| for all f X}.Hence there is some M0

Msuch that|T(f0)|

||f

0|| > M0, that is,

|T(f0)

|>M0

||f0

||, which is absurd!

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


20/26

ShouldntT(f)be |T(f)|?58=============================================================================Proof,Prop. 2, p. 157

ErratabyFitzpatrick: Replace 59 q 1 by 1 q;

CommentbyFitzpatrick: TheProof 60 does not cover the case p = 1. Forp = 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 fto be 1/m(E0) sgn(g) E0 .63

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

The function nis integrable overE since it is dominated on E by the integrable functiong., l. 14: UseThe Integral Comparison Test, p. 86.

Therefore, since n is simple, it has finite support, and hence fn belongs to Lp(E)., l. 15: If > 0is less than the smallest positive value taken by n, byChebychevs Inequality, p.80, and since n is

integrable, it follows that m

xE

n(x)=0

= m

xE

n(x)

1

En < . Hence, even

ifm(E) = , since|

fn|

p = q

nis simple but has finite support, we have E |fn|p = E0 |fn|p < withm(E0) < and fn0 onEE0.64

Errata, l. -8:||qnshould be qn. ErratabyFitzpatrick, l. -3: Remove the second comma.

=============================================================================Errata,Theo. 5, p. 159

Ishould beb

a.=============================================================================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 = nn |f|p.

||Tn|| ||T||, l. 9. In fact, ||Tn|| = inf

M |M0, |Tn(f)| M ||f||pfor all f Lp[n, n]

.

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

First paragraph after (16): Let fbe the extension of f Lp[n, n]to [n 1, n+1]that vanishes outside[n, n]. It follows

thatR(gn+1 |[n,n]) =nn(gn+1|[n,n]) f =

n+1n1gn+1 f = Tn+1(f) = T(f) = T(f) = Tn(f) =

nngn f =Rgn .

For f

Lp(R) that vanishes outside a bounded set, there is some natural number n for which f

vanishes outside[n, n]. ThenT(f) = Tn(f|[n,n]) = nngn (f|[n,n]) = Rg f. Last sentence before the last paragraph: Use PROB.5, p.162.

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

61Insert since ||T|| is not anessential upper boundfor g (see p.136), here.62As a matter of fact, the CommentbyFitzpatrickreplaces |g| > ||T||by ||g|| > ||T||.63First note that f L 1(E)and ||f||11. And now, the contradiction: Integrating both sides of |g| > 1m(E0)

E0

|g|, we haveE0

|g| >

1

m(E0)

E0

|g|

m(E0)!

64See pp.71and79.65UseAdditivity Over Domains of Integration, p. 82.


21/26

Last paragraph:

Errata: ThenTis a bounded ... should be ThenTis a bounded .... Let fbe the extension of f Lp(E) to all ofR that vanishes outside E. ThusT(f) =T(f) =

R g f = 66

Eg f =Rg.

=============================================================================ErratabyFitzpatrick, l. -8, p.161ReplaceLebesgue-StieltjesbyRiemann-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 Xwith ||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


22/26

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, useTheo. 1, p. 140.=============================================================================Lines -6, -5 and -4, p. 166

Sinceg belongs to the linear span ofF, there aren ks inRandgks in Ffor which

limn

E

fn g= limn

E

fn n

k=1

kgk= limn

n

k=1

k

E

fn gk=

n

k=1

k limn

E

fn gk=n

k=1

k

E

fgk=

Ef

n

k=1

kgk=

Efg.

=============================================================================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 inLp(E)and AE is measurable, then

limn

A

fnPROB. 10, p. 79

= limn

E

fn AProp. 9, p. 166

=

EfA

PROB. 10, p. 79 =

A

f.

On the other hand, if = ni=1ai Ei onE, where eachEi = 1(ai)(see p.71), then

limn

E

fn =n

i=1

ai limn

E

fn EiPROB. 10, p. 79

=n

i=1

ai limn

Ei

fn,

which equals, provided that limn

A fn =

A ffor every measurable subsetA ofE,

n

i=1 ai Ei fPROB. 10, p. 79

=n

i=1 ai E fEi = E f .Hence, byProp. 9(p.166), {fn} f inLp(E)

=============================================================================Errata/CommentbyFitzpatrick, p. 167

Change the font on the last line ofTheo. 11. Inthe Riemann-Lebesgue Lemma, replace ... corollary ... by ... theorem .... Also, extend the lemma

top = 1 by usingTheo. 10, the density of the simple functins in L, andTheo. 12ofChapter 2.

l. -3: Replace the second 1 by 0.=============================================================================Ex., pp.167-8

For 1 < p < , {fn} f inLp(R)also follows fromTheo. 12, p. 168. The last sentence means

limn

R

1 fn = 1 and

R1 f =0

Prop. 6, p. 163 = {fn} f inL1(R).

=============================================================================ErratabyFitzpatrick, p. 168, l. 14Replace 11 by 10.=============================================================================

Comment, pp.168-9,Proof,The Radon-Riesz Theo.At the very end, useProp. 6, p. 163.


23/26

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

no subsequence can converge inLp(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 forp = 1? , see the conclusion of theEx., pp.167-8.


PROB.17 (iii): Replace the second fnby f. PROB.19: Replace 1p < by 1 < p < .=============================================================================PROB.25, p.171

Errata: ... there is bounded linear functional ... should be ... there is abounded linear functional .... For (i), if{fn} f1,f2inX, then limnT(fn) =T(f1), T(f2)inR. HenceT(f1) =T(f2), which implies

thatT(f1 f2) =0. Therefore, since T(f1 f2) =||f1 f2||, f1 = f2.68

=============================================================================ErratabyFitzpatrick, p. 171Replace Helley by Helly, twice.=============================================================================ErratabyFitzpatrick, p. 172

l. 10: Replace the second69 ... is ... by ... in .... l. -9: Replace 6 by 7 and Helley by Helly.

=============================================================================Errata/CommentbyFitzpatrick, p. 173

l. 11: Replace 5 by 6. In the Remark, remove the comma in the first line. Also, add the assumption thatm(E) < , which is

necessary 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, withp andqinterchanged and the observation that p is the conjugate ofq, tells us that eachTnis a bounded linear functional onXand ||Tn|| =||fn||p. is where the failure takes place. (In fact, ifp = 1,then q = [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 1pby 1 < p. PROB.36: Replace Helley by Helly.

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

68

In order not to confuse the limits f1 and f2 with the first two terms of{fn}, concerning the limits, fshould be replaced by anotherletter.69As a matter of fact, the third!


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==========================================================================================================================================================PART TWO==========================================================================================================================================================13

==========================================================================================================================================================Commenton theDefinition, p. 254Is the anomalous casec1 c2 = 0 really necessary?=============================================================================PROB.1, p.255Let = S X. For the sufficient condition, supposeS is a subspace ofX. On the one hand, clearly

S+S =

x+y x,yS Sand S = x xS Swith R. On the other hand, ifx S, then

x = 1 x S for= 0, and, since x x = 0 S, x = x+0 S+S. Therefore S S and

SS+S. Now, for the necessary condition, supposeS+SS and SS,=0. Thusx+y, xS foreveryx,yS and R,=0. (Notice that 0 x= 0S since 0= x + (1) xS.) HenceS is a subspace.

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

Y+Zis a subspace ofX.In fact, let yi Yand zi Z, i = 1, . . . , n. Hence, sinceYand Z are subspaces ofX, it follows that

ni=1iyi Yand ni=1izi Z for each i,i R, i = 1, . . . , n. Therefore, ifi R, i = 1, . . . , n,

ni=1i(yi+zi)=

ni=1iyi+

ni=1iziY+Z.

Y+Zspan [Y Z].In fact, ify+zY+ZwithyYY ZandzZY Z, theny+z= 1 y+1 zspan [Y Z].

span [Y Z]Y+Z.Due toYY+Zand ZY+Z,70 ifi R,i = 1, . . . , n, andx = ni=1i xispan [Y Z]withxiYorxiZ ,i = 1, . . . , n, then eachxiY+Z, which implies thatxY+Z.

=============================================================================PROB.3, p.255(i) Consider Yi is a subspace ofXfor each index i belonging to a set Iand x1, . . . , xn iIYi. Hence

x1, . . . , xn Yi for each i I. Therefore, if1, . . . ,n R, then x = nk=1kxk Yi for each i I. Thenx iIYi.

(ii) LetiIYi be the intersection of all the linear subspaces ofXthat containsS. Hence, since S iIYi,span [S] iIYi by (i). ConcerningiIYi span [S], it suffices to show that span [S]is a subspace. In fact,considerx = nk=1kxkandx

= n

k=1kx

kare two linear combinations in span [S]withnn. Therefore, if

and are real numbers, then

x+ x =n

k=1

kxk+

kxk

+ n+1xn+1+ + n xnspan [S]

wheren+1xn+1+ + n xn occurs provided thatn+1n .(iii) Now let iIYibe the intersection of all the closed linear subspaces ofXthat containsS, argue as in (ii),

considerPROB.23, p.190, and use thatYiYifor eachiI. Therefore, on the one hand,span [S] iIYispan [S] iIYi

iIYi iIYi.

On the other hand, since span [S] is closed and contains S, span [S] = Yi for some i I.71 ThusiIYispan [S].

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

70In fact,y = y +0 andz = 0+zfor eachyYandzZ .71... the closure of a linear subspace ofXis a linear subspace., p.254, at the very end.


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l. -11: Lp(E) should be Lp(E).

l. -2: ... (1) holds forM =||T||.On the one hand, clearly, it holds ifu = 0. On the other hand, ifu= 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/||u|| ...; Prop. 2: LetXandYbe 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 alluX,74 it follows that ||T(u)|| ||T|| for alluXwith ||u|| 1. Hence

||T|| is an upper bound forT = ||T(u)|| uX,||u|| 1. As a matter of fact,||T|| = l.u.b. T. Otherwise,there is some M00 such that ||T(u)|| M0 < ||T|| for alluXwith ||u|| 1. Hence, for alluX{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, ifx,yRn, then x, y ||x||2||y||2. Now takex =(|x1|, . . . , |xn|)andy =(1 , . . . , 1).

Notice thatc1 > 0!||x|| m ||x||holds forx = 0; ifx=0, sincex/||x||S,75 it holds due tox/||x||= f(x1/||x||, . . . , xn/||x||)m.=============================================================================

Commenton ...,Yis a closed subset of the metric spaceX, ... , p.261,Proof,Cor. 6Suppose {xn} is a sequence inYthat converges to the pointxX. Hence {xn} is Cauchy (inY). SinceYis

complete, {xn} converges toxY. As the limit of a convergent sequence is unique, x = x . ThusxY.=============================================================================Errata by Fitzpatrick, p. 262

line beforeProof of Rieszs Theorem||x0 y|| should be ||x y||.

PROB.28Change compact to countable.

72See p.256.73

Ibid.74Ibid.75x/||x|| = 1.


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

Without loss of generality, consider 0 < < 1. Thus, since 0 < 1 < 1, it follows that 11 > 1. Nowargue from ||x y1|| < d1 , the new formula (8).

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

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