Title: Generalizations of the Vitali-Hahn-Saks and Nikodym theorems
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Title: Generalizations of the Vitali-Hahn-Saks and Nikodym theorems
Physical Description: v, 62 leaves : ; 28 cm.
Language: English
Creator: Jewett, Robert Stanley, 1945-
Publisher: University of Florida
Place of Publication: Gainesville, Fla
Publication Date: 1971
Copyright Date: 1971
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Subject: Measure theory   ( lcsh )
Mathematics thesis Ph. D
Dissertations, Academic -- Mathematics -- UF
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Thesis: Thesis - University of Florida.
Bibliography: Bibliography: leaves 60-61.
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Generalizations of the Vitali-Hahn-Saks
and Nikodym Theorems












By

Robert Stanley Jewett


A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA IN PARTIAL
FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY








UNIVERSITY OF FLORIDA
1971
















ACKNOWLEDGMENTS


The author would like to thank his advisor Dr. James K. Brooks

and the members of his supervisory committee, Dr. David Drake, Dr.

Gene Hemp, Dr. Steve Saxon, and Dr. Zoran Pop-Stojanovic for their

help in writing this dissertation.















TABLE OF CONTENTS



Abstract ----------------------------------------------------------iv

Introduction -----------------------------------------------------

Chapter 1: Unconditional Convergence ------------------------------6

Chapter 2: Strongly Bounded Set Functions -------------------------1

Chapter 3: Extending the Schur Theorem ----------------------------20

Chapter b: Extensions of the Vitali-Hahn-Saks and
Nikodym Theorems ---------------------------------------35

Chapter 5: A Counterexample ---------------------------------------56

References --------------------------------------------------------60

Biographical Sketch ------------------------------------------ 62



























iii















Abstract of Dissertation Presented to the
Graduate Council of the University of Florida in Partial Fulfillment
of the Requirements for the Degree of Doctor of Philosophy

GENERALIZATIONS OF THE VITALI-HAHN-SAKS
AND NIKODYM THEOREMS

By

Robert Stanley Jewett

August, 1971

Chairman: Dr. James K. Brooks
Major Department: Mathematics


The Vitali-Hahn-Saks and Nikodym theorems are two very closely

related results concerning sequences of countably additive, real

valued measures. The purpose of this dissertation is to improve

on these theorems both in the statements and the proofs. That is,

stronger theorems will be shown to be true, and the proofs will

be easier that those usually given. The author and J. K. Brooks

extend the Vitali-Hahn-Saks and Nikodym theorems to the finitely

additive vector case.

In 1969 J. K. Brooks used a result of Schur concerning uniform

convergence of a double sequence of real numbers to derive simple,

direct proofs of the Vitali-Hahn-Saks and Nikodym theorems. In

Chapter 3 of this dissertation the above mentioned theorem of Schur

is extended to the case where the double sequence is contained in

a Banach space, and then this new result is used to obtain a short

proof of the Nikodym theorem with the measures taking their values

iv









in a Banach space.

However, the Vitali-Hahn-Saks and Nikodym theorems do not

generalize directly to the case where the measures are finitely

additive, bounded, and Banach valued, and the additional hypothesis

of "strongly bounded" must be assumed. With this assumption, the

theorems may be extended, and the proofs follow the same general

line as those of Brooks'for the countably additive, scalar case.

Instead of the Schur theorem, a generalized form of a result of

Phillips is used, and the Nikodym theorem follows, but the proof

of the Vitali-Hahn-Saks theorem requires a difficult construction

with measurable sets. At the end of Chapter 4 a discussion is given

to indicate how the theorems may be extended even further to the

case where the measures take values in a locally convex linear topo-

logical space instead of a Banach space.












INTRODUCTION


This dissertation concerns the Vitali-Hahn-Saks and Nikodym

theorems (7). In 1907 Vitali (21) proved the following theorem: if


f: [o, I R are Lebesgue integrable functions converging almost

everywhere to then ., j f KcS and Jc exist and are
o 0o

equal if and only if the indefinite integrals of the fn are uni-

formly absolutely continuous with respect to Lebesgue measure.

Hahn (10) then proved in 1922 that if -: [0,1]-) f are Lebesgue

integrable functions and if ~l, fd t exists for every measur-

able set E then the indefinite integral of the fh are uniformly

absolutely continuous with respect to Lebesgue measure, and they

converge to a set function which is also absolutely contirious with

respect to Lebesgue measure. In 1933 Nikodym (12) generalized these

two results when he proved that if (S, ,/ ) is a measure space








and if n is a sequence of /u -continuous measures such that


ALM, X(E) exists for all E in 1- then the An are uni-


formly absolutely continuous with respect to / A little later


Saks (19) gave another proof of Nikodym's theorem, and it became


known as the Vitali-Hahn-Saks theorem. Using the countable addi-


tivity of the control measure, Saks defined a complete metric on the


measurable sets, and then applied the Baire Category theorem. If


the control measure had been only finitely additive, the metric


would not necessarily have been complete, and this technique could


not have been used.


Nikodym (12) proved tht if Un is a sequence of measures


defined on a 0--algebra such that /uE) : ,, (E) exists for


all E then /u is countably additive and the countable additivity


of the /u~ is uniform. This is known as the Nikodym theorem, and in


most books it is proved as a consequence of the Vitali-Hahn-Saks


theorem. Rickart (18) and Phillips (16) extended these theorems to


the case where the measures are Banach space valued. The proofs of


these results rely heavily on the fact that the set functions in









question are countably additive, and there seems to be no way to


generalize their techniques in order to prove Vitali-Hahn-Saks and


Nikodym theorems for finitely additive, vector valued set functions.


However, in 1969 Brooks (3) gave a short proof of the Nikodym


theorem and then used that result to prove the Vitali-Hahn-Saks


theorem. His proof used a difficult construction with measurable


sets along with a theorem of Schur (20) concerning the equivalence


of weak and strong convergence in Il In the same paper he gave


a short proof that the truth of the two theorems in the scalar case


would imply their truth in the vector case.


An extension of the Vitali-Hahn-Saks theorem to the finitely


additive, scalar case has been proved by Ando (1), but his theorem


was not extended to the case where the measures are vector valued.


In 1970 Brooks and the author proved Vitali-Hahn-Saks ana Aikodym


theorems for finitely additive, strongly bounded, vector valued


set functions, where strongly bounded is defined as in (17). The


strongly bounded hypothesis is especially needed in the Nikodym


theorem, because the statement would not even make sense if the









measures were only bounded; the strongly bounded hypothesis may be


dropped in the Vitali-Hahn-Saks theorem if the control measure is


assumed to be finite. The proofs of the theorems are built around


a generalization of a theorem of Phillips (15) and an extension of


a technique of Darst (6).


The first two chapters provide the background material for


the results proved later in the dissertation. However, Theorems


1.3 and 1.h are stated only for completeness, and are not used in


subsequent proofs. Although most of the theorems in these capters


may be found in Hilderbrandt (11) or Rickart (17), new proof tre


given by the author.


The purpose of Chapter 3 is to generalize a theorem of Schur


(20) and to use the new result to construct a direct proof of the


vector case of the Nikodym theorem. Since all of the measures are


assumed to be countably additive, the theorems rely only on the


results of Chapter 1, and not of Chapter 2. Chapters 3 and L, re


independent, and no theorem of Chapter 3 is used in a proof in








Chapter 4.


The most important theorems are proved in Chapter 4. Both the


Vitali-Hahn-Saks and Nikodym theorems are extended to the finitely


additive case, and it is shown how the newer version of the Nikodym


theorem implies the usual version when the measures are countably


additive. The concept of "strongly bounded" is of utmost importance


here, but Corollary 4.5 and the remark following the statement of


Theorem h.7 indicate when the condition may be dropped.













CHAPTER I


UNCONDITIONAL CONVERGENCE


Throughout this dissertation the following notation will

be used.

7 is a O'-algebra of subsets of a set 5 (l) is the

power set of the natural numbers and E and A are generic

notations for sets belonging to t and (M) respectively. R

denotes.the real numbers. X is a Banach space over the real

or complex numbers, and I)( is the norm of E 6 3 is the

conjugate space of with the norm of an element in defined

in the usual way.

Let n,n E 7 with n Define (71,>]=- ,7+I ...j7 ,


f7,oo)r ny,7+I,... j

If 4 is a real valued measure defined on a T -algebra 2 ,

then ful is the total variation of / which is defined in the








standard way.


In this chapter we define the term "unconditional convergence


and establish other conditions equivalent to it.


Definition: Let X K ,n= f,,, If t: 7r -


is a one-to-one, onto mapping, then XT(n) inI is called a


rearrangement of Xr 0 The series tn is said to be


unconditionally convergent if the sum of every rearrangement


of its terms converges.


Remark: A series that converges absolutely will converge


unconditionally, but the converse is not necessarily true. In


fact, Dvoretzky and Rogers (8) proved in 1950 that the two con-


ditions are equivalent if and only if the Banach space in question


is finite dimensional. The difference between absolute and uncon-


ditional convergence was further demonstrated by Brouxs (L) in


1969 when he showed that weakly and strongly integrable functions


with range in a Banach space correspond to unconditionally and


absolutely convergent series in the space.

00
Theorem 1.1: Let X) ,f :n 1,X,' The series Z X
,tI


II







converges unconditionally if and only if every subsequential
00
sum converges, that is, if L1 7 is a subsequence, then .-X

converges.

Proof: Let 1 XKV be a subsequence and assume the sum does

not converge. Then there exist E>o and sequences fPiI and jVi

where for each i P V

K -l = {(- X :'K:Pi', j'"7 Then form the follow-

ing rearrangement of X.I : Xn ... X7 V ,iXP ).

)()I I ) nz P^ 3 X 3 ) V3 y*tp h - jnv',

The partial sums of this sequence are not Cauchy, so the sum
00
does not exist; hence ZXn does not converge unconditionally.
n=i

Conversely, assume 19K is a rearrangement of ltI and that

the sum of {Kj does not exist. Write p 4 yv if 9p precedes

y when considered in the ordering of the sequence Xn .

There exist 6>0 and sequences -m;) and I(n where for all i ,

)< n% and /6K >) Let P, be such that for all

KrL,, h,] and all t> P , K 4 Rearrange


IYj: KI= -,) *,71 so that the elements appear in the same order







as they appear in PXnj and call this rearrangement

X5n, ,.X I Let MA >P, Again rearrange

i ( i: K ^ '"=- 'l appropriately, and label this arrangement

YlX i", X. Since 9t I5 if r ,,n,] ,

S 1 'l -n we have that fX i,", Xn~ ,Xn ). xI"* XIwz

has the same order as Xn .

In this manner, define 71 K: IK=l,. and {[k) so that
I wi*-
for every i / K Then Xh K is a subsequence

whose sum does not exist; thus the converse is proved. Q.E.D.
00
Theorem 1.2: Let Xh ,f7=I,X,* If rX converges

unconditionally, then for all > O there exists an N such

that for all L Ci',O0) IJ; I1<

Proof: Assume the conclusion is false. Then there exists

> 0 such that for all N there exists A C [J,0o) such that



Let NI= I Then there exists A C.C oo) such that

hI~I[ J> Also, there exists a finite subset of I ,

such that I > Let N = .:a There exists








SC Nr+1,00) such that I X I so there exists a finite

set rI c A such that l > Note that 1 n = 1 D where O

denotes the null set. Also, for all SE t, c 5 f

In a similar way define an infinite number of finite sets

such that for all i 1Zf ll> and if LjJ and 5E ,f j F

we have 5< .

VE Lv P.L is a subsequence of .hj If m is


any positive integer, there exists io such that for all nv E lo '
V,
SV Writing f'L [lv, i > tv we have that Xvj

I .> so is not a Cauchy sequence. Therefore


the sum of the XAv does not exist and this contradicts the uncon-
00
ditional convergence of Z h .
nr

Theorem 1.3: Let >E (X 7= ,1,2,-- Then the series

-Xn converges unconditionally if, and only if,
71=1

&(m, (X,)| = o uniformly in X where 1X The

conclusion means that if 4 >0 there exists an N so that for

all X) such that IX*1 / we have F IX()l< .
coo
Proof: Assume the series 2:- converges unconditionally,
-Az




11

and let 6 )0 By Theorem 1.2 there exists N so that n 2C

for every AC LNWo) Let X be chosen so that IX~ 1 I and define



A,:fnE[f,oo): : x)
Then A, UAI f ,oo) ,and (1/=0 Hence -X0)

ntAr, m Ez. | Cnf |

Note that for all > N /IvACi] I" therefore

ifA i [ N/ X m) -I rN1 .[ It follows
tat nCN,t~]t t ?n,6,n i .I


Likewise, it may be shown that I i)l ; consequently

IX(X) I I ) "- and the first implication of the theorem
m= N

is proved.

Conversely, assume that the sum of the X) does not converge

unconditionally. Then by Theorem 1.1 there is a subsequence PXKl

where the sum of the XHK does not exist; hence there is an >O

and sequences fPiL and fVyl where P < Vi P+I and for all ,

SCX > 6 By the Hahn-Banach theorem there exists X for

each i so that IX l I and < iXi --
K= P K P(;
^ ( ^)1.K, I








For every N there is a value of L so that
Vj
ip "'l CN, Co) therefore Jx(X )I x(Xk)\ 6 .

Consequently the limit in the conclusion of the theorem does not

converge uniformly in X where IXYI I and the second implication

is proved. Q.E.D.

Definition: Let 1E, 3X : I,,,''" and let G be the

collection of all finite sets c contained in the natural numbers.

Define a partial order on E by ", 2 if G, 2 We say
oo
the series Z(h converges to a Moore-Smith limit Xo if for


every > O there exists Oo such that for all (02 Go ,

Shf- o< ( .< Since X is complete, an equivalent condition


would be for every > 0 there exists Go so that if GC, 2

we have In K x 6
oo
Theorem 1.h: Let ZgE X =1:I,Z,-" The series ZIX

converges unconditionally if, and only if, it converges a

Moore-Smith limit.
oo
Proof: Assume ZX converges unconditionally. Let >O .

By Theorem 1.2 there exists N *so that for all LCCN,oo) ,







I t 6 Let z [oI,N] If 0 x, 2 O, we have that

J,: -_ j MJXnl itnX- ZXhj ) +
| t E, n.7 E (+ A ; Is Ego Ia '- T n I n ,-

1 + I |
I .t- <;( because both ',- o and OL-Qo are contained in

4,oo0) Therefore the series converges to a Moore-Smith limit.

Conversely, assume the series does not converge unconditionally,

so there exists a subsequence I XKI whose sum does not exist.

There exist E)o and sequences {PI iV j where P
and for each i I .K- > E Let 'o be any finite subset

of the natural numbers, and choose i so that [nPL,-"nv nV O

Letting T : oU CPi, *" Vi V we have that (T (To and


| n ec nrco and it follows that the series does not

have a Moore-Smith limit. This completes the proof. Q.E.D.














CHAPTER 2



STRONGLY BOUNDED SET FUNCTIONS



Many theorems that are true for countably additive measures


do not have extensions when the measures are finitely additive


and bounded. However, as will be seen in Chapter 4, some of


these results do carry over to the finitely additive case when


the hypothesis of "strongly bounded" is assumed. In this chapter


"strongly bounded" is defined and theorems are proved which will


be used in Chapter 4.


Definition: Let /.Z-~X be a set function. we say ,4


is strongly bounded (s-bounded) if &7iu(E,)'O whenever IEKJ is


a sequence of disjoint sets in .


This definition was first given by Rickart (17).


Theorem 2.1. Let /4:' be s-bounded and finitely


additive. Then /u is bounded.






Proof: Assume/L is not bounded. Then there exists E
such that C) >1 Suppose there exists a monotone decreasing
sequence of sets FKCE such that (FKt+I) )I> (FK))+ for all K
Then 1( -FK:,)-1K) ,and (- )IJ > /Ia) -

S1(X .(). Therefore f KF, is a pairwise disjoint sequence
of sets, whereas i does not converge to zero, which
contradicts the fact that is s-bounded.
Consequently there exists G, E such that I/A( 1 and

for all LC G, ( ) /-I())I4J There exists
such that > 1 (G)>| +2, so \ -N\- n Crl\>l
If there exists a monotone decreasing sequence of sets F iR C ,
such that /pl(F+i)f/ >/ (F1)I/1 for all K we obtain a contra-
diction as before.
Therefore there exists G6C R-G, such that /1( )I>

and for all L C&; (L1) (z )l+i i ( J D There
exists P such that ()I| )l )/I3 so 0j(P- ) UG~I

4 ( P)O- (l(Pi)\ t rf] )|> > and there exists G3C P-(G UG&)







such that \ >63) 1 and for all L C (3 ) j(L)L< |lG3)1+ 1

G3f(1 u G,)=O

Continue in this manner to define a sequence of disjoint sets

~Gi such that for all L f/(Gi)( > This clearly contradicts

the fact that A is s-bounded, hence the theorem is proved. Q.E.D.

Remark: The converse to the above theorem is false. Let Z

be the Borel sets of the real line, and let 6- 4or(R)RZ ,

the class of essentially bounded measurable functions, where

is Lebesgue measure (9). Define /uE) ~= E the characteristic

function of the set E /u is bounded because for all E in 2 ,

I E oL 1 However, letting E '=(, l) we see that

E,) 0= 1 for all 7Z so / is not s-bounded.

On the other hand, if k is the real numbers the converse

holds, so the concepts of s-bounded and bounded coincide in the

case of scalar measures.

Theorem 2.2: Let /,:. )R be bounded and finitely additive.

Then / is s-bounded.






Proof: Assume the conclusion is false. Then there exists

a sequence of disjoint sets EKJ where J / ; does not

converge to zero. Therefore there exists ( )> and an infinite

subcollection EKJ C I[EtJ such that for all L /(EK')

Let AC be such that for all 6 (EK) > 6 and for all

Ei -' /( .) 4- Either A or i is infinite.

Without loss of generality, assume L is infinite. Let /1>O.

Let r be a finite subset of A with cardinality greater than .

Then ( ) ( \ Since was

chosen arbitrarily, this proves/ is not bounded a contradiction.

Q.E.D.

Definition: Let/U: -- Define u(E)

: .\f )'E E f uis called the semi-variation

of/ .

Theorem 2.3: Let/ Z. X- ThenA is s-bounded if

and only if/ A is s-bounded.

Proof: Assume is not s-bounded. Then there exists a
/4







disjoint

to zero.


sequence EK such that I does not converge

That is, there exists 6 >0 and a subsequence IEijC E4


such that for all i I(E >} According to the defin-

ition of I for all i there exists F C E. such that

I K.. Therefore does not converge

to zero, and is not s-bounded. This proves that/i being

s-bounded implies that/ is s-bounded.

Conversely, since I/L(E) 4 (E) for every E in 2,

6m (E =o0 implies that J ([k) 0 Q.E.D.

Theorem 2.1: Let/ -- be finitely additive and

s-bounded. If IEl is a sequence of disjoint sets in Z then
o0
/ (EK) converges unconditionally.
Krl
Proof: Assume the conclusion is false. Then by Theorem 1.1

there exists a subsequence such that / 'I

does not converge and is therefore not Cauchy. Hence there exists

> 0 such that for all N there exists p > 1 where



Obtain / such that I and / ( >
irF,






and then choose p, Q such that 6 P Z Y and )> / .

In this manner we define infinite collections P and I

such that for all j, qP < ( Pi j l/(E)J >E

Let F = EK Then Fj is a sequence of disjoint sets
t L) F i Then u

and *)- (EK )|>6 so IF; does not
J LJ
converge to zero and/ is not s-bounded. This proves the theorem.

Q.E.D.














CHAPTER 3


EXTENDING THE SCHUR THEOREM



In 1969 Brooks (3) proved the Nikodym theorem by using a


result of Schur (20). In this chapter the Schur theorem is ex-


tended to the vector case, and the result is then used to obtain


a very short proof of the Nikodym theorem for countably additive,


Banach valued measures. The proof of the first the -em is a


good example of the "sliding hump" technique.


Theorem 3.1: Let in .Il, Assume that


for all i the series x is unconditionally convergent,
4=1

and also that:


(*) for all A A- rX roO .
i neA

Then the limit in (*) is uniform with respect to A That is,


for all 6 there exists (depending on 6 ) such that for


all (1 and for all A in 6 .



20







Remark: If ) is the real numbers, then Ziin converges

absolutely for every L and the conclusion is that

i n-i

Proof: Assume the limit in (*) does not exist uniformly

with respect to Then there exist 6 >o and sequences

i IuKj such that I A > To simplify notation,

let I= K so I XKn> 6 for all K Let K, Since

n r" there exists a finite set FCA such that
MiAj

'17 -- ; due to the unconditional convergence of ,

there exists R, such that for all ACFR, oo) I n )

Let T, U i(u{RT() If Ac T+l1, ) we have

r, ( D ,and 7 I > U -L -X I

For every n sK=O so there exists IJ such that
T o
for all i AI<5>T Then Zin 7fT if LI C

In thl,]} Let K: 1 then t t -+

X ad T l and X 1 I
rK -,TJ -CT,- >nr-l | xI6KgnI;T

(- A/Tn > Therefore there exists a finite
set cAKC IT s t
set 62,CAg-C/,Tj such that K- .






Since 1 XKj converges unconditionally, there exists K
?t= I
such that for all A CCR,oo) n Kh Let T=

nsa" { P2 (} ) so that for all ACT++i,o3) we have that

/PA= A = o Note that (, 0 = 0 hence =
PtPf
7 -' p A Kn when ACT+f,o) Also

Ix nX because C [I,T;] Therefore

Xi+ X-+4 1<:
P uVP uL 1 a @ I l, -
for AC T+ioo) Also, since P U a C cT +, oo) we have

that l r ,> .

In the same manner as before, obtain I such that for all

17 Iinl' and let K3= Then we have that

IfA 3- ,T > and there exists a finite set

C C 3f-D,T] such that I K >P Choose R3 such

that for all A C [R3,.) 'A 3 A and let T3

In4t(3 u {R3) If c [T3 t,o0) we then have

h1r, UpAUP 3 >-j Since G uA C T +1,oo) we have that

t P, g u.r, uAI>- and u3 UA C T +1,oo) implies that

acv uP, U r3 v 3







Continue to define an Infinite number of Pj and K. and
u J

let Then for all j | > and i> J Oi

does not converge to zero, which contradicts'the hypothesis.

Q.E.D.

Remark: Theorem 3.1 may be used to prove the equivalence

of weak and strong convergence in. iF the space of absolutely

convergent series. Let fi f E( Lr l,l,." Each element

of 11 is a sequence or real numbers, and we write fi as i injgl

and f as x 00 The sequence {fJ converges strongly

to f if i Mf i. I-, X 0 and fI converges weakly to

f if converges to F) for every 9* in We

know that the conjugate space of 1, is , the space of all

bounded sequences, and that if : i then




Theorem 3.2: Let Zf. f I ':=I,A,." Then the fL

converge strongly to f if, and only if, they converge weakly

to f.

Proof: Strong convergence always implies weak convergence,







so I will prove the converse. Assume the fi converge weakly

to f Writing f# as Xi0: and f as 4 let

i Xi -X .* For any A define y* in oo by:



01 if K E '

Then z
because the fi converge to f weakly. Therefore, by Theorem 3.1
0o
we have that A -/X ih-XhI:= and the f( converge
h'l

strongly to Note that we only needed to consider those

elements of joo which take on the value O or I Q.E.D.

Theorem 3.3: Let X. ,7:1, ... Assume that

for all in converges unconditionally, and that:

(M) for all A A, Z 'in exists.
i nEa

Then the limit in (*) is uniform with respect to A ; that is,

for every >O there exists I such that for all J I and

for all Jn 7 t L (

Proof: This proof will be divided into six parts.

Part I: If {CiK is a subsequence, then








j (XiL i~~r)- o uniformly with respect to A
K ntL
Proof of I: Let o( = XK XK+1 Then for all K

00oo
OdK 7 converges unconditionally and for all / J Kn=O .
7L= K TI0A

By Theorem 3.1, Y/A 2Ek n O uniformly with respect to ,


and the conclusion, of I follows.


Part II: For all 6> 0 there exists N such that for

P oo
all integers L and for all p N in ~ E L 6
7n1 i 7t:( I

Proof of II: Suppose that we deny the conclusion. Then

there exists 6)> such that for all N there exist I and


P2N such that:


(1) oo E 26 S


Choose 6 as indicated in the above statement. Let No be

chosen arbitrarily. Then I claim that for all I there exist

P o00
L>I and P No such that Z Xin- lIn 2 If this


were not the case we could find Io such that for all > o ,
P o o00
P No ) 2in ZDi7 For all i Xn converges,
n j4: i I T
sP oo
so there exists NL such that for all P2 #; X l TXit < .
I lh~I ? i







Let R = ty NorJ i : L' / ' o] Then for all i and all
P 00
p> R Xin i (in and this contradicts the assump-

tion (1). Therefore, since No was picked arbitrarily, we have

that for all N and I there exist L >I P N such that:

(2) i ,- n i
00 f (00
Since 6n F- in exists, IZXi7n is a Cauchy

sequence, so there exists K such that for all L, K ,

~ J Let ',:K Since -Yi,71 converges,
z I &7t: I
P o
we obtain NI such that for all P2 N, ,I -n
According to (2), there exist V>i2 V oN such that
v o0 | V V
X n Y > Therefore, since -iZ )n, +
/ l' f 7I=1 ^ '-1 1.'



have that I -i t Lin II-- -n

00,l 00 6-e_-4=_ Let A, :[,V
E I t 13
Now choose ~ such that for all 5- In xn i ^ T .

According to (2), obtain i3 > i A 2 42 such that

I n3 =3 x 6 and since ij and i, are both greater

Sa h Combining
than K we also have that z Combining
S= I l f







nI A
these inequalities, we obtain 13 > Let




Continuing this process, define { } and 7,) ,?f:,2,-"-

such that 7nAmXn %E k^ h > for all n Then


one (X 'X +) does not converge to zero uniformly


with respect to A which contradicts Part I; hence, the con-


clusion of Part II is proved.

S S
Part III: If Se ) then in X where
1:I 7:I
X),.= X(


Proof of III: This follows because the sums in question

are finite, hence the limit may be interchanged.

Part IV: For all A Xk exists.


Proof of IV: If A is finite, the conclusion holds, so

assume A is infinite. Let = IKj Letting K i nK

we see that {Z1< satisfies the hypothesis of Theorem 3.3;


consequently, we may apply the results of Parts I, II, and III

to the double sequence {~CKJ To show 2 7t exists it suffices







00
to show that 1 exists, where s/- <= .
K=:I .

In order to prove 2-6K exists, I will show that K^ Kvy


is a Cauchy sequence. Let >0 By Part II there exists an N


such that for every i and v2 N .K- l-iK .


Therefore, if V,S2 AJ we have 1 K; b i < for all .


From Part III choose I such that for all i I ,

gla = K- 3 and Yf K i-. Then for all i I ,


EVA K=1 K=l KI K K-I K-- Kl
V 700
Since 6 was chosen arbitrarily, it follows that _


is Cauchy, and the conclusion of Part IV is proved.


For all A Xn exists, and Theorem 1.1 indicates that
00
c on converges unconditionally.


Part V: For all a A i (Note that


the second sum is defined by the result of Part IV.)


Proof of V: Since the conclusion is true for finite ,


we may assume 6 is infinite.


Let n] i k X g We need to show that

on on
AwL ij- i IK and since j[Ki satisfies the hypothesis
Z K=1 K:







of Theorem 3.3, we may apply the results of the first four parts.

Let >o According to Parts II and IV, choose N so
E o0 W oa
that I IK -K Kri K. K= : 1 K=

By Part III, choose I so that for every 1 I ,

N N oo
K 1/K [C Then for every (2 I K K I



oo oo
Since E was chosen arbitrarily, n -iK = xi K .
L K=I K=I

Part VI: The proof of the conclusion of the theorem.
00
Proof of VI: Let c<'-= Xji- X By Part IV, D'n con-
00
verges unconditionally, so for all i in converges uncon-

ditionally. Also, for all L Ai -'7 = -Zt Z( n -)
Snb n 7i

A X- x = O where the last inequality results from


Part V.

Therefore, by Theorem 3.1, for all 6>0 there exists I

such that for all i21 and all A > /\ ii t

/ tEX ~- in t; I ,where the last equality


follows from Part V. This concludes the proof of the theorem.


Q.E.D.







Theorem 3.3 is the desired extension of the Schur theorem,

and will be used to prove Theorem 3.4, the Nikodym theorem.

Definition: Let / : Z/-- 6 / is countably additive

if /#(=E - (E whenever iEs} is a sequence of disjoint

sets in .

Remark: /L being countably additive implies / is s-bounded

because if/U is not s-bounded, then there exist 6>O and disjoint

sets {En such that for all n / (F I > ; hence Ku(LE|)

is not a Cauchy sequence and/U is not countably additive.

Definition: Let U1 : Z --' be countably additive, nrl,Z,- ,

and let [EiJ be a sequence of disjoint sets in Z- The /I

are uniformly countably additive if for all 6>0 there exists M

such that for all m n and for all n L:/ ti (E)I E .

Theorem 3.4 (Nikodym): Let /: u x X be countably additive,

n=: 2,-. "If/ A(E) 4~L/,( ) exists for every E then/t

is countably additive and the countable additivity of the /,,

is uniform.

Proof: Let JEg, be a pairwise disjoint sequence of sets




31


in and let (~ =/k (Fn). The double sequence ~Xi' satisfies

the hypothesis of Theorem 3.3, so by the result of Part III of

the proof of that theorem, for every 5>o there exists N

P 00
such that for every integer 4 and all P2N s> in- =

p n.)-a (E. ) Therefore the 7L are uniformly countably

additive.

On the other hand, /L =I C, = A e AE)

gin E- n the last equality following from Part V

00 00
of the proof of Theorem 3.3. Since X : T (E) we have

that / u Eln ) = I.(E) and / is countably additive. Q.E.D.
n=1

We give a proof due to Brooks and Mikusinski (5) of a result

of Banach (2).

Lemma 3.5: Let C be a separable Banach space, and let {

be a bounded sequence in Then there exists a subsequence {iK*

such that YnK,() exists for every X in

Proof: This proof consists of a standard Cantor diagonal

process. Let iX L be a dense subset of X and choose M>O

such that |






for each n so 9nix, is a bounded sequence of real numbers,

and there exists a subsequence Y,l of / i so that K YK (x,)

exists. Likewise, there exists a subsequence of

such that m. ,K(X ) exists. Continue in this manner to define

subsequences sIonI so that +(, ,K j C r i: ,KIJ KK and ,K(XL)

exists for every n For each integer K let Y .KK "

Then {YK is a subsequence of tnI and for all j ,

' K n(xj) exists.

We now show that YK( ) exists for each X k Let
S(K.

XE and 6O Since f(jj1 is dense in X ,pick Xj

so that IXj -I< M, and choose R such that for all K, 1 R

I~ K t J X) I Then for all K,~ R ,

7V 7) (Xj) x)4-

O (xj) n'(x) f + = Therefore the sequence

JIuK) is Cauchy, and A ( () exists. Q.E.D.
00
Definition: Let X h =, I, *Z, Xn is weakly
71=1 *
unconditionally convergent if for every subset A of the integers,

there exists XA such that TX A fhA









The following theorem is known as the Orlicz-Pettis theorem,


(13, 14), and the proof is due to Brooks and Mikusinski (5).


Theorem 3.6: If X is a separable Banach space, then a


series 2 t converges unconditionally if, and only if, it con-


verges weakly unconditionally.


Proof: Unconditional convergence always implies weak uncon-


ditional convergence. In order to prove the converse, I will


assume that the series converges weakly unconditionally but not


unconditionally, and then arrive at a contradiction.

00
If F-X does not converge unconditionally, there exist


an 0>o and finite disjoint sets Ai such that InAj>l for


every integer L Letting we have that ._'
-nfZ^ E'

also converges weakly unconditionally.


According to the Hahn-Banach theorem, there exist Xm such


that /I=1 and I X(-)I= I for every integer ) .

S (-yj 00
By Lemma 3.5 there is a subsequence Xi of I such


that ^-4Xi xtL(X) exists for every X in M and to simplify


notation I will assume that Mi= for all i Let 0Q = X(3).




34


For any A there exists 2a so that Z L = OA)

for all Therefore, m't, Tdi n, )

a X; n A) Since 2 Xin) exists for all \ and uncon-
L 1t6b

ditional convergence implies absolute convergence in the scalar
oo
field, it follows that X*ir()I,) ~o Therefore, the double
?tI

sequence 4inj satisfies the hypothesis of Theorem 3.3.

Letting 0d : = d we have from Part IV of the proof
oo
of Theorem 3.3 that Z 107 : I

theorem we have that din n n = uniformly in .
I 'iA C.6 ?

Choose positive integers t, and NX such that for all n 2M

da I\ and for all iNW and all A ih.-g. 1 *


Let )9n>tA Then I2J |^Hn 4 jI^l


St '- d | which is a contradiction. Hence

the theorem is proved. Q.E.D.













CHAPTER 4


EXTENSIONS OF THE VITALI-HAHN-SAKS AND NIKODYM THEOREMS


In this


are extended


and strongly


to the whole


(15).


chapter the Vitali-Hahn-Saks and Nikodym theorems


to the case where the measures are finitely additive


bounded. The first theorem to be proved is the key


chapter, and is an extension of a result of Phillips


Theorem 4.1: Let 11: P(7Z) be finitely additive

and s-bounded. If m a = c for all A then

n 2 /Un ( ): uniformly in (That is, for every


6 0 there exists N such that for all A N and for all D ,





Remark: Since the / are finitely additive and s-bounded,
00
it follows from Theorem 2.2 that Z// -(t) converges uncondi-
-t'l

tionally. Therefore (Wt) exists for every A and the
tE&A






conclusion makes sense.

Proof: Assume the conclusion is false. Then there exist

an > 0 and sequences Ii, ~ such that I / >

L= I 2 ,..... To simplify notation, assume n = .

Let ijI /I. f > so there exists N, such that

/ '(() /6 for all c N,00) Letting 7=-A lnEI N,-1

we have I I ) 1> )=--Z,)



Since j) 0O for all j we can pick L such

that 1/i () / L ) converges unconditionally,

so we can pick N,> ,I such that for all ACI ,l ) ,

IAl < /6. As a result, L Z .L( +
a 2v 6 2r',-o
+ / 2) and therefore

J~ P2 ) JE/iaf) ^) Z i >

1- 6h > .. Let ?j, lLi,a,-ij

Choose i3 such that /3 / 7 and pick 3 so

that for all A /6 Then, as before,
una0l 3 o)






/ ; 3 r > and we let ?'3 3- n .

Continue in this manner to define an increasing sequence

of positive integers JNJ such that:
-JK- I
WE nK

(2) nLJ' 76 for all A and K,3;-* ;

(3) letting K AKnld[ K- ,1)K- (No1) we have



Note: If A is a finite set disjoint from LK-IJ ,

then it follows from (1) and (2) that A / < l .

Let 6,= f.*((,tl)]t: I h)= ,,.'- For all values of ?m ,

Of is an infinite set. Also the sets iJ are disjoint

because fnl V implies that 2 ml' -I)-12~m -) for all

values of n, and Xz Since the sequence fK) is disjoint,

it then follows from the preceding statement that {u(mj l, is

a sequence of disjoint sets, where UB, is defined to be

Ulr :TE In other words, I have divided a countably

infinite collection of sets into a countable number of countably

infinite collections; and then have taken the union of each






collection. Since all the sets were disjoint, all the unions

are disjoint from each other.

Since /Ai, is s-bounded it follows from Theorem 2.3 that

/ l is s-bounded, so /A (UB,)- O Hence there exists

an no so that ,Ai(Bmo
EIA. ?" o because i,( \)> .

Let PL -.m is a countable collection of the 7? ,

so by the same process as above there exists P ,an infinite

subset of ,such that / (Up < Again ? .

Continue this process to obtain a sequence of infinite sets i2

where PmiC C and i,(Ul )/
Let r consist of the first element of r for every 7t .

P is an infinite subset of TKI and for each value of h ,

tr-f consists of a finite number of elements.

If rE then /K(, r)llI)-

/K ufr-(ru )~J) iK(fr nrK]) because r .
u fr-(rufrKU) f is a finite set disjoint from LNK>,NK41 ,

so so ufp[ Kf- t j)I)







/ UiK ([ fl '))] Combining all three inequalities we

obtain K i r) TK -- > *

Therefore / UP) if ? EP There are an

infinite number of ?' in P so /iK K)U I cannot converge

to zero, which contradicts the hypothesis. This proves the

theorem. Q.E.D.

Remark: Theorem 4.1 implies Theorem 3.1.

Corollary 4.2: Let /J,: -4~ be finitely additive and

s-bounded, n=lI,Z,-' Assume .& /',(E) exists for all E

Then if EK) is a sequence of disjoint sets,

. ~I(/ at, "-/ K) 0 uniformly in A .

Proof: Define Vj .(7)-4 X as follows: A/a)=

A( -i Ej( The V satisfy the hypothesis of Theorem

.4, so J, (,a /n)(E) -- 1 0o uniformly
A K C6 K E6

in Q.E.D.

Corollary I.3: Let /IV: be finitely additive and

s-bounded, ?=Il,Z,. Assume A/ (E) exists for all E
7?1







and let /u(E) =k /n?() Then / is s-bounded.

Proof: Assume /U is not s-bounded. Then there exist

6>0 and a sequence of disjoint sets fEg} such that for all K ,

I(EK)I>E Let iI Since /mUn()= 0( there

exists K, such that /1h (E) / However ,i /)(E)/ i

//EK) j> so there exists z > so that ) >(E-

Therefore m -~U i) ( >) I- C Because

I/LrF 1(EK)O there exists KX such that K2) 3.

&. |j (E )/= Ii(E )I > e so there exists 3 > >

so that I)3 (F 9)I and I Z-/ S 3)( 3 )3 3

Continue this process to define sequences {n Kj

where iI} is monotone increasing and for every integer ,

S/l ) > Therefore the sequence

IZ1/ ( / x)(Hj does not converge to zero uniformly

in A and this contradicts Corollary l.2. Hence the theorem

is proved. Q.E.D.

Definition: Let / -h be finitely additive and

s-bounded, 7Z~:1,I,. We say the ~ are uniformly additive









if /tm Z" X (EK 0 uniformly in A and n whenever {EK)

is a pairwise disjoint sequence of sets in 2 That is, for all

6 >0 there exists J such that for all J 2 J all n and all


'i KEJO,) n A

Remark: When the un are countably additive this is equi-

valent to the /"u being uniformly countably additive.

Theorem h.h: Let /U,.' ( be finitely additive and

s-bounded, ?=/,', ," Assume u (E) = M/n('E) exists for

every E Then the U, are uniformly additive.

Proof: Let EK be a sequence of disjoint sets and let >O .

Let V,= P(7)-4 be defined by ) )--(/ -, ( E) By

Corollary 4.3 /x is s-bounded, so the X satisfy the hypothesis of

Theorem h.1. Hence there exists an integer N such that:

(1) fJ/-) ( < E for all A2 N and all A

Since u is s-bounded, it follows from Theorem 2.4 that
o0
V- I( ) converges unconditionally, so by Theorem 1.2 there exists

I, such that:








(2) K ,) E for all

Combining inequalities (1) and (2) we have that for all t 2 N

1 ,f x(EK) M ^ 1 11E)
and for all ,,o) a n f,)

/(?nr,,-1-4 ) (Ex

Each /L is s-bounded, so there also exists 1 such that

LKEf n[r, ) I< for all 6 and all 7n N Consequently

/Mh(EK) 4< (.6 for all n and A Since E was chosen
KE6 n +I oo)

arbitrarily, this proves that the are uniformly additive.

Q.E.D.

Remark: Corollary 4.3 and Theorem 4.l are the intended

generalizations of the Nikodym theorem.

Corollary L.5: Let /h: E-~ R be bounded and finitely

additive, n =l, a,-- If /(E) ,/E ) exists for all ,

then the /, are uniformly additive.

Proof: It follows from Theorem 2.2 that the u are s-bounded,

so the conclusion follows immediately from Theorem l-.. Q.E.D.

Corollary 4.6 (Nikodym): Let /An: -9 ~ be countably







additive, I, 2, ** If / l E) = /n,(E) exists for

every E then /u is countably additive and the countable

additivity of the n is uniform.

Proof: Let > 0 and let {Egi be a sequence of disjoint

sets. By Corollary l.3, u is s-bounded, so i /U(EK) exists,
/ K=I

hence there is an I, such that for all i 1 ,

1 j (EK- 00X(E:) /X-

By Theorem h.-, the u are uniformly countably additive,

so there exists 1L such that for all n and all 2 i ,

I /E2) which means ni <)/ < because each ,u

is countably additive. Therefore:

(2) U EK) = j /A( p 0 for all 1i I .

Let I- 7 IJ11J Obviously # is finitely additive, so:

(3)

Combining inequalities (1), (2), and (3) we have K) -


= I ,ewr )h h

S.E Since E was arbitrary, this implies that


:- / .Q.E.D.
EI)= (K )







Definition: Let /~: X and V:W '- R+. We say /I

is absolutely continuous with respect to V if for every 6 > 0

there exists 6 >0 so that v(E) < implies IJE) < for all E

in i1 If / : we say the /I are uniformly absolutely

continuous with respect to d if for all 6>O there is a 6>o

such that VE) 6 implies I/r(E)k6 for all E and n .

Remark: The next theorem is a generalization of the Vitali-

Hahn-Saks theorem, and is the most important result of this

dissertation.

Theorem h.7: Let/U : .-4- X be finitely additive and s-boun-

ded, n:L, ." Assume that Am/u.(E) exists for every E

Let ) be a non-negative (possibly infinite) finitely additive

set function defined on If Uh is absolutely continuous

with respect to ) for each L then the /n are uniformly

absolutely continuous with respect to .

Remark: If ) were assumed to be bounded, then the /f

would automatically be s-bounded.

Proof: Deny the conclusion. Then there exists an >0O




45

such that for every 6 > o there exists an E, such that )(E)4

and J1/n(E)I ) for some 7 Let E, /h, be such that vE,)
and /II(E, )I Let
that /in, (E) ,

v(E)) and /In;(E) >) Then for all Fc Ez ,



Now assume that E, EK , l- nfK have been chosen. Let

SK+1 be such that for all : I ..K V(E) < CKI implies

/hE)< 3 Now choose K+l and l+ so that v(EKti) SK+1

)K+i K and I/ (E'+,) > E Then for all FC K+

and for all Z .***i , 1/ i(F) j .

Resubscripting for simplification, let /UI "-'/i We then

have that:

(1) for all I I (E)> and for all J : l*-l- and all

FcE , JF)<

Let F= : = I and assume there exists an L, >2 such

that / II E Let F2 I E L- Assume

FI'.FK have been chosen in this manner, along with i" iK *




46

Assume there exists it( > iK such that J/iK4I(FKnELK,)

Then define FK+1I F El X+ If this process continues

we obtain a countable collection of sets FK such that

VKIFKF K1)1- / ,(F)nE and



because of (1) and the fact that FK n E- zl ELK+I

Therefore |I(,, -1:)/ -(FK - -f = f-

Note that IFK-FK+I} is a sequence of disjoint sets, and

S (/i4 / f j) (FgK-.+) I} does not converge to zero

uniformly in 6 which contradicts Corollary 4.2.

Therefore the process of choosing the FK and lK has to

stop, so there exists FK lK 2., such that for all J > K,

J(K E) Let f:= K H, = F /, ( ,

and E =p, HI Then the following three results obtain.

i. (H,)

Since H, cEi it follows from (1) that Jt(H1) I <
E E
I, /6
i. >JL HI1t Y






We have that E = F, (,-F )U(F F3)U ..U(FK_,-F) U F<

| 1> 6 and / (FiO )= // (F n El ) I /

since F, n Ez C Ej and 2n< II( -FJ)h I/'F E)\ <

--< /6 because 13 > i >2 hence 3 >' Thus for

xJi -1( F +-Fj)j I+'jEi +) and
e I, ( ) I |/',-F,-F ,. .-+ I,"(F-,- FK)I+ /t i (, )|

---- +.* + i2J(H I since H,:FK Therefore

16 3Z-L +. =C

III. 'E ) --

Note that Ep +i (E+i- H,) U(Ep + n H,) H1 was chosen

so that for all J> P /Ij(EjnH,) I Hence

E ,< I 0(E, )) i(E, H,) + |it-(Ep)1,pn H,)|<

V (Et Thus E-~ IFP'

Note: From I and II we have that )(H ~ ~E '

Let FE= Assume that there exists Z > 1 such
Lt /r.^ I F' ['t r^ ')

that l/ui i 2 and let F E If

we could continue to choose i indefinitely, then we would
ob n a contradiction the same as before. So there exists
obtain a contradiction the same as before. So there exists r







and I ?.2 so that for all J > ti IF ('F n )j )


AP2 =1A / I+ 1)


8


r + p, + .-


Ep,+pt (H, uH.)


Then H, n H, = D


because Hn C l I)


(and
and E (H nH =E


(Recall the construction of

then obtained.

r'. ,1 H) 3


Since H~C EF1 C EP..l


1V (Ha)


The following results are


and P,+4 >2- we have that


2 +132


SI E-


( F2- ) (K )- F3) U'. U "

< 1 ) l -j F -


[lal"iH)


, since HI. r


and tL > I we have


U =F=' E"')


Since F(')


Sso


() c (C)C P1
E; L Ei,1C Ep,,l


that 6 (IF (F 1( E <
1/ 1 1 1 2


S+ E
P, + +Lz (


6
.2


Likewise I/ ( j -


E3


F (I) -

because i >
J4i J


for all J:1,,..K-I Hence < +- +


Let
Let Hz = F (


and
and K


= ^-^')
W .
P;L+


(1)
E,


P.,+ L + 1
.1 J+1


'+ n F +


)( F/; ) ECI <
n J+, "




49



E -
8$ 1r+ ^"^)| and it follows that I/aN I



III'. ffor all .

Since HH the sets (Ep,+p n H,) ,

(nEp pl nH ) and (EP +Pi+ (H I u ) are disjoint.

Therefore 6 / (EpP+ ) I I2)(Ep, p Lin iH') +

/2) YEP, +P ,in i E J'(2)E (f, H .1)

Note that E PE Pi", H) and |)(Ep +P+L

because of the construction of ,l and H2 Also, by definition

)I/ ,(EP +Pl- HHl) jI (i (E) Therefore



Note: From I' and II' we obtain J i >(-1 f 31

We now proceed to the general induction step.

Let K 2) /A (0)= -, E El Assume that the

following statements are true. For each J = K Hj

and Pji2 are defined and () = Ep.--Hj .' =/ EP,+i

)i'EU)> E( ( HpnHj0O if P=I---K






)j) 0J-,) IJ->


S>- j- f ,

All of the above statements are true when K= I will

now show that their being true for K implies they are true

for K+ I
(K)= (K)
Let F, I E 1: Assume there exists lZ>i-l such

that I( K)f E<) K+X and let F ) F I E .

If there exists 3 > so that (/ ( n 1Ei3 ) ,
S(K) i (K)
let F3 = F 3 If this process of choosing ij and

Fj should go on indefinitely we obtain a contradiction because

for all j Ijl -,- j J / ( n fl

(-0) (K) (K) E(
whereas -j _, E C, C p +.. +- PK + ij and Ij > imply
I (P F E (K))' zP, +-t" .+ P,+j '4
that L +Z '

Combining these two inequalities we see that for all J ,
S 1(K) II >K) K))3
P( -/4jl, (J ji -~ > + Therefore

l K' i F jj does not converge to
zero uniformly in This contradicts Corollary 4.2.







Hence there exist E(K)
S(K.) n() -- -3
J (K EJ )\I 3
2-K+


and i, a such that for all J > L ,


. Let PK+, = L.


(K)
, I-K+=, Ft
p(K') (K)
L P:Ep .


In order to show the induction step is valid, I must prove

the following four statements.


I". For all j : I\' K


Hj 1HK+I=


+ (K) (K) (K)
HK+1 C E = FK) by construction, and E,


\j 1


SHence H K+l r (J, = 0:


11 1


Since HK+1


c E +... +PK+


2P, + + + K P -.


3Z


because P > 2 .

I /.a ...(HK Z'l-

Note that E) FK)- ( ,(K) U(K) U (K)
I 3 l


Also I I(- () I fK))F : )I' n+
E ( K++) b e F(Ei.+ P K)+ .
ZK+-+I because F L CE p


and I Li


HK '- PK + PK+1 i
H+1 P +""* + Pg + Pg+, +1


FK)
UF*


(K 'P) (K)
" 7-14pt(+r+ I /'"Pj +--' + P;(+ 1P+,+


i K+1
j H


E,+...+PKt +


, I- H )/ 1


l + - PK, -IK+H







Likewise K) (KF ) 3 I ( K)-t
1:I r3)i3 +."

K1+; because F n E E + ...+P + 3 and 2 3 .
S ) (K) (K) _
In the same manner, for j :...- we have I ( rj Ij

IjP ++PK+jl.+ < +

because J j+1 Therefore E- K (E I

I++K + -+++z '"* + + )/ (HK)l ; the

first inequality follows from the induction hypothesis and the

second inequality results from the definition of HK+I and the

inequalities in the preceding sentences. Therefore,

S- ..-:+--^iT-(++i +*+ Z} <. Ir, ( 1 (K)

6 e
Since 2.+l4 +I + K + we have
6 1_5_ 6 (K)
(K+X H+1


IV''. For all ( / I (E ) > 6 +- --E

(K) K) I. (K( (K) (K) \
^..^EK~ -H,,)t+ H pK+ +PK+K++.Jf 11 H K 1^ )j >
because of the induction hypothesis.

As a result of the construction of HK+I and because PK+I+ >P +I

we have J/p+, +i( p + H2) 2 I Thus

c, PK+,^ :,-HKJ|> PO I+-^







Note: It follows from II' and III" that

J(1A (KK-0) (H I ( CE- -- C- 3-6

Therefore, by induction we may define infinite collections

[H and {/) such that for each K2 L (K-) (K-)

i 31 > 32 {HK is also

a sequence of disjoint sets. Letting % ,) =V -

O -- 1 then | +,- ) J /6- for all 72 So

(V n+iJ M+ ) K tI(H 10 does not converge to zero uniformly

in L which contradicts Corollary l.2. This concludes the

proof of the theorem. Q.E.D.

Definition: A linear topological space is said to be locally

convex if it has a base consisting of convex sets. Let

U :{UL dE AJ be such a base. Let = : U Ua:U oI

For each U, in define I( X) = a: a >O, iU and

( [)- : EII() .. Then is a pseudo-norm for the

space 6 and has the following properties.

(a) 6)0 o0

(b) + +oo00









(d) if X)Ur then a( )_ I




That is, / has all the properties of a norm except that it may be

possible to find an X such that >XO and (Y)=0

In the theorems of this chapter the fact that I=1 0 implies

X=0 was never used, and with appropriate modifications the theorems

are true if the set functions take values in a locally convex linear

topological space. First of all, one has to define s-bounded and

absolute continuity differently.

Definition: Let X be a locally convex linear topological

space, and let :({A3 be a family of pseudo-norms that determine

the topology of E Let / U-X and 'L: -- Rt be set

functions. We say / is strongly bounded (s-bounded) if

A ()F:O 0 whenever {E} is a sequence of disjoint sets. // is

absolutely continuous with respect to V if for every o( and 6

there exists 6>o such that (E6)/ implies ( ))
The topology on 6 determined by a single pseudo-norm o( is








not necessarily complete. We then let K be the completion of ,


and if /L: Z- Z is s-bounded, we consider /4 to have values in


Ej and it is then true that converges unconditionally


in )c if [Exn is a sequence of disjoint sets. The statements of


Theorem 4.1 and Corollary 4.2 do not change if 9) is locally convex,


except that An(t) is considered to be an element of for a
'tEA

fixed o Taking into consideration the new definition of absolute


continuity, the extensions of the Vitali-Hahn-Saks and Nikodym


theorems for finitely additive vector measures are valid when the


range of the measures is a locally convex linear topological space.













CHAPTER 5


A COUNTEREXAMPLE


In this chapter an example is given to show that weak con-

vergence does not imply strong convergence in the space of countably

additive set functions.


Let /t: 0(01)-9 R be countably additive, and assume that


for all A t (A): 0 Then from Theorem 4.1, we have


that V/u, (A)= 0 uniformly in Z which in turn implies

that D/-- )j/Un = O where lI is the total variation of/L .


Since Theorem 4.1 was very important in proving theorems

about set functions from a O0-algebra to a Banach space, it is

reasonable to pose the following question: if -> R

and if j/ T (E)=O for every E does it follow that


- I/, nl= 0 ? The answer to this question is, "No," as the


following counterexample shows.


56







Counterexample: This counterexample will show that if

Z/ : Z-R are countably additive and if u /-(E) =0 for

all E then it does not follow that A u-= 0 .

Construction: Let 1 be the Lebesgue measurable subsets

of (0, I) and let A be Lebesgue measure restricted to .

Let -f be defined:

\ if X is in ( ) o o < Z a, even
foO -1 if X is in ,( ) o0 <2 a. odd
0 otherwise

Let ? (E) a d A Then for all >0 A(E)<& implies

I fd)<$ which is equivalent to IL,(E)1< Note that

if K >n then /a :O

First it will be shown that if is an open interval,

then / (6) =0 Let 6>o and choose n so that 6 .

Let = -fru[i":^ 3i<9 c-frrp .f.7z: ;i f3

Then f ('-U ) (:a ) L( C and this inequality

combined with the statements in the last two sentences of the

preceding paragraph give us that I/4(6}< 2 6 for all K > .







Since E was chosen arbitrarily, the conclusion follows.

Now let V be an open set in F and let > 0 There

exists a countable collection of disjoint open intervals fl
00
such that V = U( (I am considering the null set to be an
o00o
open interval.) Then A(V) = A( and, since A(V) oo ,

there exists N such that A(V- ,)Z 6 ; hence for all 2 ,

I/x /
that for all K>T( //K k)I< Then letting

T= 7 A L: I, N1 we have that for every K >T, 1-K(V) I

IIK(v/IM -A)/ + K() ( I + N(): Therefore

we have that & /U, (V):O for all open sets V

Let E be a set in 1 and let O There exists an

open set V so that ECV and \(V-E)
/Yt(V-E ) E for all n By the result of the last paragraph

there exists T such that for all > T //kiV)I E hence

/(IE)I J /(+ V ( th(V-E)l 6 This implies that

A/6un (E)=0 for every E in .

The /- are countably additive, so they satisfy the hypothesis




59

of the counterexample. However, / f(o,) ifnl X= I for all n,

so j[jnlI does not converge to zero.














REFERENCES


1. Ando, T., "Convergent sequences of finitely additive measures",
Pac. J. of Math., 11, 395-h04 (1961).
*
/ / /
2. Banach, S., Theorie des operations lineaires, Monografje Mate-
matyczne, Warsaw (1932).


3. Brooks, J. K., "On the Vitali-Hahn-Saks and Nikodym theorems",
Proc. Nat. Acad. Sci. U.S.A.;' 6, 468-471 (1969).


4. Brooks, J. K., "Representations of weak and strong integrals in
Banach spaces", Proc. Nat. Acad. Sci. U.S.A., 64, 266-270
(1969).


5. Brooks, J. K. and J. Mikusinski, "On some theorems in functional
analysis", Bull. Acad. Pol. Sci., Math., Astron., Phys., 18,
151-155 (19707.


6. Darst, R. B., "A direct proof of Porcelli's condition for weak
convergence", Proc. Amer. Math. Soc., 17, 1094-1096 (1966).


7. Dunford, N. and J. Schwartz, Linear Operators, Part I: General
Theory, Interscience, New York (1958).


8. Dvoretzky, A. and C. A. Rogers, "Absolute and unconditional con-
vergence in normed linear spaces", Proc. Nat. Acad. Sci.
U.S.A., 36, 192-197 (1950).


9. Hewitt, E. and K. Stromberg, Real and Abstract Analysis, Springer
Verlag, New York (1965).


10. Hahn, H., "Uber Folgen linearer Operationen", Monatsh. Math.
Physik., 32, 3 (1922).














11. Hilderbrandt, T. H., "On unconditional convergence in normed
vector spaces", Bull. Amer. Math. Soc., 46, 959-962
(1940).


12. Nikodym, O. M., "Sur ler suites convergentes de functions
parfaitment additives d'ensemble abstrait", Monatsh.
Math. Physik., 40, 427-432 (1933).


13. Orlicz, W., "Uber unbedingte Konvergenz in Funktionemaumen",
Studia Math., 1, 83-85 (1930).


14. Pettis, B.
Math.


J., "On integration vector spaces",
Soc., Wh, 277-304 (1938).


Trans. Amer.


15. Phillips, R. S., "On linear transformations",
Math. Soc., 48, 516-541 (1940).


Trans. Amer.


16. Phillips, R. S., "Integration in a convex linear topological
space", Trans. Amer. Math. Soc., 47, 114-115 (1940).


17. Rickart, C. E., "Decomposition of additive set functions",
Duke Math. J., 10, 653-665 (1943).


18. Rickart, C. E., "Integration in
space", Trans. Amer. Math.


a convex linear topological
Soc., 52, L98-521 (1942).


19. Saks, S., "Addition to the note on some functionals", Trans.
Amer. Math. Soc., 35, 967-974 (1933).


20. Schur, M. J., "Uber lineare Transformationen in der Theorie
der unendlichen Reihen", J. reine u. angew. Math., 151,
79-111 (1921).


21. Vitali, G., "Sull'integrazione per series Rend. Circolo
Palermo, 23, 137-155 (1907).















BIOGRAPHICAL SKETCH


Robert Jewett was born in Portsmouth, Ohio, on August 20, 1945.


When he was eleven years old his family moved to Fort Myers, Florida,


where he graduated from high school in 1963. In the Fall of the same


year he went to the University of Florida on a golf scholarship, and


played on the golf team for two years. In 1967 he received his Bachelor


of Science degree in Math, and from that time on has been working toward


his doctor's degree. He is a member of the American Mathematical Society.


He was married to Suzanne Strobak on July 3, 1971.

/










I certify that I have read this study and that in my opinion it
conforms to acceptable standards of scholarly presentation and is fully
adequate, in scope and quality, as a dissertation for the degree of
Doctor of Philosophy.





ames K. Brooks, 'Chairman
associate Professor of Mathematics


I certify that I have read this study and that in my opinion it
conforms to acceptable standards of scholarly presentation and is fully
adequate, in scope and quality, as a dissertation for the degree of
Doctor of Philosophy.





David A. Drake
Assistant Professor of Mathematics


I certify that I have read this study and that in my opinion it
conforms to acceptable standards of scholarly presentation and is fully
adequate, in scope and quality, as a dissertation for the degree of
Doctor of Philosophy.





Zoran Pop-Stojanovic /
Associate Professor of Mathematics


I certify that I have read this study and that in my opinion it
conforms to acceptable standards of scholarly presentation and is fully
adequate, in scope and quality, as a dissertation for the degree of
Doctor of Philosophy.





Gene W. Hemp
Associate Professor of
Engineering Science and Mechanics









I certify that I have read this study and that in my opinion it
conforms to acceptable standards of scholarly presentation and is fully
adequate, in scope and quality, as a dissertation for the degree of
Doctor of Philosophy.


41' -


Stephen A. Saxon
Assistant Professor of Mathematics



This dissertation was submitted to the Dean of the College of Arts and
Sciences and to the Graduate Council, and was accepted as partial ful-
fillment of the requirements for the degree of Doctor of Philosophy.


August, 1971


Dean, Graduate School


L2_/^




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