A Study of Uniform Boundedness

CANADIAN THESES

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in the Wpartesent

R.T. SBantunga, 1983

SLW PRllSER BNFJERSI'PY

Decenber 1983 Hae:

Degree:

Title of Thesis:

_ C, Godsil ~srtsrnalgnamniner I3e-t of Hathematics Shmn Praser University 'i hereby grant to Simn Fraser University tba right to .lend ,y 'hcsis, or ~.ndadessay 4th title of which is 5taun helor) f to users ot the 5imFrasar University Library, and to mke partial or r'H.lgtrt copies oniy for such users 5r in response to a request fram the

1 Ibrary oh any other university, or other educationat institution, on its orn behaif or for one of its users. I further agree that permission

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Author: (signature)

( name f c

t,6, \98% I * f date The 76f this thesis is to review sae of the theorems

Applications Fn *t& theory af topological vector spaces and mility

------> --A Chapter I is of introduc&Xy nature- In Chapter 2, by

4 verspsce, Ln Chapter 3 we employ a simple version of Rostnthal's

> - 1- to give a proof of a remalt which is due to J. Diestel and B. P ires. i P

of rings ,of sets, namely the class of rings with property QI). '\ Among the other rehts obtained in this Chapter are enan1ir.d versions

Theorem is prmed. in several settings. At the end of'this Chapter we obtain an iatprovement of the ~rlicz-pettistheor-. I muld like to thank Dr. JJ, 5Waber for suggestbg th@ topic and far his advice is& eacauagernt in preparation of this thesis. I mfd also like to thank Ms. Sylvia Ross for the 52, Elementary proparties of mctm measures ...... 28 53. Strongly bounded =&or measures...... 32

54. Convergence and bouadedness of a sequenoe of stroqly '&ere ass i#wblra;L rsoplts cancsrning oontinuous Linsar f~~

and vactor v&Pab rhi& &rive a conclusion of dfom bom&dmasr

- ft&a+qpthe~h~~~orkrioa~~~----

p--p----ppp p------p

-- p-p--p-- discuss of t)lsarrs results and tlnir humliate appliaktioas in tbe t

2. The Vitali-WpSaks-Uzkcx&thea,r~ for finittidy additive -tor '-

3. The likeBoaadedDees tbawra 'for finitely additiw mctor.

We prove each of tlw& above results in a rore mrdl -ti-.

, In proving (1) and 12) we we priritive sliding hump argrrents of the tlypQ

possible existence of a =re general type of ring than those for which oonammd ar~regular over finites sets (3.4 Definition 2) , - 52, ~1ogicalvector spaces,

The folloving Iotation will be used thmugbout this thesis.

- set of real drs. - set of ooaaplex nmakrs, - set of positive integers.

- set- of qori-neqative real n-rs.

- power set of a given set X .

Definition 1. A subset A of a vector &ace is said to be

~------~--p ------~ ------ppppp-- -- -

(i) absorbing if for ec?ach r in X there exists a scalar a with

x € aA ; (ii) balanced if acA- forevery X vith L 1 ;

and (ivl absolutely convex if A is balanced and convex.

refinition 2. A vector space X with a topology T , which we write

- - - - 5 as IX,T), is-called a topological vector space if the oper_ations of

vector addition and salar multiplication are conthous.

Proposition 1. A vector space X with a topology T is a topological

vector space if a& only if there exists a 'fundamental neighbourhood

0 system r1(0) at the origin of X such that:

(I) Each U in qfOfa is absorbing and balanced. (2) Fur each U in ri (01 there exists V in (0) with V + V -c U .

DefinkZtion 3. A topological vector space (X,T) is locally convex in El') Each U in qfO) is absarg., Wahced, and cmvsx, !

-finition 4- bt X be a vector space. A function p: X -t Ri is

lilaa =O & XfX, n n

If, in addition, p(rf .O for every ~t#-0, then p is called an

Proposition 2. A vector space .X with a topalogy T is a tupologicaf f --&or space if and only if there exists a family F of P-seminormg

on X generating tbs= topology T on X ; also, T is Raudorff if

andonlyif F is+-. i.e., fax x EX, xfO , tbereexi~ts

p C F such that p(x) # 3 .

Definition 5. 'Let I( h a vectoi &&e. A function p: 4 + R+ is ~b,in addition, pdrf > O for every x # 0 , then p is called a , i h

Remar)E, A vector qxake X with a tapolcqy T generated by a seminorm

drespectively, nand on X is called a seminomed [respectively, no-)

space- A complete '70- space is called a Banach space.

Proposition 3. (X,Tf rs a 1~callyconvex topological vector space - --- -

if ~d only if T is generated by a'family of s-5 on X .

Definition 6. A subset 8 of a topological vector space X iscalled ------iY 4 Medif for each neighbourhood U of zero in X there exists

& f R+ such that 3 - A.u . . .

* Promsition 4. kt - X Se a topological vector hce and let A -C x .

A Fen the following statements axe equivalent. F.

1 A is bounded,

(21 For every sequence it,) of psitiye numbers with lim t = 0 n n

f and every sequence ixn f in A , limn tnn x = 0 .

1 If x is a lxally convex space, then stat~nt(lj is also

equivalent to (3) A is 'munded with respect to each continuous lz[ = z.sgnz for any z f . Proposition I. Let, (X,T1l and (Y,T 1 be two topological vector 2 spaces over the same field. The set of all continuous linear functions

wise addition and the pointwise scalar multiplication.

Proposition 2. fRt X,Y be sdormed spaces .' A linear function f

from X to Y is continuous if there exists M > 0 such that

Proposition 3. If llffl =sup{/lf (XI i[[x E X, [XI 5 13 far f E L(X,Y) , =

then 11 jl is a mxnon L(X,Y) ; moreover, it is a now if Y is -- -A

a normed space.

Theorem 1. Let f be a continuous linear mapping from a subspace A

of a topological vector space X into a complete Hausdorff topological

.vteckor space Y ; then there exists a unique continuous lineax map

F from the closure

Definition 1. Let F -c L(X,Y) where X,Y aae topological vector

spaces. Then F is called

la) pintwise bounded if { f (xf 1 f f F) is a bounded subset of

Y for each x f X ;

(b) uniformly bounded if {f (x) I f € F and x C A) is a bounded

*--r is-"---- '---"---+ax q=Gf=

X into scalars, then 'f is called a linear functional. The set of gl eont~uous- --linear - functionals- on X which is denoted by X* , is

called the dual space of X . The topology generated on X by X*

is called the weak topology of X . In case X i-s a seminormed space

X* is a Banach space with the norm defined in Proposition 3; moreover

for each x f X the linear functional 2 on X* , defined by

2(f) = f (x), Moqs to x** . Ttze locally convex topology. generated by (SIX E x) an X* is called the weak* topology on X* .

Theorem 2. (Hahn-Banach). Let X be a vector space and p a

seminorm on X , Suppose f is a linear functional defined on a vector -. subspace Y of X such that j f tx? f C p W for every x F Y . Then

f can be extended to a linear functional f on X such that

The following propositions are immediate consequences of the

Hahn-Banach theorem.

Proposition 4. Let Y be a closed subspace of a lccally cornex space X , and a C X\Y . Then there exists f f X* such that •’(a) = 1 and f(y) = 0 for y E Y . 1 pro_psition 5. Let X be a seminormed space. If 'x X with

.'xi' # 0 , then there exists f f X* such that flx) = {jx/f and

fli = 1 .

Proposition 6. Let X be a seminormed space. Then for every x X Eiausdorff 'space, tben Xf is total crpar X . i., far dach x E X with x # O there exists f € X* such that •’(XI f Q .

!t!heorer~ 3. Let X bz a no& space and r. a linear suhspaoe of X* which is also total. Then r is a weak f dense subset of X* . % Theorem 4. {Banad-ALaoglu). If X is a noaned space then the unit

- disc in X* , i-e-, If C,~*1fiffl5 11, is weak compact. If, %n addition, X is separable, then it is weak metrizable.

mrm 5. --=en6txtr&. 5F *BitSi&&--~ - =-- =------subspace of X* . lhen S is weak * closed if and only if

{f C Sl /If11 5 1) is eak closed.

We conclude this section stating sane properties of the Banach

Let w fx the set of all scalar sequences. spaces co , 8, and C 1 - men co = {(x,) t wjlh x = 0) is a 3anach space with the norm /I /L nn, for all but finitely many n} is a dense subspace of co .

a dense subspace of COD . Also note that co is a closed subapace -- - gf -4_ I m Definition 1. 'A fuW infinite series Z xi in a topological vector i=l

space (X,T) is said to be (I) convergent in (X,T) if

conwrges in (x,T) ; (ii) weakly convergent in (XrT) Xi)n 6 N 'i=l

if there exists an x C X such that Z f(xit converges to ffx) - - - iw f

for every f E X* ; iii) subseries convergent if for any increasing

(i ) series 1 - - -sequence of positive integers, the x converges ------*------i - = = - - =-=-

in (X,T) , and (iv) nnconditiodly convergent if for any permtation

o of N, theeeries L xG(i) converges to the same element x i=l

Remark: Let ACN.- Ex converges in (X,T) neans that

ia - 2% w lim x . exists in (XITI. Therefore L x . is subseries z / 1 1 n iW[l,nl* i=l

eonvergent if and only if Z x anverges for every A 5 N , i i €A ED Theorea 1. If a series xi in a locally convex s-ce X is sub- i=l

series convergent, then it is unconditio~llyconvergent.

The proof of the Wetheoran can be fowd in 19 I.

- - - --

--p Beorem 2. Let e a Cauchy sequence in a locally convex space

X If (\I mnverges weakly to x in X , then lim x = x . . 12 fl Definition 1. Let 5'l be an arbitrajcy set. A subfdly R of 2 is said to be a ring in case A LI B, A\ B 6 i7 whenevex A,B .€ ?? . m If, in addition, il A. 6 R for every sequence (Ai) in R 1 , i=l then R is called a 3-ring. %

-finition-2. A ring J? of subsets of a set R is calId an agebra

p-ring) in case for ' very sequence fA.) of pairwise disjoint sets T'> 1 (respectively, finite 4ts) in R , there exists a ce (Ai 1 -?- 7

- - - - - &finition 4. A ring R is called hereditary (or an ideal) if R is closed under subsets. If R is an ideal then f = IA~~Af R) is called a filter. {late that filters are closed under supersets.)

Example 1. A *-ring R which is not a a-ring.

Let F be a mn-principal -1 filter of subaets of N . rhis mans that 3F = 9 and if there exists a filter F' such #-at

F -c F' then F' = F . The existence of such filters is implied by

ideal satisfying the •’011owing : R contains all finite subsets of N,.

R is a QG-ring . R is not a G-ring. ,

By- the definition of a filter it is clear that A -C N , Suppose A,A' f F 1 Since F is maxisml, there exist

prove (1) let n 6 H . Since F is mn-principal In} 4 F . Henee v

%(n) 6 F vso that fn). f R , This shows that every finite set is in R . -

------pp -- --m-pj73 hi ) -Ard-w--I_8f. . ... %=-===- n

U8lara 1. Iet R be a ring of subsets of a set Q . suppose

a: R + Ft+ is a unbarnded function such that:

Then are exists a disjoint sequence (A,) of members of R such

that lim a(A,) = * . a of menbars of R such that a(~~+~)> a(Fn) + h . set A~ = ~~yF.+~.

-- ~ ~ ------p------~ p-p------... %n a(A r la(P) - a(Fn+ll by (2) and hence a(An) > n This n I .

implies that (A is a disjoint -&ce of mabers of R, such that !" t

-. case 2. Suppose for ead-i k > 0 and each E -C X with G(EI ==,

disjoint mmskrs of $2 . Hence by 1, a(P1 C a(P\ 51 + a(P fl 5).

Takingsupreraaa+r PCR-wehawe - A2) < - ; so by induction we can constmct a disjoint

(An) of members of R such that .(An) b n for n 6 N .

6 Definition 5. An &&bra R of &sets of a set 0 is said to have the

------p-pppp propert2 in pair a p--pp - -- -- terwlaatio_n_ case of sequences 1 r---even (A ) (B) , . . of msof R 'mch that An -C Bm for n,m C N , there exists 51. Introduction.

The material in this chapter is essentially contained in the

Antosik-Swartz paper [ 2 f with the exception of corollary L of

theorem. The uniform boundedness principle, one of the most important theorems in Functional Analysis, is a result which derives a conclusion

category theorera paves the way for ~amegeneralizatipn of the classical

~rsionof the theorem. As a preliminary, in section 2, we obtain a result concerning infinite matrices in a topological vector space which -m is sawwhat in the spirit of the Antosik-Hikusinski diagonal th601em 1 11.

- - - - By introducing the notion of a K-mded set, we obtain an analogous stawnt of the uniform boundedness theor* which is valid for any , arbitrary *topological vector space, We start with a lemma which can be viewed as an elementary

sliding hump type argument. - ,,

Lemma 1. Iet (Anm ) be an arbitrary infinite matrix of positive * /. numbers. Suppose (xmn) is a given infinite mktrix of non-negative * i nwnbers such that- lim x = 0 for each n and lk xm = 0 for -zi - - mn_ , each m . !t'hen there exist? a subsequence (mi) of positive integers

Proof. Set m, = 1 . Suppose ml ,m2, tm have been chusen such n I 2 < Aij for i,j = 1,2 n and i # j since/ that Xm.m ,..., . 1 j limx =O and lhx = o for i = 1,2,...,n, can choose - / P miP P mi a i xm and xm '.+I, i for m n+l > en' sud that 'i,n+l , n+lmi ' i n+l

i = 12,.. n . By induction the result follws. -

We use the above 1- to obtain our main result in this

section.

Ihwrem 1. Tet (x~)be an infinite matrix i? a topological vector

space X . Svppose (i) lin xm .= 0 for each. n and (ii) each m

subsequexe (n.1 of positive integers has a subsequence (n. ) such 3 'k m

Proof. Since eve- towlwical vector space is generated by the set of all continuazs F-SeDZiTlOrms cm X , it is sufficient to consider the

- -- A-

case when X ' is an P-seminoneed space. We shim that has a (Xmm)m~N . subsequence -which converges to zero. Since the same argmm=nt can be

applied to an arbitrary subsequence of (x ) , we b~ that ma meN A1

*tz (A. . be an infinite matrix of positive nlllabers such 13 4 that C A. . < a . Condition (ii) implies that lim x,,, = 0 for each - 2, j -13 - n s m . Thus, by iencaa 1, there exists a subsequence (nil Of -itiY

F-seminorm.) To avoid huble subscripts assume n = i Let (ik) i . be the subsequence satisfying the conclusion of condition (ii).Then for

' J" ., every k 6 W ,

:Ix. ii =/]CX. - CX. If =kik &=I 9.5 kl 'ki8 Cfk

bD ;ow note that lim if Z x 11 = 0 by fii) , and k G=l 9.5 P m lim EX. =O bythefactthat C X

- Rentark. In the above theorem it can be concluded that lim xm = 0 n uniformly for m t N . TO verify this let (mi ) and in. ) he two subsequences of --- -- positive integers. It is readily seen that the matrix bxm. n ) satisfies 1 j conditions (i) and (ii). An application of theorem 1 shsthat lim x = 0 This shas &at lira xm = 0 uniformly for rn C N i m.n . . 1 i I n The classic& unifonn bouadedness theorem states that a pointwise bounded fanily of continuous linear operators on a Banach space is uniformly bamded on bounded subsets. By introducing the- notion'of a K-bounded set we give an analegous statement of the theorem which is valid for arbitrary topological vecto

a - -A - - L - - - - - A - Definition 1. Let B be a subset of a topological vector spa& X .

3 is said to be K-hxmded if for each sequence fxn) of elements of B and each sequence of scalars (t 1 which converges to zero, the sequence - - - - LL --- (t x ) has a subsequence (tnxn 1 such that Cx. n n i i i=l' tn.xn r i < Remark. It is easy to see that every K-bounded set is bounded. Bn example of a bunded subset of a no& space, which is not K-bounded, is given at the end of this section.

Definition 2. A topological vector space X is said to- ix-- a (#)-space- - - a if every bounded subset of X is K-hunded.

-psition 1. Let X 'be an P-SaDind~gaecf space, Thgn X is a

(:c) -space if and only if each sequence (xn) in X , which wrmrges OD to zero, has a subsequence I suchthat Zx fX. "ni i=l ni

C ?roof. To prove Zecessaxy part suppose X is a (XI -Space- Let wfij+& Se a local- -base at zero in- X with V c V - nsl-* +. ')r for n C N . Since lin x = D , we can construct a @sequence ini) nn 1 of gesitive Vi for n Define #e mtqers such that x n f 2 n i

- 1 positive iitegers SJL-. <:at - (t x ) < i.. , Z x 6 X -t n. n. . n. . i=I 11 i=l 1 r,i meorem I. (The uniform boundedness principle.)

& - - -- x,et f be a family of pointwise bounded continuous linear functions of a topological vector space X to a topologicd vector space Y . Then F is unifomly bounded on every K-bounded subset

A0f X.

PKOO~. ut B = ifwif f f and x C A) . We want to show that B is a bounded subset of Let be a sequence in and (tn) a sequence o f positive n-rs with lim tn = 0 . n

4 'I Set am = tn fnit xm) for n,m = 1.2.3 ,... . Since the m - sequence ff ) of tontinaas linear functions is pointwise bounded and n t

II (1) lim a;= lint' f (t x ) = o for m = 1.2, ... . n nnmm

Since A is K-5ounded, each subsequence (mi] of bf has a

P 4 ~uhepence In. ) such t3at L t x f X Again by the facts 1. m rn . J j=l i i j j f?~t(fn) is pintvise bounded and lim t' = 0 we have n ------~~mnrk.If X is a fK)-space, then F is uniformiy bounded on every

' bounded subset of X .

Corollary I. (Banach-Steinhaus).

Let (in)be a sequence of continuous linear functions from

an F-seminormed (K)-space X to a Hausdorff topological vector space

Y . If lim fn fx) = flx) exists for every x F X , then f is a n

continuous linear function from X to Y . Moreover, this convergence

is uniform on ever). ampact subset of X .

Pmf. Since Y is Hrmsdorff, f: X -+ Y is well defined and evidently

it is linear. First we shw that f is continuous. Since X is first

countable it suffices to show that for each sequence (xn) in X with

lim f (xn) n

Construct, as in proposition 1, a sequence (t ) of positive n

numbers with lim t = OJ such that lim t x = 0 . nn n nn

{tnxn\n t N) is a bounded subset of X and, moreover. since

lim f (x) = f(x) for each x 6 X , the sequence (f is pointwise nn n

bounded. Therefore theonem 1 implies that I4 =. ifn(tmxm)'ln.m C N)

is a bounded subset of Y Since lim f (t x ) = f (t x for each . IB m nn n n

n C N , (fftnxn)fntN is a sequence in h and moreover fi is bounded.

1 Hence f (xn) = lim f (t x 1 = 0 - n n . tn subset K of X . Suppose (in-f) ,+does not mnverge to zezo 1, uniformly on K . Then there exists a sequence (nil of (n) , a sequence (x.) in K and a neighbazrhood U of zero in Y such that 1

Since X is first camtable. I[ 'Asequentially -act, and hence. perhaps by passing to a subsequence, ,we may take (xi) converging to a point x in K .

Set a. . = (f - f) (X - x . The pointwise convergence of 13 n.. j

(f,) to f implies that (2) lim aij = 0 for j = 1.2 .... . i

Since X is an F-seminormed (K)-space, proposition 1 implies that every subsequence (x. - x)~~of (I -xIjfH has - a subsequence Ik j

- x) f X and hence

(0 m 3 lim Ca = lim (f f)( Ex. -XI=O. ij - i C=l i i ~=l]kt 8

Hen= theorem d of the previous section implies that l+n tfn - ft (xi - XI = aii = O. Since 19 Ifn -f)(x) =OR 1 i 1 i we have lim (f f) (xy) = 0 which is a wntradiction to (1) I X; - - - .- 1 (K) -space, and Z a Hausdorf f topological vector space. If the bilinear map P: X x Y + Z is separately continuous, then F is jointly continuous.

~roof. Since X x Y is first countable, it suffices to show that

to zero in X and Y respectively. Consider the sequence (fn) of

f'- f'- = given f cxhtinuous linear functions of Y to Z by n (y) = F(xn,y)

- - - 7; - for-e h n . The separate continuity of F implies that lim f, (y) = 0 S n for every y in Y Since {oIy ,y is a sequentially . 12,...... compact subset of x the last corollary implies that lim fn(y? = 0 . n

'i uniformly on {O,Y ,y ...... ) and hence lim F(x y ) = li. fn(yn) = 0 . 12. n n n n

Corollary 3. If E -is a subset of a seminormed spa* X such that f (E) is bounded for every f € X* , then E is hunded. /

Proof. X* is a Banach space vith the usual norm topology. Since

A A f(E) is bounded for each f € X*, E = (xlx 6 E} is a family of pointwise bounded continuous linear functions on X* , Therefore theorem 1 The above results are usually deriv mans Qf thee

Categw theorem (see 1 . The assweption of canpletaness or barrelednessis needed there. The following is an exaaple of a nolrmed space for which the uniform Medness principle does not hold.

Let c be the space of real sequences (t,) such that 00

= 0 eventuallj. and equip c with the sup no=. The dual of tn 00

then real coo is C1 . Let en be the sequence which has value 1 in the nth place and zero elsewhere. Then is pointwise ne~

Note that the set fe,ln € N] is bounded but is not

K-bounded. ALso note that coo is neither complete nor a fK)-space. i'

An interesting but caaplicated example of a non-complete n& f ii) - space is given in LloJ . The fullowing is-a siraple example of

Let X be a Banach space, We show that X with the weak i topology is a (K)-space. bt A -c X be weakly bounded, The last corollary iiPplies that A is bounded and hen- A is K-bounded by the fact that every Banad space is a (Kf-space. TPris shows that x with the weak tppoloqy is a W-space. But in general this space is not complete. Introduction.

fie theory of vector rrreasures, in addition to its major role

in integration theory, is also important in some areas of functiok~

of vector measure theory. In doing so our strategy is to begin with

xrare basic set theoretic manipulations. This is in fact mcessary

based on the set theoretic structure of the corresponding domain space, t which is generahly a ring of subsets of a given set R . In order to ,- generalize suk#' important results, we define vector measures taking

values in an arbitrary topological vector space instead of a Banach

space. Since every topological vector space is generated by a class

of F-semiiSonns, in =st cases the resdts obtainea To=F-seminornrea I k-. \) yces can be readily generalized to topological\ fector spaces. i In section 2 ve obtain sa~basic straightforward properties

!?--% i of vector measures. Se!ction 3 is started with a simple version of thk I i msenthal's lemma. this 1- to establish a structural link

betveen t& aa~chspaces"it3 co, 8- and bounded vector measures. This

in turn beqonses a porrerful tool to obtain some important results

concerning Gs~qai+vector spaces, inbluding a generalization of

the Orlicz-pettis theorem for locally convex spaces. The materials in

sections 2 and 3, al$muc generalized to some 1extent, are essentially J.J. Uhl, ~r.1 63. Section 4 deals with convergence and boundedness

Sf sequences of vector measures. The Vitali-Hahn-Saks-likodym

theorem, which is proved in a mare general setting, plays a vital role

in this section. At the end of this section-we introduce the notion

of full dasses and disc& several applications of the previous results

in matrix sumnability,

- - - As the title indicates, we group in this section all basic

properties of vector measures which follow directly from, definitions. *. - Definition 1. Let R be a ring of subsets of a set i2 and X a

topological vector space. A function y: R +X is called a vector

called countably additive. Moreorrer, if (C(E) IE E Rl is a bounded

subset of X , then p is called bounded.

In what follows, unless otherwise stated, R 'denotes a ring

of subsets of a set fi and X denotes a seminonaed spaoe.

Definition. 2. fPt pr R + x be apector nasasnre. The variatkon

of p is the extended non-negat function 1 p 1 whose value on a

set E 5 fi is given by,

1111 (El = sup{ L ilp(l?ii) 11 In C N. El .EZI.. .,E are pairrise disjoint i=l n n rabers of R such that U Ei -C E) . i=l

If Ip] (R) < - , then is called a measure of bounded variation. varJation of the real valued measure x9 . If Ifpll(Q> < , then p is galled a measure of bounded semivariation.

!The folloving proposition is stated without a proof since its

verification involves only simple computationsf ,f' ,f'

* - t/ Pm=sition 1. a. /ul LEI Z /~I.I/~CE) for every E -c Sd and - - -

c. 1 !J 1 and IIpII are both monotone, i.e. ,

We use the follwing 1- to obtain a few other results.

- emm ma 1. If w is a finitetset of ccrnplex numbers, then there exists

Proof. Divide W into four disjoipt sets taking intersection with each quadrant of the complex plane. For at least one of these sets, call it V , we have (The last equality follows from the fact that all z € V are

%mark. As a direct consequence of this lemnra, we have the foll~g. rf sup{1 C zi 1 1 P is a finite subset of N) < , then Z 1 zi 1 < - i €F i=l

Proposition 2. A vector measure p: R + X is of bounded semivariation

proof. ~etx* C X* , l/x*!l 5 1 and let El,E 2,..., En be pairwise disjoint members of I? , Then the above lemna implies that there exists -- ~.-- -~ Consequently, if P is bounded, then JA is of bounded semivariation.

The converse is ubvious.

Emuark. In view of Proposition 2 a vector measure of bounded semivariation

''~ - is also called a bounded vector measure. tr - ---L t------7 53, strongly bounded vector measures.

Om obvious property of a countably additive vector measure y

defined on a O-ring R is that if (E is a sequence of pairuise dis- n a3 - joint members of then C u(E ) is subseries' $and unconditionally) R , n n=l convergent, Nonetheless this property is shared by many noncountably

. additive vector measures, For instance, every bounded scalar mxisure

has this property. On the other hand the vector measure V: A + co ,

&eke A is the family of all finite'subsets of H , defined by

V(A) = X, is a bounded- vector measure- - not satisfying the -eve-pro_perty.- - -

Because of its importance in theory of vector measures*we single out this

Definition 1. Let R be a ring of subsets of a set 2 and X a

tapolog+cdL vector space, A vector measure p: R -+ X is called:

(i) 'strongly additive in case I3 p iE ) csnvarrfes for each sequence n n=l

(E ) of pairwise disjoint n

( iil + strOngly bcunded in case lim p( i) = 0 for each sCquence (En) n n

of pairwise disjoint members of R .

Proposition 1. bet p: R + X be a vector measure,

(a) Suppose X is a locally mnvex space. Then if p is strongly ------bounded, Li is ?munded,

- - --- (bl If X is sequentially complete then statements (i) an?3 (ii) are

equivalent. - - -- - mf.(a1 Let 1: be a continuous seminorm on X . It is sufficient

to show that p is bounded with respect to 11 . Define a: R + R+

If A -CB then,

Since lim z(E f = (I for each disjoint sequence hn) in k? 1.5 n , n fernna 1 implies that 3 is bound&, i.e., p is bonnded.

(b) (i) always -lies (iil, To shov that (ii) implies (i) let

(En) be a sequence of paimise disjoint members of . Suppose

2 ;;!Xf dws not satisfq &&e aubng ~Kdftion. Then there exists an n=l increasing sequence (n.1 of positive integers such that 1

lin 2 u(E.) # 3 . set ? = i'u E . *en Pi) ie a sequence i j=n. I] i j-7 . j 1 1 -. f,4, theorem 2.

Z I * 2. In sCid+ieaezt iil the convergence of 7 :(En) is subseries n=l

l%.e foi',ffr'rr.rj Aefrniticn extends the earlier one to a

&finition 2. *i.e-, be a ring of subsets of a set 5 and X a

pivise &s:obr ders~f 4 , lim ;i (E j = 2 unifornly for f. n m We need the fozfowinq shpfffi;e&version of tfre Ruseni3zFs

1 le~tcea I143 to establish air Antheorem of this section. Although the proof of this lemma is simple it represents one of the most bportant

results in measure theory.

Lemm 1, Let (2) be a sequence of uniformly bounded nonnegative %a1 - n N valued- measures defined on 2 -the power set of positive integers. Then

for each E > 0 , there exists an infinite subset P of N such that lif~\fp)) < E for every p f P . P' "5 Proof. Let E > 0 . Partition N into a sequence (M,) of painvise disjoint infinite subsets of N . If there exists n € N such that

M f E for every p t Mn , our goal is achieved by setting n p

M = ? Suppose for each n , there exists pn f Mn such that n .

ut P = !pn j n c nl Then P (M~\ {pnl) = for 1 . 1 n + n = 1,2,.. . and hence

(2) (PI) + 2 (Xn \!pnj = 2(P1 8 (M,\{~,))) 5 M , where 'n Pn n the process does not stop, there is an infinite subset P2 of P 1 such that (4) v(P2) 5 M - 2€ for every p f p2 . P

Thus the process must stop before n iterations where n is the smallest positive integer such that M-n~< 0 . his completes the proof*

Now we are in a position to prwe our main theorem in this section, This theorem gives a characterization for vector measures which are not strongly bounded, Recall that coo-the space of all finitely nonzero sequences with t p norm, and m -the space of all finitely 6 0 valued sequences also wi& the sup nonn, are dense subspaces of co and Cm respectively.

Theorem 1. Let R be a ring of subsets of a set .Q and X a seminonned space. Suppose p: R -+ X is 7 bounded vector measure. Then

D is not strongly bound& if there exists a linear topological embedding T:. coo + X and a sequence (E ) of pairwise disjoint n mxnkers of R such that T(en) = p(E 1 where e denotes the n n sequence, 1 in the nth place and zero elsewhere.

If, in addition, R is a g-ring then the above st-atement r-s true if the space c is replaced by m 00 0

- -

Praof. Suppose 2: k? + X is not strongly bounded. Then there - exists a disjoint sequence (En) in R and an E > 0 such that By virtue of the Hahn-Banach theorem there is f f X* for n each n f N such that

f fJ the For n , consider variation Ifn o p 1 of the scalar

valued measure f o p Since 1 f 0 p 1 E 5 (1 for n . a

E f (Ifn !J/)nCN is a uniformly bounded sequence of nonnegative real valued measures. r

For nfN 'define p: 2N + R+ by, n

?he strong additivity of I fn 0 p 1 implies that p is a measure. n

Since for nQ? and p 5 N , il(P) = Z 1 fn E 5 p (1 , (11) is n iEP n

a uniformly bounded sequence of nonnegative real valued measures. By 1 . :emma 1 there exists an infinite subset P = {pl < p2 < ...... 1 of ,/

?i such that i

E/Z for (3) :! (P\ ipn! < every p n F P . Pn

Q) Define T: c +X by ~f(x,))= i;' xnp(E ) Sinceonly 00 . n=l 'n - - fini+ely nany tern are nonzero in the above series, it is readily seen

- - - --&at L T is linear. %reover if f C X* with •’ 5 1 , then On the other hqnd for mEN , 1

in1 c If T( (x,) ) I (since llf ! = 1) ' pm .Pm Taking supremum over m an the right hand side me have

1 (4) and (5) implies T is a linear topological enbeddkng. hall^ note that T(enl = 2 lE ) . Pn -

Moving to the case in which R is'a a-ring, we proceeq as ' - . I

, . disjoint sequence (En) of im&ers of R such that

It is readily seen that (1 is a uniformly bounded s4tguence of n mnnegative real valued nreasures . Again ~bma1 implies th& there exists an infinite subset P = ip1 < p2 < ...I of N such that.

If tx,] f ma, ue can write fxn) = C ZdA uhem m=l m

,A2 ,. . ,%< are pairwix disjoint subsets of N such that U A = N , Define Tt m + X by, = 0

The linearity of T can be'P asily verified by using same elementary set algegra. with ]If11 5 1 , then Taking supremum over 4 on the right h& side we 667

(8) and (9) implies that T is a linear to pol^@&^&. embedding of > t- m to X. 0

Finally we note that T(e 1 = p (E ) n Pn . 3

a copy of X" means that there is a linear topological enbedding T: X -t Y.

Y -- Corollary 1. Let X &-a BGchFpaicontaining no mpy of c If OD 0. , the series Z x is umrdered bounded, i.e. 1 ZX,]A is a finite n n=l n€A 00 subset of 14) is a bounded subset of X , then Z xn is subseries n=l convergi?nt.

Pmf. Let A be the ring of all finite subsets of M . Define L, L, Z p: A + X by 11 (A) = Z xn . Clearly L1 is a bounded vector measure. nfA

Since X does not mntain a copy of c theoren 1 implies that 11 0' is stronqly additive. aence C x = C ~({n)) is subseries convergent. ~ \ n ', = n=l n=l t 1 ' ~ornllaqk. bt R be a ring of subsets of a set i2 and x a locally

mnvex space. Suppose p: R -+ X is a bounded vector measure. If P lim p (E 1 exists weakly in X for every increasing sequence (E,) n n of members of R , then p is strongly bounded,

Proof We may assume that X is a seminolmed space because of the following reasons.

(1) If 11 !I is a continuous seminonn on X and if the sequence

(x,) in x weakly - anverges with 'respect to the locally convex

topology on x , then (xn) Converges weakly in (X,il 11) .

(2) p is strongly Medwith respect tothe lodally convex topology ------

if and only if is strongly bounded with respect to each - r - - - continuous semiaorm jl on X . L -

' there is a topological linear embedding T: coo +X and a sew'n

V-=-----d (Pn) of disjoint renhers of R such that p(Pn) . Set \

Em = U Fn . Then T( C en) = P(E,). n=l = n=l

First we show that lim does not exist weakly m m Suppose lim C en = (an) weakly m n=l m hence limek( C e = ek((a 1). A It m n=P

extended uniquely wer c to a member of c; . We denote this - Q extension by . m Me other hand it g i c; , then g T-le# foT 0 continuous linear functional on the subspace T(cOol of X .A virtue -1 of the Hahn-Banach theorem we can extend g T over X t a member - -7 0 P; - of X* . We denote this extension by f . It is t - foT = g . Therefore every member of c* can be 0 f - f T for some f C X* 0 . of the Hahn-Banach theorem there exists g 6 X* such that 1

and g vanishes on T(C~~)- lhis mntradicts the fact

' p(En) = x weakly in X . Hence x C T(cO04 .* lherefore

there exists a sequence (an] ' in c such that lim T(a ) = x in 00 x . n !? ~etf 6 X*; then

(2) lim f T(a ) = f Cx) o n . n

Also note that (an) is Cauchy in c since (T(a,) 1 is Cauchy in x 00 .

Consequently there is a f c such that lim an = a' in co 0 . Since - n

By (I), (2) and (3) the right hand side tends to zero as n tends n to infinity. Consequently lin Z em exists weakly. This contradiction n m=l shows that p is strongly bounded. This completes the proof,

Nuw we employ the above corollary to prove the Orlicz-~ettis theor- for locally convex spaces. This theorem was first proved by - - - -. ------arlicz for veakly sequentially cmplete Banach spaces. Kalton 18 J

------recently obtained this theorem for separable topological ,groups and then derived the result for separable locally convex spaces. For an ------alternative proof of the locally convex version of this theorem, the

reader is referred to McArthurs paper 1111.

> corollary 3. (orlicz-~ettis). Let X be a locally convex space. If

00 w 1 x is a weakly subseries convergent series in X , then C x is n n=l n=l n

subseries. convergent.

Pmof. Let A be the ring of all finite subsets of N . Define

I. lJ: -+ X I-I (A) = x*- y A by C n . Evidently is finitely additive. To n CA

sh& that p is bounded, it suffices to prove that is bounded with . . respect to each continuous seminorm 11 /I on X . Consider the subset

- h F = { .Z xn ~AIC A) of X ! 1) ; the second dual of X with respee n fA

to the seminorm topology, For every f f (x, // /I) *, f C X*; the dual

space of X with respect to the locally convex topology, and hence

m m L f (x ) is subierie; convergent so that L I f(xn) I < . COnse4uently n=l n . n=l

(XI I! * is a Banach space, the unifom boundedness princple implies

that :

that i: is bounded. hypothesis, lim p(An) = lim C x exists weakly. Therefore the m. n n ma,

last corollary implies that p is strongly bounded and hence

C x satisfies the Cauchy condition for errery A c N Since n - . nfA

C x exists weakly in X , 1.4 theorem 2 implies that - C xn is - n ~CA nfA

convergent in X . This ~letesthe proof.

. In ch-er 4 we obtain , anot3er version of the Orlicz-Pettis

theorem.

The following corollary establishes a chaxacterization of

complete seminoxed spaces not containing a copy of c 00

Corollary 4. A complete smimnned space X contains no copy of c w 00 if and only if every series L x in X , with i f (x,) < for n 1 1 n=l n=l every f f X*, is subseries convergent.

Proof. First suppose X contains no copy of c Let Zx be 00 ' n n=l a series in X with 2 i(xn) / < = for f C X*. iie define n=l

2: A + x precisely as in the proof of the last corollary and follows TO show that the converse is tme, suppose X contains a

copy of coo Then there are many nonconvergent series C x in x . n n=l

such that C f f (x)f < for f f X* . n=l

Corollary 5. Let X be a complete seninqd space. If X* does

not contain a copy of 4- then X* contains no copy of c , 0-

Proof. Let 1 fn be a series in X* such that E /p(fn)] < n=l n=l

A for F C X**. If E -c N, then Z x(f (= C f (x)) exists for x € X . na nc

By virtue of the Banach-Steinhaus theorem, .Z f converges with respect ne

to the weak* topology on X* Define p: 2N + X* by g(E) = i: f . n - n f~ weak* limit. Evidently p is finitely additive. To show that U is bounded consider the subset = f E; f -weak* limitf~c N} of X* BY F n - . n GE CO cn A the fact that E j f (x) = Z x < rn we r. / for x C X, have that F n=l n=l

is pintwise Munded. Hence the uniform boundedness pr&ciple implies Sin- X* sbot mtain a copy a•’ tco, the last part of theaem 1 cv m implies that is strongly bounded. Consequently 2 f = 2 ~({n)) n n=l n=l is su5series conveqent. Hence the last corollary implies that X* contains no copy of c . 5~sszzpletes the proof. J

The resl~ltsire obtained so far demonstrate the utility of theoren 1 in the tiieorl of topologi-a1 vector spaces. 94. Convergence and boundedness of a sequence of strongly

bounded vector measures.

The main result we obtain in this section concerning sequences of strongly bounded vector measures is the Vitali-Hahn-Saks-~ikodp :

theorem. We prove this theorem for vector measures defined on a ring with a weaker structure than of a 3-ring. The proof is a modification of the proof gimn in 1 7 I by Barbara Faires. We use this improved version of the theorem to obtain generalizations of boththe philips and

Schu lemas. Fie start with the following definition.

Definition 1. Let R be a ring of subsets of a set 3 . R is said to have property (31) if for every disjoint sequence (A,) in R and every sequence [9,i in R with Q B = @ for m,n f N , there Am n exists a subsequence (A of (A,) and C C such that: n. 1

- A c c c i J 3- = ; and c 1 = for n f fnlrn2' ...... 1. n. - .L i=l r n=l

"narks 1. Tie class of %-rings and the class of algebras with the

L?terpretaxior. ?ro-pe,rty 50th have property (QI).

2. Let 2 Se a ring of subsets of N with property (QI).

If 2 costairs all firite s.&sets of N , then R is a QG-ring. utthere is a subseweme (A 1 of (An) and c F P such #at n:

j A CC, C 7 (a\( J A) =: and C1 An = for n inl,n2 n. - + t ,...... 1. i=l 1 n=l n

03 This implies d A = C E R . Therefore R is a Qa-ring. i=ln i

Proposition 1. If R is a ring with property (PI), then R has the 1 following property (we call this property (21) ) . i For qvery disjoint sequence (An) in R and every sequence

(8,) in R with An 5 B for m,n C N there exists a subsequence m ,

(An. ) of (A~) and C F R such that: 1

Li A C C 5 lB and C ? A = 0 for n f 1x5,n2, ...... 1 n. - n n . i=l i n=l

?roof. t (A,) be a disjoint sequence in R and (B a sequence n

in R . Suppose An f Bm for n,n C N . set Dn = B1\ Bn for n t N .

Since A C B for m,n F N , A 7 D = $ for n,n f N . Since R n- m n m

.has property (QI), there exists a subsequence (A of (An) and n;

C C R such that:

(11 A c, c '1 ( = C A = n = - D n ; and 3 $ for n f {nl,n2 ,...... 1. . i=l i - n=l n The proposition follows fram (1) and (2) .

Theorem I. (Yitali-Hahn-Saks-~ik&ym).

t X be a topological vector space and R a ring of subsets of a set 2 with property (QI). Suppose (1 is a sequence of strongly n bounded X valued measures on R with lim u(E) = 0 for every E C R . n n

Then the seqlr?nce (u) is uniformly strongly bounded. i.e.. for every n disjoint seq-wnce (En) in R lim u(E 1' = 0 uniformly in m . n m n proof. Since x rs generated by a family .of F-seminarhs, we may assume that X is ar. F-semmnwd space. Suppose the contrary.. Then there exists a disjoint sequence (En) in R , an E ; 0 and a subsequence ( ) of m n

lairvise disjoint u(finite subsecs of N . Consldar the follouhg tra dlsjoint sequexes in i? :

The croperty i 21 j -Edi R implies that there exists an infinite subset been constructed such that: \

(dl) F1,F2,. ,EN- ,are pairwise disjoint. .. n-1 mnsider the following tw sequences in R .

By G Ielf the nrembers of both sequences are pairwise disjoint. Again property (QI) implies that there exists an infinite subset A' of n

2' 2' and F1 C R such that: n n

Clearly F n 3), b , c and (dl). Therefore, by 1 *=tion, we sur construct a sequence (Fk) of disjoint members of 'R - - 1 and a sequence 1 of disjoint subsets of Ii satisfying (al) (bl) i'-k , , 1 To show that Tim (Fk) = 0, let E > 0 . Then for each k there is k il 1 1 + (c R) -c Pk such that ig (%)11 > )i(F,) - E . Since p is strongly i i i I 1 1 1 mdd, lim 'ii: = 0 . Consequently lim 11 (P 1 = 0 . k i k il k 1 - 1 Choose k1 C N such that I. (P ) & and then i 2 k,

I Partition A into a sequence (If) of disjoint subsets of n k1

* \ i one the same induction prooedure to construct a sequence 2 . - -1 2 2 (F. 1 of disjoint nmLers of R and a sequence r' 1 (< 5 of K -k

&is joint

L -. - F. (E. ,r 1 =A for k6N. K 1 1 -2

, iiin d(~.1 1 = O and lim MEi ' =I0 , ii 1 ii 2

- 2 (P 1 < and then 1 'z k, 3 %, (is i2. 1-

L such +kat . , .- E ; < :j Note that E . :F . 1 - 1 1, - kl Fk2 . l7 5 2 3 sewn- (F; ) = (F ) say. in R and bn increasing oequ~ce (i of n n n positive integers such that:

(1) E CFn for n

(2) Fn 1 E = (I for 1 l k 5 n (by (bl) and (b2f). 'k

(5) . 1 > 3 for n C N . i 1n n

Let H = P L! ( Li E. ) . Then (1) implies E C H for n n k=l 5 ik - n k,n C N . Since R has property (IQ)' , there exists a subsequence

1 of. (i f and C f R such that: k

(6) dEi -C CC- 1 % and C I Ei =$I fork f {kl,k2 ,...... 1 . E=l kt k=l k

Therefore, for each p C N , J P p-1 amsepently IIui (~)fl? kfi (E 11 - lrW+ L= U!J. ( 11 1 II k 5-3 k trl kt t 4.1 %,' " P P p P P - k P By (6) an? the definition of %,Ccq =F U(U Ei). k P P p m=l m k k. P P Since F Q (U E. 1 =+ by (21, c\U E~ A~SO 1 cpk . kp m=l m m=l m P k P ' P C\U E~ =c\JE. since c n E~ =.$ for m f fkl,k2,...... 1. msl m c=1 'kc m

-CE.

'JTherefore !pi C 1 > 3 - E - E = E . This contradicts the fact k P that lim p (C) = 0 . Hence the sequence (p) is uniformly strongly i i bounded.

Corollary 1. Let R, X be as in theorem 1. Suppose 1 is a sequence n of strongly bounded X-valued measures on R such that lim p (El = p(E) n n exists for E € R . Then p is strongly boundea and, moreover, the sequence (PI is uniformly strongly bounded. n - --

If, in addition, X is complete, then for each disjoint

1 in CE p (E A =quene fEm R lim Z P m 1 = C m 1 uniformly for -c N . n mfA n mCA (E. 1 be a disjoint sequence of members of I? First we show that 1 . limLI(Ei) =p(Ei) uniformly in i . Suppose (p(Ei))nCN is not n n n

uniformly Cauchy in i . Then there exist two subsequences bk) and

(i 1 of positive integers such that: k

Since p and p are both strongly bounded - p is also stzrongly "k+l "k "k+l "k 4' bounded and, moreover, lim ( p - p ) (El = 0 for E C ?? . ~hustheorem %+l %

1 implies that lim i! p - p) (Ei) 11 = 0 uniformly in k . This

-3 rfr+l 4c 1 contradicts (1) Therefore (2) 1 p E ) = p El uniformly in i . 1 . n n

For given E > 0 , there is n f LI such that 11 p(Ei) - LI(E~)~~< E/Z for 0 n 0 i F N Since lim p (E.) = 0 , there is i C N such that . 1 i n 0 0

E E + i/ p (Ei)i; < -+-= E for i l i so that lim U(E.1 = 0 . Hence n 2 2 0 1 0 is strongly bunded.

An application of theorem 1 to the sequence (p - plnEH shows n space and fE ) 'a disjoint sequence of members of R , Then the equence - n/ y strongly additive &d p is strongly additive. NQW we

(3) for AiN, lim C P(E = C pfE 1 %is is true m m . n m€A n mEA

en A is finite, so assume A is infinite, t A = {nil < m2 < ...... 1.

m Since is unifmmly strongly additive, Z Em n = 12. . are n j=1 n j

convergent uniformly in n . merefore far given € > 0 , there exists m

n large enough to satisfy /I i U (Ern 111 < 7t Since 0 . j =n j 0

lim p (E = ;1 (Ern. for j 4 1,2,. .. ,n -1 , there exists mo E N such m 0. nn j 3 n -1 0 E that L ! 1 - Em1 fa^ n2rno. Therefore for nzm 0 j=1 n j j

This prwes (3) . define vn; zN + x by V, = E parn] for n F N m€A n

and V: 2N + X by sW = i il (Em) . Since L p(Em) is subseries rn €A m=l n

convergent for n C N , (,on) is a sequence of strongly bounded vector

measures and ,moreover, lim v CAI = vCA) for A 5 N (3) n . n

Therefore the first part of this corollary implies that (vn - is

uniformly strongly bounded. To show that lim v (A) = v(A) uniformly n n

for A -: N , suppose the contrary. Then there exists a subsequence

(V 'J)~~~of (.; t~)~~~(for notational convenience we relabel, - n - n k

('J - v) by l'ik - -.)I ) , a sequence (%I of subsets of N and an n k

E 0 such that "-3 'J) (%)'I > 5 for k i N By the definitions of k - .

'J k and there is a finite subset Fk of qc such that

i'(v 1 > E k C N Now k - u) (Fk) for . we use the induction to construct

a sequence (G.) of disjoint subsets of N and a subsequence (v, - '2) 1 i

of (Vk - 'l) such that ! (v - J G !I > 2 . This leads to a ki

contradiction since (vn - v) is uniformly strongly bounded.

Set kl = 1 and G1 = . Suppose G~,G~,..., G n disloint subsets of 9 ,and kl ( k2 < ... ': kn have been chosen such that B- for E 5 N , lim V -V (GI J G2 d U Gn) = 0 lim (vk-Y) (E)(F U every k .. . . k ' k

has only finitely many subsets.) Choose (Note that G 1 a G2 J ... 0 Gn

such that vk -v (G ii G2 d d Gn) < €/2 Set kn+l kn 1 .. . . ' n+l

= F G,+l - \(G 1 d5u... vGn) . Then kn+l 8

(by the additivity of v -v 1 ' kn+l

merefore 'lim 'i (A) = v(A) uniformly for A N n - . n

:.e., lim Z (Em) = (En) uniformly for A 5 N . n ~EAn n Fa mark 1. In the absence of +the ccmpleteness assumption the last part of the corollary can 'be modified in the following way.

R-i: - !E ) is a disjoint sequence of lierbers of i? sucn that a C p(E$ uniformly for A -c N". mCA

To see this it suffices to show that lim C u(E~)= 1 u(E~) n ~CAn m CA

for each A -C H . The remaining part of the proof runs identically.

OD t A = :m

n E R , and t3e uniform strong boundedness of (1 assure that n

;9 a Z ;1 (E C N for E > m. converges unifornly for n . Therefore given 0, j=l n 1 z P there exists f S such that & (E ) < ~/3for n f N and 0 m. 1-n I Remark 2. The last part of the cordLLary 1 may be tre&e&as a generalized version of the Phillip's lemna 1121. 'ib verify this we derive the Phillip's lemma from the last corollary.

Corollary 2. (Phillip's lemma) . Let (plWbe a sequence of bounded / n

Proof. Since 2: is Sounded and scalar valued, it is strongly bounded. n

for A -c N by the last corollary. Thus for given E > 0 , there exists n 5 N such that: 0

*, 3erefore by 3.2 lmaal, n -)5 E for n r n . i=l n 0

Z~rollar; 3. Let Le a rinq of sirjsets of 2 with pro-perty (QI)

-5 X a cscplete :;,axschrff topolqica.1 vector space. Suppose

-. ,. - : , 7; i , is a ~zu,-~taij;l; xiciitive and strongly bowtded vector .-. Proof. Let fE.1 be a disjoint sequence of members of R such that 1

w m m w U Ei C R . "Then u( U Ei) = lio U( U Ei) = lim ll (Ei) since i=l i=l n n. i=l n i=ln

1-I is countably additive for n E N . Since X is complete the last n OD CO part of corollary 1 implies that lim C p (Ei) = C u (Ei) . Hence p n i=ln i=l

is countably additive.

Uniform countable additivity of (p) follows from the fact n that (IJ) is uniformly strongly additive. n

Corollary 4. Let X be a separable Banach space and R a ring of

subsets of a set R with property (QI). If the vector measure

u: R -+ X is bounded, then is strongly bounded.

Proof. Suppose p is not strongly bounded. Then there exists a sequence

(En) of disjoint members of R and an E > 0 such that:

(1) /Iu(E~) // > E for n E N .

By virtue of the Hahn-Banach theorem, there is f n i X* with /Ifn/\= 1

such that (2) If p(En) I > E for n C N . n

By 1.3 theorem 4, the unit disc of X* is weak* conpact and since X

is separable it is metrizable with respect to the weak* topology.

heref fore-there exists a subsequence (in ) of (in)and f 6 X* with i (3) LLD. 5 3 -:': = -~~(51 for E~R. ..7- . 0 Every series in a topological vector space gives rise naturally to a

deflnltlon of a vector measure. The domaln of thls type of a vector

measure 1s determined by the nature of convergence of +he serles. In

~, ths context we use =he notlon of full classes to obtaln ce~ta~nresults concerning rnatrlx surmnabiiity. The notion of fuil classes uas

introduced by J.J. Sember and A. Freedmn ~r tneir paser 1171. t

I- numbers suck t-~ 3 x exlsts for A ;k', :her, x -=. .? n .?'A n= i

Pro_PSltlon 1. Let 2 be a •’1~11rlng and x a Banach space =ontammg

no corn of p If (x,) 1s a sequence m X such ~MK1 x L x -0 n .s. 33 nfA for A ' R , then Z x is subseries convergent. n n= 1 proof. Suppose (x,) 1s a sequence in x with C x F X for A C R n . n CA

~etf f X* . hen Z f (xn) converges for A C R . since R is full, n CA

30 a

x- I f 1 x 1 < Therefore x n is subseries convergent by Corollary 4 n= 1 n=l i Proposition 2. %t ? $= a :x-rL?g containingw all finite subsets of N axi x a emplete toplogicd MCtQr space. IP ixni is a +rice in /

convergest.

Prmf. Let J se a seqience 12 X sucz that 3 x t X for n n iA

- @ A . :or each c 5 B define i: R - x by n

Clearly (3) is a sequence of strongly bounded vector measures which

* converges setwise on i< to defined by +I (A) = Z x for A 6 R n . n fA

It follows by corollary 1 of theorem 1, that p is also strongly .P

bunded, Slnqe X is complete, p is strongly additive. Hence

C x = 1 ~({n)) is subseries convergent. II n=l n=L m Next we establish our generalization of the Schur lemma. r*

Theorem 2. Let I? be a W-ring containing all finite subsets of N and

(xmn) a infinite matrix in a complete topological vector space x .

Assume that 1 x exists for E f I? and m f N . If lim C x mn mn nfE m n€E exists for E C R , tinen (i) lim x x " exists for n f N rn mn rl

fii) lim x = 1 x uniformly for E :N . mn n - n nCE nfE

Proof. (if directly follows frm the fact that lim f x exists mn m nfE

for every E 6 R . Slnce C x exlsts for every E C !? , the previous . mn n FE i

groposLtion LnFlies L%at t? x is subseries convergent. 3efine n-1 mn

-(A) = la 1 x Since 1 x is subserles convergent, 1s a mn . mn m n6A n=1 m

strongly bounded vector measure. Now letting (E = (!n +) , we ap~ly n

?. tine last part of Z3r-liar? 1 of ~hecrem1 to have 1x11 ,(. n .) =

- Z x uniformly for n n CA

Remark. If R = ZN, in view of remark 1 after corollary 1, we can drop the completeness assumption in the above theorem.

The following example shows that R can not be replaced by any ring containing all finite sets. f~t4 be We ring of all finite subsets of ?T an& let

- men 1x1 _ x = : far every A C 4 , but clearly the conclusion of m. r, ZICA theoem 2 does not hold for the 5c

- x m=7-, are unordered uniformly convergent In the sense m' n= 1

- :?.at lf , ?, then there ex~sts n E N such that -x E, for 3 inn. n FE -- R every a whenever ??in Z 2 n . 0

Proof. 3y Theorem 2, lim x = C x uniformly for E N mn n -- . s nLE n GE

$ Therefore the sequence ! 1 x ) is uniformly Cauchy ior E = N and mn rnCN n FE -

hence there exlsts m F N such that (1) ' 1 (x,, - x ) I! < €/2 0 kn n EE for m,k 2 m and E Z N . Now we show thac for each m C N there 0 - ex1st.s pm L N such that:

(2) Ix c5/2 for ~ihEzp mn m . n EE For, qm the conttarp. men there exists a sequence (Ei) of

subsets of N such that lim Sin E = and '1 x i' 1 &/2 For i mn . 1 n CE

each 1 choose a finite subset " of E. such that " Z x > ~/3 ' i 1 mn nfF

and notice that lim %in F = = . *t = F Choose i C N such i J 1 i- i 2

that %x G .< min F and set - = 7 - Inductively we can construct 1 l 2 >Z -i2

a drslolnt sequence (5,) of f~n~tesets such that Xax 3 Hln 2 - 1 1+1

- a rid x :.I3 for i f N . Thls contradicts the fact that nFG mn 1

L.- - x 1s subserles convergent. 5.m n=l

? for Yln E i c 5y (1) and (3) .

The result follows from (3) and (4) .

We next show that theorem 2 can be viewed as a generalization of the classical verslon of the Schur lemma. c-1 fr172. ~~tf fx 1 be an infi~itematrix of mrplex nuznbers. m1 ~lppnse .Ibm/ < far every m EN. If- lim 2% exisfs, for n=l m nEE .

each E c N and if lim x = x for each n € N then - mn , m n

(i) lim Z- lxm - xn1 = o and m n=l

(ii) the series : /xm / , m = 1.2.. . . , converge uniformly in m . n=l

Proof. fi) Jiet 2 i 0 . B;' Theorem 2, there exists m f N such 0

that (xm - x )7 ' E/B for m 3m and E -N . Therefore by n o - niE

for

(ill B:: 2rrcll~:: 1, there exists c C N such that

for m E Y . This implies (ii) .

?emark. (1:) m&-:t=c that Sup - x < Ic . m n=l mn 51. Introduction.

d The subject of this chapter is one of the truly impressive

theorems of roeasure theory, the Nl~odymBoundedness Theorem, which derives

a conclusion of uniform Wndedness from a hypothesis concerning setwise

boundedness. It also has a strong impact on the theory of Banach* spaces. .

ring oq which mezwires are defmed. An algebraic characterization of such

structures is still unknown. The recent developments in this area are

largely uuntributed by the papers of G.L. sever 1161, Barbara F'aires [ 7 1 ,

RB. %rst 151 an2 corn ell.^. Constantinescu [ 3 1. i;onstantlnescu'

4eflnl~~m-I. Alt3ougF. a 23-rmg 5as a nlce algeSrarc structure, ~t ,

1s extremely dlfflcult to conqtruct such a rlng expllcltly. he am In this chapter. 1s to prove the Nikodym'Eoundedness Theorem for a moce general class of xings, namely PW-rings. gnlike the class of QO-rings, * this class contalns same well known examples of rings of sets. In thi~ chapter we also deal with tbe measures defined on substru~turesof 2* .

Tnese measures are especially important in summability theory. SO=

of the results in this chapter appear in the joint paper 1151 by

J.2. Sember and myself. 52. Defjnj,tj,~n~and qgnpe ex*leg,

The purpose of this section is to study a new class of rings

of sets intr&ced below. It will be shown in the next section that the

Nikodym Boundedness Theorem holds for measures defined on this type of

ring. One of the important features of this class is that it contains

some well-known examples of rings of sets. In what follows, unless ?ee

signified otherwise, R denotes a ring of subsets of a set R .

d Definition 1. A ring R is called a ~p-ring(respectively, an

FPg-ring) in case for every disjoint sequence (A ) of sets (respectively, Ti finite sets) in and every sequence (t of real numbers with n

lunt = * there exists a subsequence (An ) of (An) satisfying the n n i I following :

:I n n. -~OT each there ;s a partrtion i i 1 Al 'A2 ,.",A of X is. > t S. n. 1 n. 1 1 1

and such that for every seqwnce

Xemark. It is eas? to vef- f-- nat everv 27-rlng 1s a Par-r~ng and that ever:#- P2:-rmq is an YPF-ring.

5 Fwple 1. 4n increasing sequence (F.~) of psltive integers is called - .

lacunary Lf 1m (c~+~- F ) = = . We show that the rmg L of subseas n fi n of B cJenirated by lacunary sequences is ipw ,but kt-I?@. \To this i end let CA,) 5e.a sequence of pairwise disjoint finite subsets of 8 - -

- and (t a sequence of posltlve lntegers with llm t = P . n n n choose a subsequence (An 1 .of (A,) such that Max An + i i i

< Min A and then partition each A = lpl < pa < < pk] in to n n .. . i+l i

- \\ n. -. It is readily seen that i %: is Lacunary for every sequence (ki) i=l 1

with 1 L x 1 t . T& snows that L 1s FP(~. 1 r-k 1

To snow -&at L is not 2QC , let -1 = :pl T: F, ' ...... l 0 L

be an mflnlte lacunaq sequence. 3ettmg A = (Ao+n) \ (A A1 , n 3 --...... A 1, where + n = ip. + nf. , we can define inductively n-1 1 i CN

the dis2oint sequence iA j in i . n

Let (A 5e a subsequence f ;A j , Pd-r-her for each r, . n 1 n 11- n -

'1 1 1 -7 ' 1 let A ,A2 ,....-.,A 5e any flnite partition af A . ne -1 s: 3. , show that there is a seqpmce (ki), where 1 5 ki 5 s such that i'

. U %1 f L . Since An is infinite there exists 1 5 kl 5 sl such i=l .i 1

I that is infinite. Consequently =F +n for some infinite. , IjE 11 1 1

subset F of A. A similar argument shows that there exists 1 .

-al. 1 5 k2 5 s and an infinite subset FZ of F1 such that E2 .nZ 5 L 2 + .

Inductively we can construct a decreasing sequence (F. of infinite 1

subsets of A and a squefie tk.) of -gmsitiue integers such that 0 1

n i Suppose J %, = Nl i=l 1

1 7 i 5 F , are lacunary sequences. - Since F

cnere exists i .such that N'. 7 (FW1 + n is infinite. . 1 1 1 * p+l

Consequently, +&ere is an infinite .s c p such that -p+l - p+l

* - + 2 - , - Slnce F ) is a decreasmg sequence of sets, _3+L p+L - -, L

2s a:a -3 + 2 r3 , tt~ t n i --7 M, is finite. pti L - i -pi1 p i -1

--,.-.-iernfore, there exists 1; such that (G + r, -- Pa. is i F+.I P 1 - infinite. Consequently there is an infinite G c G such that .. P - p+1 ' . G +n cNi . proceeding in ;this manner we can find an infinite set P . P- 2 h

c F at the (p+l)th step such that (G1 + nl) Idi is G15 Fp+l- 1 n

. . finite for 1 5 i 5 p . This contradicti,on shows that L is not PQO . '

Example 2. Let A -c N . We denote by A(n) the number of elements of

A I{1,2, ...... n}. A is said to be a set of zero density if lim A(n) = 0 We show that the class of sets of zkro density, denoted -, . 'n n

kt (A ) M a disjoint sequence of members of q0 and (tn) n 6 a sequence of real numbers with lin t = If (A 1 has a subsequence n . n n (A 1 consisting of finite sets, then it can be easily shown that the n:

3 sequence (An) satisfies the condition given in the definition of a

PQcr-ring. So let us assume that all A 's are infinite. n

Set n = 1 . Suppose n

Now choose n. > n such that: 1+1 i Such an n exists since (1)' An ,An A are of density zero, i+l ,..., 12 ni

(2) ' 1.b Min A = , (3)' lint =m and (4)' An is infinite. n n n n i

Inductively we can construct a subsequence (A ) of (A,) n', satisfying conditions

Partition each An = {pl < p2 < ...... I into i

1 i i A1 ,A2 , ,A in the following way...... Zi+l

-i+l If n 3 n. (4) assures that A (n) > 2 Therefore, 1+1 ' n . i n by the way A is partitioned, ~~~(n)C n i

Let i be a fixed positive integer and j > i . Then for i+l lZk52 and for n5n > n we have j i

Tne'last inequaiity follows fran (1). Also fo; any n C N and

n n i+l * ,. l5k52- we "have (n) 5 1 or $i (n) 5 A (n)/ and hence qxi n i 2

Let (k,) be a sequence of positive integers such that I an i +l 1 5 ki 5 2 . We show that qri is of density zero. Let j C N and i=l i 3 n j+l . heref fore, j j+l* n.1

j+l n n. = 1 nn(the A~''.s are disjoint) i=l 1 i

The last unequality follows from (i)and (ii). The right hand side'of the inequality tends to zero as j gbes to mn i o infihity, Hence lim ( IJ qc 1 (n)/n = 0 . This shows that qg is a n i=l i i

We conclude this section with the following proposition.

~roposition'l. Every FPp-ri.ng R of subsets of N containing all finite sets is full.

Proof. Let (x,) be a sequence of positive real numbers with 1 x = w n . n=l

Choose positive integers n < n2 < < n < such that 1 ...... i ...

C -. > i for i = 1,2,...... Let a (ti) = (i) and ^k ;P n.5k

-. 0.. * (s. 5 i) of Ai there exists 15k 5 s. such that C xk 1 1 1 i 1 ." -4 . i kc%. 1 This completes the proof.

Remark. Forfiubrings of 2N containing all finite sets we have

!Full rings} -3 ~FPQU-rings}2 tpp-rings} 3 {Qo-ring} f

We have not come up with an example of a full ring which is not FEQ3 . i 53. Main results.

-- Although the Nikodym Boundedness Theorem is subject to many

generalizations, it is difficult to find one generalization that fits

the others. we therefore consider here several situations for

which the - theorem holds.

Theorem 1. (Nikodym Boundedness Theorem). b Let R be a ring of subsets of a set satisfying one of the f 01lowing :

(a) R has property (QI) . I

(b) f?, is a PQ7-ring with the hereditary property.

Also let X be a locally convex space. Suppose p: R + XI n = 1,2, ....., are bounded vector rneasures such that {p(~)In E N) is a bounded subset n of X for every X f R . Then ,;$(A) In C N and A C R) is a bounded n subset of X .

In addition, if the ,, n = 1,2, ...... , axe regular over n finite sets, then (a) and (b) can be replaced by the following :

(a') R has property (FQI) .

(b') R IS an FPW-rlng with the hereditary property.

(c') R is a full ring with the hereditary property and containing 7,

all finite sets. (In this case R = N.) .

Proof. First we establish the theorem for scalar valued measures; i.e., we assume that X = C . Define a: R -t R+ by a (A) = Sup / l-I (A) 1 . Since the sequence (u) Bs n n n

setwise bounded, a is defined and,moreover , by (1) a * is unboqded.

'\ \ We also show that: \

(3) lol(B) - a(A)l I CI(B\A) for A,B 6 R with A -CB . , Let A,B C R .

\ (2) If A IB = $, then %(A L! B) = SU~~U(AJ B) I 5 SU~~U(A)I + SU~~P(B)I n n n n n n \ \

(3) Suppose A c B . For given E > 0 there exists n N such that - 0

consequently a(B) - 'a(A) 5 a(B \ A) . Similarly C~(A) - a(B) 5 CY(B \ A) .

Now an application of 1.5 3emnz 1 to a shows that there exists a

disjoint sequence (Em) of members of R such that lim 3(E = . m m

Thus by the definition of a we can find.subsequences (p) and (Em ) i

of ( and (Em) respectively such that lim 1 p (Em.)I = m . FOT n i n 1 i simplicity we relabel the sequence (p (Em by (l-I(Ei))ifN . Then n. i i I we have - First we consider case (a) R is a ring with property (QI),

Let (t.) be a sequence of positive numbers with the limit zero such 1

that :

It is readily seen that (ti p) is a sequence of strongly bounded scalar d i valued measures (note that every bounded scalar valued measure is strongly

bounded) with lim t. p (Et = 0 for every E f R Therefore 3.3 1 . i i

Theorem 1 implies that lim t . p (E. 1 = 0 uniformly for j f N This 1 . i 'j contradicts (Sf. Henee the Hikodym Boundedness Thsrem holds when R is

a ring with property (PI).

Now we consider case (b) R is a Pp-ring with the hereditary

property. Recall (4) lim 1 .U (En) / = . Let fn = 1 !' . Since n n n R is a Pp-ring, there exists a subsequence (E of (En) and a n. 1 n n n i i i partition El ,E2 ,.....,E fs. 5t ) ofeach E suchthat s. 1 n. n 1 1 i n. 1 E R for every sequence (ki) with 1 k. 5 5. For each k. 1 1 . i=l 1 n. n 2 1 i i C N we have t n. = ; (En,)1 = ;'(El ) + il (E2 ) + ...... + 1 n. I. n. n. - 1 1 1 n n. n i' n L (Es,1 5 (E + E 2 E s. since s I n 1 1; 1 + + 1. st , n. 1 n. n n. ' 1 i 1 %-- i 1' n i there exists 1 C k s. such that / p (E I 3 tn i - 1 . n.. ki 1 1 -, n 13 i Let Ek =Ai %en (6) d A. iR and limlp (Ai)/ zrn. . 1 i i=l i n i Since R is hereditary, U A C R for P c N-. kt vi: 2N -s c P - PCP

defined by vi (PI = p ( U Ap) . Since (~1) is a sequence of bounded n pfP n i i scalar valued measures with Sup IU (E) 1 < w for every E f R , it

readily follows that (vi ) is a sequence of bounded scalar- valued

measures with sup / vi (P) i

0-algebra (hence it is a ring wit$ property (QI)), we have

I N) < . This contradicts that

I = . Hence the Nikodym Boundedness

Theorem holds when R is a PQa-ring with the hereditary property.

To prove the last part let us assume that the

;I, n- = 1,2,. .. , are regular over finite sets. Then in (3) (E can . ... n n

be rep1aced by a disjoint sequence fP f of finite sets in I? , so we have n

Now cases (a') and (b') can be treated exactly the same

way we treated cases (a) and (b). Therefore we only have to consider case