Strongly Bounded, Finitely Additive Vector Measures

And Weak Sequential Compactness

By

Harry D. Walker III

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

This dissertation probably would not have been

written without the encouragement of my wife, Patti;

and my parents.

Mathematicalj, the greatest debt is to Dr. James

K. Brooks, who directed this dissertation. Thanks are

due to all the members of my committee, but especially to

Dr. Stephen A. Saxon (see p. 79). Although not directly

concerned with this dissertation, Dr. Phillip N. Nanzetta

deserves mention for his generous guidance ir my early

graduate work.

Patricia Armstrong has done an excellent job

typing this dissertation.

TABLE OF CONTENTS

Page

Acknowledgments ii

Abstract iv

Introduction 1

Chapter

I. Basic Results 4

II. The Bart)e-Dunford-Schwartz Theorem

For Strongly Bounded Measures 34

III. Uniform Strong Boundedness And Weak

Sequential Compactness In ba(G, A) 73

Bibliography 39

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

STRONGLY BOUNDED, FINITELY ADDITIVE VECTOR MEASURES

AND WEAK SEQUENTIAL COMPACTNESS

By

Hary D. Walker III

august, 1971

Chairman: J. K. Brooks

Major Department: Department of Mathematics

This dissertation is concerned primarily with the

condition of uniform strong boundedness on a set K of vector-

valued, finitely additive measures defined on a ring. K is

uniformly strongly bounded if the images of disjoint sequence

of elements of the ring converge to zero uniformly in the

measures of K. If K consists of one measure, the measure is

called strongly bounded. We give an elementary proof of a

well known result which, when the measures are countably

additive on a sigma-algebra, frequently allows the reduction

of a problem in vector measure theory to classical (non-

negative) measure theory. Using strong boundedness, we

generalize this result to finitely additive, vector-valued

measures on rings. An extension property which we develop

enables us to derive relations between uniform strong bounded-

ness and absolute continuity with respect to a non-negative

measure. We apply these results to obtain equivalent condi-

tions for weak sequential compactness in spaces of bounded,

finitely additive, scalar-valued measures defined on an algebra

including easy proofs of several well known conditions equi-

valent to weak sequential convergence.

INTRODUCTION

In developing a theory of integration with

countably additive vector measures, Bartle, Dunford,

and Schwartz [1] found :hat a vector measure p had

associated with it a non-negative measure X such that

X is a "control measure" for P in the sense that X

and p have the same class of null sets. This enabled

them in many instances to transfer a problem in vector

measure theory to classical measure theory. We give

a short, elementary proof of the Bartle-Dunford-Schwartz

Theorem in Chapter 1 (see page 28). In the second

chapter we consider the problem of generalizing the

Bartle-Dunford-Schwartz result to finitely additive

measures. In fact, we show that such a X exists for a

given finitely additive vector measure p on a ring

precisely when p is strongly bounded. Strongly bounded

measures were introduced by Rickart [1'], where it was

shown that every such measure is bounded. We shall see

that for weakly sequentially complete vector spaces, the

converse is also true; that is, any bounded measure is

also strongly bounded. Thus this condition on finitely

additive measures is at once powerful in its implica-

tions and mild in its restrictions. The path we shall

follow investigates the question of extending countably

additive, strongly bounded measures on a ring to

countably additive measures on the generated

a-ring, and we are able to strengthen an extension theorem

of Sion [15]. Similar extension problems were considered

by Dinculeanu and Kluvanek [5], and several problems raised

in [5] will be solved in 2.20 and 2.27 (in the bounded

measure case). We shall use these results in Chapter 3

to obtain powerful results on sets of uniformly strongly

bounded vector measures. Specializing to the case of

scalar-valued measures, we obtain various useful charac-

terizations of weak sequential compactness in spaces of

bounded, finitely additive scalar measures on algebras.

Those characterizations complement those obtain ed by Dunford

and Schwartz [6], and are related to ones obtained by

Porcelli [13]. Earlier, Grothendieck [8] had obtained

similar results on weak sequential compactness for Radon

measures on u-algebras. Finally, Darst [4] was able to

simplify the theory by giving a more direct proof of

Porcelli's condition for weak sequential convergence. We

are able, using the idea of uniform strong boundedness, to

give a short proof of the equivalence of Porcelli's condition

to a related one of Hildcbrandt [10], and using our earlier

theorems, an easy proof of the equivalence of either of

these to weak sequential convergence.

Similar results on "control measures" and on uniform

strong boundedness have bjen obtained by Brooks [2j. However,

the related

the methods

different.

of a result

theorems were arrived at independently, and

of proof in this dissertation are entirely

In particular, Theorem 3.6 is a strengthening

of Brooks [2].

CHAPTER I

Basic Results

1.1 We shall be concerned in this work with various

conditions concerning vector measures. This chapter

will lay the foundation for the sequel. Theorems 1.12

and 1.15 will be of particular importance in Chapter 2.

1.2 Notation. The conventions established in this

section will be followed throughout the dissertation.

E denotes a locally convex linear topological

space with completion U. If p is a continuous semi-norm

on E, the unique continuous semi-norm on E which extends

p is denoted by p. E' is the continuous dual of E. The

topology induced on E by E' is called the weak topology.

The scalar field of E is , and may be real or complex.

The real numbers are denoted by R, the non-negative real

numbers by 2 = + U [I), and the non-negative

integers are denoted by w. For U H E, U is balanced if

aU = (au: ueU) 0 U whenever lal 1, a0c. We denote

n-dimensional Euclidian space by R or if n = 1 simply

by When E = Rn, is of course always R. If U L E

and if p is a continuous semi-norm, we say U is p-bounded

if p(U) = [p(u): ucU) is bounded in 8. U is bounded

if and only if, for each continuous semi-norm p, U is

p-bounded. We write l.c.s. for locally convex linear

topological space. A standard reference for theorems

in a l.c.s. is Kothe [11]. E will always be Hausdorff;

we will write I instead of E when E is a Banach space.

A ring R of subsets of a space Q is a set of

subsets of Q closed under finite unions and differences

of elements of R. R will denote a ring unless otherwise

specified. If 0 is required to be an element of R, we

will write A instead of R and call A an algebra. If

reR, the algebra R/r = (r'fr: r'eR] is the algebra induced

by R on r. A 6-rinq D is a ring closed under countable

intersections, and a a-ring L is a ring closed under

countable unions. The a-ring generated by a ring R is

(R). A sequence (r) c R is monotone increasing to

r = U.r. (ri r) if r. r.i+l ieJ; (r.i) is monotone

decreasing to r = nir. (ri r) if r. Q ri+l' iew. Any

function 1 on a ring R into a l.c.s. E is a measure. A

measure p:R E is countably additive if for every disjoint

sequence (ri) P R, UiricR implies

J(Uiri) = iP(ri),

where the convergence is unconditional. The measure i

is finitely additive if the above equality holder for every

finite disjoint family (r ) n=1 R. If (P e is a set

i Pd C aC

of measures, (a ) is uniformly bounded if

(M (r): aeG, rcR)

is sounded in E; if (I ) is just a single measure p, we

say that p is bounded. The set (pi ) is point-wise

bounded if for each reR, (i (r): aeG) is bounded. For

p a continuous semi-norm on E, the p-semi-variation,

p(p) : R K is defined by

p(p)(r) = sup(pp((r')): r':r, r'eR), reR.

If E is a normed linear space with norm 1 I, the

semi-variation of 1, I I] I (*) is simply the I I-semi-

variation of pi. Also, v(p)(') is the total 'arjation

of p, defined when E is a normed linear space by

v() (r)=supi( 1) (r) : (r ) CR,(r ) disjoint, and Ur r .

If p is a continuous semi-norm on E, and X: R + is

finitely additive, we write p <

lira p(p(r)) = 0,

X(r)-0

that is, given c > 0 there exists a 6 > 0 such that for all

reR such that X(r) < 6, we have p(p(r)) < e. In a normed

space, we will omit the "p" and w::ite simply p << 1.

For xEE', we say Ixl < p if jx(e)J p(e), eeE. We

will indicate the end of a proof with O .

If u: R R is a finitely additive measure, then

+ *

we define I : R R and p : R R by

p (r) = sup[p(r'): r'eR, r'r)],

U-(r) = -inf(p(r'): r'cR, r' .r].

As we shall see in Theorem 1.3, if p is a bounded measure,

we can decompose p as the sum of two non-negative, bounded

+ -

measures by p = p p

1.3 Theorem. Let p: R R be a finitely additive

measure. Then

(1) p and p- are non-negative finitely additive

measures;

(2) v () (r) (= (r) + p'(r), reR;

(3) 41l (r) = max(p +(r), p-(r)), reR.

If p is point-wise bounded, we have

(4) + = u+ p.

Proof (1) p+ and p- are obviously non-negative.

Let r,s be disjoint elements of R. Then

S+(rUs) = sup[p(t): teR, trrUs]

= sup[((t t2) : t cR t r, t t2cs)

= sup(p(tl) + 1(t2): tlt eR, t lr, t2-s]

= + (r) + p+(s).

Thus 1+ is finitely additive. Since p = (-p) -i is also

finitely additive.

(2) Set reR. Then if R' = R/r

v(p) (r) = sup[( I1-i(r ) : new, (ri) disjoint in R)

= sup(p(s)-p(t): s,tgr, s,tcR, srt=0, p (s)>02I(t)}

= sup( (s)-p(t) : s,tcr, s,teR, snt=0)

= sup[((s)-p(t): s,tzr, s,tcR)

= (r) + p (r).

The third equality above follows from: u(s), p(t) 0 imply

P(s) I(t) W P(s) = u(s) + P(0) ; P(s), p (t) < 0 imply

p(s)) (t) -p(t) = (0) pi(t); p(t) > 0 and p (s) f 0

imply P(s) p(t) f L(0) P(b). The fourth equality follows

rrom: pC(snt) 0 implies p(s) p(t) i (s-t) i(b);

p(snt) > 0 implies p(s) p(t) p (s) p(t-s).

(3) Il (r) = sup[ !I(s) : s-r, scR}

= max(sup(p (s) :scr,seR,i(s)-20),sup([-t (s) :srr, scR,p(s)O])]

= max + (r) p- (r)}.

(4) LLc reR. Then

p(r) + I- (r) = ,i(r) + sup[(- (s): scr, seR)

= sup{( (r)-p (s): sor, seR)

= sup[A(r-s): ssr, seR)

= + (r).

Since p is point-wise bounded, p (r) is finite. Thus

+ = I -

1.4 Corollary.

(1) v(): R RR

(2) p is bounded

is bounded;

(3) i+(r) = if

is finitely additive;

if and only if p (~-)

and only if p (r) = m, rcR.

Proof (1) above follows from (1) and (2) of

Theorem 1.3; (2) follows from (3) of Theorem 1.3. Since by

the proof of (4) of Theorem 1.3, 4 + P = + (3) above

follows from the fact that p is finite-valued. O

1.5 Theorem. Let P:R R+ be a bounded, countably additive

measure and = X(R). For sez, we define

S(s) = inf[ (r : (r )9R, Ur s}.

Then 3 is a bounded, countably additive measure on

into R.

Proof It is well known that i defined as above

is a countably additive extension of .. Details can be

found in Halmos [9]. Suppose i is not bounded. Then

there exists (si) s such that

F(si) > i, ieW.

By the definition of ., for ieW, there is an (r ij)jw R

such that for some N.e,

1i

N.

1= (rij) > i.

Ni

Defining r = Uj r.., we have

i

k(U= r) > i, ice,

so that i is not bounded on R. O

1.6 Corollary. In the definition of i above, the (ri)

may be required to be pair-wise-disjoint.

Proof Fix sex and (ri) 9 R such that Ur. s s.

Defining s. = r. Ui- r., we have (s.) E R pair-wise-

1 disint, Uj=l s

disjoint, U.s. 2 s and

1 1

Si-i

y.i(si) = V.(ri-U1j1 rj)

i 7.(ri) .

Thus (s) is equal to

inf[( i(ri): disjoint (r i)R; Uri s). E

The next theorem and Theorem 1.9 are basic for

our work with finitely additive measures. They frequently

allow us to reduce a problem to the case of a countably

additive measure. Theorem 1.7 can be found in Dunford and

Schwartz [6], although the proof there is stated for an

algebra rather than a ring.

1.7 Theorem. For Pny ring R on Q, there exists a ring

Rs of open and closed subsets of a compact, Hausdorff space

S such that R is isomorphic as a ring to Rs.

Remark: We call Rs the Stone ring of R, and the

isomorphism from R onto R, is denoted by r.

Proof Let A be Lhe algebra generated by RU[n).

Then by Theorems IV 6.18 and IV 6.20 in Dunford and Schwartz

[6], there is a compact Hausdorff space S such that

B(Q,A) is algebraicly, isometrically isomorphic to C(S)

under a mapping h. Suppose acA. Let a denote tne

characteristic function of a Then a) is a continuous

characteristic function of a. Then h( ) is a continuous

a

function on S, while

h(a) = h(2 ) = h2(a

a a a

implies that h(a) c (0,11. If we define

T(a) = [h( a)]-1 (1),

then by the continuity of h(a ) on S, T(a) is an open

and closed subset of S. We shall show that T is an

isomorphism, and define R = T (R). Since h is one-to-one,

s

so is T. Set aeA. Then

T (Q-a) = h(Q-a) -1 (1)

= [h(-a )]-1 (1)

= [h(%) h(ga)]-1 (1)

= [1 h(la) ]-1()

= S (a),

since h being an algebraic isomorphism implies Lhat h must

send the multiplicative identity of B(Q,A) onto the multi-

plicative identity of C(S). Similarly,

T(alna2) = T(al)nT(a2)

T (alUa2) = T (a)UT (a2)

follow, respectively, from

h(Ca a2) = h( al a2) = h( a) h(a2)

a na a a2 a2

..nd

h( alUa2) h( al a ala2

= h( al) + h( a2) h(al) h(a2).

a1 a a! a

Suppose al a2. Then T(al) S T(a2) since

T(al) = T(al a2) = T(a))nT(a2).

Then for al, a2EA,

T(al-a2) = T(al) T(alna2)

= T(a (T(1) (aa2)UT(a2-al))

= T(a1) 7(a2),

since a2-al (al-a2) implies

T(a2-al) s S T(al-a2).

Thus T is a ring isomorphism, so that R is the desired

ring. O

1.8 Remark. If (si) 9 R is a disjoint sequence of

non-null sets, then U T(s.) R .

Proof of Remark Suppose (si) S R such that

U 5. U.s., NEw.

U=1 1 i i

Then U.T(s.) is not compact, and hence not in R since

1 1 S

(T(si))i is an open cover of U.T(si) which has no finite

subcover. In fact, if for some New,

UN T (s ) = U. 'r(s.),

=l i 1 i

then for ie,,

T(Si ) c U= T(Sj),

1 j=1 ]

implies

N

s q. U s .

i j=1 j'

as in the proof of Theorem 1.7. O

One is tempted to think that Theorem 1.9 eliminates

any problem in dealing with finitely additive measures.

However, many results are valid only on u-rings, and as

Remark 1.8 shows, Rs is rarely a a-ring. We shall show

later that every ring contains a sequence (si) of disjoint

non-null sets, unless R _s c finite set. Thus Rs is a

o-ring if and only if R is finite.

1.9 Theorem. For every P: R E, the equation

-1 reR

Th(r) = H(T (r)), reR ,

defines a measure TIi: R E. The mapping T has the

s

following properties.

(1) For each finitely additive J, Tp is countably

additive;

(2) for each (r.) 9 R, lim. T(T(r)) exists in E

if and only if lim. v (r) exists in E.

1

(3) If X 2 0 is a finitely additive measure on R

such thaL U << X for each continuous semi-norm

P

p, then TX 0 is a countably additive measure

on R such that Ti << TX for each continuous

s p

semi-norm p.

Proof (1) Let r, scRs, rns = 0. Then

T-l(r)nT- (s) = T- (rs) = 0 implies

Tpi(rUs) = 2(T-l(rUs)) = P(T-1(r)UT-1 (s))

= P(T-1(r)) + -(T-I (s))

= TI (r) + Tp (s).

,Tow suppose (ri) Rs is disjoint and r = U.r.R s. Then

r is compact and covered by (r.) so that for some New,

N

r U N r..

1=1 i

In fact, since r = Uir., we have

r=U N r.

i=l ri

and hence r. = 0, i > N. Then

Ti(U.r. = T(UN r )

i Ui=1 i

N

Ni=l T(r ) = ~ Ti(ri),

so that To is countably additive.

(2) is obvious from the definition of T4.

(3) That TX > 0 is obvious; TX is countably

additive by (1). Let U be a neighborhood of 0 in E. Then

there is an E > 0 such that p(r)eU for reR whenever

X(r) < E. But if TX(r) < e for some reRs, then

(T-1 (r)) = TX (r) < e also, so that

TU(r) = p (T-l(r))U.

1.10 Definition. Let I: R E be a measure. Then

(1) a(R) = (rG,: r=U.ri for some sequence (ri)CR);

l 1

(2) p(R) = (rca: r=r' for some r'Eo(R));

(3)a) D(r,R) = (r'eR: r'-r), for rQ;

b) o(r,R) = [r'eo(R): r'nr), for rcQ;

(4)a) g(r) = lim(p(r'): r'eD(r,R)], rci,

where D(r,R) is directed by r'r" r'lr"

(assuming the limit exists).

b) (r) = lim(g(r'): r'cD (r,R)}, rOQ,

where o(r,R) is directed by r'r" r'-r"

(assuming the limit exists).

(5) M = (reR: for all seR, (s) = 4(sfr) + p(s-r)).

The next two theorems are due to Sion [15], Since

we need to examine his proofs in detail in Chapter 2, we

reproduce them here with a few small changes.

1.11 Theorem. (1) Let p: P(Q) E be such that p(0) = 0.

Then MI is an algebra and p is finitely additive on M .

(2) Let f: D E, where D is a directed

set. If for each increasing sequence S in D, f/s is a

Cauchy sequence in E, then f is a Cauchy net in 3.

Proof (1) Since p(0) = 0, it is clear that

0 EM By symmetry, if reM O-reM also. Thus if

rsEM implies rsEMP w.e have that M is an algebra

Mim r Js 1

But for t e P(0),

p(t) = p(tnr) + p(t-r)

= p(L(tnrs) + p((tnr)-s) + p(t-r)

= p(tn(rns)) + p((t-(rns))nr) + p.'(t-(rns))-r)

= p(tn(rns)) + p(t-(rns)).

If also rns = 0, then

pi(rUs) = p((rUs)nr) + p((rUs)-r)

= p(r) + p(s),

so that p is finitely additive on M .

(2) Suppose f is not Cauchy in E. Then

there is some balanced neighborhood of 0 U, such that,

for every deD, there exists d'eD with d' d and

f(d') f(d)4U.

If this were not true, then for d', d" 2 d, d',d"eD,

f(d') f(d") = f(d') f(d) (f(d") f(d))

e2U,

so that f would be Cauchy.

Choose d ED and pick a2ED such that

f(d2) f(dl) U.

n

If we have an increasing sequence (di)i=1 D such that

f(di+) f(di) 4U, <1in-l,

we can find dn+l s d using d = d in the statement shown

n+1 n n

above. Thus we have an increasing sequence (di)i ; D

which is not Cauchy. O

1.12 Theorem. Suppose the following conditions are

satisfied:

(1) P: R E is countably additive;

(2) (r ) C R monotone implies limi p(ri) eE;

(3) Oe (R).

Then p has a countably additive extension fEro,.

(R) into E.

Proof We shall break the proof into a sequence

of lemmas.

Lemma (a) g is defined on P(Q), the set of all

subsets of 0, and g/R = Pi

Proof For rEP(Q), p/D(r,R) forms a Cauchy net, by

Theorem 1.11(2) and hypothesis (2) above. That g/R = P

is obvious since reR implies rED(r,R).

Lemma (b) Let r ER, nEU, r n r. Then

n n n

g(r) = limn p (rn) .

Proof Let U be a balanced, closed neighborhood of

0. Then there exists an sER such that scr and if s'eR.

such that s'gr-s, then p(s')eU. Then for new,

pl(r) u(s,.rn) = U(rn-s) eU,

while

p(s) g(r) = limteD(rR) (p(s) p(t))

= limtD(rR)-p(t-s) EU.

Thus

lim in(r) g(r) = lim n((r n) g(r))

= lim n((rn) p(snr ) + u(snr ) g(r))

= limnn(u(rn) p(snrn)) + p(s) g(r)

e U+U = 2U.

Lemma (c) g is finitely additive on a(R).

Proof Let (rijj .) R, i = 1, 2. Then if

r = Ujr.., i = 1, 2, we have by Lemma (b), if rlnr2 =

g(rlUr2) = limn ((U=1 r1 ) U (Un r))

= lim (,i(U= rlj) + ,(U r2j))

= g(rl) -.- g'r2).

Lemma (d) Let r ec(R), neW, r r. Then

g(r) = limn g(rn).

Proof Choose s ER. ncw, such that s r a:Wd

n n

s r new. Let U be a neighborhood of 0. Then for

each n there exists by recursion a t ER such that

s t r n, tn t t P(tn ) g(r ) cU.

Since r = U t n Lemma (b) says there exists an NEw such

that for n N, g(r) t(t ) eU. But then

g(r) g(rn) = g(r) p(tn) + U(tn) g(rn)

SU + U = 2U.

Lemma (e) Let (rnn n P(), rn Then

limn g(rn) exists in E.

Proof Let U, Vn, new be neighborhoods of 0 such

n

that V. i U, new. Choose s ER so that s C rn and

ta i=0 i n n n

U(s)eVn for s C r s Then g(r ) u(s ) Vn as in

Lemma (b). Let t_ = and for new, t = fnl s... Then

t eR, tn and since (s nnti ) s. C r. s.i for

1 s i a n, we have

n

p(sn) P(tn) = p(sn.tn) = i=P(s nti-l-si)

'n V. q U.

-Li=0 1

By hypothesis (2) of the theorem limn P(t ) exists since

n1 n

t n n. Thus there is a New, such that n > N implies

n

U(t ) P(tN) EU. Therefore, for n > N,

g(rN) g(rn) = g(rN) N(sN) + (sN) (tN)

+ p(tN) t) (t(s) + (t) (s) + (s) g(r

SV'N + U+ U+ U + Vn c4U.

-N n

Lenrma (f) p is defined on P(Q), and /a (R) = g/ (R).

Proof If re c (0), then by Theorem 1.11 (2) and

Lemma (e), g/o(r,R) is a Cauchy net in E, so that F(r)

is defined in E. That /ca(R) = g/o(R) is obvious since

r e o(r,R) when r a (R).

Lemma (g) If r e P(D) and U is a closed, balanced

neighborhood of 0, then there exists an r'Ea(R) such that

rcr', g(r') l(r)eU, and for every t e Np with t C r'-r,

S(t)eU.

Proof Choose r':a(R) such that r~r' and, for

every s in a(R) with r s s S r', g(s) .(r)eU. Then

if teM- with t s r'-r, P(rUt) = I(r) + P(t), and for every

scc(R) with rUt C s 5r', g(s) F(r) EU. Thus

i(r'Jt) ](r) CU by definition of , while a(t) =

i(rUt) i(r) EU also.

Lemma (h) Let a,beo(R) and U a balanced, closed

neighborhood of 0. -If (t)cU whenever t c a or t 9 b,

then for every t such that t s aUb, p(t) e 3U.

Proof Let (an),(b n) be increasing sequences in

R with a = U a b = U b Then for any ceH with c E aUb,

P(c) = lim n I(c (a Ub )). But nEW implies

p(cnf(a Ub )) = 4(cna ) + li(cnb -a ) E 2U.

n n n n n

Thus I-L(c) e closure (2U) c 3U. But then for sea(R),

s s aUb, g(s) e 3U. therefore if t q aUb, then 3(t)e3U.

Lemma (i) R C M-.

Proof Let reR and tCscP(Q). Then given a balanced,

closed neighborhood of 0, U, there is an aao(R) so that

tir C a C r and, for a'eo(R) with tnr C a' a,

g(a') t(tnr) eU. Also choose beo(R) so that

t r C b c s r, and for every b'eo(R) with t r s b' C b,

g(b') (t-r) EU. Then for ceo(R) with t C c C aUb, we

have

tnr C cnr S a, t-r c c-r C b,

while by Lemma (c) ,

g(c) = g(cUr) + g(c-r).

Thus

g(c) (i(tnr) + i(t-r)) = g(cnr) f(tfr)

+ g(c-r) I(t-r)

E 2U,

so that r e M- since (t) = fl(tnr) + f(t-r).

Lemma (j) o(R) c M-.

Proof Let a E u(R) and t c s P(Q). Given

U a balanced, closed neighborhood of 0, choose rleR so

that rl 1 a and, for every r'eR with r' c a-rl,

U(r')eU. Thus if L' C a-rl, then g(t')eU and, since

a-r Eu(R), i(t')EU. There exists a bEo(R) such that

t-a c b c s-rl and g(b) 3(t-a)eU. Then, as in the case

of a, there is an r2cR so that r2 5 b and F(t')eU when

t' g b-r2. Then by Lemma (i),

(t) = 3(td(r- Ur2)) + i(t-(rlUr2))

= 5(tr) + (tnr2) + t(t-C(rlUr )),

since rlnr2 = 0. Since t s aUb, we have t--(r Ur2) C

(a-rl)U(b-r2), so that by Lemma (h), [(t-(rlUr2)) E 3U,

and

.(t) ( (tnr ) + (tn r)) e 3U.

By Lemma (i), since tna-rl a-r, and (t-a)-r2 C b-r2'

i(tna) i(tnrl) = i(tna-r ) e U,

I(t-a) :(tn:2) = 2 i((t-a)-r2) e U.

Thus i(t) (i(tna) + i(t-a)) e 5U.

Lemma (k) Let a e s E ?(Q), new, so that

n

a t a. Then i(a) = lim n (a ).

Proof Let U,(V.). be balanced, closed neighborhoods

n

so that i.=0 Vi s U, new. By Lemma (g), choose r ne (R), new,

so that a s r g(r ) C(a ) e V and for tcM- with

n n n n n r

t s r an, F((t)EV Let s_ = and s Ui=1 r, new.

n n n -1 n i=1 1

Then s r = Un (r -s -r ), new. Let c. = r.-si-rn,

0 < i < n. Then (c.).n is disjoint; also c. e M- by

01 0

Theorem 1.11 (1) and Lemma (j). Since c. i r.-a n ri-ai,

t e M implies it(tnci) e Vi. Thus since i is finitely

additive on M-, for any teM- we have

n

F(tn (sn-r)) =Zi0 0(tCfc )

E Zi= Vi : U.

Let a' = U.r. = U.si, so that a 9 a'Eu(R), and choose

beo(R) with a c b c a' and g(b) p(a)eN. Then for all

new,

g(bns ) g(bnr ) = -(bn(s -r )) e U.

Since an bfr nec(R), g(bnr ) i(a ) e V by the choice

of a Hence g(bns ) 1 (a ) c U + V G 2U. Since by

Lemma (f), limn g(bns ) = g(b), there is an New so that,

for n N, g(b) g(bns ) e U. But then

S(a) (an) = (0i(a) g(b)) + (g(b) g(bns;\)

+ (g(bnsN) I(an))

SU -: U + 2U = 4U.

Lemma (1) Let (a ) C M-, a t a. Then aeM-.

Proof Let t c seP(Q). Define r = a U(s-a).

Then r nP(n), neu, rnT r, and t = Un(tfir ), so by

Lemma (k)

P(t) = limrn (tnrn)

= lin. ((tna ) + i(tnrn-an))

= limn (i(tnan) + (t-a))

= j(tna) + r(t-a).

Proof of the theorem By Theorem 1.11 (1),

since I is defined on P(Q), M- is an algebra, and in

I'

fact by Lemma (1), M is a o-algebra. By Lemma (k)

p is countably additive on M-. Since / = g/ = ,

U is the desired extension. O

The following theorem, which we state without

proof (See Dunford and Schwartz [6]), is well known and

is listed for completeness. The set ca(O, ) is the space

of countably additive, scalar-valued measures u on a

o-algebra with noo- gqLven by

I I P I = v( ) (n) .

A set K s E is weakly sequentially compact if

every sequence (ki). K has a subsequence (k.ij) which

converges in E in the weak topology. We shall write

k. k if a sequence (ki)i converges weakly to k.

1.13 Theorem. A set (pC) a ca(n, ) is weakly sequentially

compact if and only if (p ) is uniformly bounded and

uniformly countably additive.

We recall that (-p ) is uniformly countably

additive when r. r 0 implies lim. p (r.) = 0 uniformly in

aeA. (See Dunford and Schwartz [6].)

One of our major interests in the next chapter is

that of generalizing the Bartle-Dunford-Schwartz Theorem

(Bartle, Dunford, and Schwartz [1]), our Theorem 1.15. The

previous proofs (Bartle, Dunford, and Schwartz [1] and

Gould [7]) of Theorem 1.15 reduce it, as we do, to the

theorem proved in 1.14. Our proof of Theorem 1.14 uses

no machinery outside of some elementary properties of the

variation of a countably additive, scalar-valued measure,

and it is both shorter and more elementary than the earlier

proofs.

1.14 Theorem. Let (a c ca(O, ). If (P ) is uniformly

countably additive, then there is a countably additive

KX 0 on such that

p << X, uniformly in a.

Proof We shall first prove that for any e > 0,

there is a 6(e) > 0 and a finite set (kii)=1 C (Pa)

such that v(Pi) (s) < 6(e), 1 < i n

implies

I (s) I < 6, aEA,

.where s is.any element of ). Suppose we deny this for

some c > 0. Choose p (a). There exists a ~i2 (u) and

an sle such that

V -( ) (s ) < -E 2(s) .

k k+1

If we have a sequence (s ) k and (4). such that

.. il i =l

(i) V("i) sj) j+1

(1) v(2)J(sl) < }+1 1 < i & j < k,

(2) i ij+l(Sj) I E, 1 j f k,

k+l

then for the sequence (i )i= (a) we can find a

Pk+2 c(ua) and Sk+1 a such that

v(ki)(sk+ +2 1 i < k,

while

IWk+2(sk+l) 2 e.

Thus (1) and (2) above hold for j a i e 1 and j 1 1

respecLively. Define t = Uin si. Then for i n

v(p i) (tn) 4 (pi) (sj)

nGn

i n =n 2i)

j^n 234+1 2n 2

while

In+l(tn) I n+(Sn) I Iln+l (tn-sn)

Se vn+ ) (tn-Sn)

Sn+l n+l n

e v(n+l) (tn+l)

/2 2

If we define t = nnt n then .i(t) = 0, i 1, so that

we have

t t 0

n

while

1n+(tn-t)I E/2, n ,

contradicting the uniform countable additivity of (p ).

For convenience, let 6 = 6(2-n) and

= l v (Uni),

kn =1 'ni

where (u )m is the set found above corresponding to

ni i=l

e = 2-n. Then defining for se ,

x s 1 __ Xj,(s)(

(s 2 1 1+ (0)

we have a countably additive measure X 0 on C. Also,

if se then

X(s) < 6n 2-n/(l + h(Q))

implies

v(i ) (s) < 6n, 1 n i < n,

and so that

i (s) <- aEA.

a 2n

Thus X is the desired measure. O

1.15 Theorem. If p: Y E is a countably additive

measure, then there exists, for each continuous semi-norm

p, a countably additive measure Xp 0 on X such that

p <

Also, p may be chosen so that for seE,

) (s) S sup(p(p(r)): r = s, rc ).

Proof Fix p a continuous semi-norm on E. Let

(Wi)aca be the set [xU: xeE', Ixl < p). Then (u ) is

uniformly countably additive since s. 0 implies for

1

e > 0 there is an New such that

I (s ) = x (p(s )) p(p(s ))

< C,

Since P is countably additive, U is bounded.

the first line in the inequality above, (pa )

i a N.

Thus by

Sc.a(n,) .

By Theorem 1.14, p exists such that

P << Xp

a p

uniformly in a. Thus for e > 0 there is a 6 > 0 such that

for all a,

lpa(s)j < e if p (s) < 6.

But then if X (s) < 6,

p(p(s)) = sup( lx(s) : xcE', |xl p)

= sup( l (s) I: aeA)

To see that we may require X (s) < p() (s), seX, consider

the choice of Xp in Theorem 1.14. For se ,

P

S (S) = 1 1 1 (p )(s

P si 2 1 + ). (Q) .j= 'ii

SsupV (pij) (s) :

ie ', 1 < j s n=n(i)]

sup(v(a) (s) : aA)

4 sup([ a (s): aeA)

= 4 sup[ u (r) l: r G s, reX, aeA}

< 4 sup(p(Gkr)): r c s, rx ].

If we define = has all the desired

p = 4 p

properties. O

Both Bartle, Dunford and Schwartz [1] and Gould [7]

proved Theorem 2.15 for a a-algebra. Dinculeanu and

Kluvanek [5] then extended the theorem to measures on

a-ring by reducing this case to the case of a a-algebra.

With our proof this extra work can be eliminated. In

fact if X.(Q) is replaced by a bound B. for ?i on the

proofs of 2.14 and 2.15 are valid for a a-ring.

CHAPTER II

The Bartle-Dunford-Schwartz Theorem

For Strongly Bounded Measures

When working with finitely additive measures,

it is frequently necessary to impose various conditions

on the measure or its range space in order to obtain

meaningful results. At the same time any such condition

should satisfy the following: (1) the condition should

not be too restrictive, and (2) it should have a natural

form. One such condition is contained in the following

definition.

2.1 Definition. A finitely additive measure p:R E is

strongly bounded if, given a disjoint sequence ri) C R,

lim. u(r.) = 0.

We note from condition (2) of Theorem 2.2 below

that a strongly bounded measure p is "almost" countably

additive. Yet as we shall see later, for many spaces,

every bounded measure p is strongly bounded: Strongly

bounded, finitely additive' measures were first considered

in Rickart [14]. Since every countably additive measure

U on a 0-ring is strongly bounded, it is perhaps under-

standable that very little work was done using the

condition until the work of Brooks and Jewitt [3], where

the condition of strong boundedness plays a fundamental

role in the study of finitely additive vector measures.

2.2 Theorem. For a finitely additive measure p: R E,

the following are equivalent:

(1) \i is strongly bounded;

(2) given a disjoint sequence (ri) C R,

Yi p(ri) E;

(3) given a rc-notone increasing sequence (r ) R,

limi p (r ) E E;

(4) given a disjoint sequence (ri) 1 R,

limn ()(Unm ri) = 0;

for each continuous semi-norm p.

Proof We shall show (1 4 3 2 1).

(1 4): Deny (4). Then for some e > 0 and some

continuous semi-norm p, there exists a double sequence

(mj, n j) such that mj+1 I nj+1 > mj, jc'm, and

p(C) (Um "r.) > c, jew. Defining s. = U." r., we have

S=nj 3 I=nj

(s.)j. R a disjoint sequence such that p(p) (sj) > e, jew.

If we choose t.eR, t. i s., such that p(U(t.)) > a, we

clearly contradict (1) with the sequence (t.).

(4 3): Let (ri) be a monotone increasing sequence

in R. Defining si+1 = ri+1 ri, iew, we have (si+l)iew R

disjoint. Then by (4), if a > 0, and p, a continuous

semi-norm, are fixed, there exists an New such that

p(p(Um=n+ s.)) < e for all m r n > N. Since

(Un+1 s) = U(rm) P(rn), we have (3).

(3 2): Let (ri) c R be disjoint. Then since

(Un= r.) is a monotone increasing sequence in R,

p(ri) = limn =1 X (ri) = lim (U =1 ri)

exists in E.

(2 1): Since the sum converges in E, the terms

must converge to 0 in E. 0

2.3 Corollary If U is strongly bounded, then

lim. i(r.) E E,

for any monotone sequence (ri) 0 R.

Proof If (ri) is monotone decreasing in R, then

(ri r.l )i is disjoint in R, so that as in (4 3) above,

p(p(rm) (rn)) = p(ri-r i+))nm 0,

m=m i 1+1 n,m

providing m n. O

2.4 Example If E is not restricted, then the only

condition on R that will insure that a finitely additive

measure p: R E is strongly bounded, is that R be finite.

In fact let R be an infinite set, as well as a ring, and

choose E = 2(R), the Banach space of bounded, measurable

functions f on 0 with the supremum norm,

Ifl supf(f(x)I: xen).

We shall show that there exists a disjoint sequence

(r.)i of non-null sets in R. Given that, define

I: R 0(R) by

U(r) = r, reR.

Then I[t(r i) L_ = 1, iEw, so that u is not strcrgly

bounded. To prove our earlier assertion, suppose such a

sequence does not exist in R. Then there is a disjoint

(rli) l C R of non-null sets such that reR implies

r U i= rl. We may assume nl a 2. Since R is infinite

at least one of [R/ i is infinite also. Say R/r

rli il /r li

n2

Then we can find a disjoint (r2i) i=1 R/r of non-null

sets such that reR/r implies r U I=l r2i Again we can

ni

take n.2 2. Continuing by induction we get (r... n.j2,

ni D "i+l

such that for all i, U n r. Uni+ r Denoting

=l1 ii j=1 i+1,j

U j: rij by r., (r. ri+l) is a disjoint sequence in

3=1 n] a 1+1 i

R since ri 2 ri+l; since ri ri+. for all i, none of the

sets are null.

Having seen that boundedness and strong boundedness

are different concepts, the natural question is under what

conditions does one of these concepts imply the other?-

The first result in this direction was proved by Rickart

[14]. Because of its importance, we state it as a theorem

without proof.

2.5 Theorem. A finitely additive measure p: R E is

bounded when it is strongly bounded.

This theorem is a corollary to Theorem 2.16 below.

Rickart [14] also noted that if E is an Euclidian n-space,

the two concepts are identical.

2.6 Theorem. Let p: R Rn be a finitely additive

measure. Then U is strongly bounded if and only if U

is bounded.

Proof The necessity is Theorem 2.5. Conversely,

if p is bounded, so is wrp, 1 i n, where rr. is the-

.th

i projection map. If we prove the sufficiency for

real-valued measures, thb.n 7i is strongly bounded for

1 < i n, so that p is also strongly bounded. Thus let

p: R R; p = p+ p-, so that if p is not strongly

bounded, at least one of I+ + is also not strongly

+

bounded. We assume + is not strongly bounded. Then

there is some E > 0 and some disjoint sequence (ri) R

such that p (r ) > e, iew. But then if s. = U r

we have (s.). s R while

S(s ) Lj= + (r ) > iE, iew.

Thus + is not bounded, so that by (2) of Corollary 1.4,

1 is not bounded. O

2.7 Corollary. If p: R R is a finitely additive

measure, then p is strongly bounded if and only if

0 and p- are strongly bounded.

Proof By (2) of Corollary 1.4, p is bounded

+

if and only if p and p- are bounded. O

Among the spaces E in which boundedness and strong

boundedness are equivalent is the space ba(Q,A), the

space of bounded, finitely additive, scalar-valued

measures p on an algebra A with norm I I = v(P) (0),

which will be considered in Chapter 3. That the two

conditions are equivalent follows immediately from the

next theorem, since ha(PA) is weakly sequentially

complete (Dunford anc Schwartz, [6]). A space E is

weakly sequentially complete if every sequence Cauchy

in E with the weak topology converges weakly in E.

2.8 Theorem. Let p: R E be a finitely additive

measure into a weakly sequentially complete space. Then

p is strongly bounded if and only if p is bounded.

Proof We need only to show that a bounded measure

is strongly bounded, L:nce the converse implication

is Theorem 2.5. Since p is bounded, xU is also bounded

and hence strongly bounded, for xeE'. Thus if (ri) Z R

is monotone, x"(ri) converges, xeE, so that p(ri)

converges weakly to some point eeE since E is weakly

sequentially complete. Therefore, if (si) S R is a

disjoint sequence,

Sp(s) = lim. P(U s)

Li i 1 j=l j

exists in E given its wek topology. Since this is also

true for any subsequence of (si), the Orlicz-Pettis

Theorem (McArthur [12]) says that, in particular,

i -(si)

converges in E. Thus 4 is strongly bounded by

Theorem 2.2(2). r

2.9 Corollary. If P: R E is a finitely additive

measure into a semi-reflexive space, then p is bounded

only when p is strongly bounded.

Proof Any semi-reflexive space is weakly

sequentially complete (K6the [11]). [

Along with the condition of strong boundedness

of a measure P is the condition of uniform strong boundedness

of a set (UP ) of measures with a common range space. This

notion will be used extensively in Chapter 3.

2.10 Definition Let (Ua) be a set of finitely additive

measures on R into E. Then (p ) is uniformly strongly

bounded if (ri)i disjoint in R implies

limi Pa(ri) = 0,

uniformly in a.

We next establish some equivalent conditions

to uniform strong boundedness. Condition (6) of Theorem

2.11 is the form of uniform strong boundedness used by

Brooks and Jewitt [3], where it is called "uniform

additivity".

2.11 Theorem. For a se-; (I ) of finitely additive

measures on R into E, the following are equivalent:

(1) (a ) is uniformly strongly bounded;

(2) given (r.) disjoint in R,

li ari)

converges uniformly in a to a point in E;

(3) if (r.) is monotone increasing in R, then

p (ri) converges uniformly in a to a point

in E;

(4) given (r.) disjoint in R,

limmn P(G )(Unm r.) = 0,

uniformly in a, for each continuous semi-norm p;

(5) for each continuous semi-norm p, if

(x ) B E' and Ix C a p, 8sB,

then (x8 U )a, is uniformly strongly bounded.

(6) given (i) disjoint in R,a a,

(6) given (r.) disjoint in R,

limn Yicn [n,-) a (ri) = 0,

n Luin[n~m) 1

uniformly in aeG and in A 0 w.

Proof We show (1 5 4 3 2 1) and

(4 6 -1).

(1 5) Let (r")i be a disjoint sequence -n X.

Then if e > 0, there exists an New such that

p(a (ri)) < E for i N and acG. Since Ixg lp(ri)I

Sp( a(ri)), (x P ) is uniformly strongly bounded.

(5 4) Deny (4). Then for some e > 0, if

icw, there are n., mi.we with n. A m. I i and p (p )

1

such that p(p') (U r.) > e. Clearly we may assume

Ei j=mi 3

1 1

mi+1 > ni. Then defining s = Ujmi r we have (s.) e R

disjoint such that p(p )(s.) > e, iew. Choosing

t iR, ti s. such that p(p (ti)) > E, we contradict (5) by

1

choosing x.eE' with Ix !i C(t.)l = p (U (t).

1 ia. 1 a. 1

(4 3 2 1) These follow as in the proof of

Theorem 2.2, allowing p to run over the p .

(4 6) If p is a continuous semi-nor:m on E,

P(iiEAn [n, ) C(ri) ) i in P() (ri)

so that the limit in (6) is uniform in A s W.

(6 1) Let A = w. Then since the sum in (6)

is uniform in a, a (ri) converges to 0 uniformly in a. O

2.12 Corollary. If P : R Rn, aeG, then the following

are equivalent.

(1) (U ) is uniformly strongly bounded;

(2) given a disjoint (ri)i R,

Yi l i(ri)

converges in 2 uniformly in a;

(3) given a disjoint (r.i)i R,

limn jieA [n,-) a(ri) = 0

uniformly in A c w and in aeG.

Proof Suppose (1) holds. We can assume that 4

is real valued since (2) will hold when (2) holds for each

i. U, 1 i < n, where ni is the ith projection map. But

for m, new there is a A a [m,n] such that

n

i--m p(ri)I = EiA H(ri) iA p(ri)

2 suptIECA, p(ri )j: A' c [m,n]),

so that (2) holds by (4) of Theorem 2.11. (2 3 1) is

obvious. 0

2.13 Corollary. If U: R E is a finitely additive

measure, then u is strongly bounded if and only if, for

each continuous semi-norr p, if (x ) 9 E' and

Ix a p, aeG, then (x p) is uniformly strongly bounded.

2.14 Corollary. If (C ) is as in the above theorem, but

E is a Banach space, then the following are equivalent:

(1) (ua) is uniformly strongly bounded;

(2) If (ri) is pair-wise-disjoint in R, then

lim mn 1 (Un" r.) = 0,

m,n a i=m i

uniformly in a;

(3) if (x g) E', Ix Ix 1 1, Sel, then

(x3 Ua ) is uniformly strongly bounded.

Proof The proof of Theorem 2.11 holds if the

p's are restricted to any set of continuous semi-norms

which generates the topology on E. 0

Having seen the relations between boundedness and

strong boundedness, one might hope for similar relations

between the corresponding uniform conditions. The following

simple examples show, hox.ever, that the condition a're not

comparable.

2.15 Example. Let (ri) be a disjoint sequence of

non-null sets, and let R be the ring generated by (ri).

We define 6.. = 1 if i = j, 6. = 0 otherwise.

(1) Let i.: R P+ be the measure determined by

i(r.) = 6... Then (Pi). is uniformly bounded

by 1, but obviously (i ) is not uniformly strongly

bounded.

(2) Let 1.: R + be the measure determined by

i (r ) = i 61j. Then (i ) is not uniformly

bounded, but it is uniformly strongly bounded.

The next theorem shows that it was not an accident

in (2) above that the sequence (Ui) was unbounded on a

single reR. Example (1) above shows that the converse of

Theorem 2.16 fails.

2.16 Theorem. Let p : R E, aeG be finitely additive

measures. If (U ) is uniformly strongly bounded and

point-wise bounded, then (p ) is uniformly bounded.

Proof Let p be a continuous semi-norm on E.

For reR, we define B(r) = sup(p(ua(r)): acG) and

S(r) = sup(B(s): seR, s c r). Since (u ) is point-wise

bounded, we have B(r) < -, reR, by definition. We

first show that (a .) is p-bounded on R/r, the algebra

induced by R on one of its elements r. Suppose not, and

let tO = r. Then S(t0) -- so that there exists

sleR, s G r such that for some p e (P ),

p(u l(sl)) > B(t0) + nl, where nI = B(t0) + 1. Then

p(u l(s)) > nl obviously, while

P(UCc(t 0-s1)) P (ki CL)) P( l (to)0

p(u i 0-s1) (s a

Sp(C (sl)) B(t0

nl,

also. We choose tl to be sl or tO sl, so that

S(tl) = -. Suppose we have a monotone decreasing

sequence (t.)m= so that defining n. = B(t i. + 1,

i i 0 i i-P

1 i m, we have for some ( )i C (P ),

(1) p(W i(ti)) > ni,

(2) S(t ) =

for 1 5 i m. Setting nm+1 = B(tm) + 1, we may choose

by (2) above, an sm+eR, sm+1 t, so that

p m+ (Sm+)) > B(tm) + nm+l.

m++l

As before, we also have p(pm+I (t sm+ )) > nm+.'

a,+l m m+1

Choose tm+1 to be sI.; on tm sm+1J so that S(tm+) =

Thus we have (U )i>. (p ) and (ti)i C R such that

(1) above holds for i > 0 when n. = B(t i_) + 1. But

then

p(k i(ti.l t)) p(i (ti)) p( a(t.il)

p (pai(t)) B(ti-)

> 1,

contradicting the uniform strong boundedness of (U )

Now suppose that (p ) is not p-bounded on R.

Then there exists (r.) s R such that p(U (r.)) T m,

1

where (U ) C (C ). We may also require that, for iew,

Pa ( a

pCai Cir.)) >

Defining s.i = ri+. (U r.), iew, we see that

1 +1 i+1 ( 1 =

P(POi+l (Si+l) ) > (ri+l)) pWC (ri+in(Ui irj)))

pP(u l (ri1) S(r )

1 +1 i+

i+l

so that we again contradict the uniform strong boundedness

of the (P ), since the (si) R pair-wise-disjoint.

Since p was an arbitrary continuous semi-norm on

E, (P ) is bounded. O

2.17 Remark. As we have seen in Example 2.15, Theorem 2.16

is false, in general, if (p ) is not required to be

point-wise bounded. This is so even if E is restricted

to be the scalars. Thus since a set K c ba(Q,A) must be

bounded if it is weakly sequentially compact, the suffi-

ciency in Theorem IV 9.12 of Dunford and Schwartz [6] is

false without the added hypothesis that K be bounded (or

at least point-wise bounded, using Theorem 2.16).

2.18 Theorem. Let (p ) be a set of finitely additive

measures on R into R. Then (p ) is uniformly strongly

bounded if and only if (p ) and (-) are uniformly strongly

a a

bounded.

Proof Since lu(r) < v (u) (r) = U+(r) + U-(r),

reR, by Theorem 1.13, the sufficiency is obvious. Suppose

(p ) is not uniformly strongly bounded. Then for so.ae

(ri) c R pair-wise-disjoint, there exist inteqers m. and

Ua. ) a ie, such that m and for some e > 0,

j ml (r) > C.

i i+1i

For each iew, if mi j < mi+1 and U. (r ) > 0, choose

i

1 +

s.eR, s. 9 r. so that p (s.) > (r.); otherwise choose

j J 1 a. j 2 a. I

s. = 0. Since (a ) is uniformly strongly bounded, there

exists an New such that n > N implies

lYj>_n ~(sj) < e/4,

for every acG. But if m. N,

Ij'imi (sj) I l .
1 i+l i

> L < U (r.) /4

2 Lmi-j mi+l ai

1 i+1 1a

> E/4,

a contradiction. 0

2.19 Example. Let R be the ring of all subsets of 5

of finite Lebesque measure. Then R is obviously a 6-ring.

If 1 is the Lebesque measure on R, then U/R is a countably

additive, non-negative, finite-valued measure on a 8-ring

which is not bounded.

Since any vector measure on a a-ring is bounded,

Theorem 2.20 below is a generalization of the Bartle-

Dunford-Schwartz Theorem, Theorem 1.15, for measures on a

6-ring. Example 2.19 shows that it is necessary in

Theorem 2.20 to assume that each p is bounded since the

X there is absolutely continuous with respect to itself

but is not strongly bounded.

2.20 Theorem. If P: R E is a countably additive

measure on a 6-ring R, then p is strongly bounded if and

only if, for each continuous semi-norm p, p has a bounded

p-control measure X .

Proof ( <- ) A basic neighborhood of 0 in E is

of the form (eEE: i (e) 1, 1 i n). Let (r.) be

disjoint in R. Then for 1 f i n there exists an N.ew

1

such that j 2 N. implies X (r.) < 6i(1). If

N max [Ni: 1 s i < n), then j N implies pi(.(j)) < 1,

1 < i < n. Thus p(r.) 0.

(-) We assume Ui 0. Then by Zorn's Lemma, there

is a set M of pair-wise disjoint elements of R for which

!i(r) # 0, reM, and such that if r'eR is disjoint froir each

reM, then J(r') = 0. Fix p a continuous semi-norm, and

define, for iew, Mi = (reM: p(u(r)) > 1/i] Since M is

pair-wise-disjoint and p is strongly bounded, M. is

1

finite for each ies. Let M = U.M. = (r.)iw. Since the

Bartle-Dunford-Schwartz Theorem, Theorem 1.15, holds for

p restricted to R/r. ieju, there exists for each i a

1

p-control X. on R/ such that X.(r.) C B, where B is a

1

bound for p on R (Theorem 2.5). Thus defining

(r) -- Xi(rnri), rER,

gives a bounded, non-negative countably additive measure

on R. By Theorem 2.2(4), there is an N such that

n > m : N implies

nP( n <

p(]p)(U. r.) < E,

where e > 0 is fixed. Thus if reR,

p(U(rnUiNri)) = limn P(P(rnU=Nri))

Slim supn P(p)(U=Nr )

Now, for 1 E.

Now, for 1 < i < N-1i, there exists a 6. such that

1

reR/r, Xi(r) < 8.

1

Set 6 -- min(6 : 1 < i

2 pl

Implies

implies p(u(r)) < /N.

N-1). Then reR, X(r) < 6

p(u(r)) = p( (rnUiri))

N-1

Si= P(u(rnri)) + p(u(rnU iNri))

< N + c = 2e.

In fact, u(r Uilri) = 0 since r Ui r.i is disjoint

from each element of M = (ri), so that P(r) = U(r U.i r.).

Thus we may set X = X. O

P

Theorem 2.20 is in fact true under much more

general conditions, but before we prove this, we need

a few more theorems. The set M4 was defined in

Definition 1.10(5).

Lemma 2.21. Let R be a ring and p a function on R into

E with u(O) = 0. Then M is a subring of R, and p is

finitely additive on My.

Proof We obviously have O~M from !1(0) = 0.

We shall first show that M is closed under relative

complements. Suppose that a, beM and teR. By set

identities we have p (tl (a-b)) + L(t-(a-b)) =

P((t-b)na) + u((t-a)U(tnbna)). But from beM we also have

P((t-a) U (tnbina)) = U((t-a)-b) + p((tnbna) J ((t-a)nb)) =

p((t-b)-a) + p(tnb) (the last equality by set identity).

Thus p(tn(a-b)) + u(t-(a-b)) = p((t-b)na) + p((t-b)-a) + "(tnb)

u(t-b) + p(tnb) = I'i) using a, bcMU We next show that

M is closed under unions, completing our proof since P

is obviously finitely additive on M By set identities

again (with a, b, t as above) U(tn(aUb)) + p(t-(aUb)) =

p((tla) U ((t-a)n (b-a))) -:- + ((t-a)- (b-a)). But

u((tna) U ((t-a)-a)(a))) = P(tna) + '((t-a)n (b-a)) since

aey Thus U(tn(aUb)) + p (t-(aUb)) =

A(tna) + U((t-a)n(b-a)) + p((t-a) (b-a)) =

p (tra) + 'p(t-a) = u(t), using b-a, aEM O

2.22 Theorem. If p: R E is a countably additive,

strongly bounded measure, then p can be extended to a

countably additive measure l on Y, the o-ring generated

by R, into E.

Proof We shall follow the proof of Theorem 1.12,

with some modifications. The proofs of Lemma (a) to

Loimna (e) are unchanged. But instead of defining p on

all subsets of 0, as in Lemma (f), we will define U only

on p(R). It is obvious that F is defined on p(R) since

Lemma (e) and Theorem 1.11(2) show that for t 9 0,

g/o(t,R)

is Cauchy whenever o(t.R) / 0. The fact that

/Va(R) = g/a(R)

follows easily as before. Since the definition of M-

depends on the domain of j, the M- we have now is changed,

but Lemma 2.21 shows that it is still a ring. If we

replace P(0) with p(R) in Lemma (f) through Lemma (m),

then the theorem follows, since in those proofs we only

used the facts that M- is a ring, that P(0) is closed

under finite intersections and countable unions, and that

each element of P(n) is a subset of a countable union of

elements of R. 0

Although we don't need this next theorem to

prove Theorem 2.25, we shall require it in Chapter 3,

and this is a convenient place prove it.

2.23 Theorem. Let and aeG, be countably additive,

strongly bounded measures on R into E. Then

(1) if ( a) is uniformly bounded, then so

is (P );

(2) if U has a bounded p-control p for some

continuous semi-norm p, then I is a bounded

p-control for j;

where p, p are the extensions of P, X to z into E and

p is the continuous extension of p to E.

Proof (1) Since a subset of E is bounded if and

only if its image under q is bounded, for each continuous

semi-norm q on E, and since each q is the extension of

V

itself restricted to E, we shall prove that for scZ and p

a continuous semi-norm on E,

sup(p(U (s)): aE) < -.

Fix e > 0 se p a continuous semi-norm on E, and act.

By Lemma (g) of 1,12, there exists an asc(R), say

a Uir., (r.) R, such that

pa(s) ga(a) e V,

where V = (ecE: p(e) e]. Since by Lemma (d)

ga(a) = limi P (V=lri)

there also exists an N ew such that

N

S(a) a(V 1 r ) E V.

Thus

p (s)) o(i(s) g(a)) + P(ga(a) a(V 'i r

N

+ p(p (Vj ri))

e + E + B = B + 2E,

where B is a uniform bound for (P(p a(*))) on R. Since

C is independent of s and of a, we see that

( (a s)) -- B, s(z, aeG.

(2) Set V = (ee'E: p(e) < 6). Since

S<< 1 there exists an E > 0 such that for reR,

p p

X(r) < e implies p(r) e V.

Now let sec such that (s) < e/3. By the definition of

I and X there exist h, h2 e o(R) such that h 9 h lh2

implies

(s) X (h) < -3, (s) U(h) e V.

Since if we set h = h-nh2, we have h e a(R), say h = U.r.,

(r.) s R. Then

1 p(s) p(h) = 1 (s) lim. Xp(U rj)I

= lim p () Xp(Uj,=~ )

and

p(li(s) (h)) = p((s) limi p(U= ir.))

= lim. p( (s) p(UJi r.)),

N

so there exists an New such that defining r = U ,rj., we

have

(s) \(r) I < |, rp((s) (1r)) < 6.

03

But then

Sp(r) < (S) +' p (s) (h) + X(h) X(r)

<-+-+ = C

3 3 3

implies P(r) e V.

Thus

i(s) = G(s) I(h) + i(h) n(r) + 4(r)

e v + v + v

shows that p(U(s)) 36. Thus << X O

P P

2.24 Theorem. If 1i: R.- E is a countably additive

measure, then U is strongly bounded if and only if,

for each continuous semi-norm p, L has a bounded,

countably additive p-control X

Proof The sufficiency follows as in Theorem 2.20.

To prove the necessity, we shall first show that U has a

countably additive, strongly bounded extension p to the

--ring D generated by R. For each rcR, denote the extension

of u to the a-algebra generated by 1R/ by 5r. This

extension 'xists by Theorem 2.22; call the g defined in

the theorem gr. We define J(d) = r(d), deD, where reR

is chosen so that d' C r. Suppose that r"eR also, with

d c r", We need to show p (d) = Ur,, (d). Define

s rfr". Since SER/r n R/r", if U is a balanced neigh-

borhood of O, we can find, using the definitions of the

functions g and P, s'Eo(R/r), s"eo(R/r,,), so that d s s',

s" s s and if t' s, t'eo(d, R/r), t" c s", t"Eo(d, R/r",)

then

r(d) g (t')eU, r, (d) gr"(t")eU.

Since gr = gr on elements of c(R/r) n o(R/r,,)

we now have

rr(d) r, = d) = d) gr(s'ns") + gr,,(s's'") r", (d)

E U U = 2U,

since s'fs" e o(d, R/r) n o(d, R/r,,). Thus we see that

u (d) = ii ,. (d), so that I is well defined. Moreover, [

is countably additive on D. Suppose (di) c D is

pair-wise-disjoint and d = U.d.ED. If we choose reR

such that d r, e ave

such that d o r, we have

(d) = r(d) = ,r(di) = i (di).

Now if we show p to be strongly bounded on D, we can

apply Theorem 2.20 to find a p-control p for U on D,

in which case the restriction of p to R will obviously

be a p-control for II. Fix (d ) c D disjoint, and U a

closed, balanced, convex neighborhood of 0 such that,

for each iEw,

i(di) U.

By the continuity of addition at 0, there exists a sequence

(Vi) of neighborhoods of 0 such that for all i

1

V. 0 U.

Zj=1 Vj 4

By Lemma (g) of Theorem 1.12, applied for each ieu, there

exists (rij)j c R such that for sane r cR,

r.i U. r.. = s. 2 di.

1 13 1 '

while (d.) I(s.) e V.,

and (d'ns. d.) E Vi, d'eD.

Thus for each iew,

(s ) U,

since otherwise,

U(di) = d(d.) i(si) + (si)

S 1 1 1

SV. + U + U U.

S 2 4 2

Now by Lemma (f) of 1.12,

(si) = g (si) = lim i (U r )

T jT k i k

Slimj p(Uk= rik)'

so for each i there exists an n. such that

1

ni 1

p(Uni r.) 4 U

k=1 ik) 2 u

1 "i

since U is closed. If we define a. = Un r.., then

2 1 j=1 *ij'

(a.) 9 R is pair-wise-disjoint, a. a s., iew, while

U(a ) 4 U, u(aind'-d ) e Vi, iew, d'eD.

i-i

We shall show that h. = a. Uj1 a. iwe, gives a

1 i ]=1 ]j

pair-wise-disjoint sequence in R whose images under p are

woundedd away from 0. it is enough to show that for each i,

ui(Uj=1 a.)) E l Vj,

since then

i

{(a) = u(h) + !(an(Ua ; a) )

1 =1 .

C p(hi) + 3 V.

1

p C(h) + U,

i' 4

implies that

U(h ) 4 U, iec,

contradicting the strong boundedness of p on R. But

u(anl(U-l a.)) = (d.na i(Ui-1 a.)) + i ((a n(Ui-1 a.)) d.)

S j=1 j=1 ] = 1 i. J ] 1

E U(d.na. n(U1 a.)) + V.,

1 1] j=l 1

while

u(dna f(U 1 a)) = 1 (d na n (a Uj1 ak)) d.)

1 = x' I i k=1

z i-V

j=1 j,

since (d.) is pair-wise-disjoint. O

The reader may have noticed that we only used the

weaker fotrm of Theorem 2.22, that is, the case when V(R)

is an algebra. This was done since Theorem 1.12 is the

only version available in the literature. In fact, however,

Theorem 2,24 is a corollary of Theorem 2.22: Given ;u

strongly bounded, j on 7 = 7(R) is countably additive,

so by Theorem 2.20 p has a p-control p ; thus p = \p/R

is the desired p-control for U on R.

The results so far have been for countably additive

measures. Using the powerful representation of R as a

Stone ring Rs, however, we can prove our most general

result on control measures for finitely additive measures.

As mentioned in the introduction, this result was shown

by Brooks [2], although we arrived at it independently

and with different techniques.

2.25 Theorem Let p: R E be a finitely additive

measure. Then U is strongly bounded on R if and only if,

for each continuous semi-norm p, P has a bounded, finitely

additive p-control X Additionally, p may be chosen so

that:

(1) x (r) sup(p(U(r')): r' 0 r, rER);

(2) p is countably additive if and only if every

p is countably additive.

Proof If p' is the measure corresponding to p on

Rs, the Stone ring associated with R, then by Theorem 1.9(2)

p' is strongly bounded on Rs, so that Theorem 2.24 gives

us a X' on R The corresponding X on R, then, is

p s p

the bounded control desired. Conversely if each p

exists, then since it is bounded, it is strongly bounded,

and one may prove p strongly bounded as in Theorem 2.20.

As for (2), if P is countably additive, Theorem 2.24 shows

that the countably additive p exist, while if they do

exist, then for (ri) 9 R pair-wise-disjoint,

P(Ujrj) (Uji r) t 0,

since for each continuous semi-norm p,

(Uj r p ) = U ) =l p(rr) i 0

from the countable additivity of p and the fact that

), is finite-valued. Finally we need to prove (1). Let

p be any continuous semi-norm. If (1) is satisfied on

Rs, then it is obviously satisfied on R also, so we assume

P to be countably additive. We found X in Theorem 2.24

P

by extending p to y on D, the 6-ring generated by R, and

using Theorem 2.20, so that in the notation of Theorem 2.20,

X p(d) = X(d) =Ci (dndi).

p 1 i (dndi)

Since by Theorem 1.15. (1) holds for each Xi on D/d.'

1

we see that (1) also holds for p on D, ie

p (d) < sup(p(F(d')): d' c d, dcD), deD.

Denote this supremum by K. Choose E > 0, and let

U be a balanced neighborhood of 0 such that 2U s V,

where

V = [eEE: p(e) < E).

Fix rcR and let d be any element of D such that d e r.

By the definition of j on D, there is a sequence

(s.) s R such that U.s. s r and

F(d) i(U.s ) e U.

Then

lim. P(U s.i ) I(Ui.s) = i(d) + (Uisi) (d)

E j(d) + U,

so that for some NEW,

I(Ui s.) e c (d) + 2U P (d) + V.

But then, if we let r' = U Nis. R, we have r' s r while

i=l1 i

p(d(r')) = p(C (r') -(d) + F(d))

> p(C(d)) p(p(r') F(d))

p(j(d)) e.

We see, then, that for reR,

sup(p(u(r')) : r'ER, r'r) -= sup(p(G(d)): deD, d G r),

and hence that (1) is satisfied. E

2.26 Corollary. Let U be as in the theorem but with E

a Banach space. Then p is strongly bounded if and only if

there exists a bounded, finitely additive measure X, 0

with U << X. Additionally X may be chosen so that

(1) (r) sup[ I Ip(r') : r'eR, r'9R), rER;

(2) p is countably additive if and only if X is

countably additive.

2.27 Corollary. For a countably additive measure

p: R E, the following are equivalent.

(1) p is strongly bounded

(2) U extends to a countably additive measure

P from (R) into E.

(3) for each continuous semi-norm p, U has a

bounded p-control X .

The corollary just above shows that in the bounded

measure case the Bartle-Dunford-Schwartz Theorem remains

valid on a ring precisely when the vector measure is

strongly bounded. This problem (for 6-rings) was raised

by Dinculeanu and Kluvanek [5]. Of course Theorem 2.20

suffices to solve this problem for a 6-ring.

The p-control, p whose existence is guaranteed

in Theorem 2.25 for a strongly bounded p, may fail to

exist if we assume only that p is bounded, even if ), is

allowed to be unbounded and R is required to be a o-algebra.

Even if p is countably additive, p may not exist when R

is a ring. Examples are given in (1) and (2) below.

2.28 Example. (1) Let L be the a-algebra of Lebesque

measurable subsets of [0, 1], and let S (L) be the Banach

space of measurable functions on Q, with the supremum

norm

Ilfl = sup{If(x) l xe [0, 1]).

Define 1: L J (L) by U(r) = ,rj reL. Let

I = (-: i 1, ice). If X is a I" 1 -control for U, and

6 > 0 is such that X(r) < 6 implies I[p(r)I < 1, then

1(( ) > 6, iel. Since I e L,

ln

(I) X((r) ) n6 ,

:i=l n

shows that X(I) cannot be finite.

(2) Let R be the ring generated by

IU(I). Then p/R is trivially countably additive. If

X is a control for p, then X(I) = as before.

2.29 Remark. If l: R E is a finitely additive measure,

then pi is strongly bounded if and only if the range of '

is weakly sequentially compact. That this is a consequence

of Theorem 2.25 has been noted in Brooks [2].

It is natural to wonder in view of the Nikodym

Theorem (see Dunford and Schwartz [6], p. 321) whether

the corresponding theorem is true for finitely additive,

strongly bounded measures. For a c-algebra, the next

theorem answers this affirmatively, and the following one

gives a condition for ic to be true on a ring. The first

theorem was mentioned in Brooks and Jewitt [3], but we

give it as a nice application of the existence of a control

measure.

2.30 Theorem. Let Pi: R E, icw, be strongly bounded,

finitely additive measures on a u-algebra R. If

limi Ui(r) exists in E, reR, then U(r) = lim. ii (r), reR,

defines a strongly bounded, finitely additive measure p.

Proof Let p be a continuous semi-norm on E. By

Theorem 2.25, each p. has a bounded p-control i.. Then

(r) = 2i XI ,

where B. is a bound on i., defines a bounded, finitely

additive measure on R such that liml (r)-0 .i(r)=0, iew.

Then by Theorem 3 of Brooks and Jewitt [3], the limit

above is uniform in i, so that i is strongly bounideC

since X is bounded. O

2.31 Theorem. Let pi: R E, ie w, be strongly bounded,

finitely additive measures. If (i ) is uniformly strongly

bounded, and if limi i(r) exists in E, reR, then

p(r) = limri i(r), reR, defines a strongly bounded,

finitely additive measure.

Proof The measure p is finitely additive since

each i. is finitely additive. Suppose (r.) is pair-wise-

disjoint in R. Then if p is a continuous semi-norm on E,

there exists an N such that for all i and all j L N,

p(Ui(r )) < 1.

But o has a unique extension to a continuous semi-norm

p on E, so that

p(p (r )) = lim. p( (rj)) < 1,

for all j > N. Thus p is also strongly bounded. O

2.32 Definition. Let p: R E be a finitely additive

measure. Then u is locally strongly bounded if, for each

reR, U restricted to the algebra R/r is strongly bounded.

As we see in the next theorem the condition defined

above is a generalization of strong boundedness which

reduces to strong boundedness on a a-ring.

2.33 Theorem. If p: R E is a finitely additive measure,

then p is locally strongly bounded if and only if (r.)

monotone decreasing in R implies lim. i(r.) exists in

E.

Proof (=) If (ri) is monotone decreasing, then

(ri) R/r and P is strongly bounded on R/r so that the

limit exists.

(H) Let reR, and (r.) R/r monotone increasing.

Then (r-r.). is monotone decreasing so that

lim. U(r.) = im. i[p(r) (p(r-r.))]

= p(r) lim. p(r-r.)

exists in E. 6

2.34 Corollary. If R is a o-ring, then p is locally

strongly bounded if and only if p is strongly bounded.

Proof By Corollary 2.3, U is necessarily

locally strongly bounded when it is strongly bounded.

Conversely if (r.) is monotone increasing in R, then

(Ujr. ri)i is monotone decreasing so that as in the

theorem above, lim. r(r.) exists in E. O

2.35 Theorem. Let p: R E be a finitely additive measure.

If, for each continuous semi-norm p, p has a p-control X ,

then p is locally strongly bounded.

Proof If reR, then each \p is bounded on R/r\

That p is then strongly bounded on R/r follows as in

Theorem 2.20. O

2.36 Remark. Local strong boundedness is actually weaker

than strong boundedness since the measure in Example 2.19

satisfies the first, but not the second condition. By

Corollary of Dinculeanu and Kluvanek [5], a locally strongly

bounded countably additive p: R E can be extended to a

countably additive measure p: D E, where D is the 6-ring

generated by R. A proof of this is also contained in the

proof of Theorem 2.24.

CHAPTER III

Uniform Strong Boundedness

And Weak Sequential Compactness In ba(., A)

This chapter is basically concerned with two

problems. The first is that of extending conditions

holding on a ring R onto the a-ring X(R). A basic tool

for this will be the extension theorem of Chapter 2,

Theorem 2.24, although most of the results of Chapter 2

will be needed. The second problem is that of investigating

conditions for weak sequential compactness in ba(Q, A),

the space of bounded, finitely additive, scalar-valued

measures on an algebra A with the variation norm:

11 I = v(p) (0).

Throughout this chapter A will denote an algebra, and

will denote the a-ring generated by R.

3.1 Theorem. If p: R -l R is a bounded finitely additive

measure, then p is countably additive if and only if p+

and p are countably additive.

Proof (C4 ) Since p+ = + p-, p is obviously

countably additive.

( ) If (r ) C R is pair-wise-disjoint,

it always follow's that

1+ (Uiri) Y ^i+ (ri),

+ + n n +

since (Uir.) 0+(Un r) = i. (ri), neu. But

since p is countably additive,

i+(Uir.) = sup(P(s): seR, s 9 Ui.r.

= sup{(Ci(si): (si)cR, U.si R, siCri, ie )

= iu(ri). E

Although we have not needed the fact so far, it is

true that the extensions of measures we dealt with in

Chapter 2 are unique. Since we shall implicitly use this

several times in this chapter, we give a short proof.

3.2 Theorem. Let U.: E, i =

additive measures. If [ = 12 on

Proof Let F = (s 7: 1 (s)

(f.) is a monotone sequence in F,

.

1, 2, be countably

R, then PI = P2'

= 2(s)]; F a R. If

then since i, i = 1, 2,

are strongly bounded,

limj ul( j) = limj u2(fj)

exists in E. Since i i = 1, 2 are countably additive

into E, the limits exist in E and

Pi(lim f ) = lim i(f ),

for i = 1, 2. Thus lim. f. EF. By the Theorem Ln

Monotone Classes (Halmos [9]), we have that F = E.

Our immediate goal, and the key theorem of this

chapter is Theorem 3.6, but first we shall prove a special

case to which it can be reduced using Theorems 3.1 and 2.18.

3.3 Lemma. Let i: j +, iew, be countably additive

measures. If New, E > 0, se7, and (bi) pair-wise-disjoint

in R with U.b. 2 s, then there exists (ai) C R pair-wise-

disjoint in R such that

(1) Uibi 2 Uai D S,

and (2) ?iPn(ai) < n(S) + E, 0 < n < N.

Proof Since each pi is an extension of itself from

R to we may use Corollary 1.6 to find (r!) C R disjoint

such that U r s and ZiUl(ri) < pl(s) + e. Then if we

let (ri) be the set (r'nbm: n, mew), we have a disjoint

sequence (r.) c R such that U.b. Q U.r. 2 s. Alro

1 1 1 311

C0(ri) = i VP 0(r1 n bj)

= .Y0(U(ri n b.))

< Li0(r') < U0(s) + e.

So we have shown the lemma for N = 0. Suppose the lemma

is valid for N-1. By Corollary 1.6 again, there exists

(di) s R disjoint such that Udi D s and i IN (di) <

UN(s) + c. Also we have (ci) c R disjoint such that

Uibi D U.c. D s, and for 1 < n < N-l, iUn(ci) < un(s) + E.

If we let (ai) be the double sequence (c idj)ijcw' then

we have Uibi U Uc. U.U.(c.nd.) = U.a. E, while (ai)

is obviously disjoint in R. Also, for 0 n -N-l,

YiPn(ai) = i,jj n(CiAd )

= iPn(Uj (ciAdj )

Slipn(ci) < Pn(s) + E'

while

YiPN(a i) = i jN(Cind j)

= ZjUN(Ui (ciN ) A)

!jpaNl(dj) < PN(S) + C.

By induction, therefore, the lemma holds for NeJ. O

3.4 Theorem. Let (p, )aE be a set of non-negative,

finite-valued, countably additive measures on X. Then

if ( a) is uniformly strongly bounded on R, (C ) is

uniformly countably additive on Y.

Proof Suppose (Il ) is not uniformly countably

additive on C. Then for some e > 0, there exists (si) ;,

and (i.) G (U ), such that s. 0, while U. (s) > e.

Fix e = e/2i+2, iew, so that e = E/2.

Since every sex is covered by a countable union of

elements of R, we can find by Lemma 3.3 a sequence (rO). G R

such that UJ.r. s and

0

Li0(ri0) < A(s0) + E'

Since ( a) is uniformly strongly bounded on R, there exists

an n0ew such that

i>n j (r)
L.~i>n0 j 1

Define r0 = U 1 rj Then for jjw,

Uj(r0sj ) = j (s ) uj(sj-r)

j (s ) i>n.j (r)

C GO.

Thus we have for n = 0, a decreasing finite sequence

(r.i)n G R such that

(1) [ (ri) < p (si) + i', 0

(2) jp(ris E E k" jai,

for 0 i s n. If we have (1) and (2) for n = m, we can

find by Lemma 3.3 (with b0 = rm, bi = i > 0) a sequence

m+l m= 1 rmr

(r 1) disjoint in R such that rm U.ri S+ n r

i m I l m+1 m

i11j (r < U ) j+ 1 (s-

for 0 < j < m+1. Since (~ ) is uniformly strongly bounded

on R, there is an nm+1 such that for iEw

>nm+ (r )

,nm+ n+l

Then defining rm+ = Uj1 r1 we have (1) and (2) above

holding for n = m+l, since for j C m+l

j (rm1 sj) = j j r) j((sj rm) rm+

S- j=0 Cj >nm+j m+

Sm+1

-Lj=0 j

Thus by induction (1) and (2) hold for all iEW with the

decreasing sequence (ri). Then by (2)

Si(ri) p.i(r.'fsi)

S- =0 EL k E/2.

Since (P.) is uniformly strongly bounded, we have by

Theorem 2.11 that limj p.i(r) exists uniformly in i,

so that by the above inequality lim pi (r.) a '/2, iEw.

But this contradicts the fact that limj Up(r ) = 0, ieW,

by (1) above. O

The proof of Theorem 3.4 above was greatly

simplified by Dr. Stephen A. Saxon. He has also made

many helpful suggestions incorporated throughout this

dissertation.

3.5 Remark. Although uniform countable additivity is a

weaker concept than uniform strong boundedness (when the

measures are countably additive), if R is a a-ring, the

two concepts are the same. Thus in Theorem 3.6, uniform

countable additivity could be replaced by uniform strong

boundedness (or uniform additivity, by Theorem 2.11).

3.u theorem. Let p : R E, acG, be countably additive

measures. If (P ) is uniformly strongly bounded on R,

then each i has a countably additive extension a from

Into E such that ( a) is uniformly countably additive.

Proof Since (p ) is uniformly strongly bounded,

each 1i is strongly bounded, so that a exists by Theorem

2.22. Suppose that ( a) is not uniformly countably additive

on L. Then we can find some (si) z and some (pi) 9 (W)

such that s. 0, while for some e > 0, and some continuous

semi-norm p,

p(Pi(si)) > e, iEW.

Define X = (xeE' :x(e) < p(e), eeE]. Since each -a is

countably additive on Y, so is xP xeX. If (ri) c R

pair-wise-disjoint, then by the uniform strong boundedness

of (p ) on R, there is an Nea such that i > N, acG implies

p(Pa(ri)) < e.

a i

Sirce I -a(r) [ p( CL(ri)), xcX, we see that (xpa) acc, XEX

is uniformly strongly bounded on R. By Theorem 2.18,

then, so are ((.4t ) ) and ((x a)-) uniformly strongly

bounded on R. Thus by Theorem 3.4 there exists Lew such

that i >- M implies for aEt, xeX,

(x )+ (s.) /2, (x )- (s ) < /2.

But by Theorem 1.3,

Ix L(si) r v(xUa)(si) = (xa)+(s)_ + (xUa)-(si) < E,

so we have a contradiction, since, for each iEW, there is

some x-X such that

iru.i(si) = p(M (si)) > c. E

One special case of Theorem 3.6 which is frequently

useful in proofs by contradiction is given in the following

corollary.

3.7 Corollary. Let the Ic be defined on X, and let

is.1 be the ring generated by (si) c N. Then if (u ) is

uniformly strongly bounded on [s 1, (l ) is uniformly

ceuntably -additive on ([s ]).

The next theorem is a very useful extension of a

result of Dunford and Schwartz [6], p. 292.

3.8 Theorem. Let P.i: R E, iew, be countably additive

measures. If (ip) is uniformly strongly bounded, and if

lim. UPi(r) exists in E, rER, then limi j.i(s) also exists

in E, seX, where i is the extension of i to .

Proof Define L = [se/: limi. i(s) exists in E)

-- 1 1

We have by hypothesis that R C L, so that if we show L to

be a monotone class, the Theorem on Monotone Classes

(Halmos [9]) will give us that L = i.e. that limit .i(s)

exists on X. Suppose that (si) s L is monotone increasing,

and let s = limi s.. Since by Theorem 3.6, (Ci) is

uniformly countably additive on X, there exists an New

such that

i (s-SN) E V, iC

where V is an arbitrary circled neighborhood of 0. Since

lim.i (sN) exists in E, there is an Mew' such that for

i, j > M,

S(sN) -- lj (sN) e V.

Thus for i, j s M,

Pi(s) lj(s) = i(s-sN) + i(sN) j(sN) Pj(s-sN)

E V + V + V = 3V.

Since for any neighborhood U of 0, there is a circled

neighborhood of 0, say V, such that 3V 9 U, we see that

lim. i.(r) exists in If (s.) L is monotone decreasing,

then

lim. (s) = lim. [-ii (s -s) + Pi(s )]

exists in S since (s,-s ) sl-s. We have sl-si e L,

iew, since for each i,

lim a (s -s ) = lim [ (s) (s) ]

exists in E. O

3.9 Corollary Suppose limi Pi(r) exists in E, reR.

Then (pi) is uniformly strongly bounded on R if and only

if each pi has a countably additive extension i to

such that limi ji (s) exists in E, se7.

Proof The necessity is Theorem 3.8. Conversely,

if the limits exist on then by the Vitali-Hahn, Saks

Theorem (Dunford and Schwartz [6]) the (pi) are uniformly

countably additive on equivalently uniformly strongly

bounded on But then (W ) is uniformly strongly bounded

on R. [

3.10 Theorem. Let (A ) be a set of finitely additive

meusuLes on R into E. If, for each continuous semi-norm

p, there is a finitely additive, bounded measure X on R

into % such that a

bounded, then the following are equivalent:

(1) (u } is uniformly strongly bounded;

(2) a <

semi-norm p.

Proof (2 1) A basic neighborhood of 0 in E is

of the form (eE: i (e) < 1, 1si'n), where each pi is a

continuous semi-norm. Let (r.) be pair-wise-disjoint in

R. Then for 1 < i n there exists an N.ew such that j r N.

1 1

implies i(r.) < 6(1). If N = max(Ni: l

implies p. ( (rj)) < 1, aeG, lsi

uniformly in a.

(1 2) Suppose that for some continuous semi-norm

p, PA p 'p does not hold uniformly in a. Then there

exists (ri) R and ( U) 0 (U ) such that for some e > 0,

S.p(ri) < -2 p(ui(r )) > G, iew.

For each iew, let xi.E' such that

and .i (e) | p(e), eeE.

Since (pi) is point-wise bounded, we may use a Cantor

diagonal process to find a subsequence (x ). of (xi)i

such that for re[ri]

limj x ni C(r)

exists and is finite, for each i w. Without loss of

generality, we may ass-me the limits exist and are finite

for the original sequence (xii ). The reader can easily

see from Ixi(e)j p(e), ecE, that for each iew,

xil-i << p In addition, we will show that the sequence

(xii ) is uniformly strongly bounded. Suppose (si) C R

is pair-wise-disjoint. Then since p (s.) 0 uniformly

in a, there exists an New such that for i N,

p(P (si)) < E, for an arbitrary fixed E > 0, and jEw.

Then obviously, for i s N, IxPj (si)I < jew.

We may now assume that all measures under

consideration are countably additive, since otherwise we

could go over to the Stone ring associated with R without

changing any relationships used below.

We may now apply Theorem 3.8, so that

lim. x.U. (s) <

1 1 1

for se ([ri]). Also, by Theorem 2.23, for iew,

x..i << p where, of course, p is the countably additive

extension of Xp to C([ri]) into R. By the V.ir.ali-Hahns-Saks

Theorem (Dunford and Schwartz [6]), we have x.ji << X

i 1 p

uniformly in iew. Thus there exists a 6 > 0 such that

scE([r ]), o (s) < 5 implies !x.U (s) I < Choose iew

such that-. < E. Then by the choice of (ri), Xp(ri) < 6.

but

IxiUi(ri) i = p(li(ri)) > E,

so that we have arrived at a contradiction. O

3.11 Corollary. Let P : R 1, acG, be finitely additive

measures. If there exists a bounded, finitely additive

measure X such that a << X, aeG, then the following are

equivalent when (P ) is point-wise bounded:

(1) ( e) is uniformly strongly bounded;

(2) Pa << \, uniformly in a.

Proof The proof of Theorem 3.10 obviously holds

if p is restricted to any family of continuous semi-norms

which generated the topology on E. O

3.12 Corollary. Let i.: R X, iew, be finitely additive

measures. If (Pi) is point-wise bounded, the following

are equivalent:

(1) (Pi) is uniformly strongly bounded;

(2) there exists a bounded, finitely additive

X:R R such that Pi << X uniformly in i.

When each pi is countably additive, X may be chosen

countably additive.

Proof We need only show (1 2). Since each

pi is strongly bounded, there exists, for iem, a bounded

control measure Xi for i".. Define

X(r) = 1 ki(r)

where Bi is a bound for i.. Then X: R R is a bounded,

1 1

finitely additive measure such that pi << X, iew. Thus

Theorem 3.10 gives the uniformity in i. If each U. is

countably additive, X may be made countably additive by

choosing each Xi to be countably additive. i

3.13 Theorem. Let t : R E, aeG be countaoly additive

measures. If (U ) is point-wise bounded, and if, for each

continuous semi-norm p, there is a bounded, countably

additive measure : R 2 such that U. << p, uniformly

p a

in a, then

Ia <

where i X, are the extensions of P ,p to .

Proof By Theorem 2.23, << X for each a. By

a p p

Theorem 3.10, (W ) is uniformly strongly bounded on R,

hence also ( ) on V by Theorem 3.6. Since (p) is

point-wise bounded and uniformly strongly bounded on R,

Theorem 2.16 gives that (U ) is bounded on R, and hence

(e ) is bounded on Y by Theorem 2.23. Then applying Theorem

3.10 again, we have that <

The next theorem could actually be proved for an

arbitrary space E, but for simplicity we do not do so.

3.1L Theorem. Let Ua: R I, aEG. be finitely additive

measures. Then the following are equivalent:

(1) (P ) is uniformly strongly bounded;

(2) there exists a bounded, finitely additive

X: R R such that a << X uniformly in a.

Proof We proved (2 1) in Theorem 3.10. To show

(1 2), we may first assume that each pT is countably

additive, since the ( a) on the Stone ring corresponding

to R would still be uniformly countably additive, and if

(2) were satisfied on the Stone ring, it would also hold

on R. It is enough to prove (1 2) with R replaced by ,

since by Theorem 3.6 (1) holds for ( ) on and (2)

holding on X trivially implies (2) holding on R. Thus

(p ) is a set of uniformly countably additive measures on

L into 1. Define (3)t = (x'I: aeG, x'el', Ix'I Il).

Then (S) o ca(Q, D) is uniformly countably additive, so

that by Theorem 1.14, there exists a countably additive

.: 3- + with << X uniformly in B. Thus e > 0 implies

there is a 6 > 0 such that if X(s) < 6 then

I a- (s) = sup x'( a s): !x'I l, x'e ')

IsE

Since 6 is independent of a, U << uniformly in a. D

3.15 Remark. The rest of this chapter is devoted to

investigating weak sequential compactness in ba(Q. A)

using the machinery developed so far. From now on we

take R = A, so that Y = R(A). We recall that by the

Eberlein-Smulian Theorem (Dunford and Schwartz [6], p. 130)

a set K in a Banach space is weakly sequentially compact

if and only if the weak closure of K is weakly compact.

3.16 Theorem. Let (U ) be a subset of ba(Q, A). Then

(Cl) is weakly sequentially compact if and only if

(1) (1 ) is point-wise bounded,

and (2) (u ) is uniformly strongly bounded.

Proof We shall first show that (1) and (2)

imply (u ) is weakly sequentially compact. As in Theorem

1.9 (3), (TY ) e ca(Qs, As) is uniformly strongly bounded;

obviously, (TU ) is point-wise bounded. Thus by Theorem 3.6,

(T! ) is uniformly countably additive on Y, while by

Theorem 2.16, (Tu ) is bounded. Thus (Ti )Cca(Qs, 7(As)

is weakly sequentially compact by Theorem 1.13. Since T

is an isomorphism, if (li) C (p ), there exists a sun-

sequence (T n.)i of (Ti.) and a I C ca(Qs (As)) such

that TI.I converges weakly to P. Since ca( 1(A )) is

a closed linear subspace of ca(s A ), we also have

n. A

Tis i, I/A

1 s

in ca(Qs, As). Since T is an isomcrphism, we then have that

x(U ) = xT-I (T ) xT- (U/ )

i "n.i 1

1 1 s

for each x E (ba (0, A))', so that n -. T-1 (/A ). And so

1 S

we see that (U ) is weakly sequentially compact.

Conversely when (u ) is weakly sequentially

compact, it is also bounded, so that (1) holds. Since T

is an isomorphism, it follows much as above that (TU )

is weakly sequentially compact. Thus by Theorem 1.13,

(T)t) is uniformly countably additive on 1(As), so that

(Tia) is uniformly strongly bounded on A Thus (U a) is

uniformly strongly bounded on A. E

3.17 Tiherem. Let (I0 I) be a subset of ca(Q, 1). Then

(lsa) is weakly sequentially compact if and only if (U )

is point-wise bounded on A, and there exists a countably

additive X: A t .such that U \ on A, uniformly in a.

Proof If (U ) is weakly sequentially compact,

then by Theorem 3.16, ( a) is uniformly strongly bounded

and point-wise bounded on A since ca(Q, 7) is a closed

linear subspace of ba(Q, A). Thus by Corollary 3.12 there

exists X: A + such that << uniformly in a. As

remarked, X may be chosen countably additive.

Conversely, i4 \ does exist with Ua << on A,

then (p ) is uniformly strongly bounded on A, so that by

Theorem 2.16, (u ) is uniformly bounded. But then (u )

is uniformly bounded and uniformly strongly bounded on

Sby Theorems 2.23 and 3.6 respectively, so that by

Theorem 3.16, (U ) is weakly sequentially compact in

ba(Q, A). Since ca(f, A) is a closed subspace of ba(Q, A),

(U ) is weakly sequentially compact in ca(Q, ). n

3.18 Convention. Let (U.i)i be a sequence in ba(O, A).

Then for acA,

P(a) -v 1 (Ui) (a)

cp(a) = 2i l+v(.i) (0)

defines a bounded, finitely additive measure on A into

+ which we call the control of the sequence (4i).

3.19 Theorem. If ( a) C ba(Q, A) is point-wise bounded,

the following are equivalent:

(1) (J ) is weakly sequentially compact;

(2) ( a) is uniformly strongly bounded;

(3) if (ai) is disjoint in A, then

limi v(U) (ri) = 0,

uniformly in a;

(/) if (ai) is disjoint in A, then

SvC(Ua) (ri)

converges in 2, uniformly in a.

(5) for each sequence (i) c (. ),

uniformly in i, where cp is the control of (Pi).

Proof Of course (1 Q 2) by Theorem 3.15. By

Theorem 3.10, a sequence (p I) ( a ) is uniformly strongly

bounded if and only if Ui << c uniformly in i, so that

(2 5) is clear, while (5 1) since by Theorem 3.15,

each subsequence (pJ ) of (~ ) is then weakly sequentially

compact. We shall show (2 -- 3 A 2) to complete the

proof. If (2) holds, then by (4) of Theorem 2.11,

li I (Um=r.) = 0 uniformly in a. Since by (2)

., M Cc n i

and (3) of Theorem 1.3, v (p ) < 211 p I, a, G, (3) of the

theorem follows. Actually we have proved more, in fact that

lim nV (u ) (U= r) = 0,

m,n a i=n i '

uniformly in a. From this it is obvious that (4) holds.

If (4) does hold, then v (! r) (r) 0 uniformly in a, so

that l a(ri) "< v(a) (ri) implies (U) is uniformly strongly

bounded. C

3.20 Remark. In (4; of Theorem 3.19,

li v(4a (ri)

may be replaced by

TLi lai(ri

by Theorem 2.11 (2). The equivalence of this to (1) in

Theorem 3.19 and the equivalence of (1) and (2) in Theorem

3.21 was done by Porcelli [13] in a very long and difficult

manner.

3.21 Theorem. For (Wi) c ba(Q, A), the following are

equivalent:

(1) Ii converges weakly to 0;

(2) (Porcelli) for (a.) disjoint in A,

1

limi. ij i'(a)I = 0;

(3) (Hildebrandt) for (ai) disjoint in A,

limit Z i.(a ) = 0;

(P) a)

b)

(u.) ;c uniformly strongly bounded, and

li (a) = 0, a

lim. a(a) = 0, aeA.

Proof (1 4) This is Theorem 3.16, since (b)

implies that (ii) is point-wise bounded and, if pi

converges weakly, i converges weakly to 0 if and only if

(b) holds.

(2 3) is obvious.

(3 4): That (b) holds is seen by taking

a1 = a, ai = 0, i > 1. Suppose (a) fails. Then by Theorem

2.11 (4), there is a double sequence (mi, ni) such that

mi m, ni mi, iEw, and for some e > 0, and some subsequence

(k. ) of (Ui),

1

lj m. k.(a)I > ,' iew.

-1 1i

Clearly we may assume mi+1 > ni, iew. Then defining

b. = U.j a., ieu, we have (b.) disjoint while Ilk (bi) I > E,

1 j'mi j 1 1

iew. For simplicity, assume Iui(bi)I > e, ifw. Since for

each jew. there is a subsequence (b ) of (bi) such that

1

IPn. (bn.) = n. (bn.) for all i a j or

S(b )I = -Pn. (bn. ) for all i B j, we may assume by a

diagonalization procedure that for each i,

I L (b ) I = i(b ), j i, or 1I (b )i = -= U (b ), j i.

Now since lim. (a) = 0, aeA, we may also assume

i-1

j=1 i(b ) < E/2, icw

In fact this clearly holds for i = 1. Suppose we have the

inequality for i = N. Then there exists an M > N such that

i N. I w 2j+la

for i M and 1 j N. If we relabel so that pi becomes