BOUNDED, FINITELY ADDITIVE, BUT NOT ABSOLUTELY CONTINUOUS

2--; 9

Ndfd

BOUNDED, FINITELY ADDITIVE, BUT

NOT ABSOLUTELY CONTINUOUS

SET FUNCTIONS

DISSERTATION

Presented to the Graduate Council of the

University of North Texas in Partial

Fulfillment of the Requirements

For the Degree of

DOCTOR OF PHILOSOPHY

By

David R. Gurney, B.S., M.S,

Denton, Texas

May, 1989

Gurney, David R., Bounded, Finitely Additive, but not

Absolutely Continuous Set Functions, Doctor of Philosophy

(Mathematics), May, 1989, 66 pp., 1 table, 3 illustrations,

bibliography, 12 titles.

Suppose ba{U)(7) is the collection of bounded, finitely

additive, real-valued set functions on a field 7 of a set JJ.

Suppose also that p € ba{R)(/) . Using refinement integrals,

( € is defined to be an element of A , that is

absolutely continuous with respect to p, if and only if for

each c > 0, there is a d > 0, so that if f € ? and

fv\ji | < i , then f y \ t | < c.

With this definition, only the non-negative elements of

ba{R)(J), i.e. elements of 6a(R)(J)+, need to be considered

and the theorem below is proved.

THEOREM The following four statements are equivalent,

i) There is a v G 6a(R){^)+ such that v £ A .

ii) There is a v € ba (R) (7) + , a sequence of sets

in 7 and d > 0 such that lia ) = 0 and for any n-ko R

tt € IN, /_ ., C• / and v (/ ) > d . »+l n n

iii) There is a sequence of sets (E J03 , € 7 such that » 8=1

lint p($n) - 0 and for any n € IN, C E^ .

iv) There is a v € 6«(IR)(/)+ such that v £ A and the r

range of v is a subset of {0,1}.

In leading up to the proof, methods for constructing

fields and finitely additive set functions are introduced

with an application involving the Tagaki function given as

an example. Also, non-absolutely continuous set functions

are constructed using Banach limits and maximal filters.

The above theorem is then used in proving a result due

to Appling and to show that, for i a Daniell integral on a

sublattice L of ba (R) (7) and p E L , if t~(/t)(I) is defined

to be i (/* | JQ) then <~(/0 € A .

Copyright by

David Robert Giirney

1989

iii

TABLE OF CONTENTS

Page

Chapter

1. INTRODUCTION 1

2. FIELD AND FINITELY ADDITIVE

SET FUNCTION CONSTRUCTION 8

3. EXAMPLES 17

4. A CHARACTERIZATION THEOREM 34

5. APPLICATIONS 51

APPENDIX 60

BIBLIOGRAPHY . 65

iv

CHAPTER 1

INTRODUCTION

The following notations and conventions will be used.

Capital letters like A,B,6 will denote sets. Sets that are

possibly empty will be marked by asterisks. Thus the sets

i * D* R*

A , B , 0 would be possibly empty. Possibly empty sets will

appear again only after Example 4.3 in the discussion,

example, lemmas and proofs before the statement of Theorem

4.8. Capital letters without asterisks will always denote

sets containing one or more elements.

Script capitals like A,B,C will denote collections of

one or more sets A. Fraktur capitals like &,$,£ will

denote collections of one or more collections A. I N , a , and

R will denote the natural numbers, whole numbers

{i.e. {0,1,2,3, ... }), and real numbers, respectively.

/ and g will denote functions with range a subset i of R.

Unless stated otherwise, lower case letters like & , b , c or

r , s , t will denote real numbers, with i through q being

reserved for the integers. The lower case Greek letters

a , 0 r i , 6 , v , t f , p will denote set functions, that is

functions with domain a collection A and range a subset i

of R or subcollection % of subsets 5 of R. Other lower

case Greek letters will denote other functions as needed.

Let R + =» {x€R: x>0>, and for any k G N, let

jfe = ^ a® a n exponent on a set, collection,

or collection of collections, as in A^, JP or will

denote all sequences of members of A, A or SI, respectively,

indexed by IN. Semicolons will be used to denote intervals.

Thus for a > b, (a*,b) = {x-. a <x<b), l&-,b) = {x: a<x<b),

(fl;oo) - {z: a<x>, etc.

The end of a proof will be marked by an open box: .

Proofs of claims within proofs or examples will be

bracketed by "proof :" and subscripted by the claim

number. Hence the proof of Claim 4.3.2 in Example 4.3

begins with "proof,,:" and ends with "•2"* Examples are

concluded with a solid box: "B".

Suppose 7 is a finite collection of sets, ( Is a

function from 7 into R, and for each A € 7, c(A) gives a

condition on A which may be either true or false. If for

each A € 7, c(A) is not true, then

Â€7 £ ^ ~ ®* c (A)

If c(3) is true for some B € 7, then for

It - [A£7: c(A) is true},

Â€7 m Â€7 * c(A)

Sets A and B will be said to "meet" or "overlap" if

they share an element. Otherwise they will be called

"disjoint" or "mutually exclusive."

Statements of Definitions 1.4, 1.6 through 1.8, l.io,

1.11 and 1.13 below can be found in Appling's 1984

expository paper [3, pp.211-14], although Appling did not

invent the concepts involved.

1.1 DEFINITION V is a subdivision of !, denoted V « {V},

if y is a set and D is a finite collection of pairwise

disjoint subsets / of f whose union is Y.

1.2 DEFINITION If £ and V are subdivisions of V, then £ is

a refinement of V, denoted £ « V, if each member of £ is a

subset of some member of V.

1.3 DEFINITION If £ and "P are subdivisions of ¥, the

intersection refinement of £ and D, denoted £KD, is the

subdivision of F each member of which is the intersection

of a member of £ with an overlapping member of V.

1.4 DEFINITION 7 is a field &£ subsets of a set V if U is

a set and 7- is a collection of subsets S of If satisfying

the following.

i) If A,£ € 7, then AUB € 7.

ii) If A € 7 and A $ II, then U\A € 7.

1.5 EXAMPLE Suppose a ,b € R such that a < b . Let 7t ,. [«;o)

be the collection of all finite unions of half-open on the

right, pairwise disjoint intervals in [«;£). A little work

wi 11 show 7 * 11 is a f ield • • [a ; o ) •

From now on suppose that 7 is a field of subsets of a

set U.

1.6 DEFINITION Suppose V € 7, "T) is a subdivision of V

with respect to 7, denoted D {If), if D « {P} and T> C 7,

1.7 DEFINITION 8 is a refinement of V with respect to 7 if

8 « D and there is a ¥ € 7 such that 8 <^ {¥} and

V <7 {?)•

1.8 DEFINITION exp {IR) (7) denotes the set of all functions

mapping elements P of 7 to subsets V of R.

1.9 DEFINITION If / is a subset of exp{U)(7), f+ will

denote the elements of / whose range is a subset of R +.

If S maps 7 into R, 6 will be regarded as equivalent

to the element 7 of exp(R)(7) given by 7(F) = {£{P)} for

any F € 7. If a € exp(R)(7), P € 7 and 8 « {V}, S is an

interpolating function of a on 8 if 6 maps 8 into R so that

for any I € 8, S(I) € a{I).

1.10 DEFINITION A«(R)(?") denotes the set of all functions

( from 7 Into a bounded subset £ of IR such that if £ and F

are disjoint elements of 7, then £(£) + £(/) = £{/U/).

The elements of ba (R)(7) are called "bounded finitely

additive set functions."

1.11 DEFINITION For a € exp(R)(7) and I € 7, the

refinement integral Of a over V, if it exists, is the real

number denoted fy<* such that for each c > 0, there is a

V {f} so that for any £ D and any interpolating

function 6 of a on 6,

| f yOl - { I) | < c .

For any ( € ba (R) {7) and any V € 7, /j,r|£| exists.

Thus the following definition makes sense.

1.12 DEFINITION For any fi € ia(R){7), ( € A if and only

if for each c > 0, there is a d > 0 so that if E € 7 and

f g | p | < 4, then /g|(j < c .

I f £ 6 , i is said to be abgoltttely continuous with

respect to p . This is sometimes called "e-S absolute

continuity."

1.13 DEFINITION For any ft 6 A«(R){^:), £ € liplp) if and

only if ( € ia{R)(/) and for some b > 0 and any f 6 7,

|nn| < wrM--Throughout this paper, let § = 0. This increases the

population of the sets defined next. A statement of this

definition can be found in Bell's dissertation [4, p.4]

among other places.

1.14 DEFINITION For /i € ba (R) (7) and p € IN,

*,(*) = {*€v X *exists}-

Note that if p = l, S ( p ) - A . t P

Also appearing in Bell's dissertation [4, p.5], if not

elsewhere, is the relation

*«>(/») c c C C

where 1 < p < q and Hu is the intersection of all for

k € IN. Each of lip (ft), S (p), #.{/»), #_{/0 and Au is a " I P f*

normed, linear, complete subspace of ba(IR){J). Having all

these subspaces of A prompted the question: Is there a r

subspace of 6«(R)(7) properly containing A ?

To answer this question, being able to construct

non-absolutely continuous set functions seemed important.

To do this, knowing more about non-absolutely continuous

set functions seemed imperative. Eventually, these last

two considerations alone provided enough material for the

present paper.

The rest of the paper is organized as follows.

Chapter 2 introduces a method for constructing fields and

setting-up finitely additive set functions on these fields,

Chapter 3 presents a few examples of how non-absolutely

continuous set functions can and cannot arise. Chapter 4

develops a characterization theorem for non-absolute

continuity, and Chapter 5 gives two applications of the

characterization theorem.

CHAPTER 2

FIELD AND FINITELY ADDITIVE SET FUNCTION CONSTRUCTION

To facilitate setting-up examples, a general method is

needed for constructing fields and finitely additive set

functions on these fields. The theorem below provides one

such method. It was originally proved by John Von Neumann

[12, pp.83-91], but the argument given is streamlined

following the example of Royden [9, pp.256-60]. First,

however, a definition is needed.

2.1 DEFINITION A collection H of subsets S of a set II is a

half-field if the intersection of any two overlapping

members of K is again in K, and the complement of each

member of K, other than is a finite disjoint union of

members of K.

2.2 THEOREM If % is a half-field of subsets of V, then the

collection of all finite disjoint unions of members of K is

a field of subsets of U, which will be called the field

generated by H.

proof: Let T C X ) = aÎN, and the S s are pairwise

disjoint members of K). The claim is that T(,K) is a field.

8

If C € H and C t 9 , by definition of H , ff\C € 7 ( 1 ) .

Now suppose A , B € 7 ( H ) . Then A = . and

B s where and are subcollections of

H , and the A - ' s as well as the £•'s are pairwise disjoint. 1 J

If A does not meet B, then Ai)B is clearly in 7 ( H ) ; so

suppose A meets B . Then 3I € N and 3 / € IN such that A . at * n t

meets B - . Let 1 = { A ^ O B j : and A - meets B - } .

Note that

JUJ - n - ui.

Suppose A.OBj £ T h e n ' * o r J ^ '' a n d s o a n d

A , are disjoint, or B • and B, are disjoint. Thus A (\B • * j i i j

does not meet A ^ O B ^ . Hence, since the members of I are in

H , A(\B is the finite disjoint union of members of H .

If A C B or B C A , AUB is in H ( 7 ) ; so suppose it is

not true that either A C B or B C A. Then A $ U $ B , A

meets U \ B which is in 7 ( H ) , and B meets U \ A which is in

7 ( H ) . Now since

AUB = ( A \ B ) U ( B O A ) U (B\A)

= ( A n ( i f \ B ) ) u (AM ) U ( B n a i \ A ) ) ,

and the last three intersections are disjoint, AUB € 7 ( H ) .

So 7 ( H ) is a field. •

If H is a half-field of subsets of U, let 7 ( H ) denote

the field generated by H . Now suppose H is a half-field of

subsets of (J.

10

2.3 DEFINITION If (i is a function from K into I R , [t will be

called additive fin 2 if for each H € K such that there is

an b € N and some pairwise disjoint subcollection

of H so that H = U^_ H •, /t(#) = £*! /»(# •). * JL I I —1 t

2.4 THEOREM If is a function from K into K which is

additive on H, then fiQ has a unique extension to a finitely

additive set function on ?(K) .

proof: For each F € T{H), there is an n € IN and a pairwise

disjoint subcollection {S-)n- from H so that / « U* .H •; ii=l t»i t'

so let p{F) => Showing ji is well-defined and

unique will prove the theorem.

Hence suppose / € J(3f) , a,n € IN, and

are pairwise disjoint subcollections from H so that

"i-i'i - u J - i V

For Vi € II , 3/ € N such that H- meets /•, so let m n I J

INff = {/€N : 3- meets Similarly, for Vj 6 W , let " I ™ * J u

= {i€N : H • meets /.•}. Now note that if i € IN , * j n t j m

= j) a n d f o r ) ^ , Jr(H/y and S^fl/k are

in H, and also if j £ k, and H .(\l ^ are disjoint.

Similarly, if j € , /. = {X fiK .); for »,/ € Nj , K- j

J J

i - n / . and Sjfl/ • are in If; and also if i # / , fl/- and J J ^ J

are disjoint. Now since

11

u!=l U/0I, «•*''» - uy-l U,€ W / <<''>>>'

(i(F) = E* = 1 /»0{iSrl)

= ^ = 1 S;€N //,0 ( #i n f; )

«

- S"=! ^ W ; >

• S°=l "0

- *(/)•

So /j Is well-defined.

If ji~ is an additive set function which is an

extension of nQ, /e~(/) = £j = 1 - E j ^ /tQ(#f0 - /i(/)

Thus ji is unique. •

2.5 DEFINITION A sequence ( ^ ) B _ 0 o f subdivisions of U is

a refinement sequence of U if 7>Q = {^} and for any a G IN,

V « 7> and D ± V . n »-l n n-1

_ _ _»

2.6 EXAMPLE ({[ (J -1) • 2 a ;i • 2 is a refinement

sequence of [0;1) which will appear again shortly. m

2.7 THEOREM If * s a refinement sequence of ff, then

U® V is a half-field. n =0 n

proof: For any two overlapping sets in , one is

contained in the other, and so their intersection is in

• Suppose D € U " a n d D £ ff. Then 3« € II such

12

that D € D . Thus since for some n € N, = {D }a- , , # » i i=1

3 j € N so that D = D- and V \ D • U{Z>-: t€N \{/}}. • m J J /Jf

Suppose CDn)n_0 is a refinement sequence of U and let

7 = 7(if° V ) . All that is needed to produce an additive u —u u

set function on 7 is to construct a set function which is

additive on • Theorem 2.4 says such a function can

be extended in a unique manner to one which is additive on

7.

2.8 EXAMPLE For any t € R, let # ( t ) be the distance from t

to the integers. Then for any x € [0;1], let

/(*> - 2"V(2Bx).

/ is known to be continuous and nowhere differentiable.

Tagaki [10] first mentioned / had these properties, and

Van der Waerden [11] later touted f for the same reason.

Figure 1 shows a few iterations toward the graph of /.

Using infinite series one can find the Riemann

integral of / from 0 to 1. By symmetry, the integral of /

1 1 from 0 to ^ equals that from ^ to 1. In fact for any

a € N, if » 6 N and i < 2n , the Riemann integral over

[(* -1}•2 B;i•2 n) can be found by analyzing what is

included and what is not included under the graph of /.

Of course the detail increases as » increases.

13

Studying the iterations toward the graph of / in

Figure 1 might lead to the conclusion that the value of /

at any dyadic rational is a finite sum. This is true since

for any a € IN, if i is odd and less than or equal to 2n,

= 2jmQ 2'^(i'2j~n)

Suppose is the refinement sequence of

Example 2.6. By putting function values of / and Riemann

integrals of f over elements of U00 J ) together in a chart n=l n

like Figure 2, where denominators are chosen to insure a

somewhat "geometric" progression from one level to the

next, the following relation becomes evident.

2.8.1 Claim Suppose n € u, i 6 N and i < 2tt. If / denotes

Riemann integration and I(a,i) = [(i-l)•2~n;i•2~n ), then

proof1: From what was shown above, /{(2t-l)»2 ) =

2~*V( {2»*-1) •2k~tt~1) . By definition,

h ( n , i ) f = ^*=1 h ( n , i ) 2

If A € and x € [0;1], let f k ( x ) - 2~k^{2kx).

As illustrated by Figure 3, if i > #, traces the

tops of 2 triangles over I ( n , i ) , where each triangle has

dx

14

as its base /(»,») and has area -|'2 = 4~^ 1

k-n -k-1

Thus

*?>m ^ = 2 "'4

k=n nk-n . -k+n 2 • 4

« 4 -tt-1

•^-0 > -k

= 2*4 -fl-1

1 - r 4 •

L

Now suppose k < n. 3 ; € II such that j < 2 and

I(n,i) C I { k , j ) . The midpoint, /» = (2;-1)•2~*_1, of I { k , j )

is such that at < (i-l)•2~,, or m > i ' - 2~ n .

CASE 1: a > I•2~B. ~n-k

Let ;> = (i -1) -2

-ifc from (;-l)*2 .

» -A

iizil = nti nK -

(J —1). Then (i-1)• 2 ~ is j>«2 units

Since the slope of /, is 1 from (/-1)*2 -k

to there is a rectangle of height p-2-'1 and width 2~n

above I(n , i ) and beneath f ^ . Above this rectangle and

still below /j. is a triangle with base 2~tt and height 2~n.

(See Fig. 3.) The rectangle may be viewed as p non-

overlapping squares of side 2~n. Each of these squares may

be viewed in turn as two non-overlapping triangles, each

with height and base 2~n as shown below.

15

For a vertical line through ( 2 i - l ) • 2 ~ n ~ 1 , each

successive segment of length 2~n~1 between 0 and f^

subtends a triangle of area -|(2~B)2 = So the area

above I(n,i) and below is just

f k ( { 2 i - l ) - 2 ""V

a-«-l •|'4 * - ( ( 2 # - l ) - 2 "

CASE 2: m < (i-1) -2'

This case is the same as Case 1 except for the fact L

that the slope of f k from a to /•2 is -1. A similar

argument, however, leads to the fact that the area above

/(»,») and below f^ is still 2~B/^((2»-1)•2~fl_1).

Thus

h ( n , i ) f = ^=0 h { B , i ) f k

[S*=0 h ( n , i ) + ^k=n h { n , i ) H

= [ C J 2-%((2 t-l).2-»-1)] + |.4-»

- 2"» [sjl* f { 21 1) •2_B_1)J + ( 2~8 • 2~R~^ )

- 2~^(2*'B~1.(2l-l))]

+ 2~®(2_0^(2B~rt~1-(2«-l)))

= 2~8[SLo ~k4(2k~tt~1'(2i-l))^

= 2~®/((21-1)•2-""1) . ox

Since a function which is additive on U00 JJ> »=0 n

determines an additive set function on 7( ) , if at € IN,

16

( V l - l C '* € N a n d '* < 2 *- and i

where the sets I{a^,i^) are pairwise disjoint,

h i ' *!-l //(»,,.,) ' " *5-1 — 1 , .

Now because for n € IN, i € II and i < 2n,

f ( ( 2 i -D-2" 8" 1) = £y = 0 2~-^((2i'-l) •2;~S~1) ,

the integral above may always be found in a finite number

of steps. m

CHAPTER 3

EXAMPLES

In this chapter, given a p in ba(R)(7), various

functions v will be constructed under certain circumstances

such that v £ A . A few fairly general construction

schemes will be given along with examples where they work

and where they do not work. No matter what elements of

&«(R)(J) are considered, Definition 1.12 has the effect of

converting them into elements of 6a{R)(J)+. Thus to make

life simpler, from now on ft and v will be assumed to be in

A«(R)(J)+.

3.1 THEOREM If there is a i^n) € 7^ such that

I in ji(E ) ~ 0, and there is an x in H such that for any B-to

» € N, x € En, then there is a v € 6e(R) (7*) so that v £ A . ft

proof: For VV € 7, let v{¥) - j j Jf J |

3.1.1 Claim v € ba(R)(7") + .

proof : Suppose £ and / are disjoint sets in 7. Then if

x £ £ and x f£ F, x £ EUF; and so ({£) + ((F) = 0+0 = 0 =

{ (EUF). If x (E £, then i )f / but x € EuF; so that

17

18

+ ((/) • 1+0 = 1 = ((I'll/). If i £ /, the argument is

similar. • 1

3.1.2 Claim v f. A . r

Pr°of2: F o r > °' 3»0

€ W such that » > »Q implies

< i , but since x € E^, v{E n) = l . • o

3.2 COROLLARY If there is an E € 7 such that p (E) = 0,

then there is a v € 6a(R)(J) such that v £ i . ft

proof: For Vs € IN, let En = E. Then (i" ) € 7^ such that ft n

lim p(E ) - o. Furthermore, for Vx € E and V» € N, x € E »-too "

and so Theorem 3.1 applies. •

Unfortunately, having some {E ) € such that alE ) n r n

approaches zero as a increases does not guarantee that all

the sequence elements Eff share an element of U.

3.3 EXAMPLE Let 7^ be the field generated by the

refinement sequence (^s)"_0 w h e r e

T>0 = { [ 0; 2 ) }

\ = {[0;1), El; 2)}

T>2 - {[0;|), [|;1), [ 1; 2) >

T>3 - < C0;-|, , [1.3) c3 ; 1 )^ [ 1 ; 2 ) )

^4 = t 1 ' 2 ^

19

T>a - {[l-2",;l-2"'~1)};!=0 U {[1-2~*;1), [ 1; 2) }

For any n € a and any [«;&) € 7>n, let ;l ([«;&)) » b - a . Then

extend } in the manner of Chapter 2 to an element of

A « ( R ) ( ^ 3 ) + -

3.3.1 filaim There Is not a (/ ) € ft such that

l i a A { E ) « 0, and for some x € U and any n € IN, x € E . »-too "

proof j: Suppose { E n ) € ft such that l i n t \ ( E ) = 0. if B-+OO N

3» € IN such that for V » € IN, Em Is not contained in m

I l - 2 ~ " ; 1), then since [l-2~*+1;l-2~*) is the set with the

smallest A value outside of [1—2—1#;2) which E could nt

contain, for Vm € IN, A { E ) > 2~".# Thus for V» € II, Iff

3/»b 6 IN such that i C [ 1-2-1*; 1). However, there is not n

an element of B in [1-2~B;1) for V« G IN. Hence there is

not an i H such that x 6 E m for V« € II, and for that a

matter, there is not an x € U such that x € E for VB € IN.

D1

To avoid this difficulty, the intersection over all

n € II of the closures of [l-2~",-l) could be considered.

[l;2) meets this intersection though, and assigning

function values would no longer be straightforward. H

20

Rather than trying to handle this problem now, the

survey of construction methods continues. First priority

will be given to general construction techniques. Any

stubborn examples will be dealt with later, if need be.

Another suitably general construction method results

from the following. Suppose ( 1 ) 6 such that n(E ) ® H

approaches zero from above as » increases, and for any

n € II, C En • For any V € T, the following sequence in

10; 1 ] can be constructed, where p ( ) = 0 if n € IN such

that V does not meet E ,

(3.4) >(WU,) -*00

- J T T J n = l

3.5 THEOREM If for any ¥ 6 7, the sequence (3.4)

converges, and

M f n o -v ( S ) = Mm ,» . ,

B-ta)

then v € ba (R) (?)* and v £ A .

proof: Suppose E and F are disjoint sets in 7.

fi (FOE ) v ( E ) + v { F ) =

^ E f ) E n ) ]

I i fit f B x

n~ko + l i " T T T J n-ka

= I im

»->oo

H(E0E ) ft (FOE )

j v r r + - J T I J

p {E0En ) + ji (FOE )

ss I J fa " ' • h . 00 * < V

21

li ((£(]£ )) == I I Iff —— r-p

• HCO '<*.>

= / in, ^

»-»Q0 P 8

= i> (JU/).

Thus v € 6«(IR)(7') + . If a € 12, however, fi(£J\B ) ft(£ )

v(S ) - — 7 7 — 5 — • I i n .n - 1 .

* t-fco " V » - * o " V Hence f j£ /i . •

3.6 EXAMPLE Theorem 3.5 can be used in Example 3.3.

Suppose {E ) € such that A($„) approaches zero from 7! U

above as n increases, and for any 1 € N, f j + 1 C

) 3.6.1 CJlalm For any ¥ € T , lint • ,« " exists.

»->oo * **'

proof : Since lia H£„) - 0, for Va € IN, 3» € IN such that 1 8-*D m

# > implies En C [ 1—2—^; 1). If V contains [l-2~/";l) for

some « € II, then for V» > n , - m

^ (rnt t ) h i „ ) = 1;

so that the limit exists and is one. Suppose that for

V/» € II, ¥ does not contain [ 1 — 2 ; 1) .

If ¥ does not contain a set of the form

[1-2 *;l-2 ! 1), ¥ does not meet any set of the form

[1-2 ,;1). If ¥ does contain a set of the form

[1-2 !;l-2 ! 1), it can contain only finitely many sets of

22

this form. In the latter case, let k be the largest

Integer such that [l-2~*;l-2~*-1) C !, and in the former

case, let k <=> 1. Thus V does not meet [l-2~*-1;l) and hence

for a > »£,

HVM ) # _ o

T T 7 ^ - = T T O = °-

So the limit exists and is zero in this case. Since these

are all the possibilities, the limit always exits. •

x (m£ n) By Theorem 3.5, if for any f € 7, v(Y) = lim , ,,, *

a-to

then v € bu(R)(7 )+ and v £ A . B O ji ^

Theorem 3.5 applies to more complicated examples where

the limits are not always zero or one, such as the

following.

3.7 EXAMPLE Let (2^ )âQ be the refinement sequence from

Example 3.3. For any » € a , let

T>2tt - <[« ;b)x[c-,d) : l * ; b ) , [c;d) G T>n }

Then CP,)fliB.0 is a refinement sequence. Let be the field

generated by F o r a nY » € « and any [a ;b)x[c ;d)

€ let ^2([a;&)x[c;i)) = (b-a)(i-c) and extend X2 in the

usual way to all of J?. Next for any n € IN, let

E n = [l-2~*;l)x[0;l>.

JN l0{V(\En) 3-7.1 Claim { E a n d for any f £ Z i'/» J » *

' 2* »'

exists.

23

proof1: Suppose n € N and {[a .;b . )x[c .;d •)}"=1 is a

pairwise disjoint subcollection of 7^ such that

y - {•[«f- -)x[c .:i •)) .

If for Vi € INb, ai > l, ^ < l, or c > l, then

ill T 7 T T T = °'

»-to 2 a

Suppose 3/ € IN such that b . » l and c • < l. Let

/ • U < C c (.) : »'€INb, and c l>. Let

a = aax(a^ : Finally, suppose /» € IN such that

a > a implies 2~a < l-a. For a > « then,

_ /I (/fl [ 0; 1)) V V " nroTTTT™"

Hence, lim . . ,, = J{/fl[0;l)). . •-to 2 a ' 1

The major problem with this current construction

method is that, given the wrong type of sequence from 7,

the limit of (3.4) may not exist at all as shown below.

3.8 EXAMPLE Consider the field 7Q generated by the O

refinement sequence (^b)^_q where

T>Q = {[ -2; 1)}

T>1 = {[-2;0) , [ 0; 1) }

"2 - U-2,-|), E-|;0) , [0;|),

24

*3 * r r n 1 4 r 1 1 | {[~2,-2)» i-2; 4)» [-}:<)).

-A. ) 2' 4' '

r-i.-i, L 4, 8J » [-§;0).

rl.15) 4' 16 '

.15 63. 16' 64

1>tt - { [-2 • 4_Ar + 1; -2 • 4~A ) , [l-4~*+1;l-4~*)}JlJ

U {[-2-4 *;0), [l-4_w;1)} -n

For [«;4) € 7^, let J([«;6)) = b-a, and extend i to all of

/ 0 in the usual manner. Suppose {E ) is the given

sequence, where for any a € u, E2jj = [-2•4~a;0)U[1-4~B;1)

and S2n+1 " t-|-4~B;0)U[1-4"B;1); so that = 3-4~B

and J(^2B+I^ 2* 4

2 A

Thus i i/r /I [E ) = 0 and for any n-ko

n € II, E^ (. En_^. For a fairly ordinary element of Jg

though, the sequence (3.4) does not converge. The sequence

for the element M - [—2;0) is constructed below.

0

1

2

3

E

[~2;1)

[ - | ; 0 )

MOE

[-~;0)U[-|; 1)

E-§;0)U[f;l)

[-2

E-i

0)

0)

0)

0)

WOE )

~ W ~ 2 3 1. 3 2 3 1. 3

25

2» [-2-4~a ;0)U[ 1-4 \ l ) [-2-4~S;0) f W

2»+l [-|-4"8;0)U[1-4~®;1) [-|. 4 - » ; 0 ) A

This problem can be circumvented through the use of

more powerful methods. Given a sequence of nested sets in

?, the Hahn-Banach theorem may be used to construct a non-

absolutely continuous set function. Banach limits, which

will now be introduced, provide the foundation for such set

functions. The development in 3.9 through 3.11 follows

that of Bhaskara Rao and Bhaskara Rao [5, pp.39-40].

Let / be the space of all bounded sequences of real

numbers. For * - € l^, let ||x|| -

3.9 THEOREM ( * j | * ) is a Banach space.

proof: Clearly, is a norm and I is linear.

Suppose is a Cauchy sequence

* n oo" F o r ^ a ^ ^ - 1 * s a C a u c^Y sequence in R and

thus converges to some SQR. Let xQ - {*0j> )®—1 and suppose

c > 0. Since is Cauchy, 3 ; € IN such that I > a > j

implies |^—||QO ."xOnl * I1).-1*.! + K.-'oal * + 3 < ¥ "

Hence c > (|«y,-*0, I )„i = H V ' o l L T l m s *,"*0 € V

xq = xj~ixj~xQ)

€ ^ and converges to i Q. •

NOW let I = [s;=1 j,.]^ € I J . If

y - (?„)„-] € Z and A > 1, then

1**1 - I f t - i ».•] - [ ^ i » , - ] |

I _i_ IY1^— 1 , y • + L • . y • 1=1 9 I I i—1 9t

< 2 fc-i ' J

00 »=1 00

Thus y € / . In fact, L is a linear subspace of tm. oo

Next, for any x = U B)® = 1 € let p ( x ) = sup {x n

Then for any x,y € I a n d c > 0, p { c x ) = c p ( x ) and

p{x) + p(y) > p(x+y).

3.10 LEMMA For any y € Z, p ( y ) > 0.

proof: Suppose 3y € L, such that p ( y ) = s u p { y # } j = b < 0,

Then E * ^ y- < nb for V» € IN. Thus is

unbounded. # •

For any y £ L, let rQ(y) = 0. Then r Q is a real

linear functional on £ satisfying TQ[y) < /?(!/) for any

y € L. By the Hahn-Banach Theorem, there is a real linear

27

functional r on / such that r(y) = Tn{y) if y 6 L, and for oo

any x € / , T(X) < p ( x ) . The functional r is called a

Banach limit.

3.11 THEOREM The Banach limit r produced above has the

following properties.

i) r{cx+dy) - CT(X) + Ar(y) for any x,y € / and

any c , d € IR.

ii) If x = {x . 6 I such that for any n € IN, x > 0, iff U JL 00 rf

then r(x) > 0.

iii) For I = (1,1,1,...), r(!) = 1.

iv) For any ( x , ) ^ E r ((*„>?-! > = r ' <x„+l' »>1>

00 v) For any x € / , |r(x)| < |jx|J

vi) If {x )00 « € / , lia inf x < Tlx) < lia sup x . 8 « = 1 oo „ , n ~ ~ ^ _ »

» oo u -* oo

vii) If jp = * s a c o n v e r 9 e n t sequence of real

numbers, then T(X) = lia x . n-to

proof: i) This is clear from the linearity of r.

ii) ~r(x) = r(-r) < p(-x) = sup{-xn}®=1 < 0,

since for V« € IN, x > 0. Consequently, T(X) > 0.

iii) -r(I) = r((-1,-1,-1,...)) < ?((-1,-1,-1,...)) = -1,

and so r(T) > 1. But since r(l) < p[T) «= 1, r(l) = 1.

lv> <*, + lC-l - - <*. + !-*. C-l- P O F V" €

l£°=i = I'.+i-'il * 2ll ( I.C=iL-

28

Thus <•<<*,+!>"_!> " '((I„)®=1) = f<<», + jC-l * <*,C-1>

" " . • r V ? - ! 1 = °- 8 i n c e < * . + r r » C - i 6

v) Suppose i -

r(i) < ,.<*> = »»><*,C.j < « ? < K l C - l " ||*||

and

,00 • * # _ % 00

00

-r(x) = r(-x) {-jr>}}jja!l -

Hence, »B/{JP >® = 1 < r(x).

-||iL - | C . j s •'»/«*,C-is r ( I ) s I N I

Thus, )*"{*) | < H Hoo*

vi) From iv) follows the fact that for VA € IN,

r((x#+A>«= L> * r ^ x » C - l * ' ProvlnSf v) gave the fact that

(or V* 6 N. <»/U, + iC-J - r ( < I»+*C=l ) - " ' ^ . t i C - r

Hence

l i a inf - I jw f| -4 00 fl-*G0

< f ( x )

< l i a (sup { x k ) ™ = n ) B-»00

* 2 Iff S U p X . # -» 00

vii) This holds since liar x exists if and only if a-to

l i a s u p x = l i a inf x . • B -» 00 FL "» 00

3.12 THEOREM Suppose (*_)® « '€ such that lia • 0 B B-*OO

29

and for any » € N, i"„ t, C £ Then there is a j* + l n

v € ba(R)(J)+ such that v A

proof: If = 0 for some n € N, the theorem is true by

Corollary 3.2. So suppose /M^„) £ 0 for Vu € II. If F 6 7

and n € IN, 0 ^ p (mi.)

) < 1 and *(P) =

7 T 7 J

00

B =1 € I .

00

Thus for VF € /, let v(¥) =

3.12.1 Claim v € fc«(R){7") + .

proof1:

0 < is/-)

00

n = l < 7" (^ ( K) ) < SUJ>

f> ( ) r r

00

8=1 < 1.

Hence v is non-negative. Suppose ¥ and V are disjoint

elements of 7. For V» € IN, the proof of Theorem 3.5 shows

p(Vf)£) ft (( VUV)f\E )

H ' t ) * *(*,) )

Consequently, <r(¥) + e (If) = <r (fUiS ) and v{Y) + v {(f) =

t {*{¥)) + r{t{V)) - r(9(V) + f i l l ) ) - r(<r(W)) »

D1

3.12.2 Claim p jf .

proof2: W > 0 , 3A € IN such that < d, but for Va > k,

= l.

Thus

V(J.) = T(HSm))

= r fi (£js(\En)

T T I 7 T "

00

1 = 1

30

00

»=1

r ( 1 )

l. n 2 •

To prove a slightly stronger version of Theorem 3.12,

the concept of a maximal filter is needed. Statements of

3.13 through 3.16 can be found in Bhaskara Rao and Bhaskara

Rao [5, pp.10-14 & 38].

3.13 DEFINITION A subcollection J of 7 is a filter in 7 if

i) ii,S € J implies € J, and

ii) A G J, B € 7 and A C B implies B € J.

A filter J in 7 is maximal if no filter in 7 properly

contains J.

3.14 LEMMA A filter J in 7 is a maximal filter in 7 if and

only if for any A € 7, either A or U\A is an element of J.

proof: Suppose A € 7 and neither A n o r d \ A is in J. If

3B 6 J such that A and B are disjoint, then B C U\A and

U\A € J.# Thus for VB € J, A meets B and one can consider

JQ = {CS.7-. 3B€J such that AHB C C}.

3.14.1 Claim JQ is a filter in 7.

proof^ Suppose G ,9 G Then 3 E ,F € such that AC\B

C C and AClF C D. Since EOF € J and is contained in

31

both C and 5, C(\B € J . Also if C G JQ and § € 7 so that

C C S, 3£ £ J such that BOA C C C S, and thus 6 G JQ.

Note that if B € J , BOA C B and BOA C A, which implies

A and B are elements of JQ. Hence J is properly contained

in JQ and J is not maximal.

For the other direction, suppose that if A £.7, A or

lf\A is in J . Suppose also that J~ properly contains J .

Then 3B G 7 such that B € J" and B f . J . Since B £ J ,

ff\B € J , and thus U\B € J". However, according to the

definition, B and ff\B cannot both be in the same filter. •

3.15 LEMMA Every filter in 7 is contained in a maximal

filter in 7.

proof: Let be the collection of all filters containing

the filter J in 7. For V ,J" € $, let J' < J" if

J% C J". Let k e a chain in <8, i.e. for V a,f} € T ,

J. < J, or < J,.

3.15.1 Claim is a filter in 7.

proof^ Suppose A ,B G • Then 3 u.fi G T such that A

G J9 and B G Since or < J 9 , A and 1 are

both in Ja or both in Hence AOB is in Ja or and

thus in . Suppose A G Vygpfy an<* B £ 7 such that

A C B. Since A G J for some ar G T, B G also and hence

9 6 urGr«V Di

32

Since for Var € T, c Ur€T^r' U?€r^7 is an upper

bound in ® to the chain Since is a n

arbitrary chain in 0, every chain in 0 has an upper bound

in <8. By 2!orn's Lemma then, <8 has a maximal element, say

J ' . Clearly, J ~ is a maximal filter containing J . •

3.16 LEMMA Suppose J is a subcollection of 7 , and for any

' 6 7- l e t " {J if I f J.

Then J is a maximal filter if and only if ( € b a { R ) { ? " ) + .

proof: Suppose J is a maximal filter. Suppose also that £

and / are disjoint members of 7 . If £ € J then F f . J , and

if / € J then £ £ J . Thus if one of £ and F is in J , then

only one is in J and i ( £ ) + ( ( F ) - 1 = £ ( £ U F ) . Suppose

neither £ nor F is in J . Then ff\£, ff\F and, consequently,

( f f \ £ ) ( \ { f f \ F ) = l f \ ( £ U F ) are in J . Thus £iiF £ / and

({£) + £(/) = 0+0 = 0 = £{£U/). Hence £ €

Next suppose £ € b a ( R ) { / ) + . Suppose A , B € J and

AOS £ J . If A does not meet 3 , A) + ( ( B ) = 2 $ ( A U B ) .

Suppose A meets B . Then A f \ B € 7 . If B C A then AMI -

B € J , and if A C B then ARB = A G J . Thus A is not

contained in B and vice versa. Hence B \ A and A \ B are both

in 7 . Since A = (irffl )U(i\ ) and AC\B does not meet A \ B ,

+ f(AJ) = { ( A ) and ((A\B) = 1. Similarly,

( ( B \ A ) = 1. Thus { ( A \ B ) + £ ( B \ A ) - 2 * ( ( ( A \ B ) U ( B \ A ) ) .

Hence AOS € J .

33

Now suppose A € J , B G 7 and A C B . If B = A, B G J .

Suppose B £ A. Then B\A € 7 and £{B) = £ (4) + ((B\A).

Hence ((i) = 1 which implies B G J and so / is a filter.

34 € J and i C f, Thus H G J . For V4 € 7 such that

B $ V, BU(ff\B) = f} and B does not meet U\B, which gives

£{B) + ({(t\B) = ({II) = 1. Hence one and only one of B and

l?\B is in J , and / is a maximal filter. •

3.17 THEOREM Suppose (E ) € ^ such that lim ) = 0 and a-to

if n G IN, E C E . Then there is a v G 6«{R)(/')+ such ii + l n that v £ A and the range of v is a subset of {0,1}.

proof: Let J - {A£7: 3»GN such that EaCA).

3.17.1 Claim J is a filter.

proof : Suppose A ,B G J , 3 »,» G N such that E^ C A and

£ C B . For I = max {a ,n), E , C Af)B. Thus AOB € J . Also fit »

if A £ J1 and B € 7 such that A C B , then B € J .

By Lemma 3.15, J is contained in a maximal filter, say

J'. For VJ G 7, let v(E) = 1 if E G and let v(E) = 0 if

E jf . The range of v is a subset of {0,1} by definition

and by Lemma 3.16, v G ba{U)(7)*. Since I in p{E ) = 0 and B - k D

i/(E ) = 1 for V B G N , v £ A . • 1} J*

CHAPTER 4

A CHARACTERIZATION THEOREM

Note that if the conditions for Theorem 3.1 hold (i.e.

there is a { i ) € ft such that I in u(E^) - 0 and for some n-ta)

a

x € ff and any n € II, x € En), and if for any n € IN, / =

them (/ ) € ft, lim ft{F„) • 0 and for any n € N, I *** JL ® » t w

»-*»

Thus the conditions of Theorems 3.12 and 3.17 8 + 1 n

hold. Also, remember that this includes the case where

/tfi1) = 0 for some E € 7.

These facts lead to the following question. Could

there be a v € ba(R)(Jr)+ such that v £ A and there is not

a (E ) € ft such that lim P(.E ) - 0 and for some c > 0 and B—>00

any n € IN, + 1 C E and V(ER) > c. To help better

understand the situation and how a sequence as described

above could possibly arise, the next theorem gives some

specific implications of non-absolute continuity.

4.1 THEOREM Suppose v € iafRH,?")* such that v 4 . There

is a (Er) €: ft satisfying the following three properties.

i) There is a c > 0 such that for any » € IN, v [E ) > c .

ii) I im p{En) = 0. n-+a>

iii) For any n ,a G IN, E meets E . ft at

34

35

proof: The approach will be to construct a sequence

satisfying the first two properties and then this sequence

will be altered so that the result satisfies all three

properties.

Since v fZ A , 3c > 0 such that for each i > 0, 3E € 7 A

so that fi(I) < d and v(E) > c. Thus for Vb G IN, 3E n € 7

such that /M^b) < \ and v {En) > c. Hence (/#) €

satisfying i) and ii).

Before continuing, a convention is needed. For any

E G 7, {E} will be said to contain one pairwise disjoint

member.

4 .1 .1 Claim 3j>Q G IN such that pQ is the maximum number of

pairwise disjoint members in t^j#=i *

proof j: If for € IN, there are p pairwise disjoint

elements of say E , E , ... >E , then ) n 1 2 p i

-Y?-.v{E ) > p c and v is unbounded.# So 3? 1 G IN such i = 1 n i

that there are not y' pairwise disjoint members of •

Let P = {jjGIN: contains p pairwise disjoint members}.

1 G P and elements of P are bounded above by p'. Thus P

has a maximum element, call it pQ, and has at most

pQ pairwise disjoint members,

For Vp G N, finite subsequences from N of length p can

be well-ordered by (» •)'_^ if where k -

p0 : »•#»'•}. Let {»•)„• , be the first subsequence of f i t I I = 1

36

length p such that E , E , ... , E are pairwise 1 2 npQ

disjoint members of { / . . For n > n , E meets E for R »-l pQ A NI

some i € N . If not, there are ?0+l pairwise disjoint P o

members in . In fact, for some i € II and » »=1 pQ

infinitely many n greater than » , E meets E . Let i 1

be the smallest such i, let /, == E and let (E1) be the

1 n ., n

subsequence: of elements of (E^) with subscripts greater

than nj , which meet / . k

Now suppose k € IN such that 1) a finite

subsequence of (^ ), so that for V i ,j € N, , E- meets E u K I j

and 2) if I is the element of IN such that E comes from the

Ith position of (#fl) as a subsequence element, there are

infinitely many n greater than I so that En meets / for k CD

€ II. and {E ) is the subsequence of elements of (E ) K U U W== 1

with subscripts greater than I which meet / for Vat € II. . IK *

At most pQ members of a r e pairwise disjoint. Thus

there is a maximum number, p^, of pairwise disjoint members

of {E^}®_., and there is, in particular, a first finite pk ?k

subsequence s o that the members of a r e

pairwise disjoint.

Again, for some i € II , there are infinitely many pk

k k a > n such that E meets E . Let J" be the smallest

*k n Ri

37

such i, let /. = Ekn and let (Sk+1)<x> , be the « + l n •„ * « l

i

k h subsequence of (t ) consisting of elements E- where i > n •„

tt j J mm | II

k it+i and Ej meets Then (^„)Basl is a finite subsequence of

{En) such that for V i,j € Nj.+1/ meets F-, Also if I is

the element of IN such that comes from the I th position

k+1

of then { ) is the subsequence of {En) consisting

of elements with subscript greater than I which meet /, K+l

Thus by induction, there is a subsequence (/ J00 , of It Jlsl

(E ) such that for V » , » 6 N, / meets / . Since I in a{F ) a » a r n »-too

• 0 and for Vb € N, ^ c, the theorem is proved. •

Returning to the previous question, with all the

interconnectedness of sets which the result of Theorem 4.1

establishes, could there be an element v of ba(R){7)+ which

is not in A^ without there also being a nested sequence in

7 whose p values converge to zero and whose v values are

bounded away from zero. The next example shows how to set

up a field 7 and a p in ft«(R)(J)+ so that there are no

nested sequences of sets whose p values converge to zero,

and yet there is a (Fn) € satisfying the following.

i) U00 J t V. »=1 a r

ii) For any n,nt € N, F meets F n nt

(4.2) iii) If k ,1 ,is, a € IN such that k $ I, at $ n and {k, 1}

£ {»,»}, then F.OFj does not meet F OF . K l an

iv) lint p{F ) = 0, u—>00 "

38

Having such a sequence around would not help in denying the

existence of a v in ba{R)(7*)+ which is not in A^ .

4.3 EXAMPLE Let U3 = N2 and let (J,)™,! be the sequence of

subsets of U- such that for any n € IN, V

En • U <(#/*)

Thus,

Ex • {(1,1), (1,2), (1,3), (1,4), ... , (1,»), ... }

E 2 = {(1,2), (2,2), (2,3), (2,4), ... , (2,»), ... }

E 3 = {(1,3), (2,3), (3,3), (3,4), ... , (3,»), ••• )

EA = {(1,4), (2,4), (3,4), (4,4), ... , (4,8), ... } 4

4.3.1 Claim 4.2 i), ii), and iii) hold for (*,),«!•

proof1: i) This is clear.

ii) If m,n € N such that m < n, then (/»,») is an

element of both E and E^.

iii) First suppose that a < n. Then EÔE^ =

<<<*.•)>!!.! u <(».*)>£=„+1) " u <«.*>>J.,+1).

Suppose (j,i) € S (\£ . It j < a, I must be a for ( j , l ) to /If 8

be in Eand I must be » for ( j , l ) to be in If / >

(j,i) is not in / ; so j = n. If I < », (j',1) is not in

£ . If I > n, j must be » for { j , l ) to be in E .# Thus ' n "

2 = n and J" fljlF = {(/»,«)). " I#

Now suppose k ,1 ,nt ,n € I N such that k ^ I , n ^ n and

{ k , l ) f {»,»}. Without loss of generality, suppose that

39

Jfc < I and m < n. Then and " {{»,*))•

Since {k,l} f {/*,»}, k $ a or I t n. Thus { k , l ) i- (/»,»)

and j does not meet ^ n • D1

Let Vn = { I f } , D* - {*,» A * . ) and f o r Vn € II such that V / 1 * •

» > 1, let

» . - U W = l V u

4.3.2 Claim For any » € IN,

and f o r any € llfl ,

v u " - i ' / - < v u " = i v u

j$k ;#k

proof : Clearly, the left-hand side contains the right hand

side in the first equation. If x € x is in ^„ + 1

or it is not. Thus the first equation holds.

The left hand side of the second equation contains the

right hand side since ^£^#+1 (*oes n o t m e e * ^ j * o r

j € N#. If x € / j t X U y . j ' y x l s e i t h e r i n £tt+1

o r At i s

not. So the second equation holds. n 2

Note that the elements of are pairwise disjoint for

any » € II. Thus for any a € IN, Dn is a refinement of T>n_1>

and (2>a)®_0 *s a refinement sequence. Let 7^ be the field

generated by {7>n)<^_0.

Next let /i(^) = 1, 81 = "§•'

= i, - |. = 5 and Mf\(/jUf 2)) - f- I'

n € IN s u c h t h a t

40

a) - 1 - 4"*.

b) for any * « «„. P j' = 2~*~"+2~2*, and

W .

c) for any i,; € WB such that k ± j , niEÔSj) - 2 * ,

then let

a ' ) > « < f \ l O t > " 1 " 4"*.

b1) " 2~*~<"+1)+2~2* for any i € « > + 1,

J #

and

c') p(EkC\Ej) - 2- i " ( 8 + 1 ) for any A € .

Having done this, for any a € N,

+ / > < w u I U v

- (1 - 2-E?!J 4"*) + 2-(»+l)-(»+l)+2-2(»

+1)

- (1 - 2-Sj:j «"*) + 2-4-'»+1)

- i - a-J?., 4"*

= /* = '

and for any A € N

+ " W i '

. (2-*-(»+D+2-2i, + 2-t-(»+i)

= 2-*"»+2"2*

>#*

Extending /t in the usual manner, then, gives an element of

ba(R)(T„)+. Also, for any n G II such that a > 1, w

41

and

' < V u * - i V u '"'-i' 1/"." 6 7

f , </s) - M * ;n V

- 2-»-» + 2-2" + mj/I*,)

= 2.2'2" + 2" ," 1(EJ^ 2-')

« 2" 2 , + 1 + 2""-' j - 2 - «

+ 1

1 - 2 " 1

= J " 2 ' " • 2~" (l-2-#+1}

= 2"»

Thus I i n p ( E ) = 0 and satisfies 4.2 iv) also. n-kn

4.3.3 Claim There is not a (/#) € ^ such that

I i n p { F ) = 0 and for any « G N, /R + 1 C /fl. »-to

proof3: Suppose { F r ) € 7 ^ . For V» € N , / # is the finite

union of sets of the form EÔE^, o r ^ o r

j f k

some k , l E N, with the restriction that k I . Since

w

I i n /i(£,\U* 1 •) = 1»« (2 m - k n j ^ - k n ~ * x >

- k - n + 2 ~ 2 k ^ _ 2 ~ 2 k

E^\\}^_Ê • contains no nested sequence in 7^ whose /« values

J j ± k

converge to zero. Finally, only contains sets of

the form ff\Uor ^/+i^j03/j' s * n c e

42

I in ft {fJ\â-.E;) = 4 "^) 53 1 - 2 (•§) = "I' flf-too 3 3 ff-to

A U * .E • contains no nested sequence in 7_ whose p values ;=i ; 3

converge to zero. Because the elements of (/#) must be

composed of elements in one of the three forms listed

above, there is no nested sequence in whose ji values

converge to zero,

Hence, if there were a way to assign set function

values to 73 such that for some v 6 6«(R)(J3)+ and some

c > 0, viE ) > c for any n € IN, then there would not be a H ~

nested sequence in whose (i values converge to zero and

whose v values are always at least c . •

Since the methods of constructing non-absolutely

continuous set functions given so far depend on the

existence of a nested set sequence whose /« values converge

to zero and whose v values are bounded away from zero, it

is fortunate that there is no way of assigning set function

values to produce the situation above. This will be

demonstrated through the use of the conventions and lemma

below.

From now on, for any ( € b&(R)(J), let £(0) = 0.

Also, for any {/fl) G ft, let /* • f 1 and for any n € IN such

that F* is defined for j € , let ? n + 1 - *

43

4.4 LEMMA Suppose (/#) € and £ € ba{R)(7) + .

i) For any n G IN, /# = 0 or F n € J.

ii) For any a € IN, Uy^/y = Uy=1/y. • •

iii) For any n,f* € N such that n f m , Fn and F are

disjoint.

iv) sup{((UnjsslFj)}^x e IR.

v> - q m l uf*k).

vi) For any c > 0, there is a k € W such that if n > k,

< c-

proof: i), ii) and iii) are clear.

iv) Since ( € &a(IR)(7) + , (UU'.j/y) )®ml is a bounded

non-decreasing sequence in R, and so has a supremum in R.

v) ,)>*_! « e(Uy = 1/y) = li" J) J~A J " a u-ta) J J /i-to J J

- '«•» s°,i ur'j) -Ji-ta J J J

vi) 3Jfc G IN such that a > A implies

c > [ >-1 « ' M - '9 -1 f , / ; '

- |'«« *>.+1 f"';*1

I a-to J J

4.5 EXAMPLE Returning to the sequence (>»*=«! f r o m E x a m P l e

4.3, suppose a ^ in ia(R)(^"3)+ can be found so that for

some i > 0 and any a € N, ) > i. Then there is a k 6 N

44

such that n > k implies S^=# < 1 ' N o W s uPP o s e

n > k + 2 . If 1 < » < j < n, E JÊ • does not meet E^fUF^, and

so i" Hi • C S* and E R E - C g*. Hence, n j j u i i

" < V - "<f*> + »(*, n

- " ( < ) - -(uj.^'.niy)i • "(u^ + 1(/„n/.))

= »</, n uj.^y) + » < 0 + sJiUi " < W

< m i , n u * . ^ . ) + s » = t + 1 ,(f*>

< -(*, n u j . ^ 0 • f

Thus,(f„ n U > , < * „ > - | > | .

Next suppose « and /» are larger than A+l and n £ a .

4.4.1 Claim fl (Uy=1^y) a*"* iffl fl (U*=1*;) are disjoint.

prooflS En fl - Uy-i<',,ni;) a n d n (Ui=lJ,;') =

U* ,(E ()£•). Since n > k, a > k and n t a, for V »,; € N., ;=1K a )' K

{/ir, i} £ {a,/} and thus does not meet

Therefore (J^ fl • ) ^ + 2 i s a s e <3 u e n c e o f Pîrwise

disjoint sets in such that if a > Ar+2,

»<i. n u*.^.) > f.

Hence v is unbounded.# Consequently, no such v exists, and

the prospects look good for non-absolute continuity with

respect to (i implying the existence of a nested sequence in

7 whose ft values converge to zero and whose v values are

bounded away from zero, a

45

Proving the existence of such a sequence requires a

couple lemmas. In the following, suppose (£#) € ft and

( € A«(R){J)+.

4.5 LEMMA For any n € IN such that n > l,

l . - elu

proof: Clearly, E n U (Uy_^ (En?[E •)) C E r . Suppose x € E n .

* * n * If x € E , there is nothing to show. So suppose x ft E .

Ji "

3; < » such that x € E - . Let be the smallest such j .

Then x € E HE . . • " 0

4.6 LEMMA If E € 7 such that for any » € N, E C J1, then

{</) - + £ y = 1 (<**>.

and as a result, ((E) > ^y = i

proof: For V» € N, f (^) * {(AUy.^y) + { {Uy^^y) » a n d s o

£(i") - {(AUy.jJy) - Thus,

^y=i HfJ) -

= * (*) - » » / {€ )

Since t ( E * ) and ((E) are in R, ib/ U (AU'^jfy) }"=1 is

also, and the desired equality holds. •

46

Now the following can be proved.

4.7 THEOREM Suppose ft, v € ba(U)(?)+, (£#) € s and c > 0

are such that lit* ~ 0 a n d f o r a nY tt e N, v ^ c* a-to

Then there is a (/ ) G and d > 0 such that I in p(F) ~ 0

n-ko

and for any a G N, ^ B + 1 C a n d ^'

proof: First note that if 3/ G 7 such that p(P) m 0 and

v{F) # 0, then, letting F « / for Va € IN, (F^) € 7^ such

that I in /*( „) = 0 a n d f o r € N, ^ B + 1 C /r and B-tao

v(Fn) => v(F) > 0 . So suppose that for V/ € 7, /»(/) = 0

implies v(F) - 0. £ C U for Va G N. Thus

0 < v ( ) < v{U) and > 0.

The proof now proceeds by induction. Let / = U and

let »Q ! = «ia{a GN: /* (^B ) < jfi(FQ)}. Then if k G IN such

that »0 has been described, let

B0,* + l = a > » Q ^ and < 2~k~2p(F0)}

Hence (aQ *s a n increasing sequence from IN such that

for Vk € N, ) < 2 * 1/t(FQ). Let ™ 0 f k

(S )? , • Then (En . ) € 7^ such that for VA: G N, a o h *=1 0,A

^ c a n d - 2~k~1P *

3«J G IN such that a > a J implies X^_n jj.) < f •

Let »^ be the smallest such a|. Then for a > a1+2,

47

£" v(E* .)

<T

Hence for V» > n1,

= " k » ny 5 ^ U i b -

Let F » U1/ • = U1/* . Then F € 7, F. C F , * J S5 1 ^

i/ (/ ) > -|c , and

< » < „ / , ) • /i[.u^.y] - J ; ^ ; . y > < £

< £® = 1 Mi 0.y> i 2"J'

< | M V -

L e t " < ' i n f o , v * » L i - T h e n e * a n d

for VA: € N, jS1 ^ C /j, j.) - anc*

w ib 1

"'i.*' S "(i0,.1+k' s /•«'») *

Now suppose I € N such that the following-two

conditions are satisfied.

i) 3(f a finite sequence in 7 such that for

V* € N r Fk C Fk_lt fi{Fk) < 2 ~ * M / q ) and

(4.7.1) v{Fk) > (1 - S* = 1 2

ii) 3(1, .)°? 1 € such that for € N, I , j j-1 " i . k c V * 11 - 2 ' ~ 1 ) c a n a

^ / jfe ^ ^ ^<'o^ # gii

3n|+1 G N such that » > #j+1 implies Sy_B j ) <

_; _ o 2 c. Let be the smallest such aj+i* Then for

48

« > » J + 1+ 2 '

V { S l n U » * " ( U " - , + 1 «

= E" 7=» !x1 + l ^

Hence for V« > n J+l

"Bi + 1 = v

'! +1

^ «-'"2 < 2 c .

• i + i . '

* <> " s ' - i

- (i - S ^ J 2"i" 1)c

-J-l -i-2 )c - 2 c .

"f+1 flJ+l

Let Z1 = U E j (, = U E U j,* Then for V« > » ^ + 1 »

> = 1

»!'/*,) * * ' 1 - s ! = i 2 " ; " 1 ' c

;=i

J+l' £ ^ 1

> (1 - I ^ = 1 2"-;"1)c = |c > 0.

Thus / j + 1 C /j € 7* and

0 < + 1

1+1 * '

U ' l i =1 l , }

ttl+1 £ J=i

< X 7 = i

= r ' - ^ i 2 " - ' = 2 " i _ W 0 ) -

l e t = ' ' i . » ( + 1n / i + i ' * = r

T h e n

( / i + i , k ' 6 ^ a n d f o r v * 6 fi+i,t c '»+i-

* ( 1 - s j = l 2 " J " 1 , C

and

49

"<Ji+1.*> * *<fi,.,+1+*» « 2 ' ° i + 1 1 ' " V

< 2 - l ' k ' 2 H f 0 ) .

Hence conditions (4.7.1) are satisfied for 1+1. So

by induction, such that for Vn G N, ^ B + 1 C #

< 2~ nHV) and > (1 - Hjml 2~j~1)c. Thus

Z»/» /»(/"„) = 0 and for Vn € IN, v ( f ^ ) > • 8-to

An immediate consequence of Theorems 3.17, 4.1 and 4.7

is this.

4.8 THEOREM The following are equivalent.

i) There is a v G ba(R)(7)* such that v £ .

ii) There is a v G ba(R)(?)* , a (^fl) € and a i > 0

such that i / < { ^ n ) = 0 and for any n G IN, B-tco

r , * i c a n d " V * J -

ill) There Is a (f) € 7^ such that lia - 0 and for /J—too

any » € N. #„+1 C I„.

iv) There is a v € S«(R)(7') + such that > )! i and the r

range of v is a subset of {0,1}.

W. Orlicz [8, 122-25] stated the major part of

Theorem 4.8, albeit in a more positive manner, back in

1968.

50

4.9 THEOREM [Orlicz] Suppose /t, v G 6«(R)(/) + . v £ A if

and only if for any (En) € ft with ^ B + 1 C En for any a € N

and It a p(E ) = 0, lint v {E ) = 0. »->oo n a-too

Orlicz's proof of the "if" part is similar to the

proof of Theorem 4.7, but no examples of non-absolutely

continuous set functions are given along with it.

CHAPTER 5

APPLICATIONS

One application of Theorem 4.8 is in another proof for

the theorem of Appling [1, p.94; 2, p.788] below. The

integral signs here denote refinement integrals.

5.1 THEOREM If £,/t € 6a{R)(?") then the following three

statements are equivalent.

i) If a € e x p (IR) { 7 ) (B) + , f y a p « 0 and exists,

then f = 0.

ii) If 7 is a function from 7 into {0,1}, /y7ft = 0 and

Jffli exists, then = 0.

iii) ( € 4^.

proof: i) 4 ii). If 7 is a function from 7 into {0,1},

then 7 € e x p {I?) ( 7 ) (B)+.

ii) 4 iii). Suppose iii) does not hold. Then £ A . r

Thus 3 c > 0 and 3(J1 ) € such that I i n p ( E ) ® 0 and for n R-ko

V» £ K, T , C g and U S ) > c . For W € 7 , let 7(F) = 1 a+1 n n

if for Va € IN, Y meets S , and otherwise, let 7 (K) = 0 . 7

is then a function from 7 into {0,1}.

5.1.1 claim !fjif' = 0.

51

52

proof^: Suppose d > 0. 3A € IN such that a > k implies

p [En) < d. If £k = U, let V = {(/}, and if £k ± U, let

V = {Efc, j.} • Next suppose £

If I € £ and I C H\£k> 7 ( I ) = 0 by definition. Thus

0 - SI€£ r( ( / ) = /*(/) - M'fc) < <*• i a k i a k

ai

Next let 5 = 1 »/{£(/' J}00 and note that b > c. J s » »=1 -

5.1.2 Claim f = b.

proof : Suppose d > 0. 31 € N such that n > I implies

{(*„) < i+i. If = V, let V = {V) and if £ U, let

7?= ff\£^}. Now suppose £ <^T).

Suppose / € £ such that ?(/) =0. 3 j € IN such that £•

does not meet I , In fact, for V» > j , since £ C £ ., £ n j n

does not meet I . Let j j = /»i»{»€IN: b>; and £ does not

meet I}.

If I € £ such that y(I) = 1, let jj = 0. Let at =

max{jj: I€£). 3 / € £ such that 7 {I) = 1. Otherwise n € IN

and no I in £ meets £ , contradicting the fact that U£ = U.

Let £~ = {I€£: ICE^} and let £" = {/€£: 7(/) = 1}. Then

b+d > f(*j) - S J € ^ {(/) > Z I £ € . 7(/)£ { /) « E J e^ 7 (/)£(/)

= ^ ( I ) Î££~ ^ ^ = ^ ^ * , /

Thus 15 - 7(1)^ (I) | < 4 and Jglt - b > 0. D2

Hence ii) does not hold, and ii) implies iii}.

53

lii) 4 i). Suppose i) does not hold. Then

3a € exp (R) (J) (B)+ such that / a/i = 0 and for some b > 0,

= b. Let « » 1 + 5tt;(U{fff/): /€/}).

5.1.3 Claim For Vi > 0 , 3£ € 7 such that £(E) > 4 — and O U

p[£) < d.

proof3: Suppose d > 0. 3 s u c h that if 8 and

a is an interpolating function of a on 8, then

S/€£ 4<i")/»U> < 3(f(J) + 1)'

Also 3Z>2 <j {U} such that if 8 an(* a *s an

interpolating function of a on 8, then

and so

|(S r e £ «(/)((/)) - i| < |.

r < Eief • ( D U D < £

Now let D be the intersection refinement of and D2<

Suppose 8 <j D. If I £8 such that inf(et{I)) <

3({(<0 + 1)' l B t ® ( I ) 6 * ( I ) S ° t h a t " ( 1 ) < 3«(*> * 1)'

For any other I € 8, just let a(I) be an element of a(I).

Then a is an interpolating function of a on 8, and thus

S/€£ < 3(£(/)b+ i) a n d < ^T£8 ®(I)((I).

If there is not an I € 8 such that

then

.»/(«(/)) > 3(i{Sl} + ir

s / e f «(/){(!) < heswrnrh-T)f(/)

= Z{i(U) + l)'S/€£ ^ { I )

54

siUh + 3(^{(Jb) + 1 } ) .

Then

Ah 3(iit) + l ) > S/€£

* S/€£* 3tf (JJ) + l)f ( / )

= m r r r + - T ) t ( ' *

Hence i > £ ( U £ + ) .

Next suppose 31 € S such that i»/(«(/)) < (lj\ + l)

Let €~ - {/€£: iaf {a ( I ) ) < jy> • A s a result,

S / € f « t 1 < S / € £- 3(f(f' + l)f ( / )

" 3(f (") + i)

= 3(f(#) +

<!•

Therefore,

< S / € £ «(/)£U> < | + £ J €£+ «(/)£(/),

and | « (i") £ (/) •

If for VI € 6, in} {a (I)) > + i) > t h en 8 = £*

and

I < f - < S7€ -

55

In either case then,

b I < S/€£* < S/€ a*(/) = b,S/€^ *(/) =

Hence < ( (U£+) . a 3

Consequently, i £ A and not i) implies not iii). • r

A final application concerns the following question.

If /t is an element of ba{R)(7r), is there an integral on

6a (R) {T) such that the integral of (i is not absolutely

continuous with respect to /»? Stating this question more

precisely requires knowing what properties make something

an integral on &«(!?)(7).

A search through the library produced two sets of

defining properties for an integral. The first set is due

to Lebesgue [7, pp.99-100] and the second set is due to

Daniell [6, p.280].

5.2 DEFINITION The integral has these properties.

i) For any a,b,h € K, f(s)is => f{s-h)is.

ii) For any € R, /J / + /J / + /J / = 0.

iii) For integrable functions / and g,

fb. </+f) " Jh, f * fl S-

iv) If / > 0 and b > a, then f\ f > 0. d

v) /* Ids = 1.

vi) If /^(x) increases toward /(x), the integral of /

tends toward that of /.

56

Before introducing the second set of properties, a

definition is needed.

5.3 DEFINITION A Set L of real valued functions on a set I

is a vector lattice l f f o r a n Y f >9 $ L and any a,b € R,

af + bg € L, ixax{f,g} € L and nin{f ,g) € L.

5.4 DEFINITION Suppose I is an arbitrary set and L is a

vector lattice of real-valued functions on X . A Daniell

integral is a linear functional i on L satisfying the

following.

i) If / € L and for any x € X, f ( x ) > 0, then <(/) > 0.

ii) If (/„) € such that for any x € X , I in f n ( x ) • 0, tt-ko

then I in i(/ ) = 0, B-+00

Now is already a linear space. If nax{(,fi)

is defined to be fnax{(,p} and nin{(,fi) is defined to be

fnin{(,/i}, where / denotes the refinement integral on

6a(R)(7"), then &«(IR)(/) is a vector lattice. If

£ € for /°f = Iy i = > /° is a Daniell

integral on b3,{\R)(7) as is easily checked.

Since Lebesgue's properties do not all hold for the

refinement integral, the Daniell integral properties are

preferred. Unfortunately, 5.2 ii) makes the Lebesgue

57

integral of a function a finitely additive set function and

Daniel1 integrals may not have this property. The

following is an attempt to rectify this situation.

For any £ € ba(R)(7) and any E € 7, let =

f ( ) for any V € 7 where ((JTII) = 0 if V does not meet E;

so that £ |.£ 6 ha{R)(J).

5.5 LEMMA If £ € ba{R)(7) and E,F are disjoint elements of

7, then if E £ 7 and E $ V, then

f I f + ( \ i u - f U - < •

proof: Suppose V € 7. Then

((inn + f(/nf) = f((Mr)U(/nrj) =

Thus if E f U, + £ | = t \ p and if V € 7,

f|,(r> - u w v ) - a n . •

5.6 DEFINITION A Daniell set function integral t is a

Daniell integral on some vector sublattice L of ba{R)(7)

such that if p € L and E € 7, then /»| E € L.

With this definition, if t is a Daniell set function

integral on some vector sublattice L of ba(U)(F) and if

/i £ £, let i~(fi)(E) = M/i j^) for any E € 7. This permits

the statement of the next theorem.

58

5.7 THEOREM If i is a Daniell set function integral on a

sublattice L of 6a{IR)(J) and /t € L, then € ba{R)(7).

proof: For V E J € 7 such that E and / are disjoint,

i~(lt)(E) + i~(p)(F) =

« i [fl\ )

- i~(fi){E[)F).

Now suppose that E, f € 7.

" Igw nax{t,0} + min{(,0}

= (Jaax{(,0})|^(V) + (fai»{t,0))\g(V)

and combined with 5.4 i), this gives

< j i ( ( f n a x { ( ' 0 >) | ) | + .0>)\g) |

= i ({faax{( ,0}) | ji) - 4 ( (J® in it »0)) |^)

< i(fnax{(,0}) - i ( f a i n { { , 0 } ) .

Thus i ~(() is bounded. •

Now the original question may be restated more

precisely. If i is a Daniell set function integral on a

vector sublattice L of ba(R)(7), is there a /J in L such

that t~(p) 0 4 ? The answer is given below.

5.8 THEOREM If i is a Daniell set function integral on a

sublattice L of ba{\R) (7) and ji € L , then t~(/1) € A .

59

proof: Suppose /t € L and *~(/t) £ . Then 3(£n)™ml €

and 3c > 0, such that lia p(E ) = 0 and for V» € IN, a-to

E _ C g and * ~ (/») (J_) > c. Then (/tin )® is a sequence J3 +1 » i9 H

in Z. If « G IN and P € 7, p\£ <K) - /i (* Of) > A(^>+1n'f) "

n

/tj„ (K) and since fn{En(\V) approaches zero as n increases, »+l

I in i [ft |n ) = 0. However, for Va € N, n-too 8

M/» | j> ) - i~(p)(Ett) > c.* • B

There is yet a final question which should be

answered. How often does it happen that I is a sublattice

of &«(R)(J), * is a Daniell integral on L, p € L and E € 7

such that ft J £ L .

Unfortunately, I do not know the answer.

APPENDIX

FIGURES

60

61

X Ln

62

00|h

Pi*

00|U>

M *

00|ui

-P-|U>

00|^ l

OJ Ui rotLn

M M M m M m

M U3|M Ln N> Ui

M 00|M k>KJ

m u>|m vo roKo

M N> U)| hO ro| i—1 M H

MN> (jalro M H M M

M OojM VO M VO

w u M v VO M vo

ro o j | ro M MH-1

M N> (j0 to M ^ M h

0 1

H H U> I-1

M^> M ^

H H OJl M M ^ M*^

M U3|M Ul M Ui

O0|F Ml"

L0| Ui M U l

ro Ml 00 4> 0^4^

oo crioo

f O H M M 00 o o

hO M Mi M oo o d o

ro M Ml M 00 O OMO

hO M Ml M 00 O ONO

00 On 00

N3 00 -F>

LO 00|00

a Ui oo|m

Ui 00| Ln

3 LO OOjOO

00| to -Pi N>

00| to 4>i N5

M m M M 1

M P H M *1 P 5 CD 03 H r t p o o OP CD cr r t p"

OP CD 03 H H H p H CD 03 QTQ CD M M M O ro

CO CD r t O - 03 hh r t P o

p* CD OP o r t CD • M 3 p" OP CD ••d CD Hi • 03

e h H hh p r t CD H* a D p*>d P P r t CD H OP O H * CD r t O hh co H* P H* CD g O H P P p <| CO r t o

fD r t co r t O M H. <! 0 H OJ O CD CD CD P P

CD f t CTQ CD

CD CD 03

O M

rt ^ P* 0

P rt CO

M <j CD CO . jd CD

M r t P" CD

3 M P M H CL £L CL^d ^ O O H M M O P 3 r t • H

CD 03 "d H

H H fD P4 fl> fD CO 05

CD r t cr P O r t r t r t co p* r t CD O ^ s . ? §

3 t r a £ ^ CO §•* o fl> Hi H H

p" r t CD P"

CD r t O

M CO r t _ pr*nd CD

& Q,

H P r t CD CTQ

H 03

< 03 M C CD CO

O H

H P* CD

HI 03 QTQ 03 ?r H*

Hj §

O r t H* O P

63

Figure 3: Exploded View Of Tagaki Function Shaded strip represents

area over typical interval.

/ \ / \ / \ / \ / \ / \ S \ fl

64

Figure 4: The Sets En From Example 4.3

Only the large dots are set elements.

3 •

2 -

msm

BIBLIOGRAPHY

1. William D.L. Appling, Some Integral Characterizations of A b s o l u t e C o n t i n u i t y , Proc. Amer. Math. Soc. 18 (1967), 94-99.

2. William D.L. Appling, Addendum to: Some Integral Characterizations of Absolute Continuity, Proc. Amer. Math. Soc. 24 (1970), 787-93.

3. William D.L. Appling, fields of Sets, Set functions, Set function Integrals, and finite Add i t i vi t y, Internat. J. Math. & Math. Sci. 7 (1984) 209-233.

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11. B.L. Van der Waerden, Sin eifaches B e i s p e i l eintr n i c k t i i f f e r e n z i e r b a r e n stetigen f u n c t i o n , Math. Z. 32 (1930), 474-5.

12. John Von Neuman, f u n d ional O p e r a t o r s , Yol. I: M e a s u r e s and Integrals , Princeton University Press, Princeton, New Jersey, 1950.

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BOUNDED, FINITELY ADDITIVE, BUT NOT ABSOLUTELY CONTINUOUS