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
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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
^A€7 £ ^ ~ ®* c (A)
If c(3) is true for some B € 7, then for
It - [A£7: c(A) is true},
^A€7 m ^A€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 ) = a^IN, 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 > ȣ,
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^ )^aQ 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)» [-}:<)).
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 J^E • 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 P^irwise
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 y : l | + 1 + l </ * . . n / I . ; > » * " ( 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 ) ^I££~ ^ ^ = ^ ^ * , /
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 + < I
contradicting the fact that / = b. Let
t = {!££: i n f ( a ( I ) ) > 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.
4. Wayne C. Bell, Characterizations of froperties of Spaces of finitely Additive Set functions in Terns of Mappings and Integrals, dissertation, North Texas State University, Denton, TX, 1975.
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