Generalizations of the Vitali-Hahn-Saks
and Nikodym Theorems
By
Robert Stanley Jewett
A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA IN PARTIAL
FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
1971
ACKNOWLEDGMENTS
The author would like to thank his advisor Dr. James K. Brooks
and the members of his supervisory committee, Dr. David Drake, Dr.
Gene Hemp, Dr. Steve Saxon, and Dr. Zoran Pop-Stojanovic for their
help in writing this dissertation.
TABLE OF CONTENTS
Abstract ----------------------------------------------------------iv
Introduction -----------------------------------------------------
Chapter 1: Unconditional Convergence ------------------------------6
Chapter 2: Strongly Bounded Set Functions -------------------------1
Chapter 3: Extending the Schur Theorem ----------------------------20
Chapter b: Extensions of the Vitali-Hahn-Saks and
Nikodym Theorems ---------------------------------------35
Chapter 5: A Counterexample ---------------------------------------56
References --------------------------------------------------------60
Biographical Sketch ------------------------------------------ 62
iii
Abstract of Dissertation Presented to the
Graduate Council of the University of Florida in Partial Fulfillment
of the Requirements for the Degree of Doctor of Philosophy
GENERALIZATIONS OF THE VITALI-HAHN-SAKS
AND NIKODYM THEOREMS
By
Robert Stanley Jewett
August, 1971
Chairman: Dr. James K. Brooks
Major Department: Mathematics
The Vitali-Hahn-Saks and Nikodym theorems are two very closely
related results concerning sequences of countably additive, real
valued measures. The purpose of this dissertation is to improve
on these theorems both in the statements and the proofs. That is,
stronger theorems will be shown to be true, and the proofs will
be easier that those usually given. The author and J. K. Brooks
extend the Vitali-Hahn-Saks and Nikodym theorems to the finitely
additive vector case.
In 1969 J. K. Brooks used a result of Schur concerning uniform
convergence of a double sequence of real numbers to derive simple,
direct proofs of the Vitali-Hahn-Saks and Nikodym theorems. In
Chapter 3 of this dissertation the above mentioned theorem of Schur
is extended to the case where the double sequence is contained in
a Banach space, and then this new result is used to obtain a short
proof of the Nikodym theorem with the measures taking their values
iv
in a Banach space.
However, the Vitali-Hahn-Saks and Nikodym theorems do not
generalize directly to the case where the measures are finitely
additive, bounded, and Banach valued, and the additional hypothesis
of "strongly bounded" must be assumed. With this assumption, the
theorems may be extended, and the proofs follow the same general
line as those of Brooks'for the countably additive, scalar case.
Instead of the Schur theorem, a generalized form of a result of
Phillips is used, and the Nikodym theorem follows, but the proof
of the Vitali-Hahn-Saks theorem requires a difficult construction
with measurable sets. At the end of Chapter 4 a discussion is given
to indicate how the theorems may be extended even further to the
case where the measures take values in a locally convex linear topo-
logical space instead of a Banach space.
INTRODUCTION
This dissertation concerns the Vitali-Hahn-Saks and Nikodym
theorems (7). In 1907 Vitali (21) proved the following theorem: if
f: [o, I R are Lebesgue integrable functions converging almost
everywhere to then ., j f KcS and Jc exist and are
o 0o
equal if and only if the indefinite integrals of the fn are uni-
formly absolutely continuous with respect to Lebesgue measure.
Hahn (10) then proved in 1922 that if -: [0,1]-) f are Lebesgue
integrable functions and if ~l, fd t exists for every measur-
able set E then the indefinite integral of the fh are uniformly
absolutely continuous with respect to Lebesgue measure, and they
converge to a set function which is also absolutely contirious with
respect to Lebesgue measure. In 1933 Nikodym (12) generalized these
two results when he proved that if (S, ,/ ) is a measure space
and if n is a sequence of /u -continuous measures such that
ALM, X(E) exists for all E in 1- then the An are uni-
formly absolutely continuous with respect to / A little later
Saks (19) gave another proof of Nikodym's theorem, and it became
known as the Vitali-Hahn-Saks theorem. Using the countable addi-
tivity of the control measure, Saks defined a complete metric on the
measurable sets, and then applied the Baire Category theorem. If
the control measure had been only finitely additive, the metric
would not necessarily have been complete, and this technique could
not have been used.
Nikodym (12) proved tht if Un is a sequence of measures
defined on a 0--algebra such that /uE) : ,, (E) exists for
all E then /u is countably additive and the countable additivity
of the /u~ is uniform. This is known as the Nikodym theorem, and in
most books it is proved as a consequence of the Vitali-Hahn-Saks
theorem. Rickart (18) and Phillips (16) extended these theorems to
the case where the measures are Banach space valued. The proofs of
these results rely heavily on the fact that the set functions in
question are countably additive, and there seems to be no way to
generalize their techniques in order to prove Vitali-Hahn-Saks and
Nikodym theorems for finitely additive, vector valued set functions.
However, in 1969 Brooks (3) gave a short proof of the Nikodym
theorem and then used that result to prove the Vitali-Hahn-Saks
theorem. His proof used a difficult construction with measurable
sets along with a theorem of Schur (20) concerning the equivalence
of weak and strong convergence in Il In the same paper he gave
a short proof that the truth of the two theorems in the scalar case
would imply their truth in the vector case.
An extension of the Vitali-Hahn-Saks theorem to the finitely
additive, scalar case has been proved by Ando (1), but his theorem
was not extended to the case where the measures are vector valued.
In 1970 Brooks and the author proved Vitali-Hahn-Saks ana Aikodym
theorems for finitely additive, strongly bounded, vector valued
set functions, where strongly bounded is defined as in (17). The
strongly bounded hypothesis is especially needed in the Nikodym
theorem, because the statement would not even make sense if the
measures were only bounded; the strongly bounded hypothesis may be
dropped in the Vitali-Hahn-Saks theorem if the control measure is
assumed to be finite. The proofs of the theorems are built around
a generalization of a theorem of Phillips (15) and an extension of
a technique of Darst (6).
The first two chapters provide the background material for
the results proved later in the dissertation. However, Theorems
1.3 and 1.h are stated only for completeness, and are not used in
subsequent proofs. Although most of the theorems in these capters
may be found in Hilderbrandt (11) or Rickart (17), new proof tre
given by the author.
The purpose of Chapter 3 is to generalize a theorem of Schur
(20) and to use the new result to construct a direct proof of the
vector case of the Nikodym theorem. Since all of the measures are
assumed to be countably additive, the theorems rely only on the
results of Chapter 1, and not of Chapter 2. Chapters 3 and L, re
independent, and no theorem of Chapter 3 is used in a proof in
Chapter 4.
The most important theorems are proved in Chapter 4. Both the
Vitali-Hahn-Saks and Nikodym theorems are extended to the finitely
additive case, and it is shown how the newer version of the Nikodym
theorem implies the usual version when the measures are countably
additive. The concept of "strongly bounded" is of utmost importance
here, but Corollary 4.5 and the remark following the statement of
Theorem h.7 indicate when the condition may be dropped.
CHAPTER I
UNCONDITIONAL CONVERGENCE
Throughout this dissertation the following notation will
be used.
7 is a O'-algebra of subsets of a set 5 (l) is the
power set of the natural numbers and E and A are generic
notations for sets belonging to t and (M) respectively. R
denotes.the real numbers. X is a Banach space over the real
or complex numbers, and I)( is the norm of E 6 3 is the
conjugate space of with the norm of an element in defined
in the usual way.
Let n,n E 7 with n Define (71,>]=- ,7+I ...j7 ,
f7,oo)r ny,7+I,... j
If 4 is a real valued measure defined on a T -algebra 2 ,
then ful is the total variation of / which is defined in the
standard way.
In this chapter we define the term "unconditional convergence
and establish other conditions equivalent to it.
Definition: Let X K ,n= f,,, If t: 7r -
is a one-to-one, onto mapping, then XT(n) inI is called a
rearrangement of Xr 0 The series tn is said to be
unconditionally convergent if the sum of every rearrangement
of its terms converges.
Remark: A series that converges absolutely will converge
unconditionally, but the converse is not necessarily true. In
fact, Dvoretzky and Rogers (8) proved in 1950 that the two con-
ditions are equivalent if and only if the Banach space in question
is finite dimensional. The difference between absolute and uncon-
ditional convergence was further demonstrated by Brouxs (L) in
1969 when he showed that weakly and strongly integrable functions
with range in a Banach space correspond to unconditionally and
absolutely convergent series in the space.
00
Theorem 1.1: Let X) ,f :n 1,X,' The series Z X
,tI
II
converges unconditionally if and only if every subsequential
00
sum converges, that is, if L1 7 is a subsequence, then .-X
converges.
Proof: Let 1 XKV be a subsequence and assume the sum does
not converge. Then there exist E>o and sequences fPiI and jVi
where for each i P V
K -l = {(- X :'K:Pi', j'"7 Then form the follow-
ing rearrangement of X.I : Xn ... X7 V ,iXP ).
)()I I ) nz P^ 3 X 3 ) V3 y*tp h - jnv',
The partial sums of this sequence are not Cauchy, so the sum
00
does not exist; hence ZXn does not converge unconditionally.
n=i
Conversely, assume 19K is a rearrangement of ltI and that
the sum of {Kj does not exist. Write p 4 yv if 9p precedes
y when considered in the ordering of the sequence Xn .
There exist 6>0 and sequences -m;) and I(n where for all i ,
)< n% and /6K >) Let P, be such that for all
KrL,, h,] and all t> P , K 4 Rearrange
IYj: KI= -,) *,71 so that the elements appear in the same order
as they appear in PXnj and call this rearrangement
X5n, ,.X I Let MA >P, Again rearrange
i ( i: K ^ '"=- 'l appropriately, and label this arrangement
YlX i", X. Since 9t I5 if r ,,n,] ,
S 1 'l -n we have that fX i,", Xn~ ,Xn ). xI"* XIwz
has the same order as Xn .
In this manner, define 71 K: IK=l,. and {[k) so that
I wi*-
for every i / K Then Xh K is a subsequence
whose sum does not exist; thus the converse is proved. Q.E.D.
00
Theorem 1.2: Let Xh ,f7=I,X,* If rX converges
unconditionally, then for all > O there exists an N such
that for all L Ci',O0) IJ; I1<
Proof: Assume the conclusion is false. Then there exists
> 0 such that for all N there exists A C [J,0o) such that
Let NI= I Then there exists A C.C oo) such that
hI~I[ J> Also, there exists a finite subset of I ,
such that I > Let N = .:a There exists
SC Nr+1,00) such that I X I so there exists a finite
set rI c A such that l > Note that 1 n = 1 D where O
denotes the null set. Also, for all SE t, c 5 f
In a similar way define an infinite number of finite sets
such that for all i 1Zf ll> and if LjJ and 5E ,f j F
we have 5< .
VE Lv P.L is a subsequence of .hj If m is
any positive integer, there exists io such that for all nv E lo '
V,
SV Writing f'L [lv, i > tv we have that Xvj
I .> so is not a Cauchy sequence. Therefore
the sum of the XAv does not exist and this contradicts the uncon-
00
ditional convergence of Z h .
nr
Theorem 1.3: Let >E (X 7= ,1,2,-- Then the series
-Xn converges unconditionally if, and only if,
71=1
&(m, (X,)| = o uniformly in X where 1X The
conclusion means that if 4 >0 there exists an N so that for
all X) such that IX*1 / we have F IX()l< .
coo
Proof: Assume the series 2:- converges unconditionally,
-Az
11
and let 6 )0 By Theorem 1.2 there exists N so that n 2C
for every AC LNWo) Let X be chosen so that IX~ 1 I and define
A,:fnE[f,oo): : x)
Then A, UAI f ,oo) ,and (1/=0 Hence -X0)
ntAr, m Ez. | Cnf |
Note that for all > N /IvACi] I" therefore
ifA i [ N/ X m) -I rN1 .[ It follows
tat nCN,t~]t t ?n,6,n i .I
Likewise, it may be shown that I i)l ; consequently
IX(X) I I ) "- and the first implication of the theorem
m= N
is proved.
Conversely, assume that the sum of the X) does not converge
unconditionally. Then by Theorem 1.1 there is a subsequence PXKl
where the sum of the XHK does not exist; hence there is an >O
and sequences fPiL and fVyl where P < Vi P+I and for all ,
SCX > 6 By the Hahn-Banach theorem there exists X for
each i so that IX l I and < iXi --
K= P K P(;
^ ( ^)1.K, I
For every N there is a value of L so that
Vj
ip "'l CN, Co) therefore Jx(X )I x(Xk)\ 6 .
Consequently the limit in the conclusion of the theorem does not
converge uniformly in X where IXYI I and the second implication
is proved. Q.E.D.
Definition: Let 1E, 3X : I,,,''" and let G be the
collection of all finite sets c contained in the natural numbers.
Define a partial order on E by ", 2 if G, 2 We say
oo
the series Z(h converges to a Moore-Smith limit Xo if for
every > O there exists Oo such that for all (02 Go ,
Shf- o< ( .< Since X is complete, an equivalent condition
would be for every > 0 there exists Go so that if GC, 2
we have In K x 6
oo
Theorem 1.h: Let ZgE X =1:I,Z,-" The series ZIX
converges unconditionally if, and only if, it converges a
Moore-Smith limit.
oo
Proof: Assume ZX converges unconditionally. Let >O .
By Theorem 1.2 there exists N *so that for all LCCN,oo) ,
I t 6 Let z [oI,N] If 0 x, 2 O, we have that
J,: -_ j MJXnl itnX- ZXhj ) +
| t E, n.7 E (+ A ; Is Ego Ia '- T n I n ,-
1 + I |
I .t- <;( because both ',- o and OL-Qo are contained in
4,oo0) Therefore the series converges to a Moore-Smith limit.
Conversely, assume the series does not converge unconditionally,
so there exists a subsequence I XKI whose sum does not exist.
There exist E)o and sequences {PI iV j where P
and for each i I .K- > E Let 'o be any finite subset
of the natural numbers, and choose i so that [nPL,-"nv nV O
Letting T : oU CPi, *" Vi V we have that (T (To and
| n ec nrco and it follows that the series does not
have a Moore-Smith limit. This completes the proof. Q.E.D.
CHAPTER 2
STRONGLY BOUNDED SET FUNCTIONS
Many theorems that are true for countably additive measures
do not have extensions when the measures are finitely additive
and bounded. However, as will be seen in Chapter 4, some of
these results do carry over to the finitely additive case when
the hypothesis of "strongly bounded" is assumed. In this chapter
"strongly bounded" is defined and theorems are proved which will
be used in Chapter 4.
Definition: Let /.Z-~X be a set function. we say ,4
is strongly bounded (s-bounded) if &7iu(E,)'O whenever IEKJ is
a sequence of disjoint sets in .
This definition was first given by Rickart (17).
Theorem 2.1. Let /4:' be s-bounded and finitely
additive. Then /u is bounded.
Proof: Assume/L is not bounded. Then there exists E
such that C) >1 Suppose there exists a monotone decreasing
sequence of sets FKCE such that (FKt+I) )I> (FK))+ for all K
Then 1( -FK:,)-1K) ,and (- )IJ > /Ia) -
S1(X .(). Therefore f KF, is a pairwise disjoint sequence
of sets, whereas i does not converge to zero, which
contradicts the fact that is s-bounded.
Consequently there exists G, E such that I/A( 1 and
for all LC G, ( ) /-I())I4J There exists
such that > 1 (G)>| +2, so \ -N\- n Crl\>l
If there exists a monotone decreasing sequence of sets F iR C ,
such that /pl(F+i)f/ >/ (F1)I/1 for all K we obtain a contra-
diction as before.
Therefore there exists G6C R-G, such that /1( )I>
and for all L C&; (L1) (z )l+i i ( J D There
exists P such that ()I| )l )/I3 so 0j(P- ) UG~I
4 ( P)O- (l(Pi)\ t rf] )|> > and there exists G3C P-(G UG&)
such that \ >63) 1 and for all L C (3 ) j(L)L< |lG3)1+ 1
G3f(1 u G,)=O
Continue in this manner to define a sequence of disjoint sets
~Gi such that for all L f/(Gi)( > This clearly contradicts
the fact that A is s-bounded, hence the theorem is proved. Q.E.D.
Remark: The converse to the above theorem is false. Let Z
be the Borel sets of the real line, and let 6- 4or(R)RZ ,
the class of essentially bounded measurable functions, where
is Lebesgue measure (9). Define /uE) ~= E the characteristic
function of the set E /u is bounded because for all E in 2 ,
I E oL 1 However, letting E '=(, l) we see that
E,) 0= 1 for all 7Z so / is not s-bounded.
On the other hand, if k is the real numbers the converse
holds, so the concepts of s-bounded and bounded coincide in the
case of scalar measures.
Theorem 2.2: Let /,:. )R be bounded and finitely additive.
Then / is s-bounded.
Proof: Assume the conclusion is false. Then there exists
a sequence of disjoint sets EKJ where J / ; does not
converge to zero. Therefore there exists ( )> and an infinite
subcollection EKJ C I[EtJ such that for all L /(EK')
Let AC be such that for all 6 (EK) > 6 and for all
Ei -' /( .) 4- Either A or i is infinite.
Without loss of generality, assume L is infinite. Let /1>O.
Let r be a finite subset of A with cardinality greater than .
Then ( ) ( \ Since was
chosen arbitrarily, this proves/ is not bounded a contradiction.
Q.E.D.
Definition: Let/U: -- Define u(E)
: .\f )'E E f uis called the semi-variation
of/ .
Theorem 2.3: Let/ Z. X- ThenA is s-bounded if
and only if/ A is s-bounded.
Proof: Assume is not s-bounded. Then there exists a
/4
disjoint
to zero.
sequence EK such that I does not converge
That is, there exists 6 >0 and a subsequence IEijC E4
such that for all i I(E >} According to the defin-
ition of I for all i there exists F C E. such that
I K.. Therefore does not converge
to zero, and is not s-bounded. This proves that/i being
s-bounded implies that/ is s-bounded.
Conversely, since I/L(E) 4 (E) for every E in 2,
6m (E =o0 implies that J ([k) 0 Q.E.D.
Theorem 2.1: Let/ -- be finitely additive and
s-bounded. If IEl is a sequence of disjoint sets in Z then
o0
/ (EK) converges unconditionally.
Krl
Proof: Assume the conclusion is false. Then by Theorem 1.1
there exists a subsequence such that / 'I
does not converge and is therefore not Cauchy. Hence there exists
> 0 such that for all N there exists p > 1 where
Obtain / such that I and / ( >
irF,
and then choose p, Q such that 6 P Z Y and )> / .
In this manner we define infinite collections P and I
such that for all j, qP < ( Pi j l/(E)J >E
Let F = EK Then Fj is a sequence of disjoint sets
t L) F i Then u
and *)- (EK )|>6 so IF; does not
J LJ
converge to zero and/ is not s-bounded. This proves the theorem.
Q.E.D.
CHAPTER 3
EXTENDING THE SCHUR THEOREM
In 1969 Brooks (3) proved the Nikodym theorem by using a
result of Schur (20). In this chapter the Schur theorem is ex-
tended to the vector case, and the result is then used to obtain
a very short proof of the Nikodym theorem for countably additive,
Banach valued measures. The proof of the first the -em is a
good example of the "sliding hump" technique.
Theorem 3.1: Let in .Il, Assume that
for all i the series x is unconditionally convergent,
4=1
and also that:
(*) for all A A- rX roO .
i neA
Then the limit in (*) is uniform with respect to A That is,
for all 6 there exists (depending on 6 ) such that for
all (1 and for all A in 6 .
20
Remark: If ) is the real numbers, then Ziin converges
absolutely for every L and the conclusion is that
i n-i
Proof: Assume the limit in (*) does not exist uniformly
with respect to Then there exist 6 >o and sequences
i IuKj such that I A > To simplify notation,
let I= K so I XKn> 6 for all K Let K, Since
n r" there exists a finite set FCA such that
MiAj
'17 -- ; due to the unconditional convergence of ,
there exists R, such that for all ACFR, oo) I n )
Let T, U i(u{RT() If Ac T+l1, ) we have
r, ( D ,and 7 I > U -L -X I
For every n sK=O so there exists IJ such that
T o
for all i AI<5>T Then Zin 7fT if LI C
In thl,]} Let K: 1 then t t -+
X ad T l and X 1 I
rK -,TJ -CT,- >nr-l | xI6KgnI;T
(- A/Tn > Therefore there exists a finite
set cAKC IT s t
set 62,CAg-C/,Tj such that K- .
Since 1 XKj converges unconditionally, there exists K
?t= I
such that for all A CCR,oo) n Kh Let T=
nsa" { P2 (} ) so that for all ACT++i,o3) we have that
/PA= A = o Note that (, 0 = 0 hence =
PtPf
7 -' p A Kn when ACT+f,o) Also
Ix nX because C [I,T;] Therefore
Xi+ X-+4 1<:
P uVP uL 1 a @ I l, -
for AC T+ioo) Also, since P U a C cT +, oo) we have
that l r ,> .
In the same manner as before, obtain I such that for all
17 Iinl' and let K3= Then we have that
IfA 3- ,T > and there exists a finite set
C C 3f-D,T] such that I K >P Choose R3 such
that for all A C [R3,.) 'A 3 A and let T3
In4t(3 u {R3) If c [T3 t,o0) we then have
h1r, UpAUP 3 >-j Since G uA C T +1,oo) we have that
t P, g u.r, uAI>- and u3 UA C T +1,oo) implies that
acv uP, U r3 v 3
Continue to define an Infinite number of Pj and K. and
u J
let Then for all j | > and i> J Oi
does not converge to zero, which contradicts'the hypothesis.
Q.E.D.
Remark: Theorem 3.1 may be used to prove the equivalence
of weak and strong convergence in. iF the space of absolutely
convergent series. Let fi f E( Lr l,l,." Each element
of 11 is a sequence or real numbers, and we write fi as i injgl
and f as x 00 The sequence {fJ converges strongly
to f if i Mf i. I-, X 0 and fI converges weakly to
f if converges to F) for every 9* in We
know that the conjugate space of 1, is , the space of all
bounded sequences, and that if : i then
Theorem 3.2: Let Zf. f I ':=I,A,." Then the fL
converge strongly to f if, and only if, they converge weakly
to f.
Proof: Strong convergence always implies weak convergence,
so I will prove the converse. Assume the fi converge weakly
to f Writing f# as Xi0: and f as 4 let
i Xi -X .* For any A define y* in oo by:
01 if K E '
Then z
because the fi converge to f weakly. Therefore, by Theorem 3.1
0o
we have that A -/X ih-XhI:= and the f( converge
h'l
strongly to Note that we only needed to consider those
elements of joo which take on the value O or I Q.E.D.
Theorem 3.3: Let X. ,7:1, ... Assume that
for all in converges unconditionally, and that:
(M) for all A A, Z 'in exists.
i nEa
Then the limit in (*) is uniform with respect to A ; that is,
for every >O there exists I such that for all J I and
for all Jn 7 t L (
Proof: This proof will be divided into six parts.
Part I: If {CiK is a subsequence, then
j (XiL i~~r)- o uniformly with respect to A
K ntL
Proof of I: Let o( = XK XK+1 Then for all K
00oo
OdK 7 converges unconditionally and for all / J Kn=O .
7L= K TI0A
By Theorem 3.1, Y/A 2Ek n O uniformly with respect to ,
and the conclusion, of I follows.
Part II: For all 6> 0 there exists N such that for
P oo
all integers L and for all p N in ~ E L 6
7n1 i 7t:( I
Proof of II: Suppose that we deny the conclusion. Then
there exists 6)> such that for all N there exist I and
P2N such that:
(1) oo E 26 S
Choose 6 as indicated in the above statement. Let No be
chosen arbitrarily. Then I claim that for all I there exist
P o00
L>I and P No such that Z Xin- lIn 2 If this
were not the case we could find Io such that for all > o ,
P o o00
P No ) 2in ZDi7 For all i Xn converges,
n j4: i I T
sP oo
so there exists NL such that for all P2 #; X l TXit < .
I lh~I ? i
Let R = ty NorJ i : L' / ' o] Then for all i and all
P 00
p> R Xin i (in and this contradicts the assump-
tion (1). Therefore, since No was picked arbitrarily, we have
that for all N and I there exist L >I P N such that:
(2) i ,- n i
00 f (00
Since 6n F- in exists, IZXi7n is a Cauchy
sequence, so there exists K such that for all L, K ,
~ J Let ',:K Since -Yi,71 converges,
z I &7t: I
P o
we obtain NI such that for all P2 N, ,I -n
According to (2), there exist V>i2 V oN such that
v o0 | V V
X n Y > Therefore, since -iZ )n, +
/ l' f 7I=1 ^ '-1 1.'
have that I -i t Lin II-- -n
00,l 00 6-e_-4=_ Let A, :[,V
E I t 13
Now choose ~ such that for all 5- In xn i ^ T .
According to (2), obtain i3 > i A 2 42 such that
I n3 =3 x 6 and since ij and i, are both greater
Sa h Combining
than K we also have that z Combining
S= I l f
nI A
these inequalities, we obtain 13 > Let
Continuing this process, define { } and 7,) ,?f:,2,-"-
such that 7nAmXn %E k^ h > for all n Then
one (X 'X +) does not converge to zero uniformly
with respect to A which contradicts Part I; hence, the con-
clusion of Part II is proved.
S S
Part III: If Se ) then in X where
1:I 7:I
X),.= X(
Proof of III: This follows because the sums in question
are finite, hence the limit may be interchanged.
Part IV: For all A Xk exists.
Proof of IV: If A is finite, the conclusion holds, so
assume A is infinite. Let = IKj Letting K i nK
we see that {Z1< satisfies the hypothesis of Theorem 3.3;
consequently, we may apply the results of Parts I, II, and III
to the double sequence {~CKJ To show 2 7t exists it suffices
00
to show that 1 exists, where s/- <= .
K=:I .
In order to prove 2-6K exists, I will show that K^ Kvy
is a Cauchy sequence. Let >0 By Part II there exists an N
such that for every i and v2 N .K- l-iK .
Therefore, if V,S2 AJ we have 1 K; b i < for all .
From Part III choose I such that for all i I ,
gla = K- 3 and Yf K i-. Then for all i I ,
EVA
K=1 K=l KI K K-I K-- Kl
V 700
Since 6 was chosen arbitrarily, it follows that _
is Cauchy, and the conclusion of Part IV is proved.
For all A Xn exists, and Theorem 1.1 indicates that
00
c on converges unconditionally.
Part V: For all a A i (Note that
the second sum is defined by the result of Part IV.)
Proof of V: Since the conclusion is true for finite ,
we may assume 6 is infinite.
Let n] i k X g We need to show that
on on
AwL ij- i IK and since j[Ki satisfies the hypothesis
Z K=1 K:
of Theorem 3.3, we may apply the results of the first four parts.
Let >o According to Parts II and IV, choose N so
E o0 W oa
that I IK -K
Kri K. K= : 1 K=
By Part III, choose I so that for every 1 I ,
N N oo
K 1/K [C Then for every (2 I K K I
oo oo
Since E was chosen arbitrarily, n -iK = xi K .
L K=I K=I
Part VI: The proof of the conclusion of the theorem.
00
Proof of VI: Let c<'-= Xji- X By Part IV, D'n con-
00
verges unconditionally, so for all i in converges uncon-
ditionally. Also, for all L Ai -'7 = -Zt Z( n -)
Snb n 7i
A X- x = O where the last inequality results from
Part V.
Therefore, by Theorem 3.1, for all 6>0 there exists I
such that for all i21 and all A > /\ ii t
/ tEX ~- in t; I ,where the last equality
follows from Part V. This concludes the proof of the theorem.
Q.E.D.
Theorem 3.3 is the desired extension of the Schur theorem,
and will be used to prove Theorem 3.4, the Nikodym theorem.
Definition: Let / : Z/-- 6 / is countably additive
if /#(=E - (E whenever iEs} is a sequence of disjoint
sets in .
Remark: /L being countably additive implies / is s-bounded
because if/U is not s-bounded, then there exist 6>O and disjoint
sets {En such that for all n / (F I > ; hence Ku(LE|)
is not a Cauchy sequence and/U is not countably additive.
Definition: Let U1 : Z --' be countably additive, nrl,Z,- ,
and let [EiJ be a sequence of disjoint sets in Z- The /I
are uniformly countably additive if for all 6>0 there exists M
such that for all m n and for all n L:/ ti (E)I E .
Theorem 3.4 (Nikodym): Let /: u x X be countably additive,
n=: 2,-. "If/ A(E) 4~L/,( ) exists for every E then/t
is countably additive and the countable additivity of the /,,
is uniform.
Proof: Let JEg, be a pairwise disjoint sequence of sets
31
in and let (~ =/k (Fn). The double sequence ~Xi' satisfies
the hypothesis of Theorem 3.3, so by the result of Part III of
the proof of that theorem, for every 5>o there exists N
P 00
such that for every integer 4 and all P2N s> in- =
p n.)-a (E. ) Therefore the 7L are uniformly countably
additive.
On the other hand, /L =I C, = A e AE)
gin E- n the last equality following from Part V
00 00
of the proof of Theorem 3.3. Since X : T (E) we have
that / u Eln ) = I.(E) and / is countably additive. Q.E.D.
n=1
We give a proof due to Brooks and Mikusinski (5) of a result
of Banach (2).
Lemma 3.5: Let C be a separable Banach space, and let {
be a bounded sequence in Then there exists a subsequence {iK*
such that YnK,() exists for every X in
Proof: This proof consists of a standard Cantor diagonal
process. Let iX L be a dense subset of X and choose M>O
such that |
for each n so 9nix, is a bounded sequence of real numbers,
and there exists a subsequence Y,l of / i so that K YK (x,)
exists. Likewise, there exists a subsequence of
such that m. ,K(X ) exists. Continue in this manner to define
subsequences sIonI so that +(, ,K j C r i: ,KIJ KK and ,K(XL)
exists for every n For each integer K let Y .KK "
Then {YK is a subsequence of tnI and for all j ,
' K n(xj) exists.
We now show that YK( ) exists for each X k Let
S(K.
XE and 6O Since f(jj1 is dense in X ,pick Xj
so that IXj -I< M, and choose R such that for all K, 1 R
I~ K t J X) I Then for all K,~ R ,
7V 7) (Xj) x)4-
O (xj) n'(x) f + = Therefore the sequence
JIuK) is Cauchy, and A ( () exists. Q.E.D.
00
Definition: Let X h =, I, *Z, Xn is weakly
71=1 *
unconditionally convergent if for every subset A of the integers,
there exists XA such that TX A*
fhA*
The following theorem is known as the Orlicz-Pettis theorem,
(13, 14), and the proof is due to Brooks and Mikusinski (5).
Theorem 3.6: If X is a separable Banach space, then a
series 2 t converges unconditionally if, and only if, it con-
verges weakly unconditionally.
Proof: Unconditional convergence always implies weak uncon-
ditional convergence. In order to prove the converse, I will
assume that the series converges weakly unconditionally but not
unconditionally, and then arrive at a contradiction.
00
If F-X does not converge unconditionally, there exist
an 0>o and finite disjoint sets Ai such that InAj>l for
every integer L Letting we have that ._'
-nfZ^ E'
also converges weakly unconditionally.
According to the Hahn-Banach theorem, there exist Xm such
that /I=1 and I X(-)I= I for every integer ) .
S (-yj 00
By Lemma 3.5 there is a subsequence Xi of I such
that ^-4Xi xtL(X) exists for every X in M and to simplify
notation I will assume that Mi= for all i Let 0Q = X(3).
34
For any A there exists 2a so that Z L = OA)
for all Therefore, m't, Tdi n, )
a X; n A) Since 2 Xin) exists for all \ and uncon-
L 1t6b
ditional convergence implies absolute convergence in the scalar
oo
field, it follows that X*ir()I,) ~o Therefore, the double
?tI
sequence 4inj satisfies the hypothesis of Theorem 3.3.
Letting 0d : = d we have from Part IV of the proof
oo
of Theorem 3.3 that Z 107
: I
theorem we have that din n n = uniformly in .
I 'iA C.6 ?
Choose positive integers t, and NX such that for all n 2M
da I\ and for all iNW and all A ih.-g. 1 *
Let )9n>tA Then I2J |^Hn 4 jI^l
St '- d | which is a contradiction. Hence
the theorem is proved. Q.E.D.
CHAPTER 4
EXTENSIONS OF THE VITALI-HAHN-SAKS AND NIKODYM THEOREMS
In this
are extended
and strongly
to the whole
(15).
chapter the Vitali-Hahn-Saks and Nikodym theorems
to the case where the measures are finitely additive
bounded. The first theorem to be proved is the key
chapter, and is an extension of a result of Phillips
Theorem 4.1: Let 11: P(7Z) be finitely additive
and s-bounded. If m a = c for all A then
n 2 /Un ( ): uniformly in (That is, for every
6 0 there exists N such that for all A N and for all D ,
Remark: Since the / are finitely additive and s-bounded,
00
it follows from Theorem 2.2 that Z// -(t) converges uncondi-
-t'l
tionally. Therefore (Wt) exists for every A and the
tE&A
conclusion makes sense.
Proof: Assume the conclusion is false. Then there exist
an > 0 and sequences Ii, ~ such that I / >
L= I 2 ,..... To simplify notation, assume n = .
Let ijI /I. f > so there exists N, such that
/ '(() /6 for all c N,00) Letting 7=-A lnEI N,-1
we have I I ) 1> )=--Z,)
Since j) 0O for all j we can pick L such
that 1/i () / L ) converges unconditionally,
so we can pick N,> ,I such that for all ACI ,l ) ,
IAl < /6. As a result, L Z .L( +
a 2v 6 2r',-o
+ / 2) and therefore
J~ P2 ) JE/iaf) ^) Z i >
1- 6h > .. Let ?j, lLi,a,-ij
Choose i3 such that /3 / 7 and pick 3 so
that for all A /6 Then, as before,
una0l 3 o)
/ ; 3 r > and we let ?'3 3- n .
Continue in this manner to define an increasing sequence
of positive integers JNJ such that:
-JK- I
WE nK
(2) nLJ' 76 for all A and K,3;-* ;
(3) letting K AKnld[ K- ,1)K- (No1) we have
Note: If A is a finite set disjoint from LK-IJ ,
then it follows from (1) and (2) that A / < l .
Let 6,= f.*((,tl)]t: I h)= ,,.'- For all values of ?m ,
Of is an infinite set. Also the sets iJ are disjoint
because fnl V implies that 2 ml' -I)-12~m -) for all
values of n, and Xz Since the sequence fK) is disjoint,
it then follows from the preceding statement that {u(mj l, is
a sequence of disjoint sets, where UB, is defined to be
Ulr :TE In other words, I have divided a countably
infinite collection of sets into a countable number of countably
infinite collections; and then have taken the union of each
collection. Since all the sets were disjoint, all the unions
are disjoint from each other.
Since /Ai, is s-bounded it follows from Theorem 2.3 that
/ l is s-bounded, so /A (UB,)- O Hence there exists
an no so that ,Ai(Bmo
EIA. ?" o because i,( \)> .
Let PL -.m is a countable collection of the 7? ,
so by the same process as above there exists P ,an infinite
subset of ,such that / (Up < Again ? .
Continue this process to obtain a sequence of infinite sets i2
where PmiC C and i,(Ul )/*
*
Let r consist of the first element of r for every 7t .
P is an infinite subset of TKI and for each value of h ,
tr-f consists of a finite number of elements.
If rE then /K(, r)llI)-
/K ufr-(ru )~J) iK(fr nrK]) because r .
u fr-(rufrKU) f is a finite set disjoint from LNK>,NK41 ,
so so ufp[ Kf- t j)I)
/ UiK ([ fl '))] Combining all three inequalities we
obtain K i r) TK -- > *
Therefore / UP) if ? EP There are an
infinite number of ?' in P so /iK K)U I cannot converge
to zero, which contradicts the hypothesis. This proves the
theorem. Q.E.D.
Remark: Theorem 4.1 implies Theorem 3.1.
Corollary 4.2: Let /J,: -4~ be finitely additive and
s-bounded, n=lI,Z,-' Assume .& /',(E) exists for all E
Then if EK) is a sequence of disjoint sets,
. ~I(/ at, "-/ K) 0 uniformly in A .
Proof: Define Vj .(7)-4 X as follows: A/a)=
A( -i Ej( The V satisfy the hypothesis of Theorem
.4, so J, (,a /n)(E) -- 1 0o uniformly
A K C6 K E6
in Q.E.D.
Corollary I.3: Let /IV: be finitely additive and
s-bounded, ?=Il,Z,. Assume A/ (E) exists for all E
7?1
and let /u(E) =k /n?() Then / is s-bounded.
Proof: Assume /U is not s-bounded. Then there exist
6>0 and a sequence of disjoint sets fEg} such that for all K ,
I(EK)I>E Let iI Since /mUn()= 0( there
exists K, such that /1h (E) / However ,i /)(E)/ i
//EK) j> so there exists z > so that ) >(E-
Therefore m -~U i) ( >) I- C Because
I/LrF 1(EK)O there exists KX such that K2) 3.
&. |j (E )/= Ii(E )I > e so there exists 3 > >
so that I)3 (F 9)I and I Z-/ S 3)( 3 )3 3
Continue this process to define sequences {n Kj
where iI} is monotone increasing and for every integer ,
S/l ) > Therefore the sequence
IZ1/ ( / x)(Hj does not converge to zero uniformly
in A and this contradicts Corollary l.2. Hence the theorem
is proved. Q.E.D.
Definition: Let / -h be finitely additive and
s-bounded, 7Z~:1,I,. We say the ~ are uniformly additive
if /tm Z" X (EK 0 uniformly in A and n whenever {EK)
is a pairwise disjoint sequence of sets in 2 That is, for all
6 >0 there exists J such that for all J 2 J all n and all
'i KEJO,) n A
Remark: When the un are countably additive this is equi-
valent to the /"u being uniformly countably additive.
Theorem h.h: Let /U,.' ( be finitely additive and
s-bounded, ?=/,', ," Assume u (E) = M/n('E) exists for
every E Then the U, are uniformly additive.
Proof: Let EK be a sequence of disjoint sets and let >O .
Let V,= P(7)-4 be defined by ) )--(/ -, ( E) By
Corollary 4.3 /x is s-bounded, so the X satisfy the hypothesis of
Theorem h.1. Hence there exists an integer N such that:
(1) fJ/-) ( < E for all A2 N and all A
Since u is s-bounded, it follows from Theorem 2.4 that
o0
V- I( ) converges unconditionally, so by Theorem 1.2 there exists
I, such that:
(2) K ,) E for all
Combining inequalities (1) and (2) we have that for all t 2 N
1 ,f x(EK) M ^ 1 11E)
and for all ,,o) a n f,)
/(?nr,,-1-4 ) (Ex
Each /L is s-bounded, so there also exists 1 such that
LKEf n[r, ) I< for all 6 and all 7n N Consequently
/Mh(EK) 4< (.6 for all n and A Since E was chosen
KE6 n +I oo)
arbitrarily, this proves that the are uniformly additive.
Q.E.D.
Remark: Corollary 4.3 and Theorem 4.l are the intended
generalizations of the Nikodym theorem.
Corollary L.5: Let /h: E-~ R be bounded and finitely
additive, n =l, a,-- If /(E) ,/E ) exists for all ,
then the /, are uniformly additive.
Proof: It follows from Theorem 2.2 that the u are s-bounded,
so the conclusion follows immediately from Theorem l-.. Q.E.D.
Corollary 4.6 (Nikodym): Let /An: -9 ~ be countably
additive, I, 2, ** If / l E) = /n,(E) exists for
every E then /u is countably additive and the countable
additivity of the n is uniform.
Proof: Let > 0 and let {Egi be a sequence of disjoint
sets. By Corollary l.3, u is s-bounded, so i /U(EK) exists,
/ K=I
hence there is an I, such that for all i 1 ,
1 j (EK- 00X(E:) /X-
By Theorem h.-, the u are uniformly countably additive,
so there exists 1L such that for all n and all 2 i ,
I /E2) which means ni <)/ < because each ,u
is countably additive. Therefore:
(2) U EK) = j /A( p 0 for all 1i I .
Let I- 7 IJ11J Obviously # is finitely additive, so:
(3)
Combining inequalities (1), (2), and (3) we have K) -
= I ,ewr )h h
S.E Since E was arbitrary, this implies that
:- / .Q.E.D.
EI)= (K )
Definition: Let /~: X and V:W '- R+. We say /I
is absolutely continuous with respect to V if for every 6 > 0
there exists 6 >0 so that v(E) < implies IJE) < for all E
in i1 If / : we say the /I are uniformly absolutely
continuous with respect to d if for all 6>O there is a 6>o
such that VE) 6 implies I/r(E)k6 for all E and n .
Remark: The next theorem is a generalization of the Vitali-
Hahn-Saks theorem, and is the most important result of this
dissertation.
Theorem h.7: Let/U : .-4- X be finitely additive and s-boun-
ded, n:L, ." Assume that Am/u.(E) exists for every E
Let ) be a non-negative (possibly infinite) finitely additive
set function defined on If Uh is absolutely continuous
with respect to ) for each L then the /n are uniformly
absolutely continuous with respect to .
Remark: If ) were assumed to be bounded, then the /f
would automatically be s-bounded.
Proof: Deny the conclusion. Then there exists an >0O
45
such that for every 6 > o there exists an E, such that )(E)4
and J1/n(E)I ) for some 7 Let E, /h, be such that vE,)*
*
and /II(E, )I Let ~~
~~
that /in, (E) ,
v(E)) and /In;(E) >) Then for all Fc Ez ,
Now assume that E, EK , l- nfK have been chosen. Let
SK+1 be such that for all : I ..K V(E) < CKI implies
/hE)< 3 Now choose K+l and l+ so that v(EKti) SK+1
)K+i K and I/ (E'+,) > E Then for all FC K+
and for all Z .***i , 1/ i(F) j .
Resubscripting for simplification, let /UI "-'/i We then
have that:
(1) for all I I (E)> and for all J : l*-l- and all
FcE , JF)<
Let F= : = I and assume there exists an L, >2 such
that / II E Let F2 I E L- Assume
FI'.FK have been chosen in this manner, along with i" iK *
46
Assume there exists it( > iK such that J/iK4I(FKnELK,)
Then define FK+1I F El X+ If this process continues
we obtain a countable collection of sets FK such that
VKIFKF K1)1- / ,(F)nE and
because of (1) and the fact that FK n E- zl ELK+I
Therefore |I(,, -1:)/ -(FK - -f = f-
Note that IFK-FK+I} is a sequence of disjoint sets, and
S (/i4 / f j) (FgK-.+) I} does not converge to zero
uniformly in 6 which contradicts Corollary 4.2.
Therefore the process of choosing the FK and lK has to
stop, so there exists FK lK 2., such that for all J > K,
J(K E) Let f:= K H, = F /, ( ,
and E =p, HI Then the following three results obtain.
i. (H,)
Since H, cEi it follows from (1) that Jt(H1) I <
E E
I, /6
i. >JL HI1t Y
We have that E = F, (,-F )U(F F3)U ..U(FK_,-F) U F<
| 1> 6 and / (FiO )= // (F n El ) I /
since F, n Ez C Ej and 2n< II( -FJ)h I/'F E)\ <
--< /6 because 13 > i >2 hence 3 >' Thus for
xJi -1( F +-Fj)j I+'jEi +) and
e I, ( ) I |/',-F,-F ,. .-+ I,"(F-,- FK)I+ /t i (, )|
---- +.* + i2J(H I since H,:FK Therefore
16 3Z-L +. =C
III. 'E ) --
Note that Ep +i (E+i- H,) U(Ep + n H,) H1 was chosen
so that for all J> P /Ij(EjnH,) I Hence
E ,< I 0(E, )) i(E, H,) + |it-(Ep)1,pn H,)|<
V (Et Thus E-~ IFP'
Note: From I and II we have that )(H ~ ~E '
Let FE= Assume that there exists Z > 1 such
Lt /r.^ I F' ['t r^ ')
that l/ui i 2 and let F E If
we could continue to choose i indefinitely, then we would
ob n a contradiction the same as before. So there exists
obtain a contradiction the same as before. So there exists r
and I ?.2 so that for all J > ti IF ('F n )j )
AP2 =1A / I+ 1)
8
r + p, + .-
Ep,+pt (H, uH.)
Then H, n H, = D
because Hn C l I)
(and
and E (H nH =E
(Recall the construction of
then obtained.
r'. ,1 H) 3
Since H~C EF1 C EP..l
1V (Ha)
The following results are
and P,+4 >2- we have that
2 +132
SI E-
( F2- ) (K )- F3) U'. U "
< 1 ) l -j F -
[lal"iH)
, since HI. r
and tL > I we have
U =F=' E"')
Since F(')
Sso
() c (C)C P1
E; L Ei,1C Ep,,l
that 6 (IF (F 1( E <
1/ 1 1 1 2
S+ E
P, + +Lz (
6
.2
Likewise I/ ( j -
E3
F (I) -
because i >
J4i J
for all J:1,,..K-I Hence < +- +
Let
Let Hz = F (
and
and K
= ^-^')
W .
P;L+
(1)
E,
P.,+ L + 1
.1 J+1
'+ n F +
)( F/; ) ECI <
n J+, "
49
E -
8$ 1r+ ^"^)| and it follows that I/aN I
III'. ffor all .
Since HH the sets (Ep,+p n H,) ,
(nEp pl nH ) and (EP +Pi+ (H I u ) are disjoint.
Therefore 6 / (EpP+ ) I I2)(Ep, p Lin iH') +
/2) YEP, +P ,in i E J'(2)E (f, H .1)
Note that E PE Pi", H) and |)(Ep +P+L
because of the construction of ,l and H2 Also, by definition
)I/ ,(EP +Pl- HHl) jI (i (E) Therefore
Note: From I' and II' we obtain J i >(-1 f 31
We now proceed to the general induction step.
Let K 2) /A (0)= -, E El Assume that the
following statements are true. For each J = K Hj
and Pji2 are defined and () = Ep.--Hj .' =/ EP,+i
)i'EU)> E( ( HpnHj0O if P=I---K
)j) 0J-,) IJ->
S>- j- f ,
All of the above statements are true when K= I will
now show that their being true for K implies they are true
for K+ I
(K)= (K)
Let F, I E 1: Assume there exists lZ>i-l such
that I( K)f E<) K+X and let F ) F I E .
If there exists 3 > so that (/ ( n 1Ei3 ) ,
S(K) i (K)
let F3 = F 3 If this process of choosing ij and
Fj should go on indefinitely we obtain a contradiction because
for all j Ijl -,- j J / ( n fl
(-0) (K) (K) E(
whereas -j _, E C, C p +.. +- PK + ij and Ij > imply
I (P F E (K))' zP, +-t" .+ P,+j '4
that L +Z '
Combining these two inequalities we see that for all J ,
S 1(K) II >K) K))3
P( -/4jl, (J ji -~ > + Therefore
l K' i F jj does not converge to
zero uniformly in This contradicts Corollary 4.2.
Hence there exist E(K)
S(K.) n() -- -3
J (K EJ )\I 3
2-K+
and i, a such that for all J > L ,
. Let PK+, = L.
(K)
, I-K+=, Ft
p(K') (K)
L P:Ep .
In order to show the induction step is valid, I must prove
the following four statements.
I". For all j : I\' K
Hj 1HK+I=
+ (K) (K) (K)
HK+1 C E = FK) by construction, and E,
\j 1
SHence H K+l r (J, = 0:
11 1
Since HK+1
c E +... +PK+
2P, + + + K P -.
3Z
because P > 2 .
I /.a ...(HK Z'l-
Note that E) FK)- ( ,(K) U(K) U (K)
I 3 l
Also I I(- () I fK))F : )I' n+
E ( K++) b e F(Ei.+ P K)+ .
ZK+-+I because F L CE p
and I Li
HK '- PK + PK+1 i
H+1 P +""* + Pg + Pg+, +1
FK)
UF*
(K 'P) (K)
" 7-14pt(+r+ I /'"Pj +--' + P;(+ 1P+,+
i K+1
j H
E,+...+PKt +
, I- H )/ 1
l + - PK, -IK+H
Likewise K) (KF ) 3 I ( K)-t
1:I r3)i3 +."
K1+; because F n E E + ...+P + 3 and 2 3 .
S ) (K) (K) _
In the same manner, for j :...- we have I ( rj Ij
IjP ++PK+jl.+ < +
because J j+1 Therefore E- K (E I
I++K + -+++z '"* + + )/ (HK)l ; the
first inequality follows from the induction hypothesis and the
second inequality results from the definition of HK+I and the
inequalities in the preceding sentences. Therefore,
S- ..-:+--^iT-(++i +*+ Z} <. Ir, ( 1 (K)
6 e
Since 2.+l4 +I + K + we have
6 1_5_ 6 (K)
(K+X H+1
IV''. For all ( / I (E ) > 6 +- --E
(K) K) I. (K( (K) (K) \
^..^EK~ -H,,)t+ H pK+ +PK+K++.Jf 11 H K 1^ )j >
because of the induction hypothesis.
As a result of the construction of HK+I and because PK+I+ >P +I
we have J/p+, +i( p + H2) 2 I Thus
c, PK+,^ :,-HKJ|> PO I+-^
Note: It follows from II' and III" that
J(1A (KK-0) (H I ( CE- -- C- 3-6
Therefore, by induction we may define infinite collections
[H and {/) such that for each K2 L (K-) (K-)
i 31 > 32 {HK is also
a sequence of disjoint sets. Letting % ,) =V -
O -- 1 then | +,- ) J /6- for all 72 So
(V n+iJ M+ ) K tI(H 10 does not converge to zero uniformly
in L which contradicts Corollary l.2. This concludes the
proof of the theorem. Q.E.D.
Definition: A linear topological space is said to be locally
convex if it has a base consisting of convex sets. Let
U :{UL dE AJ be such a base. Let = : U Ua:U oI
For each U, in define I( X) = a: a >O, iU and
( [)- : EII() .. Then is a pseudo-norm for the
space 6 and has the following properties.
(a) 6)0 o0
(b) + +oo00
(d) if X)Ur then a( )_ I
That is, / has all the properties of a norm except that it may be
possible to find an X such that >XO and (Y)=0
In the theorems of this chapter the fact that I=1 0 implies
X=0 was never used, and with appropriate modifications the theorems
are true if the set functions take values in a locally convex linear
topological space. First of all, one has to define s-bounded and
absolute continuity differently.
Definition: Let X be a locally convex linear topological
space, and let :({A3 be a family of pseudo-norms that determine
the topology of E Let / U-X and 'L: -- Rt be set
functions. We say / is strongly bounded (s-bounded) if
A ()F:O 0 whenever {E} is a sequence of disjoint sets. // is
absolutely continuous with respect to V if for every o( and 6
there exists 6>o such that (E6)/ implies ( ))
The topology on 6 determined by a single pseudo-norm o( is
not necessarily complete. We then let K be the completion of ,
and if /L: Z- Z is s-bounded, we consider /4 to have values in
Ej and it is then true that converges unconditionally
in )c if [Exn is a sequence of disjoint sets. The statements of
Theorem 4.1 and Corollary 4.2 do not change if 9) is locally convex,
except that An(t) is considered to be an element of for a
'tEA
fixed o Taking into consideration the new definition of absolute
continuity, the extensions of the Vitali-Hahn-Saks and Nikodym
theorems for finitely additive vector measures are valid when the
range of the measures is a locally convex linear topological space.
CHAPTER 5
A COUNTEREXAMPLE
In this chapter an example is given to show that weak con-
vergence does not imply strong convergence in the space of countably
additive set functions.
Let /t: 0(01)-9 R be countably additive, and assume that
for all A t (A): 0 Then from Theorem 4.1, we have
that V/u, (A)= 0 uniformly in Z which in turn implies
that D/-- )j/Un = O where lI is the total variation of/L .
Since Theorem 4.1 was very important in proving theorems
about set functions from a O0-algebra to a Banach space, it is
reasonable to pose the following question: if -> R
and if j/ T (E)=O for every E does it follow that
- I/, nl= 0 ? The answer to this question is, "No," as the
following counterexample shows.
56
Counterexample: This counterexample will show that if
Z/ : Z-R are countably additive and if u /-(E) =0 for
all E then it does not follow that A u-= 0 .
Construction: Let 1 be the Lebesgue measurable subsets
of (0, I) and let A be Lebesgue measure restricted to .
Let -f be defined:
\ if X is in ( ) o o < Z a, even
foO -1 if X is in ,( ) o0 <2 a. odd
0 otherwise
Let ? (E) a d A Then for all >0 A(E)<& implies
I fd)<$ which is equivalent to IL,(E)1< Note that
if K >n then /a :O
First it will be shown that if is an open interval,
then / (6) =0 Let 6>o and choose n so that 6 .
Let = -fru[i":^ 3i<9 c-frrp .f.7z: ;i f3
Then f ('-U ) (:a ) L( C and this inequality
combined with the statements in the last two sentences of the
preceding paragraph give us that I/4(6}< 2 6 for all K > .
Since E was chosen arbitrarily, the conclusion follows.
Now let V be an open set in F and let > 0 There
exists a countable collection of disjoint open intervals fl
00
such that V = U( (I am considering the null set to be an
o00o
open interval.) Then A(V) = A( and, since A(V) oo ,
there exists N such that A(V- ,)Z 6 ; hence for all 2 ,
I/x /
that for all K>T( //K k)I< Then letting
T= 7 A L: I, N1 we have that for every K >T, 1-K(V) I
IIK(v/IM -A)/ + K() ( I + N(): Therefore
we have that & /U, (V):O for all open sets V
Let E be a set in 1 and let O There exists an
open set V so that ECV and \(V-E)
/Yt(V-E ) E for all n By the result of the last paragraph
there exists T such that for all > T //kiV)I E hence
/(IE)I J /(+ V ( th(V-E)l 6 This implies that
A/6un (E)=0 for every E in .
The /- are countably additive, so they satisfy the hypothesis
59
of the counterexample. However, / f(o,) ifnl X= I for all n,
so j[jnlI does not converge to zero.
REFERENCES
1. Ando, T., "Convergent sequences of finitely additive measures",
Pac. J. of Math., 11, 395-h04 (1961).
*
/ / /
2. Banach, S., Theorie des operations lineaires, Monografje Mate-
matyczne, Warsaw (1932).
3. Brooks, J. K., "On the Vitali-Hahn-Saks and Nikodym theorems",
Proc. Nat. Acad. Sci. U.S.A.;' 6, 468-471 (1969).
4. Brooks, J. K., "Representations of weak and strong integrals in
Banach spaces", Proc. Nat. Acad. Sci. U.S.A., 64, 266-270
(1969).
5. Brooks, J. K. and J. Mikusinski, "On some theorems in functional
analysis", Bull. Acad. Pol. Sci., Math., Astron., Phys., 18,
151-155 (19707.
6. Darst, R. B., "A direct proof of Porcelli's condition for weak
convergence", Proc. Amer. Math. Soc., 17, 1094-1096 (1966).
7. Dunford, N. and J. Schwartz, Linear Operators, Part I: General
Theory, Interscience, New York (1958).
8. Dvoretzky, A. and C. A. Rogers, "Absolute and unconditional con-
vergence in normed linear spaces", Proc. Nat. Acad. Sci.
U.S.A., 36, 192-197 (1950).
9. Hewitt, E. and K. Stromberg, Real and Abstract Analysis, Springer
Verlag, New York (1965).
10. Hahn, H., "Uber Folgen linearer Operationen", Monatsh. Math.
Physik., 32, 3 (1922).
11. Hilderbrandt, T. H., "On unconditional convergence in normed
vector spaces", Bull. Amer. Math. Soc., 46, 959-962
(1940).
12. Nikodym, O. M., "Sur ler suites convergentes de functions
parfaitment additives d'ensemble abstrait", Monatsh.
Math. Physik., 40, 427-432 (1933).
13. Orlicz, W., "Uber unbedingte Konvergenz in Funktionemaumen",
Studia Math., 1, 83-85 (1930).
14. Pettis, B.
Math.
J., "On integration vector spaces",
Soc., Wh, 277-304 (1938).
Trans. Amer.
15. Phillips, R. S., "On linear transformations",
Math. Soc., 48, 516-541 (1940).
Trans. Amer.
16. Phillips, R. S., "Integration in a convex linear topological
space", Trans. Amer. Math. Soc., 47, 114-115 (1940).
17. Rickart, C. E., "Decomposition of additive set functions",
Duke Math. J., 10, 653-665 (1943).
18. Rickart, C. E., "Integration in
space", Trans. Amer. Math.
a convex linear topological
Soc., 52, L98-521 (1942).
19. Saks, S., "Addition to the note on some functionals", Trans.
Amer. Math. Soc., 35, 967-974 (1933).
20. Schur, M. J., "Uber lineare Transformationen in der Theorie
der unendlichen Reihen", J. reine u. angew. Math., 151,
79-111 (1921).
21. Vitali, G., "Sull'integrazione per series Rend. Circolo
Palermo, 23, 137-155 (1907).
BIOGRAPHICAL SKETCH
Robert Jewett was born in Portsmouth, Ohio, on August 20, 1945.
When he was eleven years old his family moved to Fort Myers, Florida,
where he graduated from high school in 1963. In the Fall of the same
year he went to the University of Florida on a golf scholarship, and
played on the golf team for two years. In 1967 he received his Bachelor
of Science degree in Math, and from that time on has been working toward
his doctor's degree. He is a member of the American Mathematical Society.
He was married to Suzanne Strobak on July 3, 1971.
/
I certify that I have read this study and that in my opinion it
conforms to acceptable standards of scholarly presentation and is fully
adequate, in scope and quality, as a dissertation for the degree of
Doctor of Philosophy.
ames K. Brooks, 'Chairman
associate Professor of Mathematics
I certify that I have read this study and that in my opinion it
conforms to acceptable standards of scholarly presentation and is fully
adequate, in scope and quality, as a dissertation for the degree of
Doctor of Philosophy.
David A. Drake
Assistant Professor of Mathematics
I certify that I have read this study and that in my opinion it
conforms to acceptable standards of scholarly presentation and is fully
adequate, in scope and quality, as a dissertation for the degree of
Doctor of Philosophy.
Zoran Pop-Stojanovic /
Associate Professor of Mathematics
I certify that I have read this study and that in my opinion it
conforms to acceptable standards of scholarly presentation and is fully
adequate, in scope and quality, as a dissertation for the degree of
Doctor of Philosophy.
Gene W. Hemp
Associate Professor of
Engineering Science and Mechanics
I certify that I have read this study and that in my opinion it
conforms to acceptable standards of scholarly presentation and is fully
adequate, in scope and quality, as a dissertation for the degree of
Doctor of Philosophy.
41' -
Stephen A. Saxon
Assistant Professor of Mathematics
This dissertation was submitted to the Dean of the College of Arts and
Sciences and to the Graduate Council, and was accepted as partial ful-
fillment of the requirements for the degree of Doctor of Philosophy.
August, 1971
Dean, Graduate School
L2_/^
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