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Ez 3 1 zp (En-l)I(1/2")p (F). If N L 1 is such that (1/2N)p(F)< E , then E = EN serves the purpose. 0 O. Since u << p, there exists S > 0 such that Iu(F)I < ~ / whenever 2 F E 9and Ip I(F) <S. So, if F E 9and l@l(F)<S, then v+(F)=Sup{v(B); B c F , B E % ] I E / ~ < E . Hence v+<<@. Similarly, one can show that v-<< p . (ii)+(iii). If (ii) holds, clearly u+<< lpl and V - K IpI. Hence IuI = u++u-<< IF O such that for every n 2 1, there is a set Fn in 5F such that ]pI(Fn)< 1/2” but Iv](Fn)2 E . Then the ,ret F = lim sup,,m F, has the property that lp I(F)= 0 and IvI(F)2 E . To prove this, we proceed as follows.
-
The following definition is inspired by a certain property of nonatomic measures on a-fields. 5.1.4 Definition. Let 9 be a field of subsets of a set R and p a charge on .5F. p is said to be strongly continuous on 9, if for every E >0, there exists a partition {F1,Fz, . . . ,F,} of R in 9 such that Ip1(Fi)<&for every i. Obviously, if a charge p on 9 is strongly continuous on 9, then it is bounded. One might conjecture that a charge p on 9 is strongly continuous on 9 if and only if for every E >0, there exists a partition (F1, Fz, . . . ,F,} of R in S such that ) p(Fi)l< E for i = 1 ,2 , . . . , n. But this conjecture is not true. Any non-zero charge p on 9with p (R) = 0 satisfies the later property. The following definition is inspired by yet another property of nonatomic measures on u-fields.
5.1.5 Definition. Let 9 be a field of subsets of a set R and p a charge on 9. p is said to be strongly nonatomic on 9, if for every F in 9 and 0 IC 5 lp I(F), there exists E in 9such that E c F and Ip I(E)= c. The following theorem gives the inter-relations between these concepts.
5.1.6 Theorem. Let p be a positive bounded charge on a field 9 of subsets of a set R. Then each of the following conditions implies the succeeding condition. (i). p is strongly nonatomic on 9. is strongly continuous on 9. (ii). (iii). p is nonatomic on 9.
5.
143
NONATOMIC CHARGES
If, in addition, p is a measure and 9 i s a u-field, then these conditions are all equivalent. Proof. The implications (i)J(ii)+(iii) are immediate. If p is a measure and 9 is a a-field, we show that (iii)+(i) using what is known as the “principle of exhaustion”. Let F E 9 and 0 Ic Ip (F) be given. Let %‘ = {C E 9; C c F and p ( C )Ic}. On V,we introduce the following partial order. For C1, CZin V,CIS Cz if p (Cl - Cz)= 0. We show that every chain in %’ has an upper bound in V. Let { C , ; /3 E D} be a chain in %’.Let r = Sup & ( C , ) ; /3 E D}. We can find a sequence C,, IC,, I * * such that r = p (Con).Let C = Un=l C,,. Obviously, C E V and p ( C )= r. It is easy to verify that C is an upper bound of the given chain. By Zorn’s lemma, V has a maximal element, E say, i.e. if C E %‘ and E 5 C , then p (C - E) = 0. We claim that p (E) = c. Suppose p (E)< c. Then p (F- E) = p (F)- p (E)2 c - p (E)> 0. Since p is nonatomic, by Proposition 5.1.3, there exists a set Eo in 9 such that E c c F - E and O
5.1.7 Remarks. (i). A nonatomic positive bounded charge on a field 9 of subsets of a set R need not be strongly continuous. The following is a relevant example. Let R = [0,1] and 9 the field on R generated by the collection of all intervals of the form (a, 61 c [$,f). Let p be the restriction of the Lebesgue measure to 9. Then p is nonatomic on 9but not strongly For E =$, there is no decomposition of R in 9 which continuous on 9. satisfies the required properties. (ii). A strongly continuous positive bounded charge on a field 9of subsets of a set R need not be strongly nonatomic. The following is an example substantiating this statement. Let R = [0, 1).Let 9 be the field of all sets each of which is a finite disjoint union of intervals of the type [a, 6) with rational end points and 01 a Ib I1. Let p be the Lebesgue measure restricted to 9. Then p is strongly continuous on 9 .but not strongly nonatomic on 9. (iii). If p is a strongly continuous positive bounded charge on a u-field 9 of subsets of a set R, then p is strongly nonatomic on 9. In other words, (ii)+(i) in Theorem 5.1.6 is valid if 9 is a u-field on R. See Theorem 11.4.5 for a proof. (iv). If p is a bounded charge on 9,Theorem 5.1.6 still remains valid. If p is a charge on a field 9 without necessarily being bounded, then (i) (iii) in Theorem 5.1.6. (v). If 9 is a u-field on R, (iii)+(ii) in Theorem 5.1.6 is not true even though (i)e(ii). Let R=[O, 11, 9 = B o r e l u-field on R, A =Lebesgue A (B) = 0). Let T be any 0-1 valued charge measure on 9 and 4 = {BE 9;
+
144
THEORY OF CHARGES
on 9 such that T(B)= 0 for every B in 4. Let p = A + 2 ~ Then . p is nonatomic on 9but not strongly nonatomic on 9. Now, we examine the problem of characterizing strong continuity of a charge p in terms of the positive and negative variations, p + and p - , of p, as strong continuity plays a dominant role in the decomposition theorem to be proved in the next section.
5.1.8 Proposition. Let 9 be a field of subsets of a set R. (i). A bounded charge p on 9 is strongly continuous i f and only if p + and p - are strongly continuous. (ii). I f p1 and p2 are bounded strongly continuous charges on 9,then p 1 + p2 is strongly continuous. Proof. (i) follows from the facts that p + s I p I , p - s I ~ l . 1and I p I = p + + p - - . (ii) follows from the fact that l p +pzI ~ 5 Ip11+ 1 ~ 2 1 . The following proposition is useful in establishing the uniqueness part of the decomposition theorem of Section 5.2. 5.1.9 Proposition. Let 9 be a field of subsets of a set R and p,, n 2 1 a sequence of 0-1 valued charges on 9.Let a,, n 2 1 be a sequence of real anp, is numbers such that Cnzl lanlCOO.I f the bounded charge p =CnZl non-zero, then p is not strongly continuous on 9. Proof. Assume, without loss of generality, that lull >O. If p is strongly continuous on 9, then 1pl is also strongly continuous on 9. Now, since l a l l p l s lpl =Cnzlla,Ipn, p1 must be strongly continuous. But no twovalued charge is strongly continuous. 0 5.2 SOBCZYK-HAMMER DECOMPOSITION THEOREM
In this section, we prove the Sobczyk-Hammer decomposition theorem for charges. According to this theorem, we can write every charge p as a sum of two charges, one of which is strongly continuous, the other a countable sum of two-valued charges. We need some preliminary results. 5.2.1 Definition. Let 9 be a field of subsets of a set R. A sequence p,, n 2 1 of 0-1 valued charges on 9 is said to be finitely disjoint if for every n 2 1, there exists a partition {FI,Fz, . . . ,F,} of R in 9 such that pi(Fi)= 1 for every i = 1,2, . . . ,n. This notion essentially means that any finite subcollection of {p,, n 2 1) have disjoint supports. This notion is equivalent to the sequence pn,n 2 1 being distinct. This is stated in the following proposition.
5.
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145
5.2.2 Proposition. Let 9 be a field of subsets of a set R. A sequence p,, n L 1of 0- 1valued charges on Sisfinitely disjoint if and only if the sequence p,, n 2 1 is distinct, i.e. no two charges in the sequence are the same.
Proof. “Only if” part is obvious. “If” part. This can be proved by induction. Since p1 and p 2 are distinct, we can find A, in 9 such that p1(Al)#p2(A2). Assume, without loss of generality, that pl(A1)= 1. Then {Al, A;} is the desired partition for p1 and p 2 . Suppose, for pl,p 2 , . . . , p,, there is a partition {F1, F2, . . . , F,} of R in 9 such that F ~ ( F= ~1 ) for i = 1 ,2 , . . . ,n. Since pi and p,+1 are distinct, we can find Ei in 9 such that pi(Ei)= 1 and P,+~(E:)= 1 for every i = 1,2, . . . ,n. Let Bntl = E: and Bi = Bztl n Fi, i = 1,2, . . ,n. Then {Bl,B2,.. .,B,,Bntl} is a partition of ft in 9 and p i ( B t ) = l for i = 1 , 2,..., n + l .
nL
5.2.3 Remark. (i). If p,, n L 1is a sequence of distinct 0-1 valued charges on a field 9of subsets of a set R, it is natural to ask whether the sequence is infinitely disjoint, i.e. there exists a partition {F1,F2, . . .} of R in 9such that p , ( F n ) = 1 for every n 2 1. But this is not the case as the following example shows. Let R = {1,2 ,3 ,. . .} and 9= B(O), the class of all subsets of R. Let p1 be any 0-1 valued charge on B(R) such that p l( A) = 0 for every finite subset of R. See Example 2.1.3(4). For n 2 2 , let p, on 9 be defined by @,(A) = 1, if n E A,
=0, ifngA. This sequence p,, n z 1 of distinct charges is not infinitely disjoint. (ii). If 9is a c-field on R and p,, n L 1 is a distinct sequence of 0-1 valued then pn,n L 1 is infinitely disjoint. measures on 9, The following definition is useful to express the notion of strong continuity
of a charge in a form convenient for the development of the subsequent results.
5.2.4 Definition. Let 9 be a field of subsets of a set R and p a charge Let P={FI, F2,. . . ,F,} be a partition of R in 9.Then the number on 9. p p is defined by p p = max p ( F i ) . lcisn
Let B be the collection of all finite partitions of R in 9. A positive bounded charge p on 9is strongly continuous if and only if InfpEpp p= 0. This assertion is obvious from the definition of strong continuity of p. The following two lemmas lead to the decomposition theorem.
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5.2.5 Lemma. Let 9 be a field of subsets of a set R and p a positive bounded charge on 9 which is not strongly continuous. Let a = InfpEBpp. Then there exists a set F in 9 s u c h that (i). a 5 p (F)< 2a, and (ii). a Ip (Fi)for some Fi in any partition {FI, Fz, . . . ,F,} of F in 9.
Proof. Since p is not strongly continuous, a > 0. Choose any O < E
=0, i f p ( A n F ) < a .
Then A is a charge on 9. Proof. Let A and B be two disjoint sets in 9. Case (i). p ( ( A u B ) n F ) < a . Then p ( A n F ) < a and p ( B n F ) < a . Consequently, A ( A u B )= 0 = 0+0 = A (A)+ A (B). Case (ii). p ( ( Au B ) nF) L a. Then p ( A nF ) + p ( B nF) L a. Since p (F) = p ( A n F) + p ( B n F) + p ( F - ( A uB ) n F) < 2a, p(F- ( Au B ) n F) < a. Looking at the partition { A nF, B nF, F- ( Au B ) nF} of F and in view of property (ii) of Lemma 5.2.5, either p (A nF) Ia or p ( B nF) ? a . Note that both these inequalities cannot hold simultaneously. For, if they hold, then p (F) ? p ( A nF) + p ( B nF) I2a contradicting property (i) of Lemma 5.2.5. Consequently, A ( A u B ) = a = A ( A ) + A ( B ) . This proves the lemma. 0 Finally, we prove the main theorem of this section. 5.2.7 Sobczyk-Hammer Decomposition Theorem. Let 9 be a field of subsets of a set R and p a positive bounded charge on S. Then there exists a sequence pn,n 2 0 of distinct positive bounded charges on S a n d a sequence a,, n 2 1 of nonnegative numbers with the following properties. (i). po is strongly continuous on .!F. (ii). p, is a 0-1 valued charge on 9for every n L 1.
5.
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147
(iii). Cnrl a, COO. (iv). F = p0+Cnzl anpn. Further, the decomposition (iu)is unique. u,*p,* be another Proof. First, we establish uniqueness. Let p = pg +Inzl decomposition, where pg is a strongly continuous charge on 9,p,*, n z 1 is a sequence of 0-1 valued charges on 9 and a,*, n L 1 is a sequence of nonnegative numbers satisfying CnZla: < 00. Assume, without loss of generality, that pn’s are distinct, an’spositive, pz’s distinct and az’s positive. (Either sum Cnzl a,p, and Inzl a:p,* or both could be a finite sum. The following argument carries through in these cases also.) We show that p ; = po and anpa,n L 1 is a permutation of u,*p,*,n L 1. Observe that
*
c
Po - p o =
~nllfl-
fl2l
C a:&,*. fl2l
By Proposition 5.1.8, pg - p o is strongly continuous. By Proposition 5.1.9, it follows that pg = po and CnZla,p, =Inzl a,*p,*.Next, we claim that {anpn; n 2 1)= {a,*p,*; n L 1). Suppose this is not true. If {anp,; n 1 1) is not a subset of {a ,*p: ; n L l),there is an element a l p 1, say, of {a+, ;n L 1) which is not a member of { a , * p : ; n L 1). Choose N z 1 such that Cn2N+1a,* < al. Since p 1, p p . . . , p $ are distinct (why?), we can find Fin 9such that p 1(F)= 1 and p (F)= 0 for i = 1,2, . . . ,N. See Proposition 5.2.2. Consequently,
T,
ail
C
,251
z,
T
anpn(F)=C a:p:(F)= ns1
C
nzN+1
a,*p,*(F)
This contradiction shows that {a,p, ; n 2 1)c {a,*fi,* ; n L 1). By a similar argument, we can show that {a,*p,*;n 2 l)c{a,p,; n L 1). Thus we have proved that {a,*p,*; n L 1) ={a,p,; n L 1). Since the charges in each set are distinct, it follows that u,pn, n L 1 is a permutation of a:p:, n L 1. This establishes uniqueness. Now, we prove the main part of the theorem. If p is strongly continuous, then the conclusion of the theorem is trivially true. So, assume that p is not strongly continuous. Let InfPEppp= al. Then a l > 0. Let p i be the 0-al valued charge on 9 as defined in Lemma 5.2.6. Note that p i s p . Let A 1 = p - p i. Or, equivalently, p = p i + A 1. If A is strongly continuous, we stop here and we have the desired decomposition. If A 1 is not strongly continuous, using the argument given above, we can write A l = p ; + A2, where p i is a 0-a2 valued charge with a2>0, where a2 = InfPEp( p Continuing this way, we either stop at a finite number of steps reaching the desired decomposition or obtain a sequence p:, n L 1 of charges such that each p : is 0-a, valued with a, >O. In this case, CnZla, <00 since p is bounded. Let po= p -Inzl p:. We claim that po is strongly continuous.
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THEORY OF CHARGES
Observe that
for all N. Since aN -* 0 as N -* 00, we have that InfPEp(po)p = 0, i.e. po is strongly continuous. Writing each p = a+,, where each p, is a 0-1 valued charge, we obtain the desired decomposition. 0
5.2.8 Remark. The above decomposition theorem is also valid for any bounded charge p on 9.We can work out decompositions for p + and p separately and from these we can obtain a decomposition of p in the form p0+Cflz1a,p,, where p o is strongly continuous, each p, is a 0-1 valued charge and Cnzl la,l
c Ffl,
nzl
where (i) po is strongly continuous on 9, and (ii) p, is O-a, valued with a, > O for every n L 1. (a, could be equal to 00.) (The sum Cnzl pn could be a finite sum.) Since p is unbounded, p is not strongly continuous. But for any x >0, the charge p restricted to [0, x ) n9 is strongly continuous. It follows that p,([O, x ) ) = 0 for every n L 1 and p ([0, x ) ) = po([O,x ) ) = x . This shows that po is an unbounded charge on 9. Consequently, it cannot be strongly continuous. Hence p does not admit a decomposition along the lines of Theorem 5.2.7. The above example worked because every strongly continuous charge is necessarily bounded. By relaxing the definition of strong continuity of charges, we obtain a Sobczyk-Hammer like decomposition theorem for positive unbounded charges p. For A in 9,let p / A be the charge on A n 9 defined by (p/A)(B) = p(B) for B in 9and B c A.
5.2.10 Definition. Let 9 be a field of subsets of a set fl. A positive charge p on 9 is said to be strongly continuous on Fif p / A is strongly continuous on A n 9 f o r every A in 9 with p ( A ) < a . 5.2.11 Theorem. Let 9 be a field of subsets of a set fl and p a positive unbounded charge on 9. Then there exists a strongly continuous charge po
5.
on 9 a n d a family
NONATOMIC CHARGES
149
a E r}of two valued charges on 9 s u c h that
c Pa
p=po+
asT
and {pa; a E I?} is finitely disjoint, i.e. for any distinct al,a2,. . .,a,, in r, there exists a partition {A,,, A,,, , . .,A=,,}of R in 9 such that @*,(Ai) = pai(R)for i = 1,2, . . . ,n and n 2 1. (Here r could be uncountable!) Proof. Let %' ={A E 9; p (A)< 00). %' is a ring on R. For each A in %', by Theorem 5.2.7, we can write p/A=pOA+
c
piA,
izl
where p o A is a strongly continuous charge on A n and p ; A , i 2 1 is a sequence of two valued charges on A n 9 which are finitely disjoint. pi^ could be a finite sum.) Because of the uniqueness of the decomposition in Theorem 5.2.7, the charges poA,A EV are consistent in the following sense: if A, B E %' and B c A, then p o ~ / B= poB.Define po on 9 as follows.
(cizl
POW= POA(A), if A E%', = a,
if A&%' and A€%.
a charge on 9 and is, obviously, strongly continuous in the sense of Definition 5.2.10. For i 2 1 and A in %',let fiiA(F)= piA(FnA) for F in 9. Each &A is a two-valued charge on 9. Let UAt~{fiiA; i 2 1}= { p a;a E r}.Now, it is not difficult to verify that po is
5.2.12 Remark. It is possible to show that the above decomposition is essentially unique. As a corollary of Theorem 5.2.7, we can obtain the following decomposition theorem of Measure theory.
5.2.13 Corollary. Let 9 be a a-field of subsets of a set R and p a positive bounded measure on 9. Then we can write N
R = U A , , or nzO
UA,, n=O
for some N 2 0 with the following properties. (i). An's are pairwise disjoint. (ii). A,, is a p-atom for every n # 0. (iii). The restriction of p to the a-field A. n9is a nonatomic measure.
150
THEORY OF CHARGES
Proof. By Theorem 5.2.7, we can write
where (i) p o is strongly continuous on 9, (ii) p,,, n L 1 is a sequence of distinct 0-1 valued charges on 9, and (iii) a,, > 0 for every n 2 1 with Cnrl a,,
5.3 EXISTENCE OF NONATOMIC CHARGES In this section, we examine the conditions under which there exists a .F of subsets of a set SZ. The non-zero nonatomic charge on a given fieldC following definition is instrumental in providing a solution. Definition. Let 9 be a field of subsets of a set SZ. A collection of non-empty sets {Fil,iz ,...,ik; i l , i 2 , . . . , i k is any finite sequence of 0’s and l’s, k 2 1)in is said to be a tree in 9if the following conditions are satisfied. (i). FouFI = SZ, FonFl = 0. GI. Fil.iz .....i k - 1 . 0 ~ Fii.il.....ik-l,l =Fi1.i2.....ik-, andFi,,i, ,....ik-l,onFil,i2 ,...,i k - l . l = 0 f o r a l l i i , i z , . . . , i k - i ~ { O , 1)andk 2 2 .
5.3.1
The following theorem provides a set of equivalent conditions for the existence problem.
5.3.2 Theorem. Let 9 be a field of subsets of a set SZ. The following statements are equivalent. (i). There is a non-zero positive bounded nonatomic charge on 9. (ii). $contains a tree. (iii). There is a non-zero positive bounded strongly continuous charge on 9.
5.
NONATOMIC CHARGES
151
Proof. (i)+(ii). Let p be a non-zero positive bounded nonatomic charge on 9.Since p(R)>O, we can find Fo, F1 in 9 such that FouF1=R, FOnF1= 0 , 0 < p (Fo)< p (a) and 0 < p (F1)< p (R). Applying this technique to FOand FI separately, we obtain Foo,FOI,FIO,F11. Continuing this way, we obtain a tree in 9. (ii)+(iii). Let {Fil.i2 ....,ik; i l , iz, . . . ,ik E (0, 1) and k 2 1) be a tree in 9. Let 90be the smallest field on R containing this tree. In fact, gois precisely the collection of all finite disjoint unions of sets in the tree. Define ~ o ( ..... F it)~ =~1/2k , ~foril, ~ iz,. . . , i k in(0, 1)andk 2 l.pOcanbeextendedin the obvious fashion to Soas a charge p l . Note that p l is a strongly continuous positive bounded charge on go.Let p2 be any positive charge on 9 which is an extension of p1. See Corollary 3.3.4. It is obvious that p2 is strongly continuous on 9. (iii)J(i). This is part of Theorem 5.1.6. n
The above theorem provides an interesting implication for a-fields.
5.3.3 Corollary. Let 9be an infinite a-field of subsets of a set R. Then there exists a non-zero strongly continuous positive charge on 9. Proof. In every infinite a-field, one can find a sequence A,, n 2 1of pairwise disjoint non-empty sets whose union is R. This can be proved as follows. Let B be any non-empty set in 9whose complement, B", is also non-empty. Then either B n9 or B ' n 9 is infinite. If B n9 is infinite, we can find non-empty sets B1, BZ in 9 such that B1 u Bz = B and B1nBz = 0. Then either B l n 9 or B 2 n 9 is infinite. Proceeding this way, we obtain a sequence of pairwise disjoint non-empty sets. From this, we can obtain A,,, n 2 1 with the stated properties. Write the set N = (1,2, , . .} = N1 u N2, where N1 and Nz are disjoint infinite sets. Let FO= UiEN1 Ai and F1 = UiENZ Ai. Splitting N1 into two disjoint infinite subsets as above, we obtain Fooand Fol. Applying the same technique to Nz, we obtain Flo and F11. Proceeding this way, we obtain a tree in g.Theorem 5.3.2, now, completes 0 the proof.
Theorem 5.3.2 also provides useful information on superatomic fields. Keeping in mind the definitions of an atom, atomic field and nonatomic field in Definitions 1.4.9, we define superatomic fields as follows.
5.3.4 Definition. Let 9 be a field of subsets of a set R. 9 is said to be superatomic if every sub-field of 9is atomic. 5.3.5 Remarks. (i). If R = { 1 , 2 , R, then 9is a superatomic field.
. . .) and 9 is the finite-cofinite field on
(ii). Let a be any ordinal and R=[O,a] be equipped with the order topology. This is the topology on R which has the collection of all subsets
152
THEORY OF CHARGES
C of R of the form ( a , 6 ) or [O,b)or ( a , a ]for O s a s b ICY as a base. Let 9be the field of all clopen subsets of R. Then 9is superatomic. See Pierce (1970). (iii). Let .F1 and SZbe superatomic fields on Rl and respectively. Let 9109~ be the smallest field on Rl X RZ containing {AX B; A E 9 1 and B E 95).Then S1092 is a superatomic field. (iv). If 9 is a field on a set R, then 9 is superatomic if and only if the Stone space X of 9 is scattered, i.e. no non-empty subset of X is perfect. See Sikorski (1969) especially the statement on page 35 under D. The following theorem gives equivalent versions of superatomicity and describes all the bounded charges explicitly on superatomic fields. 5.3.6 Theorem. Let 9 be a field of subsets of a set R. The following statements are equivalent. (i). 9 is superatomic. (ii). $does not contain a tree. (iii). There is no non-zero nonatomic positive bounded charge on 9. (iv). There is no non-zero strongly continuous positive charge on 9. (v). Every positive boundedcharge p on 9 h a s the following representation
where ( a ) ai 2 0 for every i and valued charge on 9.
Cizl ai <m,
and ( 6 ) each pi is a 0-1
Proof. The implications (ii)+ (iii)+ (iv)+ (v) follow from Theorem 5.3.2 and Theorem 5.2.8. (v)+(i). Suppose (v) holds and (i) is not true. We show that 9 contains a tree. Since 9 is not superatomic, there is a subfield Poof 9 which is not which does atomic. This implies that there exists a non-empty set A in 90 Then we can find A. and A1 in 90 such that not contain any atom of go. A. u A1= A, A. n A1= 0, A. # 0 and A1# 0. Repeating this procedure on A. and A1,we obtain Aoo,Aol c A. and Alo,All c A1having properties similar to the above sets. Continuing this way, we obtain a collection {Ail.iz,...,it; il, i2, . . . ,ik E (0, 1) and k 2 1)of non-empty sets in Powith the properties (a) Ai1,iZ.....ik-1.0UAil.iz.....ie-1.1 =Ail.iz,....i k - 1 and (b) Ail,iz,.... i k - 1 , o n Ai1,iz.....i k - 1 . 1 = 0 for all i l , iz, . . . ,ik-l in {0,1} and k 2 2. From this, it is not difficult to obtain a tree in 9. Consequently, by Theorem 5.3.2, there exists a non-zero
5.
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153
strongly continuous positive charge on 9. Such a charge can never be decomposed in the way condition (v) stipulates. This contradiction proves the desired implication. (i) j(ii). If 9 contains a tree, then the subfield generated by the tree is nonatomic and hence 9cannot be superatomic. This contradiction proves the desired implication. 0 We now proceed to obtain some topological conditions for the existence problem of nonatomic charges. We need some preliminary results for this purpose. Proposition. Let 2 be a u-field of subsets of a set a,9 a field on generating 2 and p a positive bounded measure on a. Then p is strongly continuous on 2 if and only if p is strongly continuous on 9. 5.3.7
Proof. “If” part is trivial. “Only if” part. Assume that p is non-zero. Let E >O. Let m be a natural number such that l / m < E . Since p is strongly continuous on a, we can find a partition {F1, F2,. . . , F,} of fl in ‘tisuch that
for every i. For each 1Ii In, by Theorem 3.5.3, we can find Gi in 9such that
Since G i c Fi u (FiAGi),we have
for every i. Further,
s
n
n
i=l
i=l
C p ( F i - G i ) s C p(FiAGi)
Defining D1=G1, D i = G i - U j l \ G j for i = 2 , 3 , . . . ,n and Dn+l= n-Ur=l Gi, we observe that {D1, D 2 , . . . ,Dn+l}is a partition of s1 in 9 0 satisfying p (Di)< E for every i. This completes the proof. The following result provides a simple necessary and sufficient condition for nonatomicity of regular measures on the Bore1 u-field of a compact Hausdorff space.
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THEORY OF CHARGES
Lemma. Let X be a compact Hausdorff space, W its Bore1 u-field and W O its Baire a-field. Let p be a positive bounded regular measure on 9.Then the following statements are equivalent. (i). p is nonatomic on 3. (ii). p ( { x } ) = 0 for every x in X. (iii). p is nonatomic on Wo. 5.3.8
Proof. (i)J(ii). This is obvious. (ii)3 (iii). Let B in Wo be a p-atom. Since p is regular, p (B) = Sup { p (C); C c B, C E $3330and C is a compact Gs subset of X}. Since p (B) >0, there exists a compact Gs set C contained in B such that p(C) >O. Let x E C. Since p is regular, 0 = p ( { x } )= Inf (II.(V); x E V, V an open F, set c X} and therefore, we can find an open F, set V, containing x such that p (V,) < p (C).{V, ; x E C} is an open cover for C. There is a finite sub-cover {V,,, V,,, . . . ,V,,,} of C. Since B is a p-atom, C is a p-atom. Since p (V, nC) 5 p (V,) < p (C), it follows that p (V, n C) = 0. Consequently, p (C) = p ((U~=I VXi)n C) = 0. This contradiction shows that there are no p-atoms in Wo. Hence p is nonatomic on Wo. (iii)J(i). Since p is nonatomic on Wo, p is strongly continuous on Wo. See Theorem 5.1.6. Hence p is strongly continuous on the bigger a-field W. So, p is nonatomic on W. O The following theorem gives some more equivalent versions for the existence of non-zero nonatomic charges, mainly topological in nature. 5.3.9 Theorem. Let 9 be a field of subsets of a set Q and X its Stone space. The following statements are equivalent. (i). There is a non-zero nonatomic positive bounded charge on $. (ii). There exists a countable nonatomic subfield 90of 9. (iii). X contains a perfect set. (iv). 9 contains an ideal 4; such that the quotient Boolean algebra 91.9 is nonatomic.
Proof. (i)+(ii). By Theorem 5.3.2, 9 contains a tree. The smallest field Soon Q containing a tree is a countable nonatomic subfield of 9. (Note that a tree is obviously a countable collection of sets.) See Corollary 1.1.14. (ii) (iii). Let so be a countable nonatomic subfield of 9.Without loss of generality, assume that gois generated by a tree. Let Y be the Stone space of so. See Theorem 1.4.10. We note that Y is perfect. In fact, Y is homeomorphic to the Cantor set {0,1}"". This follows from the fact that and goare isomorphic. For, the field V of all clopen subsets of (0, l}Ko since V is generated by the tree {{il}X {i2}X * x {in}x (0, 1)x * * ; i l , iz, * ,in ~ { 0 , 1 and } n 2 1) and gois generated by a tree, V and 9,, are isomorphic. It is obvious that (0, 1)"" has no isolated points, i.e. it is
+
-
-
5.
NONATOMIC CHARGES
155
perfect. Since the inclusion map i from goto 9defined by i(A) = A, A E $0 is a one-to-one homomorphism, there exists a continuous function f from X onto Y . See Theorem 1.4.11. Now, we claim that there exists a minimal closed subset P of X such that f ( P ) = Y. The collection 8 = {E c X; E is closed and f (E)= Y } is non-empty and is partially ordered by set inclusion, i.e. for El, EZin 8, say El IE2 if El c EZ.Let {E, ;a E D }be a chain in 8. We claim that E, E 8. Let y E Y. For each a in D, there exists x, in E, such that f (x,) = y. Since X is compact, there exists a subset of x,, a E D converging to some element x in X. Further, x p E E, whenever p Ia , since {E,; a! E D} is a chain. Consequently, since E, is closed, x E E, for every cy in D. Thus x E n,=DE, and f ( x ) = y . Hence E, is a lower bound of the chain {E, ;a E D}. So, we have proved that every chain in 8 has a lower bound. By Zorn’s lemma, there exists a minimal closed set P c X such that f ( P )=Y .
n,,D
naED
We also claim that P is perfect. Suppose not. Let x be an isolated point of P . Since P - { x } is compact, f ( P - { x } ) = Y - { f ( x ) } is compact. Consequently, f (x) is an isolated point of Y . This implies that Y is not perfect. This is a contradiction. Hence P is perfect. (iii)+(iv). Let P be a given perfect subset of X. Let V be the collection of all clopen subsets of X. Let 9’ = {CE V;P nC = 0).9’is clearly nonempty and is an ideal in %. We claim that the quotient Boolean algebra V/9’is nonatomic. Let [A] be a non-zero element in V/9’. Then A n P # 0. Further, A n P contains at least two points x, y E X , because P is perfect. Since X is compact and totally disconnected, there exists C, D in V such that X E C C A , ~ E D and ~ A C n D = 0 . Note that O<[C]<[A] and O<[D]<[A]. Consequently, %I$’has no atoms. Since 9 and V are isomorphic, (iv) follows. (iv)+(i). Since $19 is nonatomic, 919 is not superatomic. By Theorem 5.3.6, there is a non-zero nonatomic positive bounded charge p on 919. Using the quotient homomorphism from 9 to 919,the charge p can be lifted as a charge @ on 9. @ is obviously nonatomic. This completes the proof. Finally, we close this section with a result on the existence of nonatomic measures on the Borel u-field of a compact Hausdorff totally disconnected space.
5.3.10 Corollary. Let X be a compact Hausdorff totally disconnected space and W its Borel u-field. Then there exists a non-zero nonatomic positive bounded regular measure on W if and only if X contains a perfect set. Proof. Suppose p is a non-zero nonatomic positive bounded regular measure on W.Then p is a nonatomic measure on the Baire cr-field Wo of
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THEORY OF CHARGES
X. See Lemma 5.3.8. Consequently, p is strongly continuous on Wo. See Theorem 5.1.6. Note that the field %' of all clopen subsets of X is a generator See Section 1.3. By Proposition 5.3.7, p is strongly continuous on of a0. %'.By Theorem 5.3.9, X contains a perfect set. Conversely, let X contain a perfect set. Let %' be the field of all clopen subsets of X and Wo its Baire a-field. By Theorem 5.3.9, there exists a non-zero strongly continuous positive bounded charge p on %'.Every charge on %' is a measure. See Example 2.3.5(3). Since Wo is generated by %, there is a measure C; on Wo which is an extension of p. See Section 1.3 and Theorem 3.5.2. By Proposition 5.3.7, fi is a strongly continuous positive measure on Wo. Let A be the regular measure on W which is an extension of I;.. See Theorem 3.5.5. By Lemma 5.3.8, A is nonatomic. This 0 completes the proof.
5.4
DENSENESS
Let 9 be a field of subsets of a set R. Let A be the collection of all A is equipped with a topology as follows. A net probability charges on 9. pa, (Y E D in A is said to converge to a p in A if pu,(F)= p (F) for Let A 1be the collection of all nonatomic probability charges every F in 9. on 9 and A Zthe collection of all strongly continuous probability charges Obviously, A 2c A1c A.In this section, we examine the conditions on 9. under which A 2is dense in A.We need a preliminary result. 5.4.1 Proposition. Let 9 be a field of subsets of a set a. Let X be the If X is perfect, then for any given non-empty set F in 9, Stone space of 9. there exists a strongly continuousprobability charge p on S s u c h thatp (F)= 1.
Proof. The perfectness of X is equivalent to the fact that 9 is nonatomic. In the field F n9 on F, one can easily construct a tree. By Theorem 5.3.2, there exists a strongly continuous probability charge A on F n9. Define p on 9by p (E) = A (E nF), E E 9. p is the desired charge. The following is the main result of this section. 5.4.2 Theorem. Let 9 be a field of subsets of a set R and X its Stone space. Then A2 is dense in A if and only if X is perfect.
Proof. "If" part. Let p EA.We construct a net of strongly continuous probability charges converging to p. By Theorem 5.2.7, we can write
5.
NONATOMIC CHARGES
157
where (i) po is a strongly continuous positive charge on 9, (ii) ai L O for every i and po(R)+Iizl ai = 1, and (iii) p i is a 0-1 valued charge on 9 for every i L 1. Let 9i= {FE 9;p i (F)= l},i L 1.For every F in gi,choose and fix a strongly continuous probability charge pi,^ on 9such that pi,F(F)= 1. This can be done in view of Proposition 5.4.1. Consider the product set 91 X P zx with the following partial order. For (F1,FZ,. . .) and (El, Ez,. . .) in 9 1 X 9 2 x * , say (F1,F2,. . .) L (El, E2,. . .) if Fi c Ei for x sz x * . . is a directed set. For every i. Under this partial order L, S1 FZ,. . .) in S1 x s zx * * , let p(F1,~2 ,...) = p 0 + 1 + ~aipi,Fi.It is easy every (F1, to check that ~ ( F ~,...), FE Atz. ~ We claim that the net p(FI,F2,...), (F1, Fz, . . .) E S1 x 95 x * * converges to p in the topology of A. Let E in 9 be fixed. Let NI = {i L 1; E E 9i} = {il, i2, . . .} and {1,2,3, . . .}-{il, i2, . . .} = { j l , j z , : . .}. Since 95 is a maximal filter, E“Eq k for every k L 1. Define E: for each n L 1 by
--
-
-
-
ifn = i l , i z , . . . ,
E:=E, =E‘,
if
~t =jl,j2,.
...
-
Let (F1,FZ,. . .) in 9 1 x 9 z x * * be such that (F1,FZ,. . .)r(ET,E;, . . .). This implies that Fi c E for i = il, iz, . . . and Fi E” for i = j l , j z , . . . . Con= po(E)+C,,, aik= CLW. Hence the net sequently, ~(F~,F~,...)(E) ~ ( F ~,...I. F(E), ~ (FI,Fz, . . .) E X 9 2 X * * converges to p (E).This completes the proof of the “if” part of the theorem. “Only if” part. Since the perfectness of X is equivalent to the fact that 9 is nonatomic, it suffices to show that 9 has no atoms. Suppose F in 9 is an atom of 9.Take the probability charge p on 9 such that p(F) = 1. By the hypothesis, there exists a net pot a! E D in A2 converging to p in the topology of A. In particular, lirnmSD p, (F)= p (F)= 1. But p, (F) = 0 for every a! in D. This contradiction proves the result. 0 3
The following corollaries follow guite easily from the above theorem. 5.4.3 Corollary. Let 9be a field of subsets of a set R whose Stone space X is perfect. Then given any 0-1 valued charge p on 9, there exists a net pa, cu E D of strongly continuous probability charges on 9converging to p in the topology of A.
Corollary. Let 9be a field of subsets of a set R. Let X be the Stone space of 9. Then the following statements are equivalent. (i). 9is nonatomic. (ii). X is perfect. (iii). Azis dense in A. (iv) A lis dense in A.
5.4.4
158
THEORY OF CHARGES
Finally, we state a result in topological measure theory. Let X be a compact Hausdorff space and 93 its Bore1 u-field. Let A* be the collection of all regular probability measures on 93. The weak* topology on A* is described as follows. A net p,, a E D in A* converges to a p in A* if limaED5 f dp, = f d p for every real valued continuous function f on X. If, in addition, X is totally disconnected, a net pa, a E D in A* converges to a p in A*in the weak* topology of A * if and only if limaGDp, (C) = p (C) for every clopen set C c X . This follows from the fact that (f:X+R:, f= C J ~ ,for some cl, c2,. . . ,cn real, ~ 1 ~ , 2 , ...,C, clopen sets cx and n 2 1) is norm dense in the space of all real continuous functions on X equipped with supremum norm. See Section 1.3. Let A?be the collection of all regular nonatomic probability measures on 3.
xr=l
5.4.5 Corollary. Let X be a compact Hausdorff totally disconnected space and A" and AT be as defined above. Then A? is dense in A* in the weak" topology of A" if and only if X is perfect.
CHAPTER 6
Absolute Continuity
In this chapter, we formally introduce the notions of absolute continuity and singularity for charges. In Section 6.1, we study various properties of absolute continuity and singularity in the framework of charges and establish the connection with the existing notions of absolute continuity and singularity in Measure theory. In Section 6.2, we obtain Lebesgue Decomposition theorem for charges using Riesz Decomposition theorem in Vector lattices. Finally, in Section 6.3, we prove Radon-Nikodym theorem.
6.1 ABSOLUTE CONTINUITY AND SINGULARITY The following notion of absolute continuity is the main one we study extensively in this section.
6.1.1 Definition. Let p and v be two charges defined on a field 9 of subsets of a set SZ. v is said to be absolutely continuous with respect to p i f given E >0, there exists S > 0 such that Iv(E)I < E whenever E E 9 and Ip /(E)< S. If v is absolutely continuous with respect to p , we use the notation v << p.
6.1.2 Remarks. There are two other notions one could introduce related to absolute continuity. (i). v is said to be weakly absolutely continuous with respect to p if v(E) = 0 whenever E E 9 and Ipl(E) = 0. In this case, we use the notation vw<
6.1.3. Remarks. Let p and v be two charges on a field 9 of subsets of a set 0. (i). vw<
160
THEORY OF CHARGES
(iii). u < < p implies uw<<w. But the converse is not true. Let R = {l,2 , 3 , . . .}, 9=S(R),the class of all subsets of R and p on 9 be defined by
p(A)=
c
1
neA
for A in 9. Let u be any 0-1 valued charge on 9 such that u(F) = 0 for any finite subset F of R. It is obvious that U W K p. On the other hand, u is not absolutely continuous with respect to p. For E = $, there is no S > 0 such that u(F) < E whenever F E 9 and p (F) < 6. For any given S > 0, one can always find a cofinite set F such that p (F) <8. (iv). If p is a positive charge, then u << p if and only if v s << p. (v). If us<
6.1.4 Proposition. Let p and u be two charges defined on a field 9 of subsets of a set R. Then the following statements are equivalent. (i). u << p. (ii). u+<
I.
(iii) 3 (i). This is obvious. For the last part, observe that if u << p+, then u << p + + A for any positive charge A on 9.In particular, u << p + + p - = Ip 1. 0 Now, some comments are in order on the above proposition.
6.1.5 Remarks. (i). If u < < p , neither u < < p + nor v < < p - need hold. As an example, let R={1,2}, 9 = S ( R ) , p({l})=& p({2})=-$, ~ ( { l } ) ~ = $ a nu({2})=1. d Note that p+({l})= 4, ~ ~ ( ( 2=)0,) ~ ~ ( ( 1=)0,) pL-({2})= 2 and Ipl= u. Therefore, u << p. But neither u << p + nor u (< p - is valid.
6.
ABSOLUTE CONTINUITY
161
(ii). Proposition 6.1.4 still remains valid if absolute continuity is replaced by weak absolute continuity in the statement. (iii). Proposition 6.1.4 is not valid if absolute continuity is replaced by strong absolute continuity. Some simple examples based on those in (i) can be provided.
As has been pointed out earlier, absolute continuity and weak absolute continuity are not equivalent in general. However, for bounded measures on a-fields, these two notions are equivalent.
6.1.6 Theorem. Let 9 be a a-field of subsets of a set R and p and v measures on 9 s u c h that v is bounded. Then vw<
the other hand, since Ivl is bounded,
n u Fk)=!L21vl(gnFk)
IVl(F)=lvl( n z l
kzn
r l i m sup lv/(Fn)2
~
.
n+m
Thus we have IpI(F)=O and IvI(F)zE. But this is a contradiction to the assumption that v w p.~ This completes the proof. 0
6.1.7 Remark. In the above theorem, neither the assumption of boundedness of v nor the assumption that the measures are defined on a a-field can be dropped for its validity. (i). We treat the boundedness part of the above theorem. Let R = [0,1], 9= Bore1 u-field on R and p the Lebesgue measure on 9. Define Y on
162
THEORY OF CHARGES
9by v(A)=O,
ifAE9andp(A)=O,
= m , if A E 9 a n d p ( A ) > O . It is easy to check that v is a measure on 9 and that vw<
.FA 1
if A E 9 and A is finite, 1
=1- C ..AC
.(A)=
2"'
1 1neA
=2-
ifAE9andA'isfinite. if A E 9 and A is finite,
2"' 1
C -
noAC2"'
if A E 9 and A' is finite.
Note that p and v are charges on 9. Every charge on 9 is a measure. Obviously, vw<
6.1.8 Theorem. Let 9 be a field of subsets of a set fl and 2l the a-field on R generated by 9.Let p and v be two positive bounded measures on 8. T h e n v < < p o n 9 i f a n d o n l y i f v < < p o8n . Proof. If v << p on 8 , then it is obvious that v << p on 9. Now, we assume Let E > 0. There exists S > 0 such that v(E) < ~ / whenever 2 that v << p on 9, E E 9 and p (E)/S. Now, let F E 8 and p(F) < 6/2. We show that v(F) < E . By Theorem 3.5.3, there exists a set E in 9 such that p (FAE)+ v(FAE)< min { ~ / 26/2}. , Then lp (F)- p (E)I Ip (FAE)< S/2. Consequently, p (E)< p (F)+S/2 < S/2 + S/2 = 6. Therefore, v(E) < ~ / 2 .On the other hand, v(FAE)<&/2 from which it follows that Iv(F)-v(E)l<~/2. So, v(F)< v(E) + ~ / <2~ / 2 ~+ / =2E . This completes the proof of the theorem. 0
6.
163
ABSOLUTE CONTINUITY
6.1.9 Remark. The above theorem is valid for any two bounded measures and v.
p
We now study s-boundedness of charges in the light of absolute continuity. See Definition 2.1.4.
6.1.10 Theorem. Let p and v be two charges on a field 9 of subsets of a set 0. Then the following statements are true. (i). v is s-bounded if and only if 1vI is s-bounded. (ii). If v << p and p is s-bounded, then Y is s-bounded. (iii). If Y << p , p is s-bounded and v is a real charge, then v is bounded. (iv). If v << p , p is bounded and v is a real charge, then v is bounded. Proof. (i). Suppose v is s-bounded and lv( is not s-bounded. Then there exist a sequence A,,, n 2 1 of pairwise disjoint sets in 9 and E > O such that IvI(A,) > E for every n 2 1. Write IvI(A,,)= v+(A,,)+ v-(A,) for every n 2 1. Then there exists a sequence n l < n z < ' . such that v + ( A , , ) > ~ / 2 for every i 1 1 or there exists a sequence kl< k z < * such that .-(Aki)> ~ / for 2 every i 2 1. Assume that the former holds. Since v+(A,,) > ~ / 2 , there exists Bi in 9 such that Bi CAniand v(Bi)>&/2for every i L 1. Bi, i 2 1 is a sequence of pairwise disjoint sets in 9and limi+a v(Bi)f 0. This contradiction proves that Iv 1 is s-bounded. The converse is trivial. (ii). If v < < p , then v < < 11.1.1. By (i), lpl is s-bounded. Therefore, v is sbounded. (iii). Every s-bounded real charge is bounded. See Corollary 2.1.7. (iv). This follows from (iii) and the fact that every bounded charge is s-bounded. 0
-
Now, we study the countable additivity property of charges in the presence of absolute continuity.
6.1.11 Theorem. Let p and v be two charges defined on a field 9 of subsets of a set 0. If v << p, v is a real charge and ,u a bounded measure, then v is a measure.
n,,,
Proof. Let A,,, n 2 1 be a decreasing sequence of sets in 9 with A,, = 0 .It suffices to show that limn+mv(A,) = 0. See Proposition 2.3.2(2).Since p is a bounded measure, Ip I(A,,)= lp I(0) = 0. Since v << p , limn+mv(A,,)= 0. This completes the proof. 0
6.1.12 Remark. Neither the assumption that v is real valued nor the assumption that p is bounded can be relaxed in the above theorem. Suitable examples can be provided in the framework of finite-cofinite field 9 on n = { l , 2 , 3 , . . .}.
164
THEORY OF CHARGES
Now, we consider a sequence p,, n 2 1of bounded charges on a field 9 of subsets of a set s2. Let p on 9 be defined by
for F in 9.It is obvious that p is a bounded charge on 9.Further, p, << p for every n 2 1. The following theorem indicates that p is, in a sense, minimal with respect to the above property of absolute continuity.
6.1.13 Theorem. Let p,, n z 1 be a sequence of bounded charges on a field 9 of subsets of a set R. Let p be defined as above. Let u be any charge on 9 s u c h that p, << u for every n 2 1. Then p << u. Proof. Let E >O. Choose W z 1 such that 1 / 2 N + ' < ~ / 2 . Since p l , p 2 , . . . , pN are absolutely continuous with respect to u, there exists S > 0 such that Ipi(B)I < ( ~ / 2 ) ( 1 + Ipil(R)) for i = 1 , 2 , . . . ,N whenever B E 9 and IvI(B)<S. Now, if BE^ and Iul(B)<S, then
< & / 2 +E/2 = E . Hence p << v.
0
Next, we take up the study of singularity for charges.
6.1.14 Definition. Let p and u be two charges defined on a field 9 of subsets of a set 0. p and u are said to be singular if for every E >O, there exists a set D in 9 such that Ip I(D)< E and Iul(D")< E . If p and u are singular, we use the notation p I u. If p and u are positive charges, then p Iu if and only if p A u = 0. See Definition 2.5.1. 6.1.15 Remark. The following definition might seem natural as a notion of singularity. p and u are said to be strongly singular, if there exists a set D in 9such that IpI(D)= 0 = lvl(Dc).If this is the case, we use the notation pslu.
Some comments are in order about these two concepts.
6.1.16 Remarks. (i). p Iv if and only if Ip I I IuI. (ii). p Iu if and only if u I p . (iii). If p s l u, then ,u I u. But the converse is not true. Let s2= {1,2,. . .}
6.
165
ABSOLUTE CONTINUITY
and 9= P(R). Let p on 9be defined by
Let u be any 0-1 valued charge on 9 such that u ( A )= 0 for every finite subset A of R. Then p I u. But p and v are not strongly singular. Strong singularity and singularity coincide for measures on u-fields.
6.1.17 Theorem. Let 9 be a u-field of subsets of a set R and two measures on 9. Then p 1u if and only if ps 1u.
p
and u
Proof. "If" part is clear. "Only if" part. Let p I u . For each n 2 1, there exists a set D, in 9 such that Ip I(Dn)< 1/2" and IuI(DZ)< 1/2". Let D = lim sUPn+mD,. Then lp I(D) = 0. The proof is exactly similar to the one given in the proof of Theorem 6.1.6. Let E=lirnsup,,,D',. By a similar argument, it can be shown that IuI(E)= 0. Note that D" = lim infn+mD Z c lirn D', = E. So, Iul(D')=O. Hence p s 1 v . 0
As a consequence of the above theorem, we obtain Hahn Decomposition theorem for measures.
6.1.18 Corollary. (Hahn Decomposition Theorem for Measures). Let p be a measure on a u-field Yl of subsets of a set R. Then there exists a set D in 8 such that p (A)2 0
whenever A E
and A c D,
p (B) 5 0
whenever B E 8 and B c D".
and
Proof. By Lemma 2.5.5, p is either bounded below or bounded above. So, by Remark 2.5.4(ii), we have p=p+-pL-
and p + ~ p - = O .
Since p + ~ p - = O we , have p c I p - . By Theorem 6.1.17, there exists a set D in 3 such that pf(DC)= 0 = p-(D). From the definition of p + , it follows that p (B) 5 0 whenever B E 8 and B c D'. By a similar argument, it follows 0 that p (A)2 0 whenever A E Yl and A c D. This completes the proof. The following theorem is analogous to Theorem 6.1.8. This theorem also indicates that Definition 6.1.14 is a natural definition of singularity for charges in harmony with the corresponding notion used in Measure theory.
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THEORY OF CHARGES
6.1.19 Theorem. Let 2l be a u-field of subsets of a set R and p and v two bounded measures on 2L Let 9be a field on R generating M. Then p L v o n % if and only if p l v o n 9. Proof. "If" part is obvious even without the assumption of boundedness of the charges. "Only if" part. Let p Iv on a. By Theorem 6.1.17, there exists a set D in 3 such that Ipl(D)=O=Ivl(D"). Let E > O . By Theorem 3.5.3, there exists a set A in 9 such that \pl(AhD)+Ivl(AhD)<~.Now, llpl(A)~ ~ I ( D ) I I I ~ I ( A A D ) < EHence . I,uI(A)<E. Further, ~ ~ v ~ ( A c ) - / v ~ ( D c ) ~ ~ Ivl(AChD")= lvl(AhD)<~.HenceIvI(A")<E.Thiscompletestheproof. 0
We end this section with some remarks. 6.1.20 Remarks. Let p, v and T be charges on a field 9 of subsets of a set R. (i). If v << p and Y I p, then v = 0. This can be proved as follows. Let E > 0. There exists S > 0 such that lv\(A)<E whenever A E 9 and ( p\(A)< 8. One can take S < E . Since v l p , there exists a set D in 9 such that Ivl(D)<S andIpI(D")<S.Therefore, Ivl(R)=Ivl(D)+lvl(D")<S+~ < 2 ~Since& . >O is arbitrary, Ivl(R) = 0. (ii). If p << T and v L T, then p I v. An argument similar to the one above can be presented here. (iii). Some additional properties of absolute continuity and singularity will be given in Section 8.5. 6.2 LEBESGUE DECOMPOSITION THEOREM In this section, we embark on proving Lebesgue Decomposition Theorem for charges using Riesz Decomposition Theorem for Vector lattices. Recall that the space ba(R, 9) of all bounded charges on a field 9 of subsets of a set R is a boundedly complete Vector lattice. First, let us examine how far the notion of singularity introduced for charges in Section 6.1 is related to the notion of orthogonality introduced in Vector lattices. 6.2.1 Theorem. Let p, v E ba(R, 9).Then p and v are singular if and only if and v are orthogonal in the vector lattice b a ( R , m . (Consequently, the notation used for singularity of two charges is consistent with the notation used for orthogonality of two elements in the Vector lattice ba(R, 9).) Proof. Suppose p and v are singular. Let E >O. There exists a set D in 9 such that lp [(D)< E and Ivl(D") < E . Consequently, (1p I A IvI)(R) =
6.
ABSOLUTE CONTINUITY
167
Inf {lp I(B)+ lvl(Bc); B E 8< 2 ~ Since . E >0 is arbitrary, it follows that lpl A1.1 = 0. This implies that p and v are orthogonal in the Vector lattice ba(S1,9). The converse can be proved similarly. 0 Now, we characterize absolute continuity in terms of notions of Vector lattices. For the following, one has to recall the definition of the orthogonal complement of S, S', for any subset S of a Vector lattice. See Definition 1.5.7.
6.2.2 Theorem. Let p , v E ba(R, 9).Then v << p if and only if v E { { p } ' } l . Proof. From the definitions { p } l = {T E ba(Q
9); TI p}
and {{p}*}' = {A E ba(Q, 9); A IT for
every T in b }'},
it follows that v E {{p}l}' if and only if v I T whenever T I p . Suppose v << p. Let T E ba(fl, 9)and T IF. We show that v IT. But this follows from Remark 6.1.20(ii). Conversely, let v E { { p } l } * . Since {p}* = {lp I}', we assume, without loss of generality, that v and p are positive. By Theorem 1.5.12, v=
V (vAnp). n2l
Since v A n p 5 v A ( n + l ) p for all n 2 1, the set function A on 9 defined by A (F) =limn+- (v A n p ) ( F ) , F in 9 is a charge on 9. Consequently, the lattice supremum v is precisely A, i.e. v(F) = limn+- (v A np)(F), FE9. Since v A n p ~ n p v, A n p << p for every n 11. Now, we show that the sequence v A np, n 2 1 of charges converges uniformly to v. For any F in 9, lv - v A npI(F)IIv - v A n p I ( 0 )= (v - v A n p ) ( n )= v(n)- (v A n p ) ( n ) which converges to zero as n + co. Hence v A np, n 2 1 converges to v uniformly. Since v ~ n <
6.2.3 Corollary. (i). If p l , (p1+ F 2 )
168
THEORY OF CHARGES
(iv). I f p L 1 , p 2 , p € b a ( R , q , p 1 < < p a n d p 2 < < p , t h(epn ~ + p d < < p . (v). I f pn, n 2 1 is a sequence in ba(R, 9)converging to a p in b a ( R , 9 ) in the norm of ba(0, q and pn <( A for every n L 1for some A in ba(fl, 9), then p << A.
Proof. (i) and (ii) follow because {v}' is a vector sublattice of b a (Q 9 ). See (iii) follows from the fact that {v}' is a closed subspace of ba(R, 9). Theorem 1.5.19. (iv) and (v) follow from the fact that {{p}'}' is a closed vector sublattice of ba(R, 9). 0 Now, we have developed enough machinery to prove Lebesgue Decomposition theorem.
6.2.4 Lebesgue Decomposition Theorem. Let p and v E ba(0, 9). Then there exist v1, v 2 in ba(R, 9) having the following properties. (i). v = v 1 + v2. (ii). vl<< p. (iii). v z l p . (iv). I4 = b11+ I d . (v). I f v is positive, then v l and v 2 are positive. Further, any decomposition of v satisfying (ii)and (iii) is unique. Proof. By Theorem 1.5.8, { p } l and {{p}'}' are normal vector sublattices of b a ( n , 9 ) . By Riesz Decomposition theorem 1.5.10, for a given v~ ba(R, 9), there exist V I in {{p}'}' and v 2 in {p}' such that v = v 1 + v2. By Theorem 6.2.2, v 1 << p. Obviously, v 2 I p. Uniqueness of the decomposition follows from Theorem 1.5.10. By Remark 6.1.20(ii), it follows that v 1 Iv2. By Theorem 1.5.4(21) and (22),1v1= Iv1+ v21 = lvll +Iv21. Thus (iv) follows. (v) follows from the uniqueness of the decomposition and (iv). This completes the proof. 0 The above theorem gives a decomposition for bounded charges only. It is natural to enquire about the validity of Lebesgue Decomposition theorem for charges not necessarily bounded. We present some negative results. (See Notes and Comments for some positive results.)
6.2.5 Theorem. Let 9 be a field of subsets of a set R and v an unbounded real charge on 9. Then there exists a bounded charge p on 9 such that Lebesgue Decomposition theorem is not valid for v with respect to p. Proof. Let 4' = {Ft 6 ; IvI(F) < 00). 4'is an ideal in 9 Let 4 be a maximal ideal in 9containing 4'. Define p on 9by p(A)=O, if A E ~ , = 1,
if A g 4 and A E ~ .
6.
ABSOLUTE CONTINUITY
169
is a charge on 9. We claim that Lebesgue Decomposition theorem is not valid for v with respect to p. Suppose v = v1 + v 2 , where v 1 << p and v z I p . Since v 1 is real and p is bounded, v 1 is a bounded charge. See Theorem 6.1.10(iv). Since v z I p , there is a set E in 9such that 1v21(E')< 1 and p (E)< 1. Hence p (E)= 0. Consequently, E E 4;. Hence E'g 9'.Therefore, IvI(E') = m. v1 is obviously bounded on E" and v2 is also bounded on E'. Therefore, v is bounded on E' which contradicts Ivl(E') = 00. Thus the claim is established. p
6.2.6 Theorem. Let 9 be a field of subsets of a set R. Let v be any real charge on 9 such that \.\(A)= co for every non-empty set A in 9. Then Lebesgue Decomposition theorem for v is not valid with respect to any non-zero bounded charge p on 9.
Proof. Suppose there is a non-zero bounded charge p on 9 such that Lebesgue Decomposition theorem is valid for v with respect to p. Let v = v l + v 2 , where v 1 < < p and v 2 1 p . By Theorem 6.1.10, it follows that v1 is a bounded charge. As v 2 1 p , there exists a set A in 9 such that 1v21(A)< lp I(R) and Ip I(A') < lp I(R). Then A is a non-empty set on which 0 v2 as well as v l is bounded. This is a contradiction. 6.2.7 Remark. The charge given in Example 2.1.3(6)is a charge satisfying the condition imposed on v of Theorem 6.2.6. We end this section by deriving Lebesgue Decomposition theorem for measures on fields from Theorem 6.2.4. 6.2.8 Theorem. Let p and v E ca(R, 9).Then there exist v 1 and ca (a,9)having the following properties. (i). v = v l + v2. (ii). vl<
v2
in
Proof. By Theorem 6.2.4, we obtain v1 and v2 in ba(R, 9)with the above . V ~ ca(R, E 9). 0 properties. By Theorem 6.1.11, v 1 E ca(R, 9)Hence
6.3 RADON-NIKODYM THEOREM The aim of this section is to prove Radon-Nikodym theorem in the framework of charges. The proof is mainly based on Hahn Decomposition
170
THEORY OF CHARGES
thereom and involves only elementary calculations. In Chapter 7 also, we obtain Radon-Nikodym theorem for charges as a simple consequence of a result in V1-spaces. First, we establish some preliminary results.
6.3.1 Theorem. Let 9 be a field of subsets of a set R and E 7 0. Suppose p and v are two bounded charges on 9satisfying p (F) 2 - ~ / 2and u (F)2 - ~ / 2for every F in 9. Then there exists a set A in 9 and a simple function f on R having the following properties. (i). lp (E)I < E whenever E E 9 and E c A. (ii). Iv(F) - D 5, f dp I 5 E whenever F E 9 and F c A'. Proof. By Hahn Decomposition theorem (see Theorem 2.6.2), for any in ba(R, 9) and S > 0, there exists a set B(T,S) in 9such that T(E)2 -6
whenever E E 9 and E c B(7, S ) ,
7(E)I6
whenever E E 9 and E c (B(T,8))'.
7
and Let m be any positive integer
For k
= 1 , 2 , . . . , m, define
Let A = B,, A, = B,-l -Bm, A,-* = Bm-2-(Bmp1u B,), . . . , A 2= B1 -(B2uB3 u - - uB,) and Al = R-(Bl u B z u . - * uB,). Note that A, AI, A2, .. . ,A, are pairwise disjoint sets in 9with union = R. Let
(i-1)e
f = i=12()p.l(n)+&) c We show that A and f are the desired ones. Let E E 9and E c A = B,. Then
6.
ABSOLUTE CONTINUITY
171
Also, by the given hypothesis, p (E)2 - ~ / 2> -e. Consequently, Ip (E)]< e. Thus (i) is proved. Now, let E E g a n d E c A ' = A l u A 2 u . * . u A m .So, E=
m
m
i=l
i=l
u ( E n A i ) =u Ei,
whereEi=EnAifori=1,2,. . .,m.Notethatforeach2sism,EicAic Bi-l and so
On the other hand, for each 15 i 5 m, Ei c B; and so
Consequently, from the above two inequalities, we obtain
and
m
"
5 &/2+&/2 =E.
Consequently, Iu(E)-D j, f d p l s e . This proves (ii). The following theorem is a generalization of the above theorem.
0
172
THEORY OF CHARGES
6.3.2 Theorem. Let 9be a field of subsets of a set R. Let p and v be two bounded charges on 9 and E >O. Then there exists a set A in 9 and a simple function f on R such that (i). Ip (E)I< E whenever E E 9and E c A, and (ii). Iv(E)-DI,fdpI<E whenever E E 9 a n d E c A ‘ hold. Proof. In the notation of the proof of the above theorem, let B1 = B(p, ~ / 1 6 ) and B2= B(v, ~ / 1 6 )Let . pl, p2, v 1 and v2 be defined on 9by
pi(F) = p (Bi nF),
FE9,
PZ(F)= -EL (B?nF),
F E 9,
vl(F) = v(BZnF),
F€9,
vz(F)= -v(B; nF),
F E9.
and Note that p = p1 -p2 and v = vl- v2. Further, pl(F) 2 - ~ / l 6 , p2(F)2 - ~ / 1 6 , vl(F)2-&/16 and vz(F)r-&/16 for every F in 9. By Theorem 6.3.1, there exist A, in 9and simple functions fij on R such that
Ipi(E)I< ~ / 8 whenever E E g and E c Aij and
fi,. d p ; l s e / 8
Ivj(E)-D
whenever E
E and~ E c A i
E
for i, j = 1, 2. Let A = A11 u A12 u AZIu A22 and f = ( f l l - f 1 2 ) L 1 + (f21 -f22)1~e. We show that the A and f defined above are the desired ones. Let E E and~E c A . Then
E = E nA = (E n A l l ) u (E nA12) u (E nA24 u (E nA22) = Eli u El2 uEzi u Ezz
where Eij= E n A,, i, j = 1,2. Disjointizing Eii’sif necessary, we can assume, without loss of generality, that Eij’sare pairwise disjoint. In the following, we use the property that Eijc Aii for i, j = 1, 2. Now,
6.
173
ABSOLUTE CONTINUITY
Thus (i) is established. Now, let E E 9 and E c A".Then E c A; for i, J' 2. Observe that p2(En B1) = 0 = p l ( E n BE). So,
+
= 1,
v(EnBP)-D
+Ivz(EnB?)-D[ 5 4 ( ~ / 8= )~
EnBP
f~~dp21
/< 2E .
0
This completes the proof. We need a lemma before proving the main theorem of this section.
6.3.3 Lemma. Let 9 be a field of subsets of a set 0, E > O and k >O. Let .9)be such that p r O and - k p ( F ) - - E < v(F)< kp(F)+E for every F in 9. Then the following statements are true. (i). For every E ' > E , there exists a two-valued simple function f on 0 such that
p , v E ba(0,
k 2
k 2
+E
- - p (F)- E ' < v (F)- A (F)< - p (F)
for every F i n 9, where A (F)= D JF f dp, F E 5F. (ii). For every E ' > E , there exists a simple function g on s2 such that -E
< v (F)- 7( F ) < E
for every F in 9, where 7(F)= D
IF g dp, FE9.
174
THEORY OF CHARGES
Proof. (i). By Hahn Decomposition theorem, there exists a set A in 9such that ~ ( E ) ? - ( E ’ - E ) whenever E c A and E E ~ , and ~(E).(E’-E)
whenever E c A Cand E E ~ .
Let f = ( k / 2 ) I A - (k/2)IAc.Clearly, f is two-valued. Further, for any F in $,
k f d p =v(FnA)--p(FnA) 2 k 2
< - p (FnA) + E , and
k f d p = v(FnA‘)+-p(FnAc) 2
+-2k p (FnA‘).
5 ( F ’ -E )
Consequently, by adding the above two inequalities, we obtain v(F)-D
k f d p <-~(F)+E’. IF 2
One can establish, by a similar argument, the other inequality
k --p(F)-&’
k --p((F)-&*
fn
k dp< 7 p ( F ) + ~ * 2
for every F in g.The avove assertion is true when n = 1, by (i). Using (i) again for the charge v( -D I(.,f l dp, we can show that the above assertion is true for n = 2 . The general argument is similar. Now, let E ’ > E be given. Find E * > E and n 2 1such that ( k / 2 “ ) p(R) + E * < E’. The corresponding f n is the desired function. 0 9 )
The following is the main theorem of this section which we call RadonNikodym theorem for charges
6.
ABSOLUTE CONTINUITY
175
6.3.4 Theorem. Let 9 be a field of subsets of a set R. Let p, u E ba(R, 9) be such that p is positive. Then the following statements are equivalent. (i). u << p. (ii). For each E > 0, there exists a charge A in ba(ll, 9)and a nonnegative number k such that
- k p ( F )5 A ( F ) 5 k p ( F ) for euery F in 9,and I]u-A 11 5 E . (iii). For each E > 0, there exists a simple function f on R such that
for every F in 9. Proof. (i) 3 (ii). Without loss of generality, we can assume u to be positive. (We can argue separately for u c and u - . ) Let E > 0 be given. Since v << p, there exists S>O such that ~ ( E ) < E whenever E E and ~ p ( E ) < S . Let k = u ( n ) / S and A = u A k p . We show that A has the properties mentioned in (ii). Let F E 9.If p (F) < S , then u(F)- k p (F)5 u ( F )<E. If p (F)2 6, then v ( F )- k p ( F )Iu ( F ) - kS = u ( F )- u(R) 5 0 < E . In any case, we have u ( F ) - k p ( F )< E . SO,
1 1 -~ A 11 = Iu - A [(a)= Iu - u A k p I ( R )
~ ( n-(v) Akp)(R) {u(F")+ k p ( F ) ;F E .5F)
= (Y - u A k p ) ( R ) =
= u(R) -1nf
= Sup { u ( F ) - k p ( F ) ; F
E fl
5F.
From the definition of A, A 5 k p and A L 0. This proves (ii). ($3 (iii). Let E > 0. By (ii), there exists A in b a ( l l , 9 ) and a nonnegative number k such that
- k p ( F )5 A (F)5 k p ( F ) for every F in 9, and IIu -A
)I5 s/3. Then we have
- ~ / 3- k p ( F ) < A (F)< k p ( F ) + ~ / 3 for every F in 9. By applying Lemma 6.3.3(ii) for ~ ' = 2 ~ / there 3 , is a simple function f on ll such that
176
THEORY OF CHARGES
for every F in 9. Consequently, for any F in 9,
1
v (F)- D
jFf & 1
5
Iv(F) -A
(F)I+ IA (F)- D
IF 1 f dp
< ~ / +32 4 3 = E . This proves (iii). (iii)+(i). Let E >O. Let f be a simple function on SZ such that
for every F in 9.Let k = max {I f(w)l; w E R}. Take S = ~ / 2 k (Assume, . without loss of generality, that k > O . ) We show that if F E9and p(F) <S, then Iv(F)I < E . Note that
I
~ / 2 >v(F)-D
f d p LIv(F)I- D
fdp
[ F I
Consequently, Iv(F)I < E . This proves (i).
0
6.3.5 Example. Exact Radon-Nikodym derivative may not exist, i.e. there may not exist a D-integrable function f (with respect to p ) such that
v(F)=D
I,
f dp
for every F in 9, in Theorem 6.3.4. Let R = {1,2,3, . . .} and 9the finite-cofinite field on SZ. Let v and defined on 9by
p
be
v(F) = 0, if F is finite, = 1,
p(F)=
c 21
keF
=1+
T,
1 1T , kGF
Note that
v<
if F is cofinite;
2
if F is finite, if F is cofinite.
If there is a D-integrable function f on SZ such that
6. u (F)= D
ABSOLUTE CONTINUITY
sFf d p for every F in
177
$, then
f dp
v({n})=O=D[ {n)
=
fb) p ({n1)
1 = f ( n ) ; for every n 2 1. 2 SO,
I
l=u(R)=D f d p = 0 , a contradiction. For every k 2 1, by Theorem 6.3.4, there is a simple function fk on 0, which can be constructed easily, such that Iv(F) - D lFfkd p 1 < l/k for every F in 9. It can be checked that limm,n+m D Ifm -fn 1 d p = 0. But there is no function f on R such that f n , n 2 1 converges to f hazily.
CHAPTER 7
V, -Spaces
We have come across some function spaces, namely b a ( R , 9 ) and in Chapter 2, where 9 is a field of subsets of a set 0. It seems ca(R, 9), difficult to describe the duals of these spaces. However, there are some important function spaces, namely V,-spaces, related to ba(i2,9) which are more tractable and which we study in this chapter. These spaces are closely related to L,-spaces introduced in Section 4.6. In Section 7.1, we introduce V,-norms for 1r p s 00 and establish their connection with the corresponding L,-norms. V,-spaces are presented in Section 7.2 and are shown to be Banach spaces. The duals of V,-spaces for 1s p < 0O are identified in Section 7.3. In the last two sections, strong convergence and weak convergence in V,-spaces for 1s p < 00 are characterized.
7.1 L,-SPACES-AN
OVERVIEW
Let 9 be a field of subsets of a set R and p a probability charge on 9, i.e. p is apositive charge on $with p (R) = 1.In Section 4.6, we have defined L,(R, 9, p ) = {f;f is a TI-measurable function on i2 and with a pseudo-norm 1s p <00, and
l fl1,
Ifl”
= (D
is D-integrable}
If[”
dp)l’, for
f in L,(R, 9, p ) , where
L,(R, g,p ) = {f;f is a TI-measurable real valued function on R and essentially bounded}
Ifl,
for f in L,(R, g,p ) . with a pseudo-norm llfllm = essential supremum of By identifying functions which are equivalent under the equivalence p ) as the space relation f - g if f-g is a null function, we form 2?,(R, 9, p ) with a norm 11 1, unambiguously of all equivalence classes of L, (R, 9, derived from the corresponding pseudo-norm on L,(R, 9, p ) for every 1s p 5 00. These normed linear spaces (2?,(R, 9, p ) , )I*)1, are not Banach spaces in general. See Remark 4.6.8. The V, spaces to be introduced in
-
7. V,-SPACES
179
the next section are precisely the completions of these Z,,-spaces (for 1sp<0O). In this section, we give an alternative description of Z,-spaces. This description provides a natural setting for the introduction of V,-spaces. We establish some preliminary results.
7.1.1 Lemma. Let a and b be two non-negative real numbers and 1s p < 00. Then (a + b ) P s a P t l - p + b p ( l - t ) l - -for p any O < t < 1. Proof. Let us treat the case 1< p
f(t) =
For a given field 9 on a set R, recall that 9 stands for the collection of all finite partitions of R in 9. Under the notion of refinement of partitions, 9 is equipped with a natural partial order L with respect to which (9, L) becomes a directed set. If F E S , we denote the collection of all finite partitions of F in 9 by PF.
7.1.2 Proposition. Let ( R , S , p ) be a probability charge space, i.e. (a,9,p ) is a charge space and p is a probability charge on 9. Let A be any real charge on 9 s u c h that A << p. Then the net of real numbers
is an increasing net for any 1S p < 00. (If interpret 0 / 0 = 0.)
p (Fi)= 0,
then A (Fi) = 0 and we
Proof. It suffices to show that for any two disjoint sets A and B in 9,
If p (A) = 0 or p (B) = 0, this inequality is obvious. Let p (A)> 0 and p (B) > 0. In Lemma 7.1.1, let a = IA (A)I, b = IA (B)I and t = [ p (A)/p (A u B)]. Note that A is bounded. See Theorem 6.1.10(iv). Then 1- t = [ p (B)/p (A u B)]
180
THEORY OF CHARGES
Also, IA (A u B)I IIA (A)]+ IA (B)].Therefore,
0
This establishes the desired inequality.
Now, we introduce some notation. Let (a, 9, p ) be a probability charge space and A a real charge on 9such that A << p. For 15 p < 00, let
where the limit is taken over all P = {El, Ez, . . . , En}E 9. 11P
'p(Ei)]
; P ={El, EZ,. . . ,E,}E 9)
where the limit is taken over all P ={El, EZ,. . . , En}E 9.The equality of these four numbers is clear from Proposition 7.1.2. Note also that 0 IIlA Ilp 5 00. For B in 9, let A g be the charge on 9 defined by AB(A)= AfAnB), A E g.For A g , note that
Clearly, if A and B are disjoint sets in 9,then ~ ~ A +. !lABll: 4 ~ ~=~@(AuBdF.
7.1.3 Proposition. Let (a, 9, p ) be a probability charge space. Let A be a real charge on 9such that A << p. Then [[AI p = lllh Il l p for any 1I p < O3. Proof. If p = 1,then IIAllp = \A I(a)and lllA 111 = \A \(a). So, the desired assertion follows for this case. IA (F)I 5 IA I(F), it follows that IIAllp 5 Let 1< p < 03. Since for any F in 9, lllh I(Ip' It remains to be shown that lllh [[Ip IIlA \ I p . Let B E 9.From the definition of llABIIF, it follows that IA (B)/p (B)lpp(B) 5 llABll;. Consequently, IA (B)lS
7. V,-SPACES
181
I I A B I I ~ [(B)l(p~l)’p ~ = I l h ~ ~ ~ ~ [ p ( B ) ] ~ / ~l ,/ w p h+el r/ eq = 1.(Notethattheconvention 0/0 = 0 is enforced.) Thus for any partition {El,EZ,. . . ,En}€PB, we have, by Holder’s inequality for finite sums (See Theroem 4.6.2.), IA i=l
5
i]: IIAE~II~[P(E~)I~/~
i=l
(i
5 i=l
l/P IIAE~IIL)
n
( c p(Ei))
l/q
i=l
*
Therefore, from the last observation made preceding the statement of this proposition, we obtain the inequality
Taking supremum over all partitions {El, Ez,. . . ,En} in PB,we obtain [[A I(B)IP5 IIA&[p (B)]’-’. Or, equivalently,
Now, for any partition {Fl, Fz, . . . , Fm}of R in 9,
By taking supremum over all partitions {F1, FZ,. . . , F,} in 9, we obtain lllh 1; 5 IlA 1,”. From this, the desired equality follows. We now give some properties of
11 [Ip *
introduced above.
Proposition. Let (R, 9, p ) be a probability charge space. Let 1 5 p < Let A, A1, A 2 be three real charges on 9 such that A << p, A 1
Proof. (i) and (ii) are obvious. (iii) can be established using Minkowski’s inequality as follows. (See Theorem 4.6.6.)If {El,EZ,. . . ,En}is any partition of l2 in S, then
182
THEORY OF CHARGES
Theref ore,
0
Hence IIA 1 + AzIIp 5 IIA iIlp + IIAzIIp.
Now, we come to the promised alternative description of Zp(fl,$, p ) .
7.1.5 Lemma. Let (fl, 9, p ) be a probability charge space and f a simple function on fl. Let A on 9 be defined by A (F)= D f dp, F E9.Then IlA 1; = D If 1” dp for every 1 5 p < 00. Proof. Let f =I:=,aiIEi for some real numbers a l , u 2 , . . . , a,, and some partition {El, EZ,. . . , En}in P. Then
Also,
Further, if we take any partition {F1,F2,.. . ,Fm} in P finer than {El, Ez, . . . , En},then we have
Consequently, IlA 1:
=
zYXllailpp(Ei).This completes the proof.
0
7.1.6 Theorem. Let 1~p < 00. Let (a,9,p ) be a probability charge space and f E L,(n, 9, p ) . Let A on 9 b e defined by A (F)= D jF f dp, F E $. Then
D
I
Ifl”d P = IIAIIF.
Proof. First, let us treat the case p = 1. From the definition of lA111, it is clear that llA1ll = IAl(fl). By Theorem 4.4.13(xi), IA l(fl) = D If1 dp. dp. Now, we come to the case 1< p <00. We show that llAIIF I D 5 by Let the positive number q satisfy l / p + l / q = 1. Then for any E in 9, Holder’s inequality,
If[”
7. V,-SPACES
183
Let {El, E2, . . . , E,}E 8.Then
i=l
lwl”p(Ei) p(Ei)
[IAI(Ei)]P[p(Ei)]l-”
= i=l
=
: I Ifl”
i=l
D
Ei
dP = D
5 lfl”
dp.
Consequently, the inequality llAll;~DJ Ifl” d p follows. Now, we show the desired equality. Since simple functions are dense in L,(R, 9, p ) (see Theorem 4.6.15), there is a sequence f n , n 2 1 of simple D J -fl” d p = 0. For each n 2 1, let A, functions on il such that on 9 be defined by A,(F) = D JF f , dp, FE9T Then for every n 2 1, lllAnllp-IIAllpI(llhn -AllpS(DJ Iffl-fl” dp)l’”. Hence
If,
IlA 1;
=
lim Ilhnll; = lim D J I f n l p d p = D J If/” dp. n-tm
n-tm
The second equality above follows from Lemma 7.1.5. This completes the proof. 0
7.1.7 Theorem. Let (R, $, p ) be a probability charge space. Let 15 p < a. Then YP(&9, p ) = (A
E ba(il, 9); A (F)= D
I,
f dp, F
E
~
for some TI-measurable function f on R such that If 1” is D-integrable}.
0
Proof. This is now obvious from the results established above.
Now, we take up the case p = 00. Let (R, 9, p ) be a probability charge space and A a real charge on 9such that A << p. Let
5 a.We give below some properties of Obviously, 0 5 IIA )loo
11
*
IIm.
7.1.8 Proposition. Let (R, 9, p ) be a probability charge space and A, A I , A 2 real charges on $such that A c p , A l << p and A2<< p. Then the following statements are true.
184
THEORY OF CHARGES
(i). IlA llm= 0 i f and only if A = 0. (ii). llch llm= Ic 1 IlA llmfor any real number c. (iii). 1 + A2IIm 5 IIA 1IIm+ IIA2IIm. (iv)- IlA llco = 111A Illm-
Proof. (i), (ii) and (iii) are obvious. We prove (iv). IlA Ilm 5 Il1A (\loofollows from We show that l lo 5 IlA lloo.For the fact that IA (F)I5 IA I(F) for any F in 9. we have any B in 9 and {B1, B2,. . . , B,} in PB, m
m
i=l
i=l
C IA(Bt)I5 C I(h11mp(Bi)=IIhIImCL(B).
Hence 1A l(B) 5 Ilh ]lap(B), i.e., [IA l(B)/p (B)] 5 Ilh Itoo. Taking supremum we obtain Illrn5 IlA [Irn.This proves (iv). 0 over all B in 9,
7.1.9 Theorem. Let (0,9, p ) be a probability charge space and f a real valued T1-measurable essentially bounded function on R. Let A on 9 be defined by A (F)= D f dp, F E 9. Then
IF
llfllm
=Essential supremum off
= IlA
Ilm.
Proof. Note that f is D-integrable by Theorem 4.5.7 and consequently, A is well defined. First, we prove the inequality IlA llm5 Ilfllm. Let k be any positive number such that p * ( { w € 0 ;(f(w)l>k})=O. Then If15k a.e. [ p ] . Then by Theorem 4.4.13(vii),
Hence for any F in 9.
From the definition of Ilfllm, it follows that IIA llm 5 Il f l o o. To prove the reverse inequality, we proceed as follows. Let k = IlA I l m . Then \A I(F)5 k p (F) for every F in 9. This implies that D 5, -k ) dp 5 0 for every F in 9.We claim that l f l 5 k a.e. [ p ] . Define T on 9 by T(F)= D I, - k ) d p , F E9. Since T(F)5 0 for every F in 9, it follows from the definition of T + that T+(F)= 0 for every F in 9 and so, T-(F)= -T(F) for every F in 9. But T'(F) = D IF - k)+ d p for every F in 9. See k)+ is a null function. Theorem 4.4.13(xii). By Theorem 4.4.13(xiii), - k)+ - k ) - , it follows that - k 5 0 a.e. [ p ] . Since - k = Now, we claim that p *({w E R; If(w)I > k + E } ) = 0 for any E >0. Observe that k + - k)' and that - k)' is a null function. Therefore,
(If1
(If1
(If1
If1
(If1 (If1
(If1
p * ( { w E R; Jf(w)I > k
(lfl-
If\
(If1
+E})
s p * ( { w E R;
(If1
-k ) + > E } )
=O.
7. V,-SPACES
185
The claim is thus established. By the definition of Ilfllm, it follows that 11f)lm Ik + E for any E > 0. So, )Ifilm Ik . Hence 11f ) l m 5 IlA llm. This completes 0 the proof. In the above, we have indeed proved that [ f l s l l f l l m a.e. [ p ] if f is TI-measurable. Now, we provide an alternative description of Zm(R, 9, p). 7.1.10 Theorem. Let (R, 9, p ) be a probability charge space. Then Tm(fl, 9, p ) = [A
E ba(R,
.F); A(F) = D
1
f dp, F
E
~
for some essentially bounded TI-measurable real valued function f on R
I
Proof. This is now obvious.
0
.
7.2 Vp-SPACES Lk',-spaces, discussed in the previous section, are not Banach spaces in 9,p ) are not general, or, equivalently, the normed linear spaces zp(fl, complete in general. In this section, we introduce V,-spaces and show that they are the completions of the corresponding Tp-spaces for 15 p < 00. p ) be a probability charge space. For each 7.2.1 Definition. Let (R, 9, 1~p sco, let
v,(@9, I-L 1 = {A E M R , .F); 1, < 001. The following proposition is a consequence of Propositions 7.1.4 and 7.1.8. 1.2.2 Proposition. Let (R, 9, p ) be a probability charge space. Then for each 1 s p 5 co, (V,(R, 9, p ) , II-IIp) is a normed linear space. By Theorems 7.1.7 and 7.1.10, we have that Lk'JR, 9, p ) cV,(fl, 9, p) for every 1I p 5 00. Members of V,(R, 9, p ) need not admit exact RadonNikodym derivatives with respect to p. Now, we show that each V,-space is complete. 7.2.3 Theorem Let (R, 9,p ) be a probability charge space. Then for each 1Ip 5 co, (V,(R, 9, F ) , )I* \ I p) is a Banach space.
186
THEORY OF CHARGES
Proof. First, we treat the case 1s p
A (E)= lim A,(E) n+m
for E in 9.It is obvious that A is a charge on 9.Since A,,, n 2 1 converges to A uniformly over 9, A is bounded and A << p. It remains to be shown that A E V,(n, 9, p ) and that llAn -AllL,, n L 1 converges to zero. Let E > 0. There exists N r 1such that llh, - Amll; < E whenever n, m 2 N . Let P = {El, Ez, . . . , E k } E 9. Then for m 2 N,
Taking supremum over all P in 9, we obtain IlA -AmllFsE if m r N. This shows that A E V,(n, 9,p ) and that limm+m“Am -All, = 0. Hence (v,(n, 9, p ) , Il-II,) is complete for 1s p
5 IIAn -AmIImP(F)
5 IIAn -AmIIm*
This shows that A,,, n 2 l is a uniform Cauchy sequence over 9. Let A (F)= limn+, A,(F), F E 9.Then A is a bounded charge on 9and absolutely continuous with respect to p. It remains to be shown that A E Vm(n,9,p ) and that llhn-A il, n 2 1 converges to zero. Let E >O. There exists N 2 1such that ]]An -Amllm < E whenever m, n r N . Let F E ~Then . Ih(F)-A,(F)I=lim,,, Ih,(F)-hm(F)15~p(F)if m r N . Consequently, IlA - A r n l l r n s E if rn r N . This shows that A E Vm(n, g, p ) and \]A,, -All, = 0. Hence V,(n, 9,p ) is a Banach space. 0
7 . V,-SPACES
187
Now, we prove inclusion relations among V, -spaces. 7.2.4 Theorem. Let s S C O Then .
vl(n,
$9
(a,9,p ) be a probability charge space. Let
1s r 5
CL)=vr(fA 9, CL)~ V s ( 09, , P ) xVm(Q, 9, PI.
/Is
In fact, [(A111 5 IlA llr 5 IlA 5 IlA Jlmis valid for any real charge A on Babsolutely continuous with respect to p . Proof. Let 1s r <s
+ l / q = 1.
(by Holder’s inequality)
Hence IIA 11;s [llA Ils]s/p, or, equivalently, IlA [Ir 5 IlA IIS. If 15 s < co,then for any partition {El, EZ,. . . , En}in 9,
Hence IIAlls 11lA
Ilm.
This completes the proof.
0
The following is an interesting consequence of the above theorem. p ) be a probability charge space. For any real 7.2.5 Corollary. Let (a,9, charge A on 9absolutely continuous with respect to p,
lim Ilk IIp = Ilk Ilm.
P-m
Proof. By Theorem 7.2.4, limp+mIlA 1, certainly exists (may be equal to co) and is less than or equal to llAIIm. On the other hand, for any p > 1 and F in 9,
188
THEORY OF CHARGES
Now, we introduce simple charges. Recall that for any charge p on a field 9 of subsets of a set R and F in 9, p~ is the charge on 9defined by pF(E)= p ( E nF), E E9.
7.2.6 Definition. Let (R, 9, p ) be a probability charge space. A simple charge on 9 is any charge of the form aipE, for some real numbers a l , a2,. . . , a, and some partition {El,E2, . . . , En}in P. SC(R, 9, p ) stands for the collection of all simple charges on 9. SC(R, 9, p ) is a linear space and can be identified with the space of all simple functions on (R, 9) as follows. If f = aiIEiis a simple function on R for some real numbers al, a 2 , . .. , a , and some partition {El, EZ,. . . ,En}in P, then the charge A = u ~ on ~ 9Ehas~the property a i p is ~ a~simple that Ilfll, = IIA [Ip for every 1S p 5 co. Conversely, if A = charge on 9 for some real numbers al, a 2 , .. . ,a, and some partition {El, El, . . . , En}in 9, then the simple function f = I:=,u ~ I E on~R has the property that IlAll, = Ilfll, for every 1s p 5 co. It is obvious that SC(R,g, p ) c V,(R, g, p).
x:=l
Among simple charges, the charges given in the following definition are of special interest. 7.2.7 Definition. Let (R, 9, p ) be a probability charge space and A a real charge on 9 such that A << p . Let P = {El, Ez, . . . ,En}€8.Define A(Ei)
Ap=
1PEi. p(Ei)
i=l
It is obvious that
and
for any l s p
Next, we show that SC(R, 9, p ) is a dense subset of V,(R, g, p ) for any 15 p < 00. First, we need the following results.
7. V,-SPACES
7.2.8 Lemma. For any p > O and that
IX - 11’
> O , there exists a number k ( p , E ) such
E
2k (p,E)[[x
189
1’
+ p - 1 -PX]
+ IX 1’ E
for any real number x. Proof. The function f defined on the real line by f ( x ) = 1x1’ + p - 1 - p x ,
x ER
is continuous and positive for all x # 1. Further, lim~x~-,m Ix - lIp/f(x) exists and is finite. For the given E > 0, we can find 77 > 0 such that Ix - 11’ 5 E Ix 1’ whenever Ix - 11 5 77. This follows from the fact that the function g ( x )= Ix - Il”/lx 1’ for x E (0, co) is continuous at x = I . Since limlxl+ooIx - Il”/f (x) is finite, there exists 771 > q + 1 such that Ix - l l p / f ( x )is bounded on the set {x ; Ix I > q l}. Since Ix - ll”/f (x) is well defined and continuous on [-ql, 1- q ] u [ l + q ql], , it is bounded on this set. Hence we can find a constant k ( p , E ) such that
Ix - 11, 5 k ( P , E ) f ( X ) whenever Ix - 11 > q. In any case, we have
IX - 11’ 5 k ( p , E ) [ [ x 1’
+ p - 1 -PX]
+ E 1x1’
for all real numbers x.
0
7.2.9 Lemma. Let (Q, 9, p ) be a probability charge space and A E V,(Q, 9, p ) for some 1 < p < 00. Then for a given E > 0 , there exists a constant k ( p , E ) such that IIA
- APlli 5 k ( P , E
IIi - llAPII3 + E IlA IC
for any P = {El, Ez,. . . , Em}in 9’. Proof. Let P’ = {F1, Fz, . . . ,F,} be a partition in 9 finer than P. Let E > 0. Let k ( p , E ) be the constant provided by Lemma 7.2.8. Let 1 5 i 5 n be fixed. Then we can find 1 5j 5 m such that Fi c Ei. Consequently,
In Lemma 7.2.8, if we set x = A (Fi)/Ap(Fi),with the usual convention about O/O, we obtain
190
THEORY OF CHARGES
Using Ap(Fi)= [A (Ej)/p(Ej)]p(Fi) and IA (Ej)/p(Ej)l”p(Fi),we obtain
multiplying
throughout
by
By summing first over all those i E {1,2, . . . ,n } such that Fi c Ej for a given j E {1,2, . . . , m } and then summing over all j E {1,2, . . ., m } , we obtain
Since the above inequality is true for any partition P’ finer than P, we obtain
0
This completes the proof.
7.2.10 Proposition. Let (fl,9, p ) be a probability charge space and A = C:, aipEibe a simple charge on 9, for a partition P = {El, EZ, . . . Em}in
9 and real numbers a l , a2, . . . ,a,,,. Then for any partition P‘>P, IIA
-A ~
’ 1 =1 ~0
for any 1s p
0
Proof. Simply note that A - Ap’ = 0.
p ) be a probability charge space and A a 7.2.11 Proposition. Let (a,9, positive bounded charge on 9 s u c h that A << p. Then
lim IlA -(A
n-a
A
np)II1= 0.
Proof. Since ba(fl, 9)is a boundedly complete vector lattice (see Theorem 2.2.1) and A << w, A = Vnzl (A A n p ) . See Theorem 1.5.12 and Theorem 6.2.2. Note that limn+m(A A n p ) ( A )= A (A) for every A in 9, In particular, ( A -(A A n p ) ) ( R ) ,n 2 1 converges to zero, i.e. limn+m\\A -(A A np)II1= 0. This proves the result. 0 Now, we come to an important result of this section.
(a,9,p ) be a probability charge space. Then, for each 1‘ p < 00, the space SC(fl,%, p ) of all simple charges is a dense subset of V,(Q 9,p ) . More precisely, for any 1s p
7.2.12 Theorem. Let
7. V,-SPACES
191
Proof. Let us tackle the case 1< p
IIA -APll; 5 k(p, ~~Cll~llpP-ll~Pll;l+~Il~lI;. ~ = IlA A 1; p (see ~ the ~ remark ~ following Definition
Since limPEP~ > O is arbitrary, it follows that
7.2.7) and
E
Now, we come to the casep = 1.Let A be a positive charge in Vl(fl, 9, p). Then for any P in 9 and n 21, IIA -A&
5 IlA
- (A A np)II1+ ll(A A n p )- (A A n p )PI11 + II(A A n p )P - A p h
5 211A
- (A
A
np )I11 +
A
n p ) - (A A n p ) P h
(by the remark following Definition 7.2.7) 52
b - (A
A
np
)I11+ ll(A I, w )- (A
A
np ) P ~ \ z
(by Theorem 7.2.4). Since 0 5 (A A n p ) s n p and n p E Vz(fl, 9, p ) , it follows that (A A n p ) E V z ( f l , 9 , p ) . Therefore, by what we have proved in the first part, limPEBll(A A n p ) - (A A np)pll2 = 0. Hence, by Proposition 7.2.11,
p ) , write Now, it follows that limPEpIlA -Ap(ll = 0. For any A in V1(R, 9, Note that Ap=(A+)P-(A-)p. Hence limPEBIlA -APII1=O. This completes the proof. 0
A =A+-A-.
For a general bounded charge p, not necessarily positive, by defining V,-norms with respect to lpl, we can, in fact, obtain the above theorem for general bounded charges. The above theorem for p = l gives the Radon-Nikodym theorem of Section 6.3 for charges.
7.2.13 Corollary (Radon-Nikodym Theorem). Let $be a field of subsets of a set R. Let p and v be two bounded charges on 9 s u c h that v << p. Then for each E > 0, there exists a simple function f on R such that
for every F in 9.
192
THEORY OF CHARGES
We also obtain from the above theorem that V,(R, 9, p ) is the complep). tion of Lfp(R,9, Corollary. Let ( R , 9 , p ) be a probability charge space. Then V,(R, 9, p ) is the completion of Lfp(R,9,p ) for every 1 ' p
7.2.14
Proof. This is a consequence of Theorem 4.6.15 and Theorem 7.2.12. 0
7.2.15 Remark. We have refrained from making any statement about Lfm(R,9, p ) in the above corollary. It is not true that Vm(R,9, p ) is the p ) . The following example justifies this point. completion of Zm(R,9, We recall the example given in Remark 4.6.8. Let R = {1,2,3, . . .}, F the finite-cofinite field on R and p the charge on 9defined by 1 if A is finite, p(A)= n o A 2"'
c
1 2"'
1-
=2 -
n.AC
if A is cofinite.
We state a number of interesting facts about this ( R , 9 , p ) . 1. Lfl(R, 9, p ) is not complete. See Remark 4.6.8. 2. Any bounded charge v on 9 is absolutely continuous with respect to p. (Let&>O.Findm ~lsuchthatCn,,Ivl({n})<~.TakeS =C,,,p({n}).) 3. Vl(R, 9, p ) = ba(R, 9). 4. A real valued function f on R is T1-measurable (Tz-measurable) if and only if f ( n ) , n 2 1 converges. 5 . A real valued function f on R is essentially bounded if and only if f is bounded. 6. If a real valued function f on R is bounded, then \lflloo= Sup{If(n)l; n E W . 7. Lfm(R,9, p ) = c, the space of all convergent sequences of real numbers p ) is complete. equipped with the supremum norm. Thus Zm(R,9, 8. A bounded charge v on 9is in V,(R, 9, p ) if and only if v ( { n } ) / p ( { n } ) , n 2 1 is a bounded sequence of numbers. 9. Let v on 9be defined by
n even
n odd
For this v, there is no essentially bounded T1-measurable function f on R
f d p for every F in 9. J , 10. Z m ( R , 9 , p ) is a proper closed subspace of
such that v(F) = D
v,(R, 9, p ) is not the completion of 9 m ( R , 9, p).
Vm(fl,9,p). Thus
7. V,-SPACES
193
We need an analogue of Proposition 7.2.11 for V,-norms to establish some results on strong convergence in V,-spaces. This is acheived in the following.
7.2.16 Lemma. Let (R, 9, p ) be a probability charge space and v a simple Then positive charge on 9. lim (Iv - (v A np)llp= 0
n+a3
for any 1' p
=EL,
Proof. If v aipEifor some nonnegative real numbers a l , a 2 , .. . , a,,, and some partition {El, EZ,. . . , Em}in 8,then v 5 n p for sufficiently large n. Hence the above limit is zero. 0 7.2.17 Theorem. Let (R, 9 , p ) be a probability charge space and 1s p < 00. Let v be a positive charge in V,(R, 9, p ) . Then
Proof. Let P={El, EZ,. . . ,En}be any partition in 8.Then for any n L 1, IIv - (v
A
n p Ill, 5 IIv - vpll,
+ llvp - ( v p A np Ill, + ll(vP A np
- (v A n p
l,.
By Theorem 1.5.4(29), I(vp A n p ) - ( v A n p ) /5 Ivp- vI. Consequently, for n L 1, llv-(v ~ n ~ ) I l , ~ 2 l l v - ~ ~ J ~ , + J J v ~ -ByLemma7.2.16,OS (~~~np))),. lim sup,,a3 IIv - (v A np)llp5 21(v- vpllp+ 0. Since this inequality is true for all P in 8,by Theorem 7.2.12, we have limn-,m IIv -(v A np)llp= 0. This 0 completes the proof.
7.3 DUALS OF V,-SPACES In this section, we show that the dual of V,(Q .F,p ) for any 1s p < m is V,(R, 9, p ) , where l / p + l / q = 1.
7.3.1 Theorem. Let (R, 9, p ) be a probability charge space. Then the dual of V,(Q 9, p ) for any 15 p <00 is isometrically ismorphic to V,(R, g, p), where l / p + l / q = 1. Proof. Let l < p < 0 0 and l < q < 0 0 be such that l / p + l / q = l . Let v E V,(R, 9, p ) . We define a linear functional T,, on V,(R, 9, p ) as follows. Let A ~ s C ( R , 9 , p ) .Then A =ELlaipEi for some real numbers a l , az, . . . , a , and some partition {El,Ez, . . . , E m } in 8.Define T,,(h)= EL, aiv(Ei). T, is well defined on SC(n,9, p ) . Further, T, is a linear
194
THEORY OF CHARGES
functional on the linear space SC(n, 9, p ) . Observe also that
(by Holder’s inequality)
5 Ilk llPIlv114.
T,, can be extended uniquely as a continuous linear functional to V,(Q .!F, p ) since SC(Q g,p ) is dense in V,(R, 9, p ) . Let us denote this extension again by T,. T,, has the property: ITu(A)I
4I~114lIAllP
for any A in V,(R, .!F, p ) . Let us compute the norm of T,,. Obviously, IlT,lls IIvIIq.We shall establish there exists a simple the reverse inequality by showing that for any P in 9, charge A such that lT,(A)I = IIvpllqllhIlp. Then it would follow that I l ~ P l l q J l ~ I l= P
lTu(A)I s l l T v 1 l l i 4 l P
from which we obtain lIvpllq 5 llT,,ll for any P in 9. Hence IIvIIq 5 llTulland so, the equality ensues. Let Now, let P = {El, El,. . ,En}E 9.
.
for i = 1,2,,
. , ,n
and A =Cy=1 c i p ~ ,Then .
Let us calculate IlA .,1
Consequently, we have IT&
)I = 1(vplI,IlA [Ip’
7. V,-SPACES
195
Thus we have associated a continuous linear functional T, on V,(n, 9, p) for any given v E V,(Q, 9, p ) such that (IT,IJ= IIvl/,. The map v + T,, from V,(n, 9, p ) to V:(R, 9, p ) is a linear map and is norm preserving. This map is obviously one-to-one. In order to show that this map is onto, we proceed as follows. Let T be any nonzero continuous linear functional on V,(n, 9, p ) . Define We enumerate the properties a set function vT on 9by vT(F) = T ( ~ FF )E,9. of vT. (i). Since T is a linear functional on V,(n, 9, p ) , V T is a charge on 9. (ii). Since T is a continuous linear functional, VT is bounded. For, for any F in 9, IvT(F)I= IT(PF)I5 ~~T~~IIPFII~ s l l T b ( F ) SllTll. (iii). vT<< p. Let E >O. Take S = ~/11T11.If F E9 and p(F)<S, then IYT(F)I <E. (iv). v T ~ V q ( n , g , pLet ) . P={El,Ez,. . . ,E"}EP.Then
i=l
Hence llvT[[:1llT1F. Thus v.T E V,(n, 9, p ) . For this VT, TVT = T.Hence llvTll = IlTll. This completes the proof that V:(n, 9, p ) and V,(n, 9, p ) are isometrically isomorphic for 1< p
i=l
Thus T, is a bounded linear functional on S C ( n , 9 , p ) and SO can be p ) . We extended uniquely as a continuous linear functional on Vl(S1, 9, denote this extension again by T,. Then I T u ( ~ ) l Is I ~ l l m l l ~ I l ~
for every A in V1(R, 9, p ) . Consequently, llT,ll 5 Ilvllm. To prove the reverse ~ p(F)>O. Let ul= inequality, we proceed as follows. Let F E and
196
THEORY OF CHARGES
sign(v(F))/p(F) and az=O. Let A have
= U I ~ F + U ~ ~ F C Since .
llA1ll= 1, we
Since this inequality is true for any F in 9with p (F)> 0, we have
Hence llvllm I~ ~ T Thus v ~we ~ have . proved that llT,,ll=IIVllm. Hence, the map v + T, from V,(R, 9, p ) to V ? ( R , 9 , p ) is a linear map such that IlT,,ll= IIvllm. So, this map is one-to-one. In order to show that this map is an isometric isomorphism, it suffices.to show that for any p ) , there exists VT in given continuous linear functional T on Ll(R, 9, Vm(R,9, p ) such that T = T,,, The existence of vT can be established along the same lines as that of the case 1< p
7.3.2 Remark. Let (R, 9, p ) be a probability charge space and p > 1, q > 1 be such that l / p + l / q = 1. For each v in V,(R, 9, p ) , T,, defined p ) and for each A in above is a continuous linear functional on V,(n, 9, p ) ,T A is acontinuous linear functionalon V,(R, 9, p ) . Theduality V,(R, 9, between V,(R, $, p ) and V,(n, 9, p ) can be expressed by T u ( A = TA( v )
for every A in V,(R, g, p ) and v in V,(R, 9, p). For any parition P = {El, Ez. . . . , Em}in 9,
Thus we have shown that Tv(Ap) = TA(vP).Since limpEp"AP- Allp = 0 = lirnpEpIIvp- vll,, it follows that Tv(A)= TA(v) for any A in V,(R, 9, p ) and v in V,(n, 9, p ) . This argument also shows that
exists and is equal to T,(A) = TA(v) for every A in V,(n, 9, p ) and v in v m , 9, PI.
7 . V,-SPACES
197
7.4 STRONG CONVERGENCE In this section, we give some conditions for strong convergence in V,-spaces for 15 p < 00. First, we need the following result.
7.4.1 Proposition. Let (Sz,%,p) be a probability charge space and v a bounded charge on 9absolutely continuous with respect to p . Then for any l % p , pl, ~ Z < O O , t l , t z 2 0 with tl+t2= 1 a n d p =tlp1+t2p2, the inequality
11~11;-= l l ~ l l ~ ~ f 1 1 1 ~ l l ; ; ~ 2 holds. Proof. Let P = {El, Ez, . . . , En}E 9 and t l , t2> 0. Then
Applying Holder’s inequality to the numbers l / t l and l / t z , i.e.,
we obtain
5 IIvI12:111vI12z.
Since this inequality is true for every P in 9, we obtain
1.”,
5
The case t l = 0 or tz = 0 is trivial.
ll~Il,”:f111~ll~f2. 0
The following theorem provides a sufficient condition for strong conp ) for p > 1 in terms of strong convergence in vergence in V,(Q 9, V l ( Q %, p ) .
7.4.2 Theorem. Let 1< r 5 00. Let (a,%, p ) be a probability charge space p ) satisfying the following two and v, v,, n L 1 a sequence in Vr(Sz,g, conditions. (i). v,, n 2 1 converges to v in V1(Sz,%, p ) . (ii). Supnzl llvnllr Then v,, n 2 1 converges to v in V,(Sz,9, p ) for any 15 p
198
THEORY OF CHARGES
Proof. In view of Theorem 7.2.4, it sufficies to prove for the case r < m . Assume, also, without loss of generality, that v = 0. In Proposition 7.4.1, if we take p1= 1, p 2 = r, tl = (r - p ) / ( r - 1) and t 2 = ( p - l)/(r - l), then for every n L 1, 1l v 11p v, :-p)/(r-l) r(P-l)/(r-l) n p-Il
II
IIvnllr
I ~ ~ v , ~ ~ ~ - ~ ) / k - l )r ( p - l ) / ( r - l )
[SUPIIvnllrl nzl
0
IIv,llp = 0. This completes the proof.
By (i), it follows that
The following therorem is analogous to the Lebesgue Dominated convergence thereorem proved in Section 4.6. This result also gives a sufficient condition for strong convergence in V,(R, 9, p ) for p > 1in terms of strong convergence in V1(R,9, p). 7.4.3 Theorem. Let (i2,9,p) be a probability charge space and v, v,, n 2 1 a sequence in Vl(R, 9, p ) having the followingproperties. (i). v,, n 2 1 converges to v in V,(R, 9,p ) . (ii). lv,I S A for every n 2 1 for some A in Vp(i2,9, p ) with 1 s p
Proof. For the case p = 1, there is nothing to be proved. Let us assume p > 1. By (ii), it follows that each v, eVp(R,5F, p ) . By (i), it follows that v,(F) = v(F) for every F in 9. It also follows that limn+m(v,l(F) = IvI(F) for every F in 9.(Use the fact that 1 ~ ~ 1n, 2 1 converges to 1.1 in Vl(R, 9, p ) . ) Consequently, IvI < A . Hence v E V,(R, 9, p ) also. Assume, without loss of generality, that each v, is positive. From the lattice properties of ba(R, 9), for any m, n 2 1, we have (A A mp)(v, A m p )S A - v,. See Theorem 1.5.4(29).Therefore, from this inequality, we obtain 0 5 v, -(v, A m p )s A -(A A mp).
Hence llv, - (v, A mp)IlpI l(A - (A ll v - v n l l p s l l v - ( v
A
mp)llp Thus for any m,n 2 1,
~ m l . ~ ) l l p + l l ( v ~ ~ m l ~ ) - ( v n ~ m ~ ~ ) I I~pm+ ~ l l )( v- v , nllp
5 IIv - (v A
mp
)]Ip
A
m p ) - (v, A mp )11p
+ Ilk - (A
A
m p )IIp
Observe that for each fixed m 2 1, (vn A mp),n 2 1 converges to (v A m p ) in V1(R,5F, p ) . This follows from the inequality II(v A m p )- (v, A m p ) l l l s ((v- v,(I1 for every n L 1 and (i). Further, (v, A m p )Im p for every n 2 1. Consequently, by Theorem 7.4.2, (v, A mp), n 2 1 converges to (v A mp) in V,(R, 9, p ) for every fixed m 2 1. Therefore, for every fixed m L 1, 0 s lim sup I(v- v,IIp 5 I1v - (v A mp)llp+ 0 +(\A - (A n-m
A
mp
)IIp
7. V,-SPACES
199
Now, by letting m + 00, by Theorem 7.2.15, we have OsIim sup IIv - vnllp < O + O . n-W
Hence limn,, ))v- vnllp = 0. This completes the proof.
0
The following theorem gives necessary and sufficient conditions for strong p ) for 1< p <03. convergence in Vp(n,9, 7.4.4 Theorem. Let 1< p
Proof. (i). This follows from the inequalities
I I I ~ n I I p - I I ~ I l p l ~ I-vIIp I~n and (p--l)/p
lvn (F) - v (F)I 5 llvn - vllp[c~(F)I
5
lbn - vllp
for every F in 9. (ii). v is obviously a charge on 9.We show that u < < p and that V E V,(R, 9, p ) . Since ilvnllp, n 2 1 is convergent, there is a positive constant M such that llvnllps M for every n 2 1. From the definition of ~ ~ v nit ~ ~ , , follows that for any F in 9 and n I1,
Equivalently, we have the inequality Iv, (F)I IIIv,IIp[p(F)](p-l)'p5 M[p(F)]'p-l)'P. Letting n + CO, we obtain lv(F)(sM[p(F)(p-l)'p.From this, it follows that v << p. For any partition P = {El, E2, . . . ,Em}in 8,
Taking supremum over all partitions P in 8, we obtain IIvII;sMP, or v E Vp(R, 9, PI.
200
THEORY O F CHARGES
Now, for any n 2 1 and for any given E > 0, by Lemma 7.2.9,there exists a constant k such that for any partition P ={El, Ez, . . . , Em}in 9, we have IIv-v,llp
4lv-vPIIp +llvP-(vfl)Pllp +ll(v,)P-~flIlp
Observe that
and this converges to zero as n + 00 since limfl+mv,(F) in 9. Therefore, lim sup I I -~vfl1lp5 1 1 v- ~
= v(F)
for every F
+~+[~(II~II::-II~PII~+&II~II~I~’~
P I I ~
fl-47
for every P in 9.By taking limPEp,we obtain limsupIlv-v,llp~o+o+&IIvllp. n+m
Since & > 0 is arbitrary, we conclude that limfl+mIIv - vnllp= 0. This com0 pletes the proof. The following theorem gives a necessary and sufficient condition for p). strong convergence in V1(R, 9,
7.4.5 Theorem. Let (0,9, p ) be a probability charge space and v,, n 2 1 a sequence in Vl(R, 9, p ) . Then v,, n 2 1 converges strongly in Vl(fl, 9, p) if and only if v, (F),n 2 1 converges uniformly over F in 9. 0
7.5
WEAK CONVERGENCE
Weak convergence in V1(R, 9, p ) will be considered quite extensively in Chapter 8. In this section, we prove just one result about weak converp ) for 1< p COO. gence in V,(R, 9,
7.5.1 Theorem. Let 1 < p < CO. Let (a,9, p ) be a probability charge space p). and v,, n 2 1 a sequence in V,(Sl, 9, (i). If v,, n L 1 is weakly convergent to a v in V,(R, 9, p ) , then v(F) = lim,,+mv,(F) for F in 9 a n d sup,,^ llvnllp
7. V,-SPACES
201
Proof. (i) If v,, rz 2 1 converges to v weakly in V,(n, 9, p ) , then TA(v,), n 2 1 converges to T A ( v )for every A in V,(Cl, 9, p ) , where l/p + l/q = 1. For F in 9, let A = p~ which obviously belongs to V,(n, 9, p ) . Consequently,
v(F)=T,(A)= TA(v)=limn+mTA(v,) = lim Tv,(h)= lim vn(F). n+m
n+m
See Remark 7.3.2.It is well known that every weakly convergent sequence in a Banach space is norm bounded. See Thoerem 1.5.16. (ii). As in the proof of (ii) of Theorem 7.4.4,it follows that Y E V,(Cl, .9,p ) . Now, we show that v,, n 2 1 converges to Y weakly in V,(Q, 9, p). Let A E V,(Cl, 9, p ) and Thbe the continuous linear functional on V,(n, 9, p) induced by A, where l/p + l/q = 1. TA is determined on SC(Cl, 9, p ) by k k uipEifor some real = uiA(Ei),where 7 = the following rule. TA(7) We show numbers ul, uz, . . . , a k and some partition {El,E2, . . . ,Ek}in 9. . would then imply weak converthat T A ( v n n) , 2 1 converges to T A ( v ) This gence of Y,, n rl to v in V p ( f l , 9 , p ) . Observe that for any partition P = {F1,F2, . . . ,F,) in 9,
ITA ( v n
I I
-T A(v)
TA ( v n ) -
+ ITA
I
TA
T A ( ( v n )P)l
( ( v n )PI
( v n ) - TA
First, we let n + CO. Then we obtain
-T A (vP)l ((vn)P)l
+ IT A
(vP)
- TA
I
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THEORY OF CHARGES
Since this inequality is true for any P in 9, by taking the limit over P E P and using Theorem 7.2.12, we obtain
This completes the proof.
0
7.5.2 Remark. Given a probability charge space (R, S,p ) , one can find a charge space (X, %, A ) in which % is a cr-field on X and A is a probability 8 such that V,(fl, 9 , p ) is isometrically isomorphic to measure on 5 9,(X, %, A ) for every 1S p S 00.
CHAPTER 8
Nikodym Theorem, Weak Convergence and Vitali-Hahn-Saks Theorem
Nikodym theorem for measures gives a necessary and sufficient condition for norm boundedness of a set of measures on a cT-field. Vitali-HahnSaks theorem for measures characterizes uniform absolute continuity in the space of all bounded measures on a cT-field. These results are useful in the study of weak convergence in the space of all bounded measures on a a-field. The exact analogues of Nikodym and VitaliHahn-Saks theorems are not valid for charges on every field. In this chapter, we seek the exact analogues of these theorems for charges by considering more restrictive fields and yet general enough to include a-fields as a special case. The presentation of this chapter is somewhat at variance with that of the previous chapters in the following sense. In the earlier chapters, we have obtained weaker versions of some results of Measure theory for charges valid on any field. In this chapter, we obtain exact versions of the above theorems by considering some special class of fields. Section 8.1gives the relevant background information about Nikodym and Vitali-Hahn-Saks theorems. Section 8.2 gives some examples to demonstrate that Nikodym and Vitali-Hahn-Saks theorems are not valid for charges on fields. Section 8.3 presents Phillips’ lemma which plays an important role in our quest in extending Nikodym and VitaliHahn-Saks theorems for charges on some special type of fields. Phillips’ lemma is also used in characterizing weak convergence of charges. Section 8.4 presents the desired extension of Nikodym theorem. Section 8.5 studies norm boundedness of a set of bounded charges in the light of uniform absolute continuity. Section 8.6 gives a decomposition theorem for a given set of bounded charges in terms of norm bounded sets and finite dimensional sets. Section 8.7 is mainly concerned with characterizations of weak convergence in the space of bounded charges. Finally, Vitali-Hahn-Saks theorem is extended to special fields in Section 8.8.
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THEORY OF CHARGES
8.1 NIKODYM AND VITALI-HAHN-SAKS THEOREMS IN THE CLASSICAL CASE One of the important problems in the study of the space ca(R, %) of all bounded measures on the o-field % of subsets of a set R is the identification of weakly sequentially compact subsets M of ca(R, %). A set of necessary and sufficient conditions for m to be weakly sequentially compact involves norm boundedness of m and uniform absolute continuity of m with respect to some A in ca(Cl,%). Nikodym theorem characterizes norm bounded subsets of ca(R, %) and Vitali-Hahn-Saks theorem deals with uniform absolute continuity.
8.1.1 Nikodym Theorem. Let % be a o-field of subsets of a set Cl. Let M be a subset of ca(R, 3).Then Sup {lp (A)I;p E M}< 00 for every A in % if and only if Sup {lip 11; p E M} < 00. For a proof of this result, one may refer to Dunford and Schwartz (1964, Theorem IV.9.8, p. 309). We will obtain this result as a corollary of a more general result we prove in Section 8.4 in this direction. It is worth noting that if Nikodym Theorem 8.1.1 is valid for every countable subset of M, then it is also valid for M. For Vitali-Hahn-Saks theorem, we need the following definitions.
8.1.2 Definition. Let 9 be a field of subsets of a set R and M c ba(Cl,9). Let A E ba(R, 9).M is said to be uniformly absolutely continuous with respect to A if for every E > 0, there exists 6 > 0 such that Iv(A)I< E for every v in M
whenever A E 9 and IA I(A)<6.
8.1.3 Definition. Let be a o-field of subsets of a set R and M c ca(R, a). M is said to be uniformly countably additive if for any sequence A,, n 2 1 of pairwise disjoint sets in % and e > 0, there exists a natural number N L 1 such that
12,
P (Ad
I<
&
for every n 2 N and p in M. Now, we state Vitali-Hahn-Saks theorem.
8.1.4 Vitali-Hahn-Saks Theorem. Let % be a o-field of subsets of a set R and pn, n 2 1 a sequence in ca(R, %) such that p,(A), n L 1 converges to a real number for every A in %. Define p on % by p(A) = n-m lim p,(A),
A E 3.
8.
NIKODYM A N D VITALI-HAHN-SAKS
THEOREMS
205
Then the following statements are valid. (i). M = {p,, n 2 1) is uniformly absolutely continuous with respect to A whenever A E ca(R, '21) and p , c A for every n 2 1. (ii). M is uniformly countably additive. (iii). p is a bounded measure on '21. The non-trivial part of this theorem is the conclusion (i) from which (ii) and (iii) follow easily. For a proof of this result, one may refer to Dunford and Schwartz (1964, Theorem 111.7.2, Corollary 111.7.3 and Corollary 111.7.4, pp. 158, 159 and 160). We also obtain this result as a corollary of a more general result we present in Section 8.8.
8.2 EXAMPLES In this section, we present two examples which illustrate the point that extensions of Nikodym and Vitali-Hahn-Saks theorems fail in the framework of charges on fields. The first example we give here is concerned with Nikodym theorem.
8.2.1 Example. Let R be any infinite set and 9 the finite-cofinite field on SZ: Fix a sequence x,, n 2 1 of distinct points in R. For each n L 1, define pn on 9by n, if A is finite and x n E A, pn (A)=
I
0, if A is finite and xn& A,
-n,
if A is cofinite and xn& A,
0, if A is cofinite and x,
E A.
It is easy to check that each p n is a bounded charge on 9 and that (p,,(A)I
A@)=
1 nd(A)2
c
7,
1 n c I ( A ) 2"'
= 1+
C
-
if A is finite, if A is cofinite,
206
THEORY OF CHARGES
where I ( A ) = {n L 1; x, E A}. A is a charge on 9and each p, << A. The later statement can be verified as follows. Fix n 2 1 and let E > 0. Take 0 < 6 < 1/2". If A E 9 and A (A)< S , then A is finite and x,& A. Consequently, p,(A) = 0 < E . However, {p,, n 2 1)is not uniformly absolutely continuous with respect to A. The above examples also demonstrate that Nikodym and Vitali-HahnSaks theorems fail to hold in the space ca(f2,9), where 9 is a field of subsets of f2.
8.2.3 Remark. Nikodym theorem for positive charges on any field trivially holds whereas Vitali-Hahn-Saks theorem may not hold. For example, let f2= {1,2,3, . . .}, 9the finite-cofinite field on f2 and p,(A) = 1, if n E A, =0, ifn@A,AE$ for every n 2 1. Let p (A) = CnsA1/2", A E 9. Then p,(A) exists for every A in 9and p, << p for every n 2 1.But {p,, n 2 1) is not uniformly absolutely continuous with respect to p.
8.3 PHILLIPS' LEMMA In this section, we prove a lemma due to Phillips (1940, p. 525) which is basic in our study of Nikodym and Vitali-Hahn-Saks theorems. First, we need the following concept of semi-variation of a charge. 8.3.1 Definition. Let For F in 9,let
p
be a charge on a field 9 of subsets of a set f2.
C; (F)= SUP {Ip (E)\;E c F, E E 9).
C; is called the semi-variation of p. We give some properties of C;. 8.3.2 Proposition. Let p be a charge on a field 9 of subsets of a set f2 and C; its semi-variation. Then the following statements are true. (i). C;(E)5 C; (F) whenever E, F E 9and E c F. (ii). C; (Eu F) IC; (E)+ C; (F) whenever E, F E 9. (iii). C; (E)5 lp ((E)5 2C; (E)for any E in 9.
8.3.3 Phillips' Lemma. Let f2={l, 2 , 3 , . . .} and S=P(f2),the class of all subsets of f2. Let p,, n 2 1 be a sequence of bounded charges on 9. In the following, ( i ) j (ii)@(iii).
8.
NIKODYM AND VITALI-HAHN-SAKS THEOREMS
207
(i). limn+mp,(A) = 0 for every A in 9. (ii). limn-tmSup {Ip,(A)[;A c R, A finite}= 0 , i.e. p,, n 2 1 converges to 0 uniformly over all finite subsets of R. (iii). Iimn+mC k Z l lpn({k})l= 0. In particular, if (i) holds, then p,({n}) = 0.
Proof. (i)j (ii). This is proved in the following steps. 1". Given any integer m o r 1 and E > 0, we can find n o 2 1 such that lp,,({k})l<E whenever n 2 no. For this, use limn+mIp,,({k})l=0 for each k = 1 , 2 , . . . ,mo, from (i). 2". Suppose (ii) is not true. Then there exists 77 > 0 such that for any given m r 1, there is an m' 2 m and a finite set A c R such that Ip,,,,(A)Ir 277. 3". We develop some notation. If A is a finite subset of R, let min (A) stand for the smallest element in A and max (A) for the largest element in A. 4". We claim that for any given integer p 2 1, 7 given above in 2" and any integer 41 2 1, there is an integer 4 r q 1 and a finite set A c R such that p 5 min (A) and lpq(A)\2 77. The above claim obviously follows when p = 1 from 2". Suppose the above claim is false for some p > 1. Then there exists an integer q 2 2 1 such that for any 4 z q 2 and every finite set A c R with p ~ m i (A), n IpLq(A)I < 77 holds true. Since there are finitely many sets B c R having the property that max (B) < p and limn-tmp,,(B)= 0, we can find q3 2 1 such that Ipq(B)I< 77 .for every finite set B c R with max (B) < p and q 2 q 3 .Now, let q be any integer 2 max {q2,q3}. Let A be any finite subset of R. Let B = { i E A ; i < p } and C = { i E A ; i z p } . Then B n C = 0, B u C = A , max (B) < p and rnin (C)>p. Further, lpq(A)l= [pq(B)+pq(C)/ 5 lpq(B)I+ Ik,(C)l< 77 + 77 = 277. This contradicts 2". Hence the claim is established. 5". Now, we obtain two sequences n1 < n 2 < . and ml < m 2 < * * of positive integers and a sequence A,, n 2 1 of finite subsets of R having the following properties. (a). mi 5 min (Ai)5 max (Ai)< rni+l, i = 1 , 2 , 3 , . . . . (b). (pni(Ai)( r 7, i = 1 , 2 , 3 , . . . . (c). 121 Ipni({k})/<77/8, i = 2 , 3 , . . . . First, we obtain m l , A l , n1 and m2. Take m l = 1. In 4", take p = 1 and q1 = 1. We find an integer nl r q l and a finite set A1 t R such that m l = p 5 min (A,) and (pnl(A1)l 2 77. Let m 2= max (Al)+ 1. Using m2,we proceed to obtain A,, n2 and m3 satisfying (a), (b) and (c). In lo, take mo=m2 and E = q / 8 . We can find nk 2 1 such that Ip,,({k})l<77/8 for every n r n ; . In 4", take p = m 2 and q1= m a x { n ~ , n l } + l . We can find n 2 z q 1 and a finite set A2cR such that L 77. Note also that Cr2, lp,,&k})/< ~ / 8 . min (A,) 2 m 2 and Ipn2(A2))
x:Zl
-
208
THEORY OF CHARGES
Using m3, as in the above, we can obtain A3, n3 and m4 satisfying (a), (b) and (c). Proceeding this way, we obtain the desired sequences. Observe that A,, n 2 1 is a sequence of pairwise disjoint sets. 6". Now, we find a sequence S,, n 2 1of infinite subsets of R and a sequence p1 < p z < * * * of positive integers with the following properties, (a)'. P k is the smallest element in s k , k 2 1. (b)'. Sk+lC s k -(Pk}, k 2 1. (c)'. f i n p , ( U i a S k + 1 Ai) 5 77/89 k 2 1. First, we obtain S1, p1 and Sz. Write R = Ti, where T,, n 2 1 is a sequence of pairwise disjoint sets and each Tj is infinite. We claim that there is a j 0 2 l such that f i n l ( U i e ~ , , A i ) < ~Suppose /8. not. For every j' 2 1, iinl(uis~, Ai)> 77/8. We can find, for each j 2 1, a set Bi C U I ~Ai T, such that Ipnl(Bj)l< q / S . Since pnl is a bounded charge on 9and Bj, j 2 1 is a sequence of pairwise disjoint sets in .F,limjem ,unl(Bj)= 0. This contradiction establishes the claim. Let S1 = Ti,. Let p1 be the smallest element in S1. Then p 1 # 1. For, if p1= 1, then q / 8 Aj) 2 lFn,(A1)l2 77, by 5"(b). This contradiction shows that p1 # 1. By decomposing S1 -(PI} into a countable number of pairwise disjoint infinite sets, as above, we can find an infinite subset Sz c S1-(PI} such that finp,(UiES2 Ai)5 q / 8 . Hence the desired S1, p1 and S 2 are obtained satisfying (a)', (b)' and (c)'. If p 2 is the smallest element in Sz, then p 1 < p z . Continuing this way, we obtain the desired sequences having properties (a)', (b)' and (c)'. 7". Let A = Ui2lA , . For any fixed i 2 1, write
ujzl
2fi,l(ujo~1
i-1
u A , u u A,. By 5"(b), Ipnpi(APi)177. Note that max ( u;.:APi) : < mpi-l+lI m p ( ,by Sob). A = A, u
j=l
B
Consequently, by 5"(c),
Since
jzi+l
8.
NIKODYM AND VITALI-HAHN-SAKS
THEOREMS
209
by 6"(c)'. Consequently,
3
>q-q/8-~/8=zq.
-
Since P I < p 2 < * np,< npz< * .Therefore, pnp,,pnpz,. . . is an infinite pnPi(A) = 0. The above subsequence of p l , p z ,. . . . So, by (i), inequality is a contradiction to this limit. Hence (ii) is valid. (ii)+ (iii). Note that a,
m
0 ~r lirn sup n+m
~r 2
C lpn({k})I= limn-msup rn-mk-1 lim 1 kzl
( p , ({k})I
lim sup Sup {lp,,(A)l;A c R and A finite} = 0, n-rm
by (ii). Hence (iii) follows. (iii)+ (ii). Note that 0 5 lirn sup Sup {(pn(A)I; A c n and A finite} n+m
5 lirn sup n-m
C
Ipn( { k } ) (= 0,
by (iii).
k z l
Hence (ii) follows.
8.3.4 Remark. In the above lemma, (iii)+(i) is not true. Take any sequence pn, n 2 1 of distinct 0-1 valued charges on P(R) each of which vanishes on finite sets.
8.4 NIKODYM THEOREM We develop this theorem in the framework of Boolean algebras. The various notions connected with Boolean algebras are given in Section 1.4.
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THEORY OF CHARGES
8.4.1 Definition. A Boolean algebra B is said to have Seever property if for every two sequences a,, n 2 1 and b,, n 2 1 in B satisfying the following conditions (i). a l s a z s a 3 * ~, ~ ~ (ii). bl? bz 2 b3 L * and (iii). a, b, for every m, n 2 1, there exists c in B such that a, s c 5 6, for every m, n 2 1.
-
Boolean cr-algebras and cr-fields of subsets of any set constitute an important class of Boolean algebras with Seever property. We give some examples.
8.4.2(i) Example. Let R = {1,2,3, . . .} and 9 the finite-cofinite field on $2.The Boolean algebra 9does not have Seever property. Take A, = { 2 , 4 , 6 , .. . ,2m},
m 21,
and
B , = R - { l , 3 , 5 ,..., 2n+1},
n ~ l .
There is no C in 9such that A, c C c B, for all m, n L 1. The following is an example of a Boolean algebra with Seever property which is not a Boolean cr-algebra.
8.4.2(ii) Example. First, we prove the following result. Let B1 and Bzbe two Boolean algebras and h onto homomorphism from B1 to Bz. If B1 is a Boolean algebra with Seever property, then Bz also has Seever property. Let a,, n 2 1 and b,, n L 1 be two sequences in BZsuch that a, 5 u,+l5 bm+l5 b , for all m, n 2 1. Let c,, n L 1 and d,, n L 1 be two sequences in Bl such that h(c,) = a n for every n L 1 and h(d,) = b, for every m L 1. Define e l = c1 A dl, f l = c1 v d l , f n t l = f , A (cnClv d,+l v e,) and e,+l = e , v ( c , + l ~ d , + l ~ f , ) ,n s l . Then e l 5 e z " * * " f 2 5 f l , h(e,)=a, and h ( f m )= b, for all m, n L 1. Since B1 has Seever property, there is an x in B1 such that e, s x sf, for all m, n L 1. Then a, Ih ( x )5 6 , for all m, n 2 1. This completes the proof. Now, we give the desired example. Let R = {1,2, . . .} and 9 the ideal of all finite subsets of 0. Since the natural homomorphism from P(R) to the quotient Boolean algebra P ( R ) / 9 is onto and "(0) has Seever property, by what we have proved above, P(R)/,9 has Seever property. We show that P(Q)/9 is not a Boolean a-algebra. Let B,, n 2 1 be a sequence of pairwise disjoint subsets of R each of which is infinite. Then Vnrl [B,] does not exist. This we prove as follows. Let [C]z[B,] for every n 2 1. Then we exhibit [D] in P ( R ) / 9 such that [C] > [D] 2 [B,] for every n 2 1.
8.
NIKODYM AND VITALI-HAHN-SAKS THEOREMS
211
Since B, - C is finite for every n 2 1, we can find a point x , in C n B , for every n L 1. Let A = {x,; n r 1)and D = C - A. This D serves the purpose. [B,] does not exist. Consequently,
v,
The most important result about Boolean algebras with Seever property is the following.
8.4.3 Theorem. Let B be a Boolean algebra with Seever property. Let p be any bounded charge on B. Let X = {b E B; lp 1(b)= 0). Then the quotient Boolean algebra B / N is complete.
Proof. Since B/X satisfies the countable chain condition, i.e. any family of pairwise disjoint elements in B / N is at most countable (because there cannot be more than n pairwise disjoint elements b in B such that Ip I(b)> l k l ( l ) / n ) ,it suffices to show that B/X is a Boolean c+-algebra.See Theorem 1.4.8. Let [a,], n 2 1 be a sequence of equivalence classes in B / N . We show that Vn,l [a,] exists in B/X. Without loss of generality, assume that a l r a 2 1 - * - . Let C = { b € B ; b r a , for every n r l } . Let r = Inf {Ipl(b);b E C}. Then there exists a sequence b,, n 2 1 in C such that r= Ip](b,).Without loss of generality, assume that bl 2 b22. * . Since B has Seever property, there exists c in B such that a, 5 c I b, for every m, n z 1. It is now obvious that V,,I [a,] = [ c ] . Thus B/X is a 0 Boolean c+-algebra.
-
Now, we are ready to prove Nikodym theorem for charges on complete Boolean algebras.
8.4.4 Theorem. Let B be a complete Boolean algebra and p,,, n 2 1 a sequence of bounded charges on B. Suppose for every b in B, Sup,,, Ipn(b)l< 00. Then Sup,,, llpnll< 00. Proof. Suppose Sup,,,, Ilp,ll= 00, i.e. Sup,,l SUPbeB Ip,(b)l = CO. First, we find a sequence c,, n L 1of pairwise disjoint elements in B and an increasing sequence mk,k r 1 of positive integers such that limk+mIpmk(ck)l= 00. For each a in B, define t, = Sup,,, SupbSa Ikn(b)l.Note that for each a in B, either t, = co or tl-, = co, where 1 is the unit element in B. For, if t, < 00 and tl-, < 00, then tl = Sup,,, SUpbeB Ip,(b)l < 00. Similarly, if t, = 03 for a in B, then either t b = co or t a - b = co for any b in B satisfying b 5 a. From the supposition that Sup,,,, Ilp,,ll=m, we can find m l z l and dl in B such that Ipml(dl)(>Supnrl Ip,(1)1+2. If tdl=co, t a k e e l = l - d l . We find that Iprn1(Cl)l=
I~rnl(1-dl)lz Ikrnl(dAl - I p m l ( l ) l
2 Ipm1(dl)l-SUPnz,
I/~n(1)1>2.
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THEORY OF CHARGES
If tl-& = 00, take c1= dl. In this case, we find that lprnl(cl)l= Iprnl(dl)l>2. In any case, we have Ipml(cl)l> 2 and tl-cl = 00. Since tl-E1= co,we can find dZI1 - c1 and m2 > ml such that (prn2(d2)1 > Sup,,l Ip,(1-cl)l+3. If t&=co, takecz=(1-c1)-d2. Note that Iprnz(cz)l= Iprnz((1
-cl)-dZ)I LIprnz(dz)I-Ipm,(l
-cI)I
~ I ~ r n z ( d 2 ) l -I pSn ~( 1~- ~ 1 ) 1 > 3 . nzl
If t(l-cl)-dz = 00, take c z = dZ.Now, Ipm2(c2)I = Iprn2(d2)I > 3. In any case, we observe that m2 > ml, C I A cz = 0, I p m l ( C 1 ) 1 > 2, IKrnz(Cz)l> 3 and tl-(clvc2)= 00. Continuing this procedure, we obtain a sequence c,, n L 1 of pairwise disjoint elements in B and an increasing sequence mk, k L 1 of positive integers such that limk--tmI/Arnk(Ck)l = a. For each k L 1, we define A k on P(R), where R = {1,2,3, . . .}, by
for A c R . Note that each Ak is a bounded charge on P(R). Further, limk,m &(A) = 0 for every A c a. This follows from limk+mI&,,(Ck)l= and SUpkzl Iprnk(ViE~ cj)l < 00 for any A c R. By Phillips' lemma 8.3.3, we conclude that limk,m Ak({k}) = 0. But hk({k}) = 1 for every k 1. This contradiction proves the result. 0 Now, we generalize Theorem 8.4.4 to Boolean algebras with Seever property.
Theorem. Let B be a Boolean algebra with Seever property and p,, n 2 1 a sequence of bounded charges on B. Suppose for every b in 5, Supnzl Ipn(b)l<m. Then Supnz1 I l p n l l < a .
8.4.5
for b in 5. Then A is a bounded charge on 5 and p, << A for every n 2 1. Let X = {a E B; A (a) = 0). Then, by Theorem 8.4.3, the quotient Boolean algebra BIN is complete. For each n z 1, define fin on 5 / N by fi,([bI) = p , ( b ) , b E B. Since pn << A, fin is well defined on B / N for all n L 1. Note that fin, n 2 1 is a sequence of bounded charges on 5 / N satisfying the hypothesis of Theorem 8.4.4. The conclusion of Theorem 8.4.4, now, gives the result. 0 The following is the main result of this section which gives Nikodym theorem for sets of bounded charges on Boolean algebras having Seever property.
8.
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THEOREMS
213
8.4.6 Corollary. Let M be any collection of bounded charges on a Boolean algebra B having Seever property. Suppose for every b in B, Sup {lp( b ) ] ; p E M} < 00. Then Sup {Ilpll;p E M} < a. Proof. Suppose Sup {IlpII;p EM}=00. Then we can find a sequence p,, n L 1in M such that sup,,^ I(p,ll = 00. By the given hypothesis, it is obvious that for every b in El, Sup,,, Ip,(b)l<m. This contradicts the validity of Theorem 8.4.5. 0 Of course, the classical Nikodym Theorem 8.1.1 follows.
8.4.7 Corollary. Let M be any collection of bounded charges on a u-field
9l of subsets of a set 0. Suppose for every A in a, Sup {Ip (A)I;p E M} < 00. Then Sup {lip 11; p E M} < 00. In particular, Theorem 8.1.1holds. 0
8.5
NORM BOUNDED SETS IN THE PRESENCE OF UNIFORM ABSOLUTE CONTINUITY
Nikodym theorem (Corollary 8.4.6) gives a set of necessary and sufficient conditions under which a given subset M of b a ( R , 9 ) is norm bounded when 9 is a field of subsets of a set R having Seever property. Now, we give a set of sufficient conditions based on the notion of uniform absolute continuity and strongly continuous charges, for a given set M c ba(R, .F)to be norm bounded without any conditions on 9. We need some preliminary results on absolute continuity, singularity and strongly continuous charges. We say that a bounded charge v on a field 9of subsets of a set R is atomic if v is a countable sum of two-valued charges on 9.
8.5.1 Proposition. Let 9be a field of subsets of a set 0. (i). I f p , v and T are bounded charges on 9such that p << v + T and p 17, then p v. (ii). I f p , v and T are bounded charges on 9such that p << v +T, then we can write p = p 1 + p 2 such that v and p2-x T. (iii). If p, v and T are bounded charges on 9such that p << Y + T and v I T , then we can write p = p l + p 2 such that p l << v and p2<< T , and such a decomposition is unique. (iv). If v is a two-valued bounded charge on 9 and p is any charge on 9 such that p << v, then p is two-valued. (v). If p is a strongly continuous charge on 9 and v is a bounded two-valued charge on 9, then p l u .
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(vi). I f p is a strongly continuous charge on 9and v is a bounded atomic then p Iv. charge on 9, (vii). I f p and v are two bounded charges on 9 s u c h that v is atomic and p << v, then p is atomic. (viii). I f p and v are two distinct two-valued charges on 9, then p Iv. In fact, there exists A in $such that Ipf(A)= Ipl(R) and IvI(A') = Ivl(R). More generally, if vi, i L 1 is a sequence of distinct two-valued charges on 9,then p IA, where p = CisDaivi and A = pjvj, with D and E being disjoint subsets of {1,2,3,. . .}, C i s D Iail
xjeE
Proof. Without loss of generality, we assume that all the charges involved are positive. (i). Let E > 0. There exists S > 0 such that p (B) < ~ / whenever 2 B E 9and (v +r)(B) < 8. Assume, without loss of generality, that S < E . Since p 17, there is a set D in 9such that p (D) < S/2 and 7(DC)< S/2. Now, we show that p << v. Let C be any set in $such that v(C) < S/2. Note that p(C nD) < 6/2 < ~ / 2and (v + 7)(Cn D') = v ( C nD') + r ( C n D") < S/2 +S/2. So, p (C nD') < ~ / 2 Thus . we see that p (C) < E . (ii). By Lebesgue Decomposition Theorem 6.2.4, we write p = p1 +p~z such that p l<< Y and p 2 1I/.Clearly, p1 and p2 are positive. Now, p 2 5 p1+p~2
(iii). By (ii), we can write p = p 1 + p 2such that pl<
.
=O<E
and p ( (
5 Ai))=p(Aio)<E.
i=l
Hence p I v. (vi). Since v is atomic, positive and bounded, we can write v aivi such that ai > 0 for every i L 1, CiZlai < co, vi is 0-1 valued for every i 2 1 and vi, i L 1 is a distinct sequence of charges. Let E > 0. Choose N 1 1 N such that CizN+lai < ~ / 2 .By (v), aivi. So, we can find D in 9 such that p(D) < ~ / 2 and ZEl aivi(D') < s/2. Now, v(D') = N aivi(DC)+Ci,N+l a i v i ( D ' ) < ~ / 2 + ~= / 2E . This completes the proof. (vii). By Sobczyk-Hammer Decomposition Theorem 5.2.7, we can write p = p l + ~ : !such that p l is strongly continuous and p2 is atomic. Since
lxi=l
xi=1
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NIKODYM AND VITALI-HAHN-SAKS THEOREMS
215
u. But by (v), p l lu. Hence p1 = 0. Therep1+p2<
Proof. We prove only (a) of (ii). Without loss of generality, we assume that all charges involved are positive. Let E > 0. Since M is uniformly absolutely continuous with respect to u, there exists 8 > 0 such that p (A) < e/2 for every p in M whenever A E 9 and u ( A )<S. We take S < E . We show that p1(A) < E for every p1 in M1 whenever A E 9 and ul(A) < S/2. Let A E 9and ul(A) < S/2. Let ~1 E MI. Note that, by Proposition 8.5.l(vi), p 1 1 u 2 . So, there exists D in 9 such <S/2. Note that pl(A nD) <6/2 <~ / 2 . that pl(D) < S/2 and v2(DC) Also, u(A nD") = vl(A nD") + u2(AnD") < S/2 + S/2 = S. Consequently, p(AnD')<&/2. So, p1(AnD")<&/2. Therefore, p1(A)= p l ( A n D ) + p1(A nD') < ~ / +2~ / =2E . This completes the proof. 0 Now, we prove the main theorem of this section. 8.5.3 Theorem. Let 9 be a field of subsets of a set R and M c ba(R, 9) be uniformly absolutely continuous with respect to a v E ba(R, S). (i). If u is a strongly continuous charge on P, then M is norm bounded, i.e. SuP{llPll;ELEM}<~. (ii). If every p in M is a strongly continuous charge ofz 9, then M is norm bounded. Proof. (i). Without loss of generality, assume that all charges involved are positive. Since M is uniformly absolutely continuous with respect to u, for E = 1, there exists S > 0 such that p(B) < 1 for every p in M whenever B E 9 and u(B) < 8.Since v is strongly continuous, there exists a partition (B1, B2, . . . ,Bk} of fl in 9 such that v(Bi) < S for i = 1,2, . . . , k. So, for k any p in M, p (a)= C,=l p (Bi)< k. Hence M is norm bounded. (ii). By Sobczyk-Hammer DecompositionTheorem 5.2.7, write u = u1 + u2, where ul is a strongly continuous charge on 9 and v2 is an atomic charge
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THEORY OF CHARGES
on 9. By Proposition 8.5.2(ii), M is uniformly absolutely continuous with respect to v l . (i) completes the proof now. 0
8.6 A DECOMPOSITION THEOREM Let 9 be a field of subsets of a set R and M c ba(R, 9) be uniformly absolutely continuous with respect to a v in b a ( R , q . In this section, we make an attempt to isolate that part of M which is responsible for the failure of M to be norm bounded. More precisely, we obtain a decomposition of M into a norm bounded part and a finite dimensional part. We start with a definition. N
8.6.1 Definition. Let A api,where v l , vz, . . . , vN are distinct 0-1 valued charges on 9 and a1, a z ,. . . , aN are any non-zero real numbers. A, is defined to be the collection of all bounded charges on 9 absolutely continuous with respect to A. Any subset of A, is called a finite dimensional set. 8.6.2 Remark. One can show that any p in A, is of the form pivi for some real numbers PI, p2, . . . ,pN. One can use Proposition 8.5.1 to establish the veracity of this statement. This is the reason for the use of the term “finite dimensional set”. The following proposition is instrumental in proving the main result of this section.
8.6.3 Proposition. Let 9be a field of subsets of a set R and M a subset of ba(R, uniformly absolutely continuous with respect to an atomic charge v in b a ( R , 9 ) . Then there exist two subsets MI and Mz of ba(R, 9)satisfying the following properties. (i). M1 and MZ are both uniformly absolutely continuous with respect to v. (ii). M c MI + M2 = { p + 7 ;p E MI, 7 E Mz}. (5).M1c A, for some finite linear sum A of 0-1 valued charges on 2F, i.e. MI is a finite dimensional set. (iv). MZis norm bounded.
Proof. The proof is carried out in the following steps. 1”.Without loss of generality, we can assume v to be positive. Since v is atomic, we can write v = Cizl aivi,where ai > 0 for every i 2 1,Cizl ai < CD and vi, i 2 1 is a sequence of distinct 0-1 valued charges on 9. (Iizl aiui could be a finite sum.)
8.
NIKODYM A N D VITALI-HAHN-SAKS
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217
2". Since M is uniformly absolutely continuous with respect to v, for E = 1, there exists S > 0 such that Ip [(B)< 1 for every p in M whenever B E 9 and v(B) < 6. N 3". There exists N 2 1 such that CizN+la; <S. Let A aivi and T = CizN+laivi.Note that ~ ( 0< S) and v = A + T . 4".Let (A1, A2, . . ,AN} be a partition of R in 9such that vi(Ai)= 1 for i = 1 , 2 , . . . ,N and vj(Ai)= 0 for every i Zj,i, j = 1,2, , , ,N. 5". Since the charges vi, i 2 1 are distinct, A IT. See Proposition 8.5.l(viii). 6". Let p EM. We can write p = p1+ p 2 , where p l and p2 are charges on 9, p1<
.
.
N
Ic~zl(R)=C IpZI(Ai) i=l N
=
N
C Ip2I(Fi)+i C= l i=l
1~21(Ai-Fi)
N
Note that for each i = 1 , 2 , . . . ,N, N
<0+6=S. By 2", IpL((Ai-Fi)
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THEORY OF CHARGES
there exist two sets M1 and M2 contained in ba(R, 9)having the following properties. (i). M1 and M2 are both uniformly absolutely continuous with respect to v. (ii). M c M1+ M2. (iii). M1 c A, for some finite linear sum A of 0-1 valued charges on 9, i.e. M1 is a finite dimensional set. (iv). M2 is norm bounded. Proof. Assume, without loss of generality, that v is positive. Write v = v1 + v2, where v 1 is a strongly continuous charge on 9 and v2 is an atomic Write each p in M also as p 1 + p 2 , where p1 is a strongly charge on 9, continuous charge on 9 and p2 is an atomic charge on 9.Let H I = { p l ;p EM} and H2 = { p z ;p EM}. H1 and H2 are both absolutely continuous with respect to v and M c H1+&. By Proposition S.S.Z(ii>,HI is uniformly absolutely continuous with respect to v1 and H2 is uniformly absolutely continuous with respect to v 2 . By applying Proposition 8.6.3 to H2 and v2, we obtain two sets H3 and H4 contained in ba(R, 9)such that H2c H3 + H4, H3c A, for some finite linear sum A of 0-1 valued charges on 9and H4 is norm bounded. By Theorem 8.5.3(i), H1 is norm bounded. Take M1 = Hs and M2 = H1+ H4. MI and Mz are the desired sets. 0
8.7 WEAK CONVERGENCE Let 9 be a field of subsets of a set R. The principal object of study in this section is the space ba(R, 9)in its weak topology. It seems difficult, i.e. the space of all in general, to characterize the dual space ba*(R, 9), continuous linear functionals on the Banach space ba(R, In this section, we characterize weak convergence in ba(R,9). Recall that the weak topology on ba(R,9) is the smallest topology on ba(R, with respect to which all functions in ba*(R, 9)are continuous. The classical results on weak convergence in the space of bounded measures on a cr-field follow as simple corollaries of the results of this section. First, we give two results which are useful in characterizing weak convergence.
a.
8.7.1 Theorem. Let 4, pn, n L 1 be a sequence of bounded charges on a field $of subsets of a set R satisfying the following conditions. (i). {pn,n 2 1)is uniformly absolutely continuous with respect to 4. (ii). Supnal Ipn(F)I<W forevery F i n 9. Then Supna1 l l p n l l < 00.
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THEOREMS
-
Proof. Suppose Supnzl Ilpnll= 00. We exhibit a sequence n l < n2 < * of positive integers and a sequence A,, n L 1 of pairwise disjoint sets in 9 such that Ipnk(Ak)l 2 1 for every k 2 1. Let r =Supnzl Ip,(R)l. Then r
-
The above result can be labelled as Nikodym theorem for charges on fields. There are no conditions on the field but we impose an extra condition on the sequence F,, n L 1 to conclude that the set {p,, n L 1)is norm bounded. If p,,, n L 1 satisfies condition (i) only, it is not true that {p,, n 2 1) is norm bounded. As an example, let R = {1,2,3, . . .}, 9= 9(R)and p any 0-1 valued charge on 9such that p (A) = 0 for every finite subset A of R. Then the sequence { n p , n L 1) is uniformly absolutely continuous with respect to p but Sup,,l I(npII = 00. The argument presented in the proof of the above theorem is similar to the one used in the proof of Lemma 2.1.5.
Theorem. Let ( R , 9 , p ) be a probability charge space and p,, n L 1 a sequence in V1(R, 9,p ) such that {p,,, n 2 l} is uniformly absolutely continuous with respect to p and limn+, p,(F) = 0 for every F in 9. Then p,, n 2 1 converges to 0 weakly in V1(R, 9,p ) .
8.7.2
Proof. Let A E V,(R, 9, p ) .We show that limn.+, T A ( p , ) = 0. (See the proof of Theorem 7.3.1 for the definition of T A . ) Let E >O. There exists S > O such that lp,l(F)<~for every n L 1 whenever F E and~ p(F)<S. Since A E V,(R, 9, p ) , A E V1(R, 9, p ) . By Theorem 7.2.12, there exists Po = {F1,Fz, . . . ,F,} in 9 such that
{ C IA (Ej)- Ap,(Ei)l;P k
IlA
-Apolll
= SUP
i=l
= {El, Ez, .
I. ..
. . ,Ek} E 9 < SE.
Let r = Supnzl Ilpnll.By Theorem 8.7.1, r
220
THEORY O F CHARGES
Hence
Observe that
as n + CO, and
Therefore, limsup ) ~ ~ ( p n ) J s l i m JsTuAp- A p o ( p n ) J + O n +m
n+m
5 E [ r + 211A
llool.
Since E > 0 is arbitrary, it follows that limn-,.mT A ( p , ) = 0. This completes the proof. 0 The followingtheorem gives a comprehensive list of equivalent conditions for weak convergence of a sequence in ba(fl,9). 8.1.3 Theorem. Let S be a field of subsets of a set $2 and sequence in ba(fl,m. Let 4 on 9be defined by
pn, n 2
1a
8.
NIKODYM AND VITALI-HAHN-SAKS THEOREMS
22 1
for A in 9. Then the following statements are equivalent. (i). p,, n L 1 converges to 0 weakly in ba(S1, .9). (ii). C k z l p,(Ak) = 0 for every sequence Ak, k L 1 of pairwise disjoint sets in 9. (iii). limn-tmx k z l Ipn(Ak)l= o for every sequence Ak, k L 1 of pairwise disjoint sets in 9. (iv). limn-tmp, (A) = 0 for every A in 9and limn,, p, (A,) = 0 for every sequence Ak, k 2 1 of pairwise disjoint sets in 9. (v). limn-tmp,(A) = 0 for every A in 9 and {p,, n L 1) is uniformly absolutely continuous with respect to 4. (vi). lim,,,p,(A)=O for every A in 9 and { p n , n~ l is }uniformly absolutely continuous with respect to v whenever v E ba(S1, 9)and pn << v for every n L 1. (vii). p,, (A) = 0 for every A in 9and there exists a v in ba(S1, 9) such that {p,, n L 1) is uniformly absolutely continuous with respect to v. (viii). p,, n L 1 converges to 0 weakly in Vl(S1, 9,*). (ix). p,, n L 1converges to 0 weakly in Vl(S1, 9, v ) whenever v E ba(S1, .9) and p, << v for every n L 1. (x). There exists a v in ba(S1,9) such that p, << v for every n L 1 and p,, n 2 1converges to 0 weakly in V1(S1,9, v). Proof. (i)*(ii). Let Ak,k Define T on ba(S1, .9) by
L 1 be
a sequence of pairwise disjoint sets in 9.
for p in ba(R,9). Since IT(p)IsCkzlIpl(Ak)llpl(S1)=IIp1I,T is a continuous linear functional on ba(S1, .9). Since p,, n 2 1converges to 0 weakly in ba(S1, 9), lim T ( p n )= T ( 0 )= 0 = lim
1 p,(Ak).
n+05 k z l
n-tm
(ii)+ (iii). Let Ak,k L 1be a sequence of pairwise disjoint sets in 9. Define, for each n L 1, A, on P(N) by An(D)=
1
koD
pn(Ak)
for D c N, where N = { 1 , 2 , 3 , . . .}. Note that each A, is a bounded charge A,(D) = 0 for every D c N. By Phillips' lemma on P(N). By (ii), limn-+m 8 . 3 . 3 (i) (iii),
+
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THEORY OF CHARGES
(iii)+(iv). Let A E ~ Take . Al = A , A z = 0, A3= 0, . . . . So, by (iii), limn-,wC k 2 1 Ipn(Ak)l= 0 = limn+wlpn(A)I. NOW, let A,, n L 1 be any sequence of pairwise disjoint sets in 9’ By .(iii),
This completes the proof of (iv). (iv)+ (v). Assume that (iv) holds and f i n , n L 1 is not uniformly absolutely continuous with respect to 4. This means that there exists E > O such that for every S > 0, we can find A in 9 and n 2 1 such that $(A) < S , but lpn(Nl2 28. For S = 1, we find A1 in 9 and n l L 1 such that lpnl(A1)l22.5 and $(Al) < 1. Since pl,p 2 ,. . . ,pnlare absolutely continuous with respect to $, there exists S1> O such that Ip,(A)I < .5/2’ for j = 1,2, . . . , nl whenever A E and ~ $(A)<S1. For S = S 1 , we find Az in 9 and n 2 ? l such that Ipn,(AZ)(2 2.5 and *(A2)< S1. Obviously, n2 > nl. Also, Ipnl(B)I< .5/2’ whenever B E 9 and B c A2, since $(B) 5 $(A2) < 81. Since PI, pz, . . . ,pnz are absolutely continuous with respect to 4, there exists SZ>O such that 2 =~ 1,2, . . . , nz whenever A E 9and $(A) < S2.For S = SZ, Ipj(A)I< ~ / forj we find A3 in 9 and n 3 2 1 such that IpCLn,(A3)]>2~ and $(A3)<82. 2 Ipn2(B)I ~ < &/Z3 whenever B E 9 Obviously, n3 >nz. Also, lpnl(B)I< ~ / and and B c A3, since $(B) 5 $(A3) < SZ. Proceeding this way, we obtain a sequence Ak, k L 1 of sets in 9 and a sequence n l < nz < * * * of positive integers having the following properties. (a). Ipnk(Ak)l2 2.5, k 2 1. (b). lpni(B)I< for j = 1,2, . . . , k - 1, k 2 2 whenever B E 9 and B C Ak. We relabel pnl,p n z , .. . as p l , pz,. . . . This is admissible since the property (iv) is hereditary, i.e. any subsequence of p,, n 2 1 has property (iv) . Now, we construct a sequence Hk,k 2 1of pairwise disjoint sets in 9and find a sequence 0 = p o < p l
(Hj+l)l> E
for every j L 0. This then gives a contradiction as property (iv) fails to hold for the sequence p l , ppltl, pp,+l,. . . , and the property (iv) is hereditary. First, we give H1 and p l . Let F1 = Al. If there is an integer i > 1 such that Ipi(F1nAi)l > ~ / 2 let , il be the smallest positive such integer and F2=F1-Ail. If there is an integer i > i l such that Ipi(FznAi)I>E/2, let i2 be the smallest positive such integer and F3 = Fz -Ai,. If this process does not stop at any finite stage, we get a sequence F1n Ail,F2nAi,, . . . of pairwise disjoint sets in 9such that ]pi*(FknAc)I > ~ / for 2 every k 2 1.This contradicts the validity
8.
NIKODYM A N D VITALI-HAHN-SAKS THEOREMS
223
of property (iv) for the sequence pi*, k z 1. So, let us assume that the above procedure stops at some finite stage. This means that there is a positive 2 every i >pl. Take H1 = Fpl.We integer p1 such that pi(Fpln Ai)5 ~ / for show that Ipl(H1)I2 2s - ~ / 2 . = IPl(Fp1-1 -AiplL1)I IPl(H1)I = Ip~(Fpi)l
= ICL1(Fp1-2-Ai,l-2-Ai.1-1)/
1~ 2 IP =
1(A1- Ail - Ai2 - * 1
-*
* *
-A
ipl-l
)I
(AllI - IP 1 (A1 Ail)I - IP 1 (A1 nAiJl * *
- Ipi(A1nAipl-,>l
- &/2’2- . . . - & / y P , - 1
2 2&- &/24
12E - & / 2 > & . Look at the sequence AI1)=Apl+l -FPlri r 1 and the sequence pI1)= ppl+i,i r l of charges. We claim that these sequences have properties similar to those of (a) and (b) with 2~ replaced by 28 - ~ / 2Note . that
IpL1)(AI1))I = IFpl+i(ApI+i-Apl+i nH1)1 2 Ippl+i(Apl+i)I - Ippl+i(Apl+in
z 2.5 - &/2 for every i r 1.
To prove (b), let B E 9 and B c A!’), i z 2. Then for j = 1,,2, . . . ,i - 1,
ICLjl)(~)I 5 &/2plCi I 3&/zi+l= 3 1 2 E F
1 2
= (2E - &/2)7.
Thus the sequences A!’), i 2 1 and p !I), i z 1 have the following properties. - ~ / 2for every i 1 1. (a)’. Ipi1)(A!1))122~ (b)’. Ipjl’(B)II(2.5 -&/2)(1/2’) for j = 1 , 2 , . . . ,i - 1, whenever B E 9 and B c A!” for i z 2. By using the argument given after (a) and (b) for the sequences figuring 2 ~ / 2 we ~ ,obtain an integer p 2 > p 1 , a in (a)’ and (b)’ and replacing ~ / by set H2 in 9 and sequences A!”, i z 1 and p?), i z 1 having properties ~ ( H2 ~~~) ~/ 2 -) ~ / 2 ’ . similar to (a)’ and (b)’. Further, lp!‘)(H2)I= I , U ~ ~ + 2 Observe that H2 c A\*)= APl+l-HI and so, HI n Hz = 0. Continuing this procedure, we get the desired sequences Hk, k 1 1 and pk, k r 0. Hence (iv)+ (v).
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THEORY OF CHARGES
+
(v) (vi). It suffices to show that $ << v. But this is evident from Theorem 6.1.13. (vi)+ (v). This is obvious. (v) .$(vii). This is also obvious. ( v ) j (viii). This follows from Theorem 8.7.2. +) is a subspace of V1(R, 9,v). (viii)j(ix). By Theorem 6.1.13, V,(n, 9, Since p,, n 2 1 converges to 0 weakly in V1(R, 9, $), it follows that p,, n L 1 converges to 0 weakly in V,(n, 9, v). ( i x ) j (viii). This is obvious. (viii) (x). This is also obvious. (vii)+(x). This follows from Theorem 8.7.2. (x) (i). This is analogous to the proof given for the implication (viii)j(ix) because V1(Q 9,v ) is a subspace of ba(R, 9). This completes the proof that all the ten statements are equivalent. 0
+ +
Now, we formulate a theorem on weak convergence of a sequence of bounded charges analogous to the above theorem without specifying the limit. 8.7.4 Theorem. Let 9 be a field of subsets of a set n and p,, n 2 1 a sequence in ba(fl, Let $ be the charge defined on 9by
m.
for A in 9. Then the following statements are equivalent. (i). p, L 1 is weakly convergent in b a ( l 2 , a . (ii). p,,(A), n L 1 converges to a real number for every A in 9and {p,, n L 1) is uniformly absolutely continuous with respect to $. (iii). p,(A), n 2 1 converges to a real number p (A) for every A in 9, p E ba(R, 9)and {p, - p , n 2 1) is uniformly absolutely continuous with respect to $. Proof. (i)+ (ii). Let p,, n 2 1 converge to p weakly in ba(@ 9). Then p, - p , n L 1converges to 0 weakly in ba(n, By Theorem 8.7.3(i) .$(vi), lim,,oo ( p , -p)(A) = 0 and {p, - p , n L 1) is uniformly absolutely conwhenever p, - p << v for every n ? 1. tinuous with respect to v E ba(& 9) Note that {p,, n L 1)c Vl(n, 9,$) and Vl(n, 9, $) is a norm closed subspace of ba(R, 9). Consequently, Vl(s2, 9, $) is a weakly closed subspace See Theorem 1.5.17. So, p EV,(R, 9,$ Thus ).we observe of ba(R, By Theorem 8.7.3, that p, - p , n L 1converges to 0 weakly in V,(R, 9,$). {p, - p, n 2 1) is uniformly absolutely continuous with respect to sr/. Since p << $, {p,, n L 1) is uniformly absolutely continuous with respect to $. (ii) (iii). Let p (A) = lim,,oo p, (A), A E 9. Since {p,,, n 2 1) is uniformly
a.
a.
+
8.
NIKODYM AND VITALI-HAHN-SAKS
THEOREMS
225
absolutely continuous with respect to 4, p << 4. So, by Theorem 6.1.10(iv), is bounded. Consequently, {p, - p , n L 1) is uniformly absolutely continuous with respect to rfi. (iii)J(i). By Theorem 8.7.3(v)+(i), p, - p , n z 1 converges to 0 weakly . p,, n 2 1 is weakly convergent in ba(R,9). 0 in ba(n, 9)Consequently, p
Now, we show that if the field 9 has Seever property, characterization becomes very simple. of weak convergence of sequences in ba(R,
m
8.7.5 Theorem. Let 9 b e a field of subsets of a set n having Seever property and p,, n L 1 a sequence in ba(l2,m. Then p,, n z 1 is weakly convergent in b a ( Q 9 ) if and only if p,(A), n L 1 Converges to a real number for every A in 9. Proof. If p,, n I1converges weakly in ba(R, 9), by Theorem 8.7.3, p,(A), n 2 1 converges to a real number for every A in 9. This proves the “only if” part of the theorem. Conversely, let p(A) = limn+m@,(A), A E$. It is obvious that p is a Further, by Theorem 8.4.5, p is bounded. For each n 2 1, charge on 9. let v, = p, - p . For this sequence v,, n 2 1, we check the validity of (iii) of Theorem 8.7.3. Let Ak, k 2 1 be a sequence of pairwise disjoint sets in 9. For each n I1, define T, on P(N) by
for D c N, where N = {1,2,3,. . .}. Each T, is a bounded charge on S(N) and it follows that limn+ooT,(D) = 0 for every D c N. By Phillips’ lemma 8.3.3(i)+ (iii), lim
1
17,({k))l= 0.
n+m k z l
But ~ k ~ l 1 T , ( { k }=Ikzl )l Ivn(Ak)l. Hence v,, in b a ( n , 9 ) .
It
z 1 converges to
o weakly 0
Now, we obtain as a corollary the classical result on weak convergence of sequences of bounded measures on u-fields.
8.7.6 Corollary. Let 2l be a u-field of subsets of a set R and p,, It h 1 a sequence in ba(fl, 2l). Then p,, n 21 converges weakly in ba(R, 2l) if and only if p, (A),n I1converges ?o a real number for every A in 2l. In particular, if {p,, n 2 l }ca(R, ~ a), then p,, n 2 1 is weakly convergent in ca(R, a) if and only if @,(A),n 2 1 converges to a real number for every A in 2l. Proof. Note that every u-field has Seever property.
0
226
THEORY OF CHARGES
The results proved in this section give a new characterization of uniform absolute continuity.
8.7.7 Theorem. Let 9 be a field of subsets of a set R and p,, n I1 a sequence in ba(Q 9)Let . $ be the charge on 9 d e f i n e d by
for A in 9. Then {p,, n 2 1) is uniformly absolutely continuous with respect to if and only if for every sequence A,, n L 1 of pairwise disjoint sets in 9, limn+mpn(An) = 0. Proof. “If” part of this theorem is essentially the proof of the implication (iv)+ (v) of Theorem 8.7.3. We prove the “only if” part. Suppose for some sequence A,, n 2 1 of pairwise disjoint sets in 9, limn+mp,, (A,) # 0. Let E > 0 be such that lim Supn+mlpn(A,)[> E . Since pn, n L 1 is uniformly absolutely continuous with respect to $, for the E above, there exists 6 > 0 such that lpn(A)I< E for every n 2 1whenever A E9and $(A) <6. Since $ is a bounded charge, limn+m$(A,) = 0. So, there exists N L 1 such that $(A,) < 6 for every n IN . Consequently, ~p,,(A,,)~< E for every n 2 N . This implies that lim Supn+mIp,,(A,)I5 E . This contradiction proves the result.
8.8 VITALI-HAHN-SAKS THEOREM In this section, we obtain Vitali-Hahn-Saks theorem in ba(R, 9) for fields 9having Seever property. For this purpose, we introduce the notion of uniform additivity.
8.8.1 Definition. Let 9be a field of subsets of a set R and M c ba(R, 9). M is said to be uniformly additive if for any sequence A,, n I1of pairwise disjoint sets in 9, lim Sup 1 [p(A,)I=O. m+m
EM n z m
The following theorem gives equivalent versions of uniform additivity.
8.8.2 Theorem. Let 9 be a field of subsets of a set R and M c ba(R, 9). Then the following statements are equivalent. (i). M is uniformly additive. (ii). For any sequence A,,, n L 1 of pairwise disjoint sets in 9,
8.
NIKODYM A N D VITALI-HAHN-SAKS
THEOREMS
227
(iii). For any sequence A,,, n 2 1 of pairwise disjoint sets in 9, lim s u p s u p m-rw
&EM D c N
Ic
I
p ( ~ , = ) 0.
nzm
noD
Proof. It can be seen easily that each of the properties (i), (ii) and (iii) is hereditary, i.e. if M has any of the properties (i), (ii) and (iii), then any subset of M has the same property. (i).$(ii). This is obvious. (ii) (i). The proof is carried out in the following steps. Suppose (i) is not true. We show that (ii) is not true. 1". Since (i) is assumed to be false, there exist E >0, a sequence E,,, n 2 1 of pairwise disjoint sets in %, a sequence p,,, n P 1 of charges in M and a sequence jl<j z < * of positive integers such that
+
1 lp,,(Ej)l>~,
n = l , 2,....
i ?in
Since for any bounded charge p on 9, limm+mC,,,, Ip(E,,)I = 0, we can take p,,, n 2 1 to be distinct. 2". Look at p l . Since Ipl(Ej)I>E, we can find m l > j l such that x,2jlIpl(Ej)I> E . We can also find P I > ml such that Ipl(Ej)l< ~ / 4 . Now, for pp17 we can find m2 >j,, such that Ipp,(Ej)l> E and p z > m2 such that lppl(Ej)l< ~ / 4 . Continuing this procedure we obtain two sequences ml, m 2 , .. . and pl, p z , . . . of positive integers satisfying the following conditions. (a). jl< m l< p l I j,, < m z < p z s jm< m3
cizjl
cjzp,
cizp,
-
+
.
I
k"+l
with the understanding that ko = 0.
I
228
THEORY OF CHARGES
4".We show that property (ii) fails to hold for the sequence ppl,p p z , .. . and the sequence F1, FZ,. . . of pairwise disjoint sets from g.This would prove the implication (ii)j(i). For every n L 0, note that
z
> €12- €14= €14. This follows from the elementary inequality la + b I L la1 - 161. Since k, -+ 00 as n + CO, property (ii) fails to hold for the sequence ppl,p p 2 ,. .with respect to the sequence F1, FZ,. . . of pairwise disjoint sets in 9. (i)j(iii). This is obvious. Let (iii)j(i). Let A,, n L 1 be a sequence of pairwise disjoint sets in 9. E >O. Since (iii) is assumed to be true, we can find m o r 1 such that for every m L mo,
.
SUP SUP
1c
D C N ~ E Mn z m nED
p(An)J< E / 2 *
Now, if m L mo and p E M, we have
c
nzm
c
c
nED
nsE
Ip(An)I=)n z m F ( A n ) l + /n z m p(An)l <E/2+
€12 = E ,
whereD={n rl;p(A,)rO}andE={n rl;p(A,)
8.
NIKODYM A N D VITALI-HAHN-SAKS
229
THEOREMS
Proof. (iii). It is clear that p is a charge on 5. Since p n ( b ) ,n 5 1 converges to a real number, Supnzl (p,,(b)l< CO. So, by Theorem 8.4.5, Sup,,, Ilpnll< 00. Hence p is bounded. (i). Since B has Seever property, by Theorem 8.7.5, F,,, n 2 1 is weakly convergent to some bounded charge T on B. Obviously, T = p. Define I,$ on B by
for b in B. By Theorem 8.7.4(ii), {p,,, n 2 1) is uniformly absolutely continuous with respect to $. By Theorem 6.1.13, 4 << u. Hence {pn,n ? 1) is uniformly absolutely continuous with respect to V. (ii). Since p,,, n L 1 converges to p weakly, p n - p , n L 1 converges to 0 weakly. By Theorem 8.7.3(i)J (iii),
for any sequence ek, k L 1 of pairwise disjoint elements in 5. Let There exists m l L 1 such that
2 ml.
>O.
b n ( e k ) - & (ek>l<E/2
k z l
whenever n
E
There exists m2L 1 such that
c
kzm
Ip (ek)l< E l 2
whenever m L m2. Let mo= max {ml, m2}.There exists N 2 1 such that
1
kzm
Ipi(ek)- (ek)l< E/2
whenever m L N and i = 1 , 2 , . . . , mo. Now, let m 2 N . Then kzm
Ipn(ek)15
C Ipn(ek)-p(ek)l+
kzm
c
kzm
Ip(ek>l
< E / 2 + & / 2= E for every n
5 1. This
shows that {p,,, n L 1) is uniformly additive.
0
The classical Vitali-Hahn-Saks theorem follows as a corollary. 8.8.5 Corollary. Let "I be a u-field of subsets of a set s1 and p,,, n L 1 a sequence in ba(s1, "I).Suppose p,,(A),n L 1 converges to a real number p (A) for every A in '$ Then I. the following statements are true. (i). {p,,,n zl} is uniformly absolutely continuous with respect to u whenever p,, << v for every n 2 1 for any v in ba(s1, fl),
230
THEGRY OF CHARGES
(ii). b,,, n 2 1) is uniformly additive. (iii). p is a bounded charge on 8. In particular, if bn, n 2 1) c ca(Cl, ? then, I)under , the above condition, the following statements are true. (i). b,,, n 2 1 ) is unifarmly absolutely continuous with respect to v whenever v E ca(S1, a) and p,, << v for every n 2 1. (ii). {p,,, n L 1)is uniformly countubly additive. (iii). p is a bounded measure on 8.
Proof. Every a-field has Seever property.
0
CHAPTER 9
The Dual of b a ( Q 9) and The Refinement Integral
One of the important problems in the study of the Banach Space ba(R, 9), the space of all bounded charges on a field 9 of subsets of a set R, is to This problem seems to be difficult and no describe its dual ba*(R, 9). is known. The satisfactory representation of the elements of ba*(il, 9) purpose of this chapter is to present one of the attempts made to describe using what is known as refinement integral. In Section 9.1, we ba*(n, 9) present some basic ideas on refinement integral. In Section 9.2, for superatomic fields 9, we show that refinement integral representation exists for We also show that the later property characterizes all elements of ba*(R, q. superatomic fields. We also obtain refinement integral representation for elements of VT(R,9, p ) in this section.
9.1 REFINEMENT INTEGRAL
9.1.1 Definition. Let 9be a field of subsets of a set R and p a bounded charge on 9. Let f be any real valued function defined on 9 and A E 9. Then the refinement integral off with respect to p over A is said to exist if there is a real number a with the property: for any E >0, there is a partition Po in PAsuch that for any partition P = {El, EZ,. . . , Em}of A in 9finer than Po,
If
i=l
f(Ei)p(Ei)-al
<E*
We write (Y =IAfp. (Recall that PAstands for all finite partitions of A in 9.) By the statement ‘‘IA fp exists”, we mean that the refinement integral off with respect to p over A exists. I f f = 1, then 5, fp exists for any A in 9 and any bounded charge p on 9andSafp=pCA). We give some basic properties of refinement integrals.
232
THEORY OF CHARGES
9.1.2 Proposition. Let 9 b e a field of subsets of a set Q and f a real valued function on 9. Let p and A be two bounded charges on 9 and A E 5F. (i). I f JA f p exists and p is a real number, then IA f/3p exists and JA fop =
p IAfp. (ii). I f
JA f
p exists and
I,f A
exists, then
IAf p + A
exists and
I,
f p +A =
I*fP +I*fA. (iii). I f f is a bounded function on 9 a n d J Afp exists, then
(iv). Let f be a bounded function on 9. Let p,, n L 1 be a sequence of bounded charges on 9 and a,, n 2 1 a sequence of real numbers such that Cnzl l a n l l p n l ( a ) < ~ . L e t p=Cnzl a,p,.If JAfp,exiStSforeueryn 21, then JA f p exists and JAfp=
c [ fpn. an
nzl
A
Proof. (i) and (ii) are obvious from the definition of the refinement integral. (iii). Let JA f p = a . Let E >O. Then there exists a partition P = {El,Ez, . . . ,En}of A in 9such that f (Ei)p(Ei)- a I < E . This implies that
Since E > 0 is arbitrary, we obtain
aipi,n L 1. By the given condition, A,, n 2 1 converges (iv). Let A, = to p in the norm of ba(R, 9). Note that, by (iii),
c
n z l
Let M = SupBEsI f (B)I and
E
I a n 1A f p n l < a *
> 0. We can find N
2
1 such that I(AN -pII <
9.
THE DUAL OF
b a ( R , 9 ) AND
REFINEMENT INTEGRAL
233
Further, we can find a single partition Po in PA such that for any patition P = {F1, F z , . . . , F,} of A in 9 finer than PO,
< ~ / +3N ( & / 3 N+) ~ / =3E . Hence J Af p exists and IAfP =
C
nz1
an
A
fpn.
0
We now show that every bounded real valued function on 9 for which the refinement integral with respect to p over R exists for every p in ba(R, 9)defines a continuous linear functional on ba(Q 9).
9.1.3 Theorem. Let 9be a field of subsets of a set R and f a bounded real valued function on 9.Suppose jn f p exists for every in ba(R, 9).Let T be defined on ba(R, 9) by
for p in ba(R, 9).Then T is a continuous linear functional on ba(R, 9). Proof. This follows from Proposition 9.1.2(i), (ii) and (iii).
0
The main problem considered in the next section is to characterize those fields 9 for which every T in b a * ( R , q admits a representation of the above type.
234
THEORY OF CHARGES
9.2 THE DUAL OF ba(Cl,F) Recall the definition of a superatomic field from Definition 5.3.4 for the following theorem. 9.2.1 Theorem. Let 9be a field of subsets of a set R. Then the following statements are equivalent. (i). Forany continuous linearfunctional Ton ba(R, 9), i.e. T E ba*(S2, 9), there exists a real valued bounded function f on 9such that ,5 f p exists for all p in ba(S2,9) and
w=[ nf P for all p in ba(S2, 9). (ii). 9 is superatomic. Proof. (i) j (ii). The proof is carried out in the following steps. 1". Suppose 9is not superatomic. By Theorem 5.3.6, there exists a strongly continuous probability charge A on 9. We define T on b a ( R , 9 ) by T ( p )= pl(S2)for p in ba(S2, 9), where p = p1+ p 2with p 1<< A and p21A. See Lebesgue Decomposition Theorem, i.e. Theorem 6.2.4. Since this decomposition is unique, T is a well defined linear functional. Continuity of T also follows from Theorem 6.2.4 if we observe that IT(p)l=IPCLI(WI5 IPll(W 5 IPllm + IP2l(n)
(I.
5 IP I(R) = 1 1 F
2". Since we are assuming that (i) holds, there exists a bounded function f on 9 such that T(p) = J n f p for all p in ba(S2, 3". For F in 9, let A F on 9 be defined by AF(E)=A(EnF)for E in 9. Then for any F in 9,
a.
I.
A(F)=AF(S2)=T(A~)= J fhF= n
r
JF f A .
(The Lebesgue decomposition of A F with respect to A is AF+ 0.) 4". Define 93 = {FE 9; every finite partition of F in 9 has a set B in 9 such that A (B) > 0 and f (B) 2 i}. 5". We claim that for every A in 9 with A (A)> 0, there exists F in 9such that F c A and F E 93. Suppose the claim is false. Then there exists A in 9 with A (A)> 0 such that F !ii93 whenever F c A and F E 9, i.e. for every either A (Bi)= 0 or f (Bi)< $ for all i. partition {B1, Bz, . . . , B,} of F in 9, By 3", since A (A) = JA f A , we can find a partition P = {Fl,F2, . . . ,F,,,}of A in 9such that
9.
THE DUAL OF
ba(fl, 9) AND
REFINEMENT INTEGRAL
Note that f(Fi)A(Fi)I$A (Fi) for all i, so that Therefore, A( ~ ) /5 2
If
i=l
235
f(Fi)A(Fi)IA (A)/2.
I
f(Fi)A( F i ) - A (A) < A ( ~ ) / 2 .
This contradiction establishes the claim. 6". Since A is a strongly continuous probability charge on 9, we can find Bo and B1 in .Fsuch that Bo nB1 = 0, A (Bo)> 0 and A (BI)>0. By 5", we can find A. c Bo and Al c B1 such that Ao, Al E 93. By the same argument, we obtain Aoo,Aol cAO; Alo,All CAI; AoonAol= 0; AlonAll= 0 and Aoo, Aol, Alo,All E 93. Continuing this way, we obtain{A,,,,,, ...,,,, ;~i = 0 or 1 for all i, n 2 l},a subcollection of 9 with the following properties. 6). A,,,,, ,..., 3AEI.E2 ,..., for all n 2 1. (ii). A,,,,, ,_... nA6,,6,.___. 6, = 0 if ( € 1 , E Z , . . . ,E , ) f (SI,SZ,. . . ,S,), n 2 1. 7". Let E = ( E ~ , E ~. ,. .) be a sequence of 0's and 1's. Let 8, = {A,,, A,,,,,, AE1,E2.E3, . . .}. Let 9,be a filter in $containing 8, and maximal with respect to the following property: (*) "For every A in 9€, there is a B in 93 n9, such that B c A." Existence of 9, can be established using Zorn's lemma as follows. We look at the collection % of all filters in 9 containing 8, and having the property (*), This collection is non-empty since the filter in 9generated by 8, (which exists by remark 1.1.23(4)) has the property (*). In the usual partial order of inclusion, one can show that every chain in % has an upper bound. Hence by Zorn's lemma, there is a filter 3, in 9containing 69€ and maximal with respect to the property (*). 8". Now, we claim that 9,is a maximal filter in 9. Suppose the claim is . 9;be the false. There exists C in 9 such that neither C nor C'E 9,Let filter in 9 generated by Seand {C} and 9f the filter in 9 generated by .!FEand {C?. In fact, En
9,' = { D E ~D; I C n B for some B in 9,} and
9:= {D E 9; D 3 C'n B for some B in 9,}. By the definition of 9,, 9;and 5Ff do not have the property (*). So, there are sets B1, B2 in gesuch that C n B1 and C'nBz do not contain any member of 93.We can assume that B1 = B2 = B, say, by considering BI nBz. By 5", A (C nB) = 0 = h(C'n B). This implies that A (B) = 0. But, since B E 9,, A (B) > 0. This contradiction shows that 9, is a maximal filter in 9. 9". We claim that if E = ( E ~ E, ~ .,. .) and S = (Sl, S2, . . .) are two sequences of 0's and 1's such that E # 8, then 9, and 9 6 are distinct. Suppose 9, =9 6 . Since E # S , there exists n 2 1 such that ( & I , E Z , . . . , E , ) # (61, SZ, . . . ,6,).
236
THEORY OF CHARGES
,,
So, A,,.,, n A,,.,, ..... = 0 E 9,a, contradiction. Thus the claim is valid. 10". We claim that for every n 2 1, { E E (0, l}Ko; A (A) 5 l / n for every A has at most n elements. (Recall that (0, l}Ko is the space of all in sE} sequences of 0's and 1's.) Suppose the above set has more than n elements. Pick up any n + 1 distinct elements E " ) , E " ) , . . . ,E ( n + l ) from this set. Since FEq 9,9. . . ,9,(n+1) are distinct, we can find Bi in Sc(:), i = 1,2, . . . , n + 1 such that B1, Bzr. . . ,B,,' are pairwise disjoint. Since A (Bi)5 l / n for every i = 1,2, . . .,n + 1,h(Urf: Bj)~ (+ ln) / n .But A (0)= 1.This contradiction establishes the claim. 11". Since
h (A) 2 l / n for every A E PE},
the set on the left is countable. Since (0, l}Nois uncountable, there exists q E {O,l}"o such that InfA,F,, A (A) = 0. 12". Let u on 9be defined by u ( A ) = 0, if A &9,,, A E 9,
=1, i f A E g V . u is a 0-1 valued charge on 9. 13". Observe that u LA. See Proposition 8.5.l(v). Hence the Lebesgue decomposition of u with respect to A is 0 v. Therefore, T ( u )= 0. 14". On the other hand, we show that J,fu>:. It suffices to show that
+
given P={E1, EZ,. . . , E m } in 9,there exists a finer partition P = (F1, Fa, . . . ,F,} in 9 such that I:=,f(Fi)v(Fi) 2;.For P = {El, Ez, . . . ,E m } in 9,there exists exactly one i, say i = 1, such that u(Ei)= 1, i.e. El E g,,. We can find B e F , , n B such that B c E 1 . Take P '= {B, EI-B, Ez, E3,. . * Em}. Thus 0 = T ( u )# J,fu 2 .; This contradiction shows that 9is superatomic. Now, we prove (ii)+ (i). This is carried out in the following steps. 1". We, first, collect some basic facts about derived sets in topology. Let X be any topological space and A c X. Set A' = A, A' (the derived set of A') = {x E A'; x is an accumulation point of A'}, if (Y is a limit ordinal, set A" = ADand for any ordinal a,set A"+'= (A")'. Then A", a 2 0 is a decreasing net of sets and each A" is a closed subset of A. For what follows, we assume that X is a scattered compact Hausdorff totally disconnected space. Then, there exists an ordinal (YO such that X"O is a non-empty finite set and Xn0+l = 0. This can be proved as follows. Let p be the least ordinal such that Xp = Xp+l. Then X p = 0.For, if Xp # 0 , then Xp+' is a proper subset of X p . (Since X is scattered, the 9
no<"
9.
THE DUAL OF
ba(R, 9)AND
REFINEMENT INTEGRAL
237
closed set Xp is not perfect and hence has isolated points in it.) Now, we claim that p = a O +1 for some ordinal ao. If p is a limit ordinal, then Xp = Xy= 0 and this implies that Xy= 0 for some y
nv
f (V) = SUP { T ( p1; p E V'}. If V = 0 ,let f (V) = 0. f is obviously a bounded function on 9. 4". First, we show that Jxfp = T ( p ) for any 0-1 valued charge p on 9. By lo,there exists a clopen set C c X and an ordinal po such that Cpo= { p } . Consequently, f(C) = T ( F ) C . also has the property: if E E 9and p E E c C, then EPo= E n Cpo= { p } .Now, for the partition {C, C"},we havef(C)p (C) + f (Cc)p(C") = T ( p )+ 0 = T ( p ) . Further, if P = {El, Ez, . . . , E,} is a refinement of {C, C'}, then CE,f(Ei)p(Ei)= T ( p )from the property of C mentioned above. Hence Jxfp = T ( p ) . 5". Let p be any bounded charge on 9. Since 9 is superatomic, p = Cnzl anpnfor some sequence I*.,,, n 2 1 of 0-1 valued charges on 9 and for some sequence an,n 2 1 of real numbers satisfying lan/<m. See the comment following Theorem 5.3.6. The conditions of Proposition 9.1.2(iv) are met and so lxfg exists and
IXfF= C
nzzl
an
I
x
fpn
=
C
nzl
anT(pn>
238
THEORY OF CHARGES
as C Z , aipi, n 2 1 converges to p in the norm of b a ( R , 9 ) and T is a continuous linear functional on ba(fl, 9). This completes the proof. 0 Now, we obtain a refinement integral representation for continuous linear p ) . For functionals on V1(R, 9,p ) for a probability charge space (a,9, relevant information on V1(fl, 9,p ) , see Chapter 7. 9.2.2 Definition. Let 9 be a field of subsets of a set R and p a positive bounded charge on 9. A real valued function f on 9is said to be convex with respect to p if
for every A, B in 9with A n B = 0. (Recall the convention that 0/0 = 0.) 9.2.3 Theorem. Let ( R , 9 , p ) be a probability charge space and T a continuous linear functional on V1(R, 9,p ) . Then there exists a bounded real valued convex function f on 9 such that T ( v )=In f v for every v in Vl(Q, 9, r*).
Proof. By Theorem 7.3.1, there exists a bounded charge A on 9 in Vm(R,9, p ) such that T ( v )= TA(v) for every u in V*(R,9, p ) . From Remark 7.3.2,
where P = { E l , E Z ,. . . , E m } is a generic element of 9. For F in 9, define f (F) = A (F)/p(F). Since A E Vm(R, 9, p),
Now, we check that f is a convex function with respect to p. For A, B in 9withAnB=0,
9.
THE DUAL OF
ba(R, 9) AND
REFINEMENT INTEGRAL
239
This shows that f is a convex function with respect to F. Now, for any in VI(R, 9, F),
Y
m
1 f(Ei)v(Ei)= PEPi=l
T ( v )= Th(v) = lim
n
fv,
where P = {El, E2, . . . , Em}is a generic element in 9.This completes the 0 proof. 9.2.4 Remark. We can obtain a refinement integral representation for F) for any 1 < p
CHAPTER 10
Pure Charges
A charge on a field may not be pure in the sense that there may be a part of it which is countably additive. If we isolate successfully a maximal countably additive part of a given charge, what is left may be termed as purely finitely additive. This chapter is devoted to the pursuit of these ideas. After presenting the preliminaries in Section 10.1, we prove a decomposition theorem due to Yosida and Hewitt in Section 10.2. In Section 10.3, we look at pure charges on a-fields. In Section 10.4, we discuss some examples which illuminate some aspects of pure charges. Finally, in Section 10.5, pure charges on Boolean algebras are studied.
10.1 DEFINITIONS AND PROPERTIES In this section, we introduce pure charges and give some of their properties.
10.1.1 Definition. A positive charge p defined on a field 9 of subsets of a set SZ is called a pure charge if there is no non-zero positive measure A on 9 satisfying A IF. More generally, a charge p on 9 is said to be a pure charge if lp 1 is a pure charge. Before characterizing bounded pure charges, we recall some results from Section 1.5 and Chapter 2. b a ( R , 9 ) stands for the vector lattice of all ca(R, 9) stands for the collection of all bounded bounded charges on 9. measures on 9 and is a vector sublattice of ba(SZ,9). We have already seen that b a ( Q 9) is a boundedly complete vector lattice and that ca(SZ, 9) is a normal vector sublattice of ba(SZ, 9). See Theorems 2.2.1 and 2.4.2. ca(R, 9)L stands for the set {A E ba(S1, 9); A 17 for every T in ca(0, Note that A 1~if and only if IAl A 171= 0. The following is a characterization of bounded pure charges.
m}.
10.1.2 Theorem. Let p E ba(R, 9).Then p E ca(n, 9)l.
if
p
is a pure charge if and only
10.
24 1
PURE CHARGES
Proof. Suppose p is a pure charge. Let 7 E ca(R, 9). Note that Ip I A 171Ilp I and Ip I A 17)I171. Since 171 is a bounded positive measure on 9, it follows that lp I A 171 is a measure. Since p is a pure charge, Ip I A 171= 0. Consequently, p 17.Hence p E ca(0,S)'. Conversely, let p E ca(fl, 9)l and A E ca(n, be such that (A 1 5Ip I. Since Ip I IIA I, i.e. Ip I A ( A I = 0, it follows that A = 0. Hence p is a pure charge. 0
a
The following is a simple characterization of a pure charge in terms of its positive and negative variations. 10.1.3 Corollary. Let p E ba(Q, 9).Then if p + and p - are pure charges.
p
is a pure charge if and only
Proof. Note that ca(R, 9)lis a sublattice of ba(0, 9). See Theorem 1.5.8. So, if p E ca(n, 9)', p + and p - c~ a ( R , 9 ) l . Conversely, if p + and p - ~ c a ( n , 9 ) * , then p = p + - p - ~ca(R, 9)'. 0 10.1.4 Corollary. Let p l , p 2 E ba(R, 9)and a any real number. I f p~ and p2 arepurecharges, s o a r e p 1 + p 2 , p l v p 2 , p1 ~ p a2n d a p l .
Proof. ca(R,9)' is a vector sublattice of ba(R, 9). See Theorem 1.5.8. 0 The following corollary shows that purity of charges is preserved under passage to limits. 10.1.5 Corollary. Let p, pn,n 2 1 be a sequence in ba(R, 9).Suppose for each n 2 1, pn is a pure charge and pn,n 2 1 converges to p under the total i.e. limn+.mlpn -pl(R) = 0. Then p is a pure variation norm on ba(R, 9), charge .
a'
a,
Proof. Since ca(R, is a normal vector sublattice of ba(R, it is closed. ConSee Theorem 1.5.19. Since each p,, is a pure charge, pn E ca(R, 9)'. sequently, the limit p E ca(R, 9)'. Hence p is a pure charge. 0
10.2 A DECOMPOSITION THEOREM The results developed so far yield the following decomposition theorem.
a
10.2.1 Theorem. A n y p in ba(fl, can be written in the form p = p p+ p,, where p p is a pure charge in ba(R, 9)and p , E ca(R, 9).Further, such a decomposition is unique.
Proof. This follows from the Riesz Decomposition Theorem, i.e. Theorem and S = ca(0, From Theorem 1.5.10 in which we take L = b a ( Q 9) 2.4.2, it is clear that S is a normal vector sublattice of L. 0
a.
242
THEORY OF CHARGES
We give another description of the countably additive part p, of p in the following theorem. 10.2.2 Theorem. Let p be a positive bounded charge on a field S o f subsets of a set a.Then for any A in 9,
p(A,); A,, n L l t A , A, €9, n21 = Inf
[1
p (A,); A,,
n 2 1 is a sequence of
nzl
pairwise disjoint sets in
with
U A,
nzl
=A
I
.
Proof. The equality of the two expressions on the right above is clear. For A in 9, let T(A)= Inf
1 p (A,); A,, n 2 1is a sequence of (nz1
pairwise disjoint sets in 9with
u A,
nz1
=A
I
.
Suppose v is any positive bounded measure on 9such that v 5 p. Obviously, v 5 T . Now, we show that T is a charge. Let A, B E 9be such that A nB = 0. Let E >O. We can find two sequences A,, n 2 1 and B,, n L 1 of pairwise disjoint sets in 9such that UnzlA, = A , UnzlB, = B and
Consequently,
7(AUB)s
C
P(An)+
C
p(Bn)
n2l
fl2l
+ T(B)+ E .
5 T(A)
Since E > 0 is arbitrary, it follows that T ( AuB) 5 T(A)+T(B).To prove the reverse inequality, we proceed as follows. Let E >O. We can find a sequence C,, n I1of pairwise disjoint sets in 9such that C, = A u B and p ( C n )~ T ( A B) u +E. nr1
10.
PURE CHARGES
243
But
Since E > 0 is arbitrary, it follows that T(A)+T(B)5 T(AUB). ConNext, we show that T is a measure. Let A,, sequently, T is a charge on 9. n I 1be asequence of pairwise disjoint sets in 9 s u c h that Unz1A, = A E 9. Since T is a positive charge on 9, C n , l ~ ( A n ) s ~ ( ASee ) . Proposition 2.1.2(viii). For any E > O and n 2 1, we can find a sequence Ani,i L 1 of Afli= A, and pairwise disjoint sets in 9such that
uizl
C
izl
P (Ani)I T(An) + ~ / 2 ~ *
Consequently,
T Wn1z l~iCz l p(Ani)snCz l T(An)+E. Since E > O is arbitrary, we obtain T ( A ) ~ 7(A,). C ~ Hence ~ ~
This implies that T is a measure on 9. Since r I p , we can write p = T + ( p - T ) . We claim that p - T is a pure charge. If v is any positive bounded measure on 9 such that v 5j.t -7, then v + T 5 p. Since v + T is a measure, by what we have remarked above, v + T IT. Hence v I0. Since v is positive, v = 0. This shows that p - T is a pure charge. Since, in the Riesz Decomposition Theorem, the decomposition is uniquely achieved, it follows that bc= T. This proves the result. 0
10.3 PURE CHARGES ON U-FIELDS In this section, we obtain further characterizations of pure charges defined on cT-fields. The following theorem implies that any positive bounded pure charge and any positive bounded measure are singular.
10.3.1 Theorem. Let p be a positive bounded charge on a field S o f subsets of a set R. Then EL is a pure charge if and only if for every positive bounded measure h on 9, A in s a n d a > 0, there exists B in 9 s u c h that B c A , A(B)
and
p(A-B)
244
THEORY OF CHARGES
Proof. By Theorem 10.1.2, p is a pure charge if and only if p E ca(a, 9)*. Equivalently, p is a pure charge if and only if p A A = 0 for every positive bounded measure A on 9. 0 The above theorem can be strengthened if 9 is a u-field.
10.3.2 Theorem. Let 9 be a u-field of subsets of a set fl and p a positive bounded charge on 9. Then p is a pure charge if and only if for every bounded measure A on $and E > O , there exists B in 9 s u c h that
and
p (B) = 0
A (B')
<E .
Proof. Suppose p is a pure charge. Let A be a positive bounded measure on 9. Let E >O. By Theorem 10.3.1, there exists a set B, in 9 such that p (B,)
< ~ / 2 " and A (BE) < ~ / 2 "
nn2,
B,. Note that B E 9.For every n L 1, p(B) 5 for every n 2 1. Let B = p (B,) < ~ 1 2 " . Consequently, p (B) = 0. Also, A (B') = A (Unzl B:) I CnzlA (B:) IC,,?, ~ / 2 ,= E , as A is countably subadditive. The converse is a consequence of Theorem 10.3.1. Here is another characterization of pure charges on (+-fields useful in constructing examples.
10.3.3 Theorem. Let A be a positive bounded measure on a u-field 9 of subsets of a set a. Let p be a bounded charge on 9 such that p w A,~i.e. p (A)= 0 whenever A E 9 and A (A) = 0. Then p is a pure charge if and only if there exists a decreasing sequence A,, n L 1 of sets in 9 such that A (A,), n L 1 converges to zero and lp [(A:) = 0 for every n L 1.
Proof. The necessity part of this theorem follows from Theorem 10.3.2. To prove the sufficiency part, let (I, be any positive bounded measure on 9 satisfying (I, 5 lp I. From the given hypothesis on p and A, it follows that (I, << A. Let A,, n 2 1 be the given decreasing sequence of sets in 9 such that lim,,,A(A,)=O and lpI(A:)=O for every n r l . Since (I,<
(I,(nnZl
(I,(nnzl
10.4
EXAMPLES
The aim of this section is to construct some interesting examples of pure charges.
10.
245
PURE CHARGES
10.4.1 Example. Let 9 be the Borel c+-field on the real line, R,and let p be a bounded charge on W such that p (B) = 0 for any bounded set B in 9.Then p is a pure charge. (Indeed, if p vanishes for all bounded Borel subsets of R, so does lp 1.) Proof. Let (I, be any positive measure on 9 such that (I, 5 Ip I. Then
Hence (CI = 0. Using the result that any ideal in W is contained in a maximal ideal in 9, one can construct non-trivial charges p of the above type. One simply has to look at the ideal 4 of all bounded Borel subsets of R and then find a maximal ideal in W containing 4. See Section 1.2. It seems impossible to construct a pure charge on a c+-field without taking recourse to some axiom related to the Axiom of Choice.
10.4.2 Example. Let 9 be the field of all clopen subsets of the Cantor set (0, l}Ko. There is no non-zero pure charge on 9.More generally, let 9 be the field of all clopen subsets of a compact Hausdorff space s1. There is no non-zero pure charge on 9.Much more generally, there is no non-zero pure charge on a field 9 of subsets of a set s1 if and only if 9 is a compact A,, = 0 class, i.e. for any decreasing sequence A,,, n 2 1of sets in 9, implies that A,, = 0 for some n 2 1.
n,,
Proof. We will prove the most general statement. If 9 is a compact class, every charge on 9 is a measure. Hence there is no non-zero pure charge If 9 is not a compact class, we can find an infinite sequence A,, on 9. n 2 1 of pairwise disjoint non-empty sets in 9 whose union is R. Let So={A c 0; A is a finite union of sets from {A,,, n 2 1) or A" is a finite union of sets from {A,,,n 2 1)). gois a subfield of 9. On So,let p be defined by p (A) = 0, if A is a finite union of sets from {A,,, n 2 l}, = 1,
otherwise. p is a 0-1 valued charge on Poand is certainly not a measure on go. By Corollary 3.3.5, there is a 0-1 valued charge fi on 9which is an extension of p from go. It is obvious that fi is a non-zero pure charge on 9. 0
10.4.3 Example. Let 9 be a field of subsets of a set R. Any 0-1 valued charge p on 9 is either a pure charge or a measure on 9. Indeed, if (I, is a non-zero measure on 9 satisfying 0 I (I, s p, then p is a measure. 0 10.4.4 Example. Let R = [0, 1) and 9 the field of all sets each of which is a finite disjoint union of sets of the form [ a ,b ) with 0 I a 5 b 5 1. For t
246
THEORY OF CHARGES
in (0, 11, define p, on 9 by pr(A) = 1, if there exists a S > 0 such that [t - 8, t ) c A, = 0,
otherwise.
pr is a 0-1 valued charge on $. Since p.r is not a measure, pr is a pure charge. Conversely, if p is any 0-1 valued pure charge on 9, then p = pr for
some t in (0,1]. This can be proved as follows. Since p is a 0-1 valued pure charge, we can find a decreasing sequence A,, n 2 1 of sets in 9 such A, = 0 and p ( A , ) = 1 for every n 2 1. For each n 2 1, we can that findaninterval[a,,b,)cA,suchthatp{[a,,b,)}= land[al, b ~ ) ~ [ ba 2z) ,3 . * . Obviously, [a,, b,) = 0. Let t = a, = b,. Since [a,, b,) = 0, b, = t for some n 2 1. Consequently, p =PI. Now, we obtain explicitly the decomposition of a given positive bounded charge on 9 as a sum of a pure charge and a measure. Let p be any positive bounded charge on 9.Define a real-valued function m on [0, 1) by m ( t )= p{[O,t ) }for 0 5 t < 1. Then m is a monotonically non-decreasing function on [O, 1) and limrpl m(t)
nnzl
-
nn21
nnz,
Since each ptn is a 0-1 valued pure charge, CnZlanpr,is a pure charge. The m function (as defined above for p ) of p -Cnzl anprnis continuous, and consequently, p - C n z l a , p t , is a measure on 9. The above is the desired decomposition of p into a pure charge and a measure. 0
10.5 PURE CHARGES ON BOOLEAN ALGEBRAS The notion of a measure on a field 9 of subsets of a set R is not the same when 9 is viewed as a field of sets and when 9 is viewed as a Boolean algebra. If B is a Boolean algebra and b,, n 2 1 is a sequence of elements in B, Vnzl b, denotes the supremum of {b,, n 2 I}, if it exists. If 9 is a field of sets on a set R, B,, n 2 1 is a sequence of sets in 9 satisfying the B,. If Unzl B, fails to condition that UnzlB, E g,then Vnzl B, = Unal it is possible that Vnzl B, exists in 9 when 9 is viewed as a belong to 9, Boolean algebra. This is the feature that distinguishes the notion of a measure on a field and the notion of a measure on a Boolean algebra. We give an example to amplify this point. Before that, let us formalize the notion of a measure on a Boolean algebra.
10. PURE
CHARGES
247
10.5.1 Definition. Let B be a Boolean algebra or 9 a field of subsets of a set 2 ! viewed as a Boolean algebra. An extended real-valued function p on 9 (or on B) is said to be a measure if for every sequence B,, n L 1 of p(VnrlB n ) = C n r l p(B,) pairwise disjoint sets in 9 with Vnrl B, in 9, holds. The following example demonstrates the difference in the outlook that persists when dealing with measures on fields as well as on Boolean algebras.
10.5.2 Example. Let n={l,2 , 3 , . . . ,a}, and 9, the collection of all finite subsets of {1,2,3,. . .} and their complements. 9 is a field on R. 9 is also a compact class. (See Example 10.4.2.) Consequently, every charge on 9 is a measure. Now, let us view 9 as a Boolean algebra. Consider the following set function p on 9. p ( A )= 0, if A is a finite subset of {1,2, 3, . . .}, = 1, otherwise.
is a charge on 9 but p is not a measure on 9 when 9 is viewed as a Boolean algebra. For, let B, = { n } ,n L 1.B,, n 2 1is a sequence of pairwise disjoint elements in 9 with VnZlB, = R. But p(R) = (Vnrl B,) = 1 and Cn21P(Bn) = 0. p
The decomposition theorem, developed above for a bounded charge defined on a field 9 of subsets of a set 0 as a sum of a pure charge and a measure, takes the following form when 9 is viewed as a Boolean algebra. Can we write p = p l+ p 2 , where p1 is Let p be a bounded charge on 9. a pure charge on 9 and p2 is a measure on 9?We will show subsequently that such a decomposition theorem is possible. Before that, we characterize measures and pure charges on 9 when 9 is viewed as a Boolean algebra.
10.5.3 Theorem. Let %be a field of subsets of a set 0. Let Y be the Stone space of 9, % the field of all clopen subsets of Y, T a n isomorphism from 9to % and Wo the Baire u-field on Y. Let p be a positive bounded charge on 9and b the positive bounded measure on % given by /-z (C) = p (T-'C), C E %. Let 6 be the extension of @ from % to aJ0 as a measure. (See Theorem 3.5.2.) Then the following statements are true. (i). p is a measure on 9when 9is viewed as a Boolean algebra if and only if h ( C ) = O for every nowhere dense closed G s subset C of Y. Equivalently, p is a measure on 9when g i s viewed as a Boolean algebra if and only if fi (D) = 0 for every first category Baire F, set D c Y . (ii). p is a pure charge on 9 when 9is viewed as a Boolean algebra if and only if there is a &st category Baire F, set DOc Y such that 12 (DG)= 0.
Proof. In the argument given below, we view 9 as a Boolean algebra. (i). Let p be a measure on 9. Let C be any nowhere dense closed Gs
248
THEORY OF CHARGES
nn2%
subset of Y. We can write C = C,, where each C, is a clopen subset of Y and C1 1 C 2 x C 3 1.* * . This is possible because in any compact Hausdorff totally disconnected space Y, for any closed set C contained in an open set U, one can find a clopen set V such that C c V c U. Since C is a nowhere dense set, Anrl T - k , = 0 in 9.Since p is a measure on .F, 1 p ( T - C,) = 0. Consequently, limn+m& (C,) = 0. Hence & (C) = 0. The converse can be proved by retracing these steps. (ii). Suppose p is a pure charge on 9. Let r =Sup &(D), where the supremum is taken over all first category Baire F, subsets D of Y. It is obvious that r < a.Let D,, n 2 1 be an increasing sequence of first category Baire F, subsets of Y such that &(D,) = r. Let DO= Un21 D,. Then &(Do)= r and DOis a first category Baire F, subset of Y. Further, Do has the following property. If D is any first category Baire F, subset of Y, then I;. (D -Do) = 0. Let $ on Wo be defined as $ (A)= & (A)- & (A nDo) $ is a measure on 30. Let T be the charge on 9 whose for A in 30. corresponding measure on Wo as explained in the statement of this theorem is the $ given above. If C is a first category Baire F, set, then $(C)= 0. By (i), T is a measure on 9.Since T 5 p and p is a pure charge on 9, T = 0. This shows that for any A in Wo,
&(A)= & (A nDo). Consequently, & (Dg) = 0. For the converse, we proceed as follows. Let p be a charge on 9 such that & (D‘o)= 0 for some first category Baire F, subset Do of Y. Let T be a measure’on 9 such that 0 5 7 s p . By (i), $(C)= 0 for any first category Baire F, subset of Y. In particular, i(Do)=O. Consequently, ;(Y)= $(Do)+?(DG) = 0. Hence T = 0. This shows that p is a pure charge. 0 Now, we give the desired decomposition theorem for charges on Boolean denote the algebras. Let 9 be a field of subsets of a set R. Let Z ( R , 9) collection of all bounded measures on 9 when 9 is viewed as Boolean algebra. Let P(R, 9)denote the collection of all bounded pure charges on 9with 9being viewed as a Boolean algebra. 10.5.4 Theorem. Let 9 be a field of subsets of a set R and p a bounded charge on 9. Then there exists p l in P(R, 9)and p2 in b ( R , 9) such that EL = p t + p z .
Further, this decomposition is unique. Proof. A proof of this theorem can be modelled on the one for Theorem 10.2.1 by establishing that G ( R , 9 ) is a normal vector sublattice of 0 ba(R, 9).
CHAPTER 11
Ranges of Charges
The range of a measure on a cr-field of subsets of a set s1 is a very well understood phenomenon. For example, the range of a real measure p on a cr-field of sets is a closed subset of the real line and further, if p is nonatomic, then its range is an interval. In this chapter, we study ranges of charges. In Section 11.1, we make some general comments on the ranges of bounded charges on fields. In Section 11.2, we show that the range of a bounded charge on a a-field is either a finite set or contains a perfect set. In Section 11.3, we examine the cardinalities of the ranges of charges. In Section 11.4, the validity of Liapounoff’s theorem for strongly continuous charges on fields is examined. In Section 11.5, we construct a positive bounded charge on a field such that its range is neither Lebesgue measurable nor has the property of Baire.
11.1 RANGES OF BOUNDED CHARGES ON FIELDS To begin with, we introduce some basic definitions and establish notation.
11.1.1 Definition. Let p be a charge defined on a field 9 of subsets of a set n. The range of ,u is denoted by R(p) and is defined to be the set
11.1.2 Definition. A charge p defined on a field 9 of subsets of a set s1 is said to be finitely many valued if its range R(p) is a finite set. p is said to be infinitely many valued if R(p) is an infinite set. The range of a finitely many valued charge is easy to describe. For this, we need a lemma.
11.1.3 Lemma. Let p be a finitely many valued real charge defined on a field 9of subsets of a set s1. Then we can find pairwise disjoint sets AI, non-zero real numbers a l , a 2 , . . . , a,, 0-1 valued charges A2,. . . , A, in 9,
250
THEORY OF CHARGES
p1, p2, . . . ,pn on 9 such that pi(Ai) = 1for i = 1,2, . . . ,n and n
p =
C aipi. i=l
Proof. Let R(p) consist of k points. Let
0 = {{BI,Bz, . . . , B,,,}; Bi's are pairwise disjoint, Bi's are in 4, p (Bi)# 0 for every i, and m 2 1). Note that if {B1,B2, . . . , B,}E 0,then m 5 k. Let {Al, AZ,. . . , A,} be a maximal family in 0, i.e. if {B1,B2,. . , ,B,}E 0,then m s n . Each Ai has the following property. If B E 9and B c Ai, then p (B) = 0 or p ( B )= p (Ai). If this is not true, then there is a B in 4 such that B c Ai and p (B) # 0, p(Ai -B) # 0. Then {Al, A2,. . . ,Ai-l, B, Ai -B, A i t l r . ..,A,}€ 0 contradicting the maximality of {Al, Az, . . . ,A,}. Let a i = p(Ai) and p j on 9 be defined by pi(B)= (l/ai)p(Ain B ) for B in 9 and i = 1 , 2 , . . . , n. It Further, is clear that each pi is a 0-1 valued charge on 9. " CL = C aipii=l
For this, use the fact that if B E 9 and B c (A1u A Z U p (B) = 0.
* *
u A,)', then
0
From the above lemma, the following proposition is obvious.
11.1.4 Proposition. Let p be a finitely many valued charge defined on a field 9of subsets of a set 0 having the representation n
1-1. =
C atpi, i=l
where p i ' s are 0-1 valued charges sitting on disjoint sets in 4 and a l , az, . . . ,a, are non-zero real numbers. Then r k
c { l , 2, . . . , n } and 1sk
5n
I
.
Now, we look at infinitely many valued bounded charges defined on fields. We need a preliminary result.
11.1.5 Proposition. If p is any bounded charge defined on a fieId 9 of subsets of a set il and if there is a real number c > O such that Ip(A)I>c whenever A E 9 and p (A)# 0, then p is finitely many valued.
11.
RANGES O F CHARGES
25 1
Proof. We attempt to write p in the form I:=,aipi in the spirit of Lemma 11.1.3. Let SupFEsIp(F)I= k. Since p is bounded, k < 00. Find the largest positive integer N such that Nc 5 k. Let {A, ;a E r}be any family of pairwise disjoint sets in 9 such that p(A,) # 0 for every a in r. We show that r is a finite set and in fact, the cardinality of r is 52N. Suppose the cardinality of r is >2N. Let rl = {a E r; p (A,) > 0 ) and r2= {a E;'I p (A,) < O}. Then, either the cardinality of rl is > N or the cardinality of r2is >N. Assume, without loss of generality, that the cardinality of rl is >N. Select N + 1 distinct sets A,,, A,,, . . . ,A,,+, with a1, a 2 , .. . ,aN+l in rl. Let A = Ami.Then p (A)> ( N + 1)c > k. But p (A) 5 k. This contradiction establishes the claim. As in the proof of Lemma 11.1.3, let
UL'
0 = {{B1, B2, . . . ,B,,,}; Bi's are pairwise disjoint, Bi's are in 9, p (Bi)# 0 for every i, and m L 1). Let {Al, A2, . . . , A,} be any maximal family in 0 with n being the largest possible integer. Following the argument in the proof of Lemma 11.1.3, we can write n
CL =
C i=l
aipi,
where al, a2,. . . ,a, are non-zero real numbers and p l , p 2 , . . . ,p, are 0-1 valued charges on 9sitting on disjoint sets in 9. Hence p is finitely many 0 valued.
As a consequence of the above proposition, we prove the following result which tells us about the nature of R(p) when p is infinitely many valued,
11.1.6 Theorem. Let p be an infinitely many valued bounded charge on a field 9 of subsets of a set R. Then 0 is an accumulation point of R(p). More strongly, R ( p ) is a dense-in-itself set, i.e. R(p) has no isolated points.
Proof. From Proposition 11.1.5, it follows that 0 is an accumulation point of R(p), i.e. in every open interval (-c, c ) with c >0, there exists A in 9 such that p (A) # 0 and p (A)E (-c, c). To prove the second part, let A E 9 be such that p (A) # 0. We show that p (A) is an accumulation point of R(p). Now, we note that either p is infinitely many valued on A or p is infinitely many valued on A". Case (i). p is infinitely many valued on A. By the first part of this theorem, since p is also bounded on A, we can find a sequence B,, n 2 1 in 9such that each B, c A, p (B,) # 0 and limn+cop (B,) = 0. Then p (A-B,) = p (A)- p (B,), n L 1 converges to p (A). Note that p (A - B,) # p (A) for all but a finite number of n's. Hence p(A) is an accumulation point of R(p). Case (ii). p is infinitely many valued on A'. By an argument similar to
252
THEORY OF CHARGES
the above, we can find a sequence B,, n r l of sets in 9 such that p (B,) = 0. Note that B, c A', p (B,) f 0 for every n 2 1 and lim,,,p(AuB,)=p(A)+lim,,,p(B,)=p(A). Also, p ( A u B , ) # p ( A ) for all but a finite number of n 's. This shows that p (A) is an accumulation point of R(p). 0 The results established so far can be summed up in the following theorem.
11.1.7 Theorem. Let p be a bounded charge on a field 9of subsets of a set 0. Then either every point of its range R ( p ) is an isolated point in which case R(p) is a finite set or every point of R(p) is an accumulation point of Rb). Now, we take up the case of real charges (not necessarily bounded) defined on fields of sets. Theorems 11.1.6 and 11.1.7 are no longer true for such charges. We present a couple of examples to amplify this point.
11.1.8 Example. Let C! = (1,2 ,3 , . . .}, 9= Finite-cofinite field on C! and p on 9be defined by p ( A ) = #A, = -#A',
if A is finite, if A is cofinite,
where # A denotes the number of points in A. p is a real charge on 9. p is also infinitely many valued. But every point in the range R(,u) = {. . ., -3, -2, -1,O, 1 , 2 , 3 , . ..) is an isolated point of
Rb). 11.1.9 Example. Let C! = ( 0 , 1 , 2 , 3 , . . .}, 9 the finite-cofinite field on 0 and p on 9be defined by
= -p(A"),
if A is cofinite.
(If n = 0, 1+ l / n is interpreted as equal to 1.) p is a real charge on 9 taking infinitely many values. 0 is not an accumulation point of R(p). Note that 1 is an accumulation point of R(p).
11.2 RANGES OF CHARGES ONU-FIELDS Ranges of bounded charges on c+-fieldsexhibit some additional properties over what we have established in the previous section. The main result of
11. RANGES
253
OF CHARGES
this section is that the range of any bounded charge on a u-field is either a finite set or contains a perfect set. We need some preliminary results.
11.2.1 Definition. Let p be a charge defined on a c+-field % of subsets of a set R. Let A,, n z 1 be a sequence of pairwise disjoint sets in a. We say that p is u-additive across A,, n 2 1 if for any B in 8,B c UnzlA,,
F(B)= C PLAnnB) nzl
holds. The following proposition characterizes this notion.
11.2.2, Proposition. Let p be a bounded charge on a a-field 8 of subsets of a set SZ. Let A,, n z 1 be a sequence of pairwise disjoint sets in a. Then the following statements are equivalent. (i). p is u-additive across A,, n z 1. (ii). p + is u-additive across A,, n z 1 an d p - is c+-additive across A,, nzl. (iii). Ip I is u-additive across A,, n 2 1. (iv). I~l(Unz1 An)=Cnzl IpI(An)* (v). Ipl(B,), n z 1 converges to 0, where B, Am,n 2 1.
=urn,,
Proof. (i)+ (ii). Let A E % and A c U n z l A,. We show that p+(A)= C n z l p + ( A n A , ) . Let B E % and B c A . By (i), p(B)=C,,, p ( B n A , ) s C n z lp + ( A nA,). Since this inequality is true for every B in $ with ?I B c A, it follows that p+(A)
+ +
lpl(umzn
C
mzl
IpI(Am)=IpI( mUz l A m ) = I p I (mZ= l Amu mU z n Am)
m=l
m=l
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THEORY OF CHARGES
Hence, equality should prevail throughout, and consequently,
Ip I(B,) =
Cmz, l ~ l ( A ~By) .(iv), it follows that (v)+(i). Let B E % and B c U n Z 1 A , . Then for any n r l , I C m z n p ( B n F(U,,,~,, B AAm) = 0. Hence Am11 5 Ip C IU ,, Am). By (v), P(B)=
C
mzl
W(BnAm).
0
Thus p is u-additive across A,, n I1.
The following proposition gives another property of the above notion. 11.2.3 Proposition. Let p be any charge on a u-field 2l of subsets of a set 0. Let A,, n 2 1 be a sequence of pairwise disjoint sets in 2l such that p is u-additive across A,, n 2 1. Let B,, n r 1 be any sequence of sets in 2l such that B, c A,, for every n 2 1. Then p is u-additive across B,, n 2 1.
0
Proof. This follows directly from Definition 11.2.1.
In the following proposition, for any given bounded charge p which is not finitely many valued, we exhibit a sequence of pairwise disjoint sets across which p is u-additive. 11.2.4 Proposition. Let p be a positive bounded charge on a a-field 2l of subsets of a set a.Suppose the range of p is not finite. Then there exists a sequence W,, n 2 1 of pairwise disjoint sets in % such that p (W,) > 0 for every n I1 and p is u-additive across W,, n 2 1.
Proof. First, we exhibit a sequence Z,, n I1 of pairwise disjoint sets in 8 such that p (Z,) > 0 for every n 2 1. Since R(p) is not a finite set, we can and the range of p on A1 is an find A1 in 2l such that 0 < p (Al) < p (a) infinite set. By similar argument, we can find AZ in %, A 2 c A 1 such that O
-
11.
255
RANGES OF CHARGES
union of sets from {Z,, n 2 1). It follows that limn+oo p(Y,)=O. Consequently, Y, = 0. Define W, = Y, -YflCl, n ? 1. W,, n z 1 is a sequence of pairwise disjoint sets in %, p (W,) > 0 for every n z 1 and p is a-additive across W,, n z 1 by Proposition 11.2.2.
n,,
The following theorem is the main result of this section.
11.2.5 Theorem. Let p be a bounded charge on a cr-field % of subsets of a set R. Then the range R ( p ) of p is either a finite set or contains a perfect set. Consequently, R(p) is eirher finite or has the power of the continuum.
Proof. Suppose the range of p is not finite. We prove the theorem, first, when p is positive. By Proposition 11.2.4, we can find a sequence W,, n z 1 of painvise disjoint sets in % such that p (W,) > 0 for every n 2 1 and p is cr-additive across W,, n z 1. Then (c,,cp(W,); C c ( l , 2 , 3 , . . .}}c R(p). But {InEC p (W,); C c {1,2,3, . . .}} is a perfect set. Now, let p be any bounded charge on %. Since R(p) is infinite, R(Ip1) is infinite. For, if R(Ip1) is finite, since IpI = p + + p - , then p + takes finitely many values and p - takes finitely many values. Consequently, p = p c - p L takes finitely many values which is a contradiction. Since R(1p I) is infinite, by Proposition 11.2.4, we can find a sequence W,, n 2 1 of pairwise disjoint sets in 8 such that 1p I(W,) > 0 for every n z 1 and IpI is cr-additive across W,, n ? 1. Since lp I(W,) > 0, we can find B, in M such that p (B,) # 0 and B, c W, for every n 2 1. By Propositions 11.2.2 and 11.2.3, p is cr-additive across B,, n z 1. Either we can find an infinite subset N1 c { 1 , 2 , 3 , . . .} such that p(B,) > O for every n EN^ or we can find an infinite subset N Z c { l ,2 , 3 , . . .} such that p(B,)
{xncC
Now, we strengthen the above theorem under the same conditions.
11.2.6 Theorem. Let p be any bounded charge on a a-field % of subsets of a set R. Then either every point in the range R ( p ) of p is an isolated point of R ( p ) in which case p is finitely many valued or every neighbourhood of any point a in R(p) contains a perfect set C c R(p).
Proof. Assume that p is not finitely many valued. For the point 0 E R(p) and the neighbourhood ( - E , E ) with E > 0, we exhibit a perfect set C c R(p) such that C c ( - E , E ) . Since R(1p.I)is infinite, as in the proof of Theorem 11.2.5, we can find a sequence C,, n z 1 of pairwise disjoint sets in % such C,,,), n z 1 converges to that p is cr-additive across C,, n z 1, zero and p(C,) > 0 for all n 2 1 or p(C,) < O for all n 2 1. Ignoring the C,) < E . It follows first few sets if necessary, we can assume that ( pl(Um21
lpl(u,,,
256
THEORY OF CHARGES
that for any subset D c { l , 2 , 3 , . . .}, Ip(UmED Cm)l< E . Thus we find that the perfect set (CmaDp (Cm);D c {1,2,3, . . .}} c ( - E , E ) . Now, let a be a non-zero point in R(p). Let A E % be such that p (A) = a. Suppose p when restricted to A is infinitely many valued. Then, by what we have proved above, for a given E > 0, there exists %' c A n9 such that {p( C ) ;CE %'} is a perfect set contained in (-e, E ) . Note that {p(A-C) = p(A)-p(C); C E ~is} a perfect set contained in ( ~ ( A ) - E~, ( A ) + E ) = (a - E , a + F ) . If, on the other hand, p happens to be finitely many valued on A, then p when restricted to A"is infinitely many valued. By what we have established above, we can find 9 c A'n % such that { p (D); D E 9} is a perfect set contained in (-E, E ) . Now, note that { p (A u D) = p (A)+ p(D); D E ~ is}a perfect set and is contained in &(A)-&,~ ( A ) + E ) = (a - E , a + E ) as well as in R(p). This completes the proof of the theorem. I7
11.3 CARDINALITIES OF RANGES OF CHARGES
In this section, we make some remarks on the cardinalities of ranges of charges. In Section 11.2, we saw that if p is a bounded charge on a v-field of sets, then its range R(p) is either a finite set or has the cardinality of the continuum. If p is a bounded charge on a field 9of sets, then its range R(p) can be of any cardinal number. It is an amusing and instructive exercise for the reader to construct a charge on a suitable field of sets so that its range is of cardinality n, a given integer >1. The following theorem goes beyond finite cardinals. 11.3.1 Theorem. Let K be any infinite cardinal less than or equal to the cardinality of the continuum. Then there is a set R, a field 9 of subsets of R and a real charge p on 9such that the cardinality of the range R(p) of p is K. Proof. Let X be any subset of R having the following properties. (i). Cardinality of X = K. (ii). If x, y E X and a , p are rational numbers, then a x +by E X . Such a set X can be constructed as follows. Let B be any subset of the real line R with cardinality K. Let X = ( ( ~ 1 x +1 a 2 ~ 2+ *
* *
+(Y,x,;
X I ,~
2
. ., . ,X ,
E B,
al, a2,. . .,a , rational numbers and n 2 I},
Then the set X has the above properties (i) and (ii).
11.
257
RANGES OF CHARGES
uy=l
[ai,bi), Take R = R. Let 9 be the collection of all sets A of the form where [al, b l ) , [a2,b2),. . . , [anr6,) are pairwise disjoint intervals, a l 5 b l , a 2 s b Z,..., a , 5 b n , a l , a z , ...,a , E X , 61, b2,..., b , ~ X a n d n r l ,and their complements. 9is clearly a field on R. Define p on 9by n
@(A)=
n
1 (bi- a i ) , i=l
= -@(AC),
p
if A is of the form
u [a;, bi),
i=l
if A' is of the above form.
is a real charge on 9 and R ( p )= X. This shows that R(p) has cardinality
K.
0
11.3.2 Remarks. (i). One can construct a bounded charge p with cardinality of R(p) = tc in Theorem 11.3.1. Further, one could have I.L to be positive as well. (ii). If p is allowed to take infinite values, the construction of a positive charge p with cardinality of its range being a prescribed infinite cardinal number could be made much simpler.
11.4 CHARGES WITH CLOSED RANGE If p is a bounded charge on a field 9 of subsets of a set R, then its range R(p) need not be a closed subset of the real line R. See Theorem 11.3.1. In the following, we give an example of a bounded charge p on a a-field '% of subsets of a set R such that its range R(p) is not a closed set.
11.4.1 Example. Let R = {1,2,3, . . .} and '% = P(R), the class of all subsets of R. Let po be any probability charge on '% such that pO(A) = 0 for any finite subset A of R. For each n r 1, let p, on '% be the measure defined by p,(A)=O, if n EA , = 1,
if n E A and A c R .
Let p =CnrO(1/2,+')pn. Note that $ & R ( p but ) $ is an accumulation point of R ( p ) .Hence R ( p ) is not a closed set. In view of the above example, it is of interest to derive a set of sufficient conditions under which R ( p ) is a closed set. Sobczyk and Hammer Decomposition theorem (See Theorem 5.2.7.) provides a basis for further exploration in this direction. We need a definition, to begin with.
258
THEORY OF CHARGES
11.4.2 Definition. Let 9be a field of subsets of a set fl. A sequence p,,, n L 1 of 0-1 valued charges is said to be discrete if for any given positive integer n, there exists a set A in 8 such that @,(A)= 1 and pm(A)= 0 for every m # n. Let us state a lemma about discrete sequences of charges on fields.
11.4.3 Lemma. If p,,, n L 1 is a discrete sequence of 0-1 valued charges on a field 9 of subsets of a set fl, then there exists a sequence A,, n 2 1 of pairwise disjoint sets in 8 such that p,,(A,) = 1 and p,(A,,) = 0 for all m and n such that m # n. Proof. If B,, n L 1 is a sequence of sets from 9 such that p,(B,) = 1 and p,(B,) = 0 for m # n, then the sequence A,, n 2 1 defined by
u B,,
n-1
A1=B1, and
A, =B,-
n22
m=l
serves the purpose of the lemma. The notion of discreteness introduced above is weaker than the notion of infinite disjointness. See Remark 5.2.3(i). The following is an example amplifying this point. Let f l = { 1 , 2 , 3 , . . . , 00) and 8 the collection of all finite subsets of {1,2,3, . . .} and their complements. On the field 9,for each n 2 1, define pn by @,(A)= 1, if n E A,
=0, ifngA. This sequence p,, n 2 1 of distinct 0-1 valued charges is discrete but not infinitely disjoint. This is because for any countable partition {Fl, Fz, . . .} of R in 9, all but a finite number of sets among {Fl, Fz, . . .} are empty. The following is an instance when the range is a compact set.
11.4.4 Theorem. Let a,,, n L 1 be a sequence of real numbers such that be a a-field of subsets of a set fl and p,,, n 2 1 a discrete sequence of 0- 1 valued charges on 8.Let
Cnzl la,]
p=
C
anpn.
nrl
Then the range R ( p ) of
p
is a compact subset of R and hence is closed.
Proof. By Lemma 11.4.3, we can find a sequence A,, n 1 1 of pairwise disjoint sets in '2l such that p,(A,) = 1 for every n L 1 and pm(A,)= 0 for all m # n. Define a real valued map h on the Cantor set {O,l}Hoby h ( x l ,~
2 , .*)
=
C
nzl
xnan
11.
RANGES OF CHARGES
259
for (xl, XZ, . . .) in (0, l}'". ((0,l)'" is equipped with the usual product topology.) Since Cn2,( a , ( < c ~h, is a continuous function on (0, I}~O. Let D be the range of the function h. Since (0, is a compact space, D is a compact subset of R. Now, we claim that D = R(p). Let a E D . Then there A,,, exists (XI,XZ, . . .) in (0, l}'" such that h(x1, XZ, . . .) = a . Let A = UnEC where C = ( n 2 1; xn = 1). Since % is a cT-field, A E 5%. Further,
= h(x1, x,,
. . .) = a.
Consequently, a E R(p). Conversely, let b f R(p). Then there is a set A in 9l such that p(A) = b = C n z l a,,p,,(A)=Inrl a,p,,(AnA,). Define the sequence x,,, n 2 1 by x,,
= 1,
if p,,(AnA,) = 1,
=0,
if pn(AnA,,)= 0 , n 2 1.
Consequently, ~ ( x I , x. .~.)= ,
C
anxn
n2l
=
C
nr,
aflpfl(AnAn)
=p(A)=beD. This proves that D = R(p).
0
The following result is an analogue of one dimensional Liapounoff's theorem for nonatomic measures on a-fields. See Theorem 5.1.6.
11.4.5 Theorem. Let p be a positive bounded strongly continuous charge defined on a cr-field Vl of subsets of a set a. Then the range R(p) of p is a closed interval. Proof. Let a E (0,@(a)). We show that there exists a set A in 9l such that p(A) = a . Since p is strongly continuous, for E =;, we can find a finite partition of R in such that each set in the partition has p-value <$.Let A1 be the largest possible union of sets from this partition satisfying the Let B1 be any set condition that p (A,) 5 a. Al # R because 0 < a < p (a). in the partition which is not in this union. Then 0 < p (B1)<$and p (A1u B1) > a. If p (A,) = a, the desired claim is established. Suppose p (Al) < a. Since p restricted to B1 is strongly continuous, we can find a finite partition of B1 in 9l such that each set in the partition has p-value<1/22. Let A2 be the largest possible union of sets from this partition such that p (A,) 5 a -p(A1). Since A2# B1, we can find a set BZ from this partition which is
260
THEORY OF CHARGES
not in this union A,. Then 0 < p (Bz)< 1/22 and p (A, u Bz) >a - p (Al), or equivalently, p (Al) p (Az)+ p (B2)> a. If p (A,) = a - p (Al), then Al u A, is the desired set. Otherwise, we proceed to obtain A3 and B3 as above using Bz. If this procedure terminates at a finite stage, we obtain a set A in 3 with p-value equal to a. Otherwise, we get a sequence Al, BI; A,, Bz; . . . of sets in 3 having the following properties for every n 2 1. (i). A , , n B , , = 0 . (ii). A,,+1u B,,+l c B,,. (iii). p (Ad + p (Ad + + p (A,,) 5 a. (iv). pLA1) + F (A,) + * * * +p(An) + p @,)>a. (v). 0 < p (B,) < 1/2". It is now clear that p (A,,) = a . We now show that p(Uflz1 A,,) = CnZlp(A,). Note that
+
---
for every n 2 1. Consequently,
But the inequality
is always true for positive charges. Hence
This proves the result.
0
Now, it is natural to attempt to extend Theorem 11.4.5 to cover the case when we have a general strongly continuous bounded charge on a v-field. We give an example to show that Theorem 11.4.5 fails in general. Before that, we prove a proposition which will be useful in the construction of the desired example.
11.4.6 Proposition. Let p be a bounded charge on a field 9of subsets of a set fl and R(p) its range. Let a = Sup,,~p((B) and p = InfAEsp(A). Then the following statements are equivalent.
11. RANGES
261
OF CHARGES
(i). p admits a H a h n set, i.e. a E-Hahn set for E =O. (See Definition 2.6.1. (ii). a E R(p). (iii). p E R(p). Consequently, R(p) is not a closed set if p does not admit a Hahn set.
Proof. We show that (i) and (ii) are equivalent. (i) 3 (ii). If A is a Hahn set for p , then p (A) = a. (ii) +(i). If A E 9and p (A) = a , then A is a Hahn set for
p.
0
Now, we construct the desired example. In view of Proposition 11.4.6, it suffices to construct a strongly continuous bounded charge on any given infinite u-field not admitting a Hahn set.
11.4.7 Example. Let % be a given infinite cT-field of subsets of a set 0. Let u be any strongly continuous probability charge on a. See Corollary 5.3.3.Using Theorem 11.4.5, we obtain a tree {A61,62 ,...,6,; 81,SZ, . . . ,a, a finite sequence of 0’s and l’s, n 2 1) having the following properties. ....,6,,1= 0 for any n 2 1 and any finite sequence (i). A,,,,, .....6,.0nAS1,62 S1, 82,.. . , S , of 0’s and 1’s. (ii). AS,,^, ....,6,.0uAs,,~, .....6,,1= A61.62....,6, for any n 2 1 and any finite sequence 81, Sz, . . . , S , of 0’s and 1’s. (iii). A o nA, = 0. (iv). A0 u A1= R. and any (v). V(AS1.62....,8 , ) =al(Sl)aZ(S2) * ’ & , ( a n ) for any n sequence 81,SZ,. . . , 8, of 0’s and l’s, where a, (0)= l / ( n + 1)and a , (1)= n / ( n + l ) , n 2 1. Let X = {A E 8 ;v(A) = 0). Look at the quotient Boolean algebra %/Af. For A in [A] denote the equivalence class in % / X containing A. Note .....6,]; S1,Sz, . . . ,a, a sequence of 0’s and l’s, n 2 1) is a tree that {[A6162 in %/X.As in the proof of (ii)+(iii) of Theorem 5.3.2, one can construct a strongly continuous probability charge 7 on 2i/N such that
a,
?([A61,62,....
~ , ] ) = ( Y ~ ( ~ - S ~ ) ( Y Z ( ~’ -* S’ an(l-Sn), Z)
for any finite sequence S1,&, . . . ,S , of 0’s and l’s, and for any n 2 1. Now, define T on 8 by T(A)=?([A]) for A in %. T is a strongly continuous probability charge on % because for any finite sequence S1,Sz, . . . ,a, of 0’s and l’s, 7(A61,62.....~n) 5 l/n, for every n 2 1. v
AT
= 0 because for any n 2 1,
uAs,,s,.___. and uA61,62...., S,*O
where both the unions are taken over all S I , & , .
6
,
~
~
. . , 6, in (0, l), are disjoint
262
THEORY OF CHARGES
a,
with union equal to v-value of the first set and .r-value of the second set are each equal to l / ( n +2). 7 also has the property that if A E 8 and v(A)= 0, then 7(A) = 0. Now, define p on 8 by p = v -7. Since v A T = 0 , p + = v and p - = T. See Theorem 2.2.2(4). It is clear that v and T are distinct and p is a strongly continuous bounded charge on a. We note that p does not,admit a Hahn = 0. set. If p admits a Hahn set A in a, then pf(AC)= v(A') = 0. So, 7(AC) Also, p-(A) = T ( A = ) 0. This implies that T(R)= 0 which is a contradiction. Theorem 11.4.5 says that the range of a positive bounded strongly continuous charge defined on a u-field is convex and closed. The convexity part can be generalized to any finitely many positive bounded strongly continuous charges defined on a c+-field whereas the closedness cannot be generalized.
11.4.8 Example. Let v and T be positive bounded strongly continuous charges on an infinite u-field % as in Example 11.4.7. R(v, T ) = {(.(A), T(A)); A E a} is not a closed set because (1,O)a R(v, T ) but belongs to the closure of R(v, T ) . 11.4.9 Theorem. Let pl,p2, . . . , p,, be positive bounded strongly continuous charges defined on a u-field % of subsets of a set R. Then the range R(~l,~Z,...,~")=~(~l~A)r~z(A),...~~"(A));A~~~Of~
a convex subset of the n-dimensional Euclidean space R". Proof. For n = 1, the result was already established in Theorem 11.4.5. Let us assume the result to be true for n = k and prove the result for n = k + 1. Let p l , pz, . . . ,pk+l be positive bounded strongly continuous charges on a. The proof is divided into several steps as follows. 1". Let ~ ~ = p ~ + p ~ + l - t - . *for * +i p =1 ~ ,+ 2 ~, ..., k + 1 . Then it is easily seen that R(pl, p z , . . . , p k + l ) is convex if and only if R(71, T ~. ., . , T k + 1 ) is convex. So, we prove that R(TI,T ~ . ,. . ,T k + 1 ) is convex. 2". If we show that for every A in a, there exists a set B in % such that B c A and T ~ ( B=)$ T ~ ( for A ) i = 1 , 2 , . . . ,k + 1, then it would follow that R ( T ~T ,~. ., . , 7k+1) is convex. Let us see why. For any given set A in %, by repeated application of the above assertion, for every dyadic rational r between 0 and 1, we can find a set A, with the following properties. (i). A. = 0. (ii). A1 = A. (iii). A, c A, whenever the dyadic rationals r and s satisfy 0 5 r <s I1. (iv). T~(A,) = ~ T ~ for ( A i) = 1,2, . . . ,k + 1 and for any dyadic rational Olrsl.
11.
RANGES OF CHARGES
263
(v). For any real number a between 0 and 1, if we define A, = U A , , where the union is taken over all dyadic rational numbers r 5 a, then A, E fl and T~(A,)= aTi(A)for i = 1 , 2 , . , , , k + 1. Let C, D E %and O 5 a 5 1. Then UTi (C)+ (1 -a)Ti(D) = Ti((C-D)a u (C nD) u (D - C)i-,)
for every i = 1,2, . . . ,k + 1. Using the properties (i), (ii), (iii), (iv) and (v) given above, one can establish the above equality. Consequently, R ( T ~Q, , . . . , Tk+1) is convex. Thus it suffices to exhibit a set B in '%,for a given A in 8, such that B c A and T ~ ( B ) = $ T ~for ( A )every i = 1 , 2 , . . . , k + l . 3". Let C E '% be a set obtained by the induction hypothesis satisfying C c A and T~(C) = $T~(A) for i = 1 , 2 , . . . ,k. If T k + l ( c ) = &k+l(A), then we have finished. If not, let T k + l ( C ) < : T k + l ( A ) < T k + l ( A - C ) . NOW, look at the Sets C,, 0 Ia 5 1 and (A - C),, 0 5 a i 1 obtained as in Step 2" with respect to T I , 7 2 , . . . ,Tk. The existence of these sets is assured by the induction hypothesis. Since T k + l s T k , W e have that Tk+l(Ca- c b ) 5 Tk(Ca- c b ) = (a - b)Tk(C)if a 2 6. Hence Tk+l(C,) is a continuous function of a. Similarly, Tk+l((A-C),) is also a continuous function of a. Hence for O s a 5 1, T k + l ( c a u (A-C)l-,) is a continuous function of a taking the value ? - k + l ( c ) at a = 1 and the value ?-k+l(A-C) at a = 0. Hence there exists a real number a0 in [0, 11 such that ?-k+1(Caou ( A - C L J = $n+l(A). However, for l r i s k , T~(C~U(A-C)~-~)=U~T~(C)+(~-~&~(A-C)
+ (1 - u ~ ) $ T ~ ( A )
=aOh(A) = $ T ~(A).
Thus Ca0u(A-C)l-,=
B serves the purpose.
0
The analogue of Theorem 11.4.9 for bounded strongly continuous charges which are not necessarily positive is also true and is an easy consequence of Theorem 11.4.9.Recall that a bounded charge p is strongly continuous if lp is strongly continuous.
I
11.4.10 Theorem. Let p l , pz,. . . ,pn be a finite number of bounded strongly continuous charges defined on a u-field fl of subsets of a set 0. Then the range R ~ IPZ,, . . . , pn) ={(,UL(A), PZ(A),. . . ,p n ( A ) fA ; = %I of P I , p2, . . . ,p,, is a convex subset of R". Proof. Using the Jordan Decomposition Theorem, write pi = p: - p ; for i = 1,2,. . . ,n. Then all the p:'s and pL;'s are strongly continuous. So, R ( p l, p L, p l, p y, . . . ,p :, p i ) is a convex set by Theorem 11.4.9. From 0 this, it follows easily that R(p1, pz, . . . ,p,,) is convex.
264
THEORY OF CHARGES
11.5 CHARGES WHOSE RANGES ARE NEITHER LEBESGUE MEASURABLE NOR HAVE THE PROPERTY OF BAIRE In this section, we examine whether the range of any probability charge on a a-field is a Borel subset of the real line R. Given any infinite cr-field 8 on a set a,we will exhibit a family of probability charges on 8 the range of each of which is neither Lebesgue measurable nor has the property of Baire. For this purpose, we need some notions and results from Topology and Measure Theory. Let 8 be a a-field of subsets of a set R and 7 a positive measure on 8. The charge space (R, 8,7)is said to be complete if B E 8 whenever B c A for some A in 8 with 7(A) = 0. If (R, %, 7 ) is not complete, we can enlarge 8 so as to make the resultant triplet complete. More precisely, let
& ={B
A A; B c C for some C in 8 with 7(C)= O and A E 8}.
One can show that $l is a cr-field on R and contains %. Define ? on $l by
;(B A A) = 7(A), = 0 and A E 8. F is unambiguously where B c C for some C in 8 with T(C) defined on %, agrees with T on 8 and is indeed a measure on 8.The charge space (R, fl,?) is complete. Sets in 3 are usually called measurable sets. The following is a particularly interesting case of the above. Let 93' be the Borel cr-field on the real line R and A Lebesgue measure on B. Then (R, 93,A ) is not complete. Let 8 be the completion of 9 with respect to A as described above. The sets in 8 are called Lebesgue measurable sets. Obviously, 8 contains the Borel cr-field W properly. Let X be a topological space. Recall that a subset A of X is said to have the property of Buire if there exists an open subset U of X such that A A U is a set of first category. Let W* be the collection of all subsets of X each of which has the property of Baire. Then W*is a cr-field on X and contains the Borel a-field of X, i.e. the smallest cr-field on X containing all open subsets of X. Let X be a complete separable metric space and B its Bore1 o-field. Let XNobe the product space X x X x . . * ={(xl, xz, . ..); x, E X for all n L 1). Let Bmbe the product o-field on XKo,i.e. the smallest a-field on XKocontaining all finite dimensional cylinder sets {Al x AZx * * * x A, x X x X x * * ; A I ,Az,. . . , A , E 9,n 2 1). Let p,, II 2 1 be a sequence of probability measures on BI-Then there exists a unique probability measure p on Bm with the following property.
-
p ( ( A l x A z x . . * x A , x X xX x . . . ) = ~ l ( A l ) p * . 2 ( A 2 ) . . ' ~ n ( A n )
11.
RANGES OF CHARGES
265
for any A I ,A2, . . . ,A, in and n 2 1. p is called the product prot Tbility p,. measure of p,, n 2 1 and is denoted by In the above framework of product spaces, we introduce the followlng notion. A set D c X N o is said to be a tail set if ( y l , y 2 , . . . , Y k , xk+l, X k + Z , . . . ) E D whenever ( X I , X 2 , . . . ,xk, X k + l r X k + z , . . .) E D for some X I , X Z , . . . ,xk in X, for any y l , y 2 , . . . ,Y k in X and for any k 2 1. Now, we state, in this connection, a useful result.
n,,,
Kolmogorov’s 0-1 Law. Let D be a p-measurable tail subset of XHo. T h e n p ( D ) = O o r 1. Now, we look at the space XKoin its product topology, where X is a complete separable metric space. The following result is a topological analogue of the above law.
Oxtoby’s Category Analogue of Kolmogorov’s 0-1 Law. Let D be a tail subset of XKowith the property of Baire. Then either D is of first category or D“ is of first category. Now, we are in a position to construct the desired family of probability charges.
11.5.1 Definition. Let 9 be a field of subsets of a set R and p,, n 2 0 a sequence of 0-1 valued charges on 9. p o is said to be an accumulation point of {p,, n 2 1)if for every A E 9 with p o ( A )= 1,there exists an infinite subset D of {1,2,3, . . .} such that p , ( A ) = 1 for every n E D . The above notion has a simple interpretation in terms of the Stone space Y of 9.Each p n can be identified as a point in Y and the above definition is equivalent to saying that po is an accumulation point of the subset {p,, n 2 1)of Y in the usual topological sense.
11.5.2 Theorem. Let % be a a-field of subsets of a set s2 and p,, n 2 0 be a sequence of 0-1 valued charges on such that p,, n 2 1 is a discrete sequence and p o is an accumulation point of {p,, n 2 1).Define p on % by P=
C
1 p
p
n
-
n 20
Then the range R(p) of p is neither Lebesgue measurable nor has the property of Baire. Proof. Let Z = {0,1} and v the measure on the discrete u-field on Z given ) Let C = ZHoequipped with the product u-field and by v({O})= ~ ( { l=}4. 7 the product probability measure v x v x v . . . on this u-field. Sets in this product u-field are called Bore1 subsets of C . (C is the Cantor set.) Define
266
THEORY OF CHARGES
a real valued function h on C by
hbi,~
.I= nCz l
Xn
2 , .
for ( x l , x z , .. .) in C. This function has many interesting properties. It is one-to-one except for a countable set of points; it is a homeomorphism except for a countable set of points; h(A) is a Borel subset of [0,1] if and only if A is Borel subset of C; h preserves the measures 7 and the Lebesgue measure A on [0, 11, i.e. 7(hP1(B))= A (B) for every Borel subset B of [0,1]; h(A) is a Lebesgue measurable subset of [0,1] if and only if A is a 7-measurable subset of C; h(A) has the property of Baire in [0, 11 if and only if A has the property of Baire in C. See Kuratowski (1966). In view of these properties of h, if we show that F = {(@*(A), p1(A),pz(A), . . .); A E?l} is neither 7-measurable nor has the property of Baire in C , then it will follow that R(p) is neither Lebesgue measurable nor has the property of Baire in [0,1]. This is because h(F)= N P 1. Let D = {(pl(A),p2(A), . . .); A E8,pO(A)= 1). So, if we show that D is neither 7-measurable nor has the property of Baire in C, the desired conclusion about F follows. First, we show that D is not 7-measurable. We prove that D is a tail set. Since pn, n 2 1 is a discrete sequence, by Lemma 11.4.3, we can find a sequence A,, a 2 1 of pairwise disjoint sets in 8 such that p,(A,) = 1 for every n L 1and p,,,(A,) = 0 for all rn # n. Since pois an accumulation point of {p,,, n 2 I}, po is distinct from all p,,, n 2 1. One can assume, without = 0 for every n 2 1. This follows from the loss of generality, that pO(An) fact that if 6 and 7 are two distinct 0-1 valued charges on 91, then there is a set A in 8 such that e(A)=O=q(AC).Let (x1,x2,.. .)= (pi(A),pZ(A),. . . ) E D for some A in ?l with pO(A)= 1. Let k be any positive integer and y1, y 2 , . . . ,Yk be any finite sequence of 0’s and 1’s. Let E l = { l ~ i ~ k ; y i = O }and E Z = { l ~ i ~ k ; y i = l } .Let B = (A-UIEEI Ai)u (UIEE2 A;). It is obvious that pO(B)= 1, pl(B) = yl, Pz(B) = Yz, . . . pk(B) = Ykt Pk+i(B)= @k+l(A), P k + z ( B ) = pk+Z(A),, . . Consequently, (yl, y2, . . . ,yk, &+I, &+2, . . .) E D. Hence D is a tail set. Suppose D is 7-measurable. By the Kolmogorov’s 0-1 law, T(D)= 0 or 1. Let us look at the map 4 from C to C defined by 4(x1, x 2 , . . .) = (1-xl, 1-xz, . . .) for (xl,xz, . . .) in C. We claim that 4(D)nD = 0 and $ ( D ) u D = C . Suppose $ ( D ) n D # 0 . Let (xI,x2,.. . ) ~ l j l ( D ) n DThen . we can find two sets A and B in ?l such that po(A)= 1= pO(B)and 9
(xi, XZ, . . .>= (pi(A),FAA), . . .I,
(1-xi, 1- X Z , . . .) = (pi(B), pz(B),
.I.
11.
RANGES OF CHARGES
267
Note that p o ( A n B ) = 1 and ( p l ( A n B ) , p z ( A n B), . . .) = (0’0, . . .). This is a contradiction to the fact that po is an accumulation point of {p,,,n L l}. Therefore, +(D) nD = 0. To show that 4(D) u D = C, let (xl, x2,. . .) E C . Let E = {n 2 1;x,, = 1) and A = UnEE A,,. Then (pl(A), pZ(A),. . .) = (x1,x2,.. .). If p o ( A ) = 1, then (x1,x2,.. . ) E D . If pdA)=O, then (xl,x2, . . .) E $(D). This shows that 4(D) u D = C. Note that t,b preserves = T(G) for every 7-measurable set G contained the measure 7 , i.e. ~((tr(G)) = 1, then T(~(D)) = 1 and consequently, 7(C) = 2 which in C. Now, if T(D) = 0, then T ( $ ( D ) )= 0 which is again a contradicis a contradiction. If T(D) tion since it works out that T(C)= 0. Thus D is not 7-measurable. To prove that D does not have the property of Baire, one can repeat the above argument and use Oxtoby’s category analogue of Kolmogorov’s 0 0-1 law and Baire Category theorem.
11.5.3 Remark. If 3 is an infinite u-field on a set R, one can always find a sequence p,,, IZ 2 0 of 0-1 valued charges on satisfying the conditions imposed in Theorem 11.5.2.
CHAPTER 12
On Lifting
The purpose of this chapter is to present some ideas on lifting in the setting of charges on fields. In the following, we elucidate the concept of lifting. For the Let 9 be a field of subsets of a set R and 9 an ideal in 9. following definition, recall the definition of the quotient Boolean algebra 919 and the natural homomorphism h from 9 to 919 which takes each (See Section 1.4.) set A in 9to its equivalence class [A] in 9.
12.1 Definition. The natural homomorphism h from 9 to 919 is said to admit a fifting if there exists a subfield Poof 9 such that the map h restricted to Sois an isomorphism from to 9/9. A lifting is tantamount to selecting one set from each equivalence class of 9 so that the resulting collection of sets becomes a field on Q and is isomorphic to 919.This notion of lifting can be reformulated as follows. Suppose the natural homomorphism h from 9 to 9/9admits a lifting. Let II, be the induced isomorphism from 919 to 90.
z
949/9
90.
Let p be the composition of h and II,, i.e., p = II, h. Then this map p has the following properties. (i). p is a homomophism from 9to 9. (ii). p is onto Po. (iii). If A, B E 9 and A B, then p (A) = p (B). (iv). A p (A) for every A in 9. For ease of expression, we call p the lifting of 9 with respect to the ideal 9. If n={l,2 , 3 , . . .}, 9=9’(R) and 4 the. ideal of all finite subsets of Q, it is not difficult to see that 9does not admit a lifting with respect to 9. A well-known result of von Neumann and Maharam establishes the existence of a lifting in the case when (Q,@,p) is a complete measure space, i.e. 9is a c+-field of subsets of R, p is a positive bounded measure on 9 with the property that B E 9 whenever B c A, A E 9 and p (A) = 0, and 9 ={A E 9; p (A) = 0 ) the ideal of all sets in 9with p-measure zero. See Maharam (1958). 0
-
-
12.
269
O N LIFI'ING
It is natural to enquire about the existence of a lifting of 9with respect to 9 in the above if we assume p to be a charge only. But lifting fails to exist in the generality mentioned above and the following theorem due to Maharam-Erdos amplifies this point. (Recall the notion of a density charge from Example 2.1.3(10).)
12.2 Theorem. Let a={1,2,3, . . .}, 9= P(a)and p any density charge on 9. Let 9 = {A E 9; p (A) = 0). Then there is no lifting of 9with respect to the ideal 9. Proof. The proof is carried out in the following steps. 1". Suppose there exists a lifting p of 9with respect to 9. 2". Let 1< p l < p 2 < . * be a sequence of positive integers such that pi and pi are mutually prime for every i # j and Cizl l/pi < 00. It follows that this sequence has the following properties. (i) lim,,m n / p , = 0. (ii) (1- l / p i ) converges (to a non-zero real number). 3". F o r l s j s p i a n d i = 1 , 2 , 3,..., let
nizl
A i j = { j j, + p i , j + 2 p i , . . .}.
Note that, for each i r 1, Ail,A i 2 , .. . , Aipiare pairwise disjoint sets with union R. So, p (Ai1),p (Ai2),. . . ,p (Aipi) are pairwise disjoint sets with union R. Let Ri be the set among {p(Ail),p(Ai2),. . . ,p(Aipz)} containing i and denote the corresponding set Aii by Si for each i = 1 , 2 , 3 , . . .. 4". Obviously, p ( S i )= Ri = 0. Since Si) contains S j for every j 2 1, it follows that p ( u i z l S i ) =a. 5". Let Ti be the set obtained from Si by removing the first element of Si, i r 1.LetT* =UizlTi. Sincep(T*)~p(Ti)=p(Si)foreveryjL1,itfollows that p(T*) =a. This implies, by the definition of lifting, that T* si, i.e. p (T*) = 1 . We show that
uizl
uirl
p(uizl
-uizl
which will then lead to a contradiction. 6". We use the notation # A for the number of elements in A. Let Let r be a positive integer such that
1 1 -<&I6
i and -<&I6 i>rpi Pi Let ko be a positive integer such that ko)p1p2*
foreveryilr.
2' p r and - < ~ / 6 . ko
E
> 0.
270
THEORY OF CHARGES
We show that for any integer k > ko,
(Since p is a density charge on 9, we will then have #T* n { l , 2 , . . . , k } p(T*)= lim
k
k-fw
= 1 - n (1-;).I is1
7". Let k > ko be fixed. Let s be the smallest integer i such that pi > k, i.e.
(ulzl r
n { l , 2 ,..., k } =
C # T i n { 1 , 2,..., k }
i=l
r
r
- Ci = l j C= l # T i n T j n { l , 2 , . . . , k } i <j I
I
I
+I C C # T i n T j n T u n { 1 , 2 ,. . . , k } - * * . j=l u = l j = l
i<j
+(- l ) ' - ' # T 1 n T Z n . . . n T r n { l , 2 , . . . , k } .
Since Ti is of the form {ri +pi, ri +2pS . . .} for some 15 ri
k Tin{1,2, . . . , k } = -
Pi
for every i B 1. We now show that, for i # j ,
+
Ti nTi c {rij +pipi, rij 2pipj,rij + 3pipj, . . .} for some 15 rij
#Ti nTi n { l , 2 , . . . ,k}---. PiPi
12.
27 1
ON LIFTING
Pursuing this argument, we obtain
r
r
k
r
-1 1 1 -+*.*+(-1)' PiPiPU
i = l i=l u = l
P1P2 '
* *
Pr
i<j
n{1,2,
. . . ,k}-k
lr+(~)+...+(:)=2'-1<2'. 9". Using the same argument as above, we note that k #TiA{1,2, ..., k } l l + Pi for every r + 1s j 5 s - 1. 10". Thus, finally, we have
1
#T*n{1,2,. k
. . ,k}-
k
[' #
UielTi n { l , 2 , . . . , k} )k
2' < - + ~ / 3 + ( - +s - 1 k k
1 -), 1
i>rP;
by 8", 6" and 9" respectively. s-1
< ~ / 6 ~+ / +-+ 3
~ / 6
Ps-1
< ~ / 6 4+ 3 -k ~ / 6 ~+ / 6 E<. This completes the proof.
(1-3)
APPENDIX 1
Notes and Comments
CHAPTER 1 Sections 1.2, 1.3 and 1.4 offer fairly standard, though cursory, treatment of the topics covered. For set theoretical notions dealt in Section 1.2, one could refer to Kamke (1950) or Kelley (1955) or Dunford and Schwartz (1954) for more details. Kelley (1955) is a standard reference for topological concepts covered in Section 1.3. Halmos (1963) and Sikorski (1969) are excellent sources for ideas on Boolean Algebras. Some parts of Section 1.1 may be found in Halmos (1950). Theorem 1.1.9(1) is due to Pettis (1951). Corollary 1.1.12 is inspired by Section 4 of Sikorski (1969). A substantial part of Section 1.5 on Functional Analytic concepts is inspired by the fundamental paper of Bochner and Phillips (1941).
CHAPTER 2 Bochner and Phillips (1941) identify the space ba(R, S)of all bounded of charges on the field 9 of subsets of the set R with the space ca(R’, 9’) all bounded measures on a suitable cr-field 9’of subsets of a suitable set R‘. From this it follows that ba(R, 9)is a boundedly complete vector lattice. Theorem 2.2.1 arrives at the same conclusion more directly using a bank of ideas from vector lattices. Construction of invariant charges using Banach limits is a standard practice in Ergodic Theory. See Example 2.1.3(8). We have used Banach limits to show the existence of Density charges. See Example 2.1.3(10). The comprehensive Jordan Decomposition Theorem proved in Section 2.5 is new. Existence of &-HahnDecomposition for charges for every E > 0 was first established by Darst ( 1 9 6 2 ~ ) . Topsae (1979) gave some conditions under which a charge becomes a measure. This result is a generalization of Theorem 2.3.4. First, we need a definition.
APPENDIX
1
273
Definition. A collection % of subsets of a set R is called a monocompact class if it has the following property: if C,, n 2 1 is a decreasing sequence of sets in % with C, = 0,then there exists m 2 1 such that C, = 0.
n,,,
Theorem A.l. Let 9be a field of subsets of a set R and % a monocompact class of subsets of R. Let p be a positive bounded charge on 9having the following approximation property: for any F in B a n d E > 0 there exist C in % and G in 9such that G c C c F and p (F- G )< E . Then p is a measure on S. Christensen (1971) gave some conditions under which a charge becomes a measure. We present some of his results. Let 9 be a a-field of subsets of a set R. For v in ca(R, S),let h, be the map on .F defined by
h,(F) = v(F),
F E9.
Let % be the smallest a-field on 9with respect to which each of the maps h , is measurable for v in ca(R, 9).
Theorem A.2. Let p be a real charge on 9, If p is measurable with respect to the a-field % on 9, i.e.
h i ' (B) = {FE9; p(F) E B}E % for every Borel subset B of the real line R, then p is a measure on 9. Another result in the context of Polish spaces, i.e. complete separable metric spaces, can be described as follows. Let R be a Polish space and 9 its Borel a-field, i.e. the smallest a-field on R containing all open subsets of R. Let p be a probability charge on 9. Let R* be the collection of all closed subsets of R. The gist of the following result is that if p restricted to R* is a decent function on R", then p is a measure on 9. We now elaborate this statement. We can introduce a suitable metric d" on St" so that (a*,d " ) becomes a separable metric space. Let d be a metric on R compatible with its topology. Since R is separable, one can always choose d to be a precompact metric on R. Define a metric d* on R* by d * ( A , B) =Sup {max { d ( a ,B), d ( A , b)}},
A, B E R".
aeA beB
Then
(a*,d * ) is a separable metric space. Let 9"be the Borel a-field on
R*. Theorem A.3. If the map p restricted to R* is measurable with respect to 9*, then p is a measure on 9.
274
THEORY OF CHARGES
Rao (1971) gave a sufficient condition under which a charge becomes a measure. Let p be a positive real charge defined on a field 9 of subsets of a set SZ. A subfield goof 9is said to be p-pure if the following conditions are met. (i). p (AN)= 0 for some N L 1 wheneve; A,, n L 1 is a sequence in 90 satisfying Al 3 A2 3 A3 3 . . and A, = 0. (ii) p(A)=Inf(C,,,p(A,); { A f l , n Z 1 } c 9 o and U n ~ l A , ~ Aforl every A in 9. (This condition means that the Caratheodory measure induced by p on gocoincides with p on 9.)
-
n,,
Theorem A.4. If there exists a p-pure subfield 90 of 9, then p is a measure on 9. Rao (1971) stated that the converse of the above theorem is true. This is not correct, however, as the folIowing discussion demonstrates.
Theorem AS. Let 9 be a a-field of subsets Q f a set s2 and p a nonatomic probability measure on 9. Let 90be a p-pure subfield of 9 and 91the smallest u-field on SZ containing 90. Then p is nonatomic on 9 1 . (For the definition of a nonatomic measure, see Chapter 5.) Proof. Obviously, 9 1 c 9. We show that 9 and g1are p-equivalent, i.e. given A in 9 there exists B in slsuch that p (A A B) = 0. To begin with, given E > 0 we show that there exists a set B, in g1such that p (A A B) < E . Since 90is a p-pure subfield of 9, there exists a sequence En,n 2 1 in 90 such that UnZlE, 3 A and Cnzl p(E,) < p (A) + E . Clearly, UnZl E, E gl. Take B, =UnzlEn. Thus for each n 2 1, we can find B, in sl such that B,. Since A A (lim p ( A A B,) < 1/2". Take B = Iim B,) c lim sup,,oo (A A B,) and p (lim sup,+m (A A B,)) = 0 (Borel-Cantelli Lemma), it follows that p ( A A B) = 0. Finally, since 9 and 9 1 are pequivalent, p is nonatomic on S1. 0 Theorem A.6. Let 9be a a-field of subsets of a set s2 and p a nonatomic probability measure on 9. Let g o be a p-pure subfield of 9.Then p is strongly continuous on 90. Proof. This is a consequence of Theorem A5 above and Proposition 5.3.7. Theorem A.7. Let 9 be a a-field of subsets of a set SZ and p a nonatomic probability measure on 9. Let 90 be a p-pure subfield of 9. Then there exists a set A in 9of cardinality greater than or equal to the cardinality of the coniinuum c such that p (A) = 0.
Proof. By Theorem A6, p is a strongly continuous probability charge on 90. So, there exist two sets Bo and B1 in .F0 such that Bo n B 1 = 0 , O < p (Bo) < 1/1(2) and O
APPENDIX
1
275
p0300)<1/2(2*),O
u
n,,,,
n,,,,
Sierpiiiski showed with the aid of continuum hypothesis the existence of a set R, a cr-field 9 on R and a nonatomic probability measure p on 9 such that p (A) = 0 if and only if A is at most countable. In this case, in of 9. For a discussion view of Theorem A7, there is no p-pure subfield 90 on the above type of measure, see Marczewski (1953,7(iv), p. 123). In fact, continuum hypothesis is not needed at all for an example. Any non-compact measure on a countably generated cr-field would serve the purpose. See Bhaskara Rao, Bhaskara Rao and Rao (1982) and Frolik and Pachl (1973). Theorems A.l, A.2, A.3 and A.4 stated above are analogous to Proposition 2.3.2 and Theorem 2.3.4 in spirit.
CHAPTER 3 The results of Sections 3.1 and 3.2 are due to Tarski (1938) and Horn and Tarski (1948). Our treatment is slightly different from the one given in Horn and Tarski (1948). The results of Section 3.3 are due to Los and Marczewski (1949). Section 3.4 contains some ideas of Los and Marczewski (1949) in its development. Some of the results of Section 3.5 are due to Pettis (1951). For the results of Section 3.6, see Guy and Maharam (1972). Extending some of the results of this chapter to group-valued charges, Bhaskara Rao and Aversa (1982) have proved the following theorem.
Theorem A.8. Let G be an algebraically compact abelian group. If p is any G-valued charge defined on a subfield %’ of a field 9 of subsets of a set R, then there is a G-valued charge fi on 9 which extends p. Carlson and Prikry (1982) have proved that the above result is true for all groups using a result on Specker groups. For Specker groups see Fuchs (1970).
276
THEORY OF CHARGES
CHAPTER 4 The notion of a.e. [ p ] introduced in Definition 4.2.4 is slightly different from the standard one adopted in Measure Theory. For example, we say f s g a.e. [ p ] if there exists a null function h such that f s g + h . The definition prevalent in Measure Theory is that f s g a.e. [ p ] if and only if JpI*({wE R ; f ( w f > g ( w ) } ) = O . Of course, if IpI*({w ER; f ( w ) > g ( w ) } ) = O , then f s g a.e. [ p ] according to Definition 4.2.4 though the converse is not true. If a D-integrable function f has the property that D
I
f d p 2 0 for every E in 9,
E
thenf? 0 a.e. [ p ] in the sense of Definition 4.2.4 but Ipl*({w E R;f(w)
Example. Let R = {1,2,3, . . . , co}, 9 ={A c R; A or A' is a finite subset of { 1 , 2 , 3 , . . .}} and p on 9be defined by A is finite, p ( A ) = 0 if A E 9, = 1 otherwise. Let % be the smallest c+-fieldon R containing 9. Indeed, % = P(R). Let @ be the extension of p from 9 to % as a measure. In fact, for A c R, @(A)=1 if c o ~ A , =O
ifco&A.
APPENDIX
1
277
It can be observed that Ll(s1, 9, p ) and L ~ ( f la, , C;) are both complete but Ll(s1, 9,p ) f Llfs1, 8,C;). For, obviously, rim)€ Li(s1, a, @) but I { m ) a L1(sl, 9, p ) . (Observe that is not T2-measurable in the framework of
(a,9, @).I CHAPTER 5 The main decomposition theorem presented in Section 5.2 is due to . results in this chapter are due to Sobczyk and Hammer ( 1 9 4 4 ~ )Other the authors (1973) and (1978).It is possible to derive Theorem 5.2.7 using Riesz Decomposition Theorem on vector lattices. See Section 1.5.
CHAPTER 6 The treatment of absolute continuity, singularity and Lebesgue Decomposition Theorem given here follows closely that of Bochner and Phillips (1941).Darst (1962a),(1962b)and (1963) has worked extensively on extensions of Lebesgue Decomposition Theorem in the unbounded case. The following are some of his results.
Theorem A.9. Let p and u be two charges on a field 9of subsets of a set s1 such that one of p and u is bounded. Then u admits a Lebesgue Decomposition with respect to p if and only if there exists a sequence E,, n 2 1 of decreasing sets in 9 s u c h that Ip [(E,) = 0 and lim,,m IuI(E;) isfinite.
A special case of this theorem is the following result. Theorem A.lO. Let p and u be tw5 charges on a field 9of subsets of a set s1 such that u is bounded. Then u admits a Lebesgue Decomposition with respect to p. The Radon-Nikodym Theorem presented here was originally formulated and proved by Bochner (1939). Later, it was proved again by various people at various times: Bochner and Phillips (1941),Dunford and Schwartz (1964), Fefferman (1967), Darst and Green (1968), Dubins (1969), Darst (19706)and Pachl(l972). The proof presented here is due to Pachl(l972). This proof, even though it looks lengthy, is elementary and in our opinion, the simplest. Another proof of Radon-Nikodym theorem appears in Chapter 7 as a simple consequence of a result on V1 spaces. This proof is due to Bochner and Phillips (1941). Maynard (1979) gave necessary and sufficient conditions for the existence of exact Radon-Nikodym derivatives.
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THEORY OF CHARGES
Dunford and Schwartz (1964) use Stone Representation Theorem for Boolean algebras in their proof of Radon-Nikodym theorem.
CHAPTER 7 V,-spaces were introduced by Bochner (1939). Leader (1953) studied these spaces in great detail. Our presentation follows closely that of Leader (1953) with simplifications. Remark 7.5.2 follows from Fefferman (1968) and Bhaskara Rao and Halevy (1977). In the later paper, the results on V,-spaces were obtained using Stone Representation Theorem on Boolean Algebras. Green (1970/71) gave necessary and sufficient conditions for the comp ) for a given charge space (R, 9,p ) . pleteness, in particular, of 21(R, 9, Let 9, p ) be a probability charge space. For A, B in 9, say A - B if p (A A B) = 0. is an equivalence relation. Let 9*be the collection of all equivalence classes of 9. 9*is a Boolean Algebra. Let R‘ be the Stone 9’the class of all clopen subsets of R‘ and 3’the Bore1 space of 9*, cr-field on R’. There is a natural probability measure p ’ on 3’which corresponds to the probability charge p on 9via the correspondences
(a,
-
9+ 9” f* 9’3’. Green’s necessary and sufficient conditions for the completeness of
LZ1(R,9, p ) are that R‘ be extremely disconnected (i.e. the closure of every open subset of R’ is open) and every open subset of R’ is equivalent to its closure under p ’ . These conditions are not correct. An example can be constructed from the following theorem of Bhaskara Rao and Aversa (1982).
Theorem A . l l . Let p be a positive bounded measure on a field 9of subsets of a set R. Let % be the smallest a-field on R containing 9 and fi the positive bounded measure on 2l which extends p. Then
2 l ( R , 9, p )=21(R,
%9
6)
if for any E > 0 and for any A in 2l there exist B and C in 9such that B c A c C a n d p(C-B)<€. This theorem gives the following two examples.
Example 1. Let R be any set and 9be the finite-cofinite field on R. Then for any positive bounded measure p on S,2?l(R, 9, p ) is complete.
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279
Example 2. Let 9 be the field generated by all the open subsets of the g 1 ( R , 9, p) real line R. Then for any positive bounded measure p on 9, is complete.
CHAPTER 8
With the exceptions of Sections 8.5 and 8.6, the treatment presented here follows that of Phillips (1940a), Darst (1966), Seever (1968), Porcelli (1960), Brooks and Jewett (1970) and Leader (1953). In Sections 8.5 and 8.6, we have included theresultsof Bell (1979). The proof of Phillips’Lemma presented here is essentially the original proof of Phillips (1940a). The results on Nikodym theorem and Vitali-Hahn-Saks theorem use essentially the ideas of Seever (1968). We now make some observations on Theorem 8.7.3 which gives equivalent conditions for weak convergence. Condition (ii) is due to Hildebrandt (1934). Condition (iii) is due to Porcelli (1960) The proof of (iv)J(v) is due to Darst (1966). Condition (v) is due to Leader (1953). The results on weak convergence in ba(fk, F)are scatteed in the literature and we made an attempt to bring some semblance of order by unifying them in our Theorem 8.7.3. Some equivalent conditions in Theorem 8.7.3 and quite a few results in this chapter are new.
CHAPTER 9
The basic ideas in the development of refinement integrals are due to Kolmogoroff (1930). The main result (Theorem 9.2.1) in Section 9.2 is due to Keisler (1979). Our proof of this result is slightly different from the one presented by Keisler (1979) in that, we avoid using a deep result in refinement integrals due to Kolmogoroff. CHAPTER 10
The decomposition theorem presented in Section 10.2 is due to Yosida and Hewitt (1952) (and Kakutani, as Yosida and Hewitt mention). Their original proof is a working-out of the proof of Riesz Decomposition
280
THEORY OF CHARGES
a.
Theorem for the vector lattice ba(R, 9)and the normal sublattice ca(R, Some alternative proofs, extensions and generalizations of this decomposition theorem can also be found in the literature. See Ranga Rao (1958), Plachky (1971), Traynor (1972) and Huff (1973). Pure charges on Boolean algebras are characterized by Lloyd (1963). See Theorem 10.5.3.
CHAPTER 11 A substantial part of these results is from a paper of K. P. S. Bhaskara Rao (1981). Theorem 11.2.5 is due to Sobczyk and Hammer (19446, Theorem 3.3, p. 849) when p is nonnegative. They gave an example of a bounded charge on a a-field of subsets of a set R whose range is countably infinite, thus negating the validity of their Theorem 3.3 for general bounded charges on a-fields. See Sobczyk and Hammer (19446, Theorem 3.4, p. 850). This example is incorrect. Their Theorem 3.3 is true for any bounded charge. See Theorem 11.2.5.See also K. P. S. Bhaskara Rao (1981). Here is an amusing example of a real charge whose range is the set of all rational numbers. Let R = {1,2,3, . . .}, 9 the finite-cofinite field on R and p on 9 is defined by p(N=
c n1 -3
if A is finite,
neA
= - I -,n1
if A is cofinite, A c R.
noAc
The range R ( p ) of p is the set of all rational numbers. This follows from the Egyptian Fraction Theorem in Number Theory, that, every positive k l/nj, where n l , n z , . . . , n k are rational number can be written as distinct positive integers. We now give an example of a bounded charge p taking positive and negative values such that 0 is not a two-sided accumulation point of R (p ). Let R = {1,2,3, . . .}, 9the finite-cofinite field on R and p on 9is defined by 1 if A is finite, F (A) = -YA2R) = 1 - p (A'),
if A is cofinite.
is a bounded charge on 9 and 0 is not a two-sided accumulation point of R h ) . p
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281
Theorem 11.4.5 is due to Sobczyk and Hammer (1944a, Theorem 5.1, p. 843). It was also rediscovered by Maharam (1976, Theorem 2, p. 49). According to Theorem 11.4.4, if p,,, n 2 1 is a discrete sequence of 0-1 valued charges on a cT-field % of subsets of a set n, then the range R ( p ) of p (1/2")p,, is compact. Erdos (see Maharam (1976) posed the problem of whether the range R(p) of p =CnZl (1/2,,)p, is a non-Bore1 set when {pl, p 2 , . . .} is not a discrete set. The answer to this question is affirmative in some special cases as pointed out in Theorem 11.5.2. If the answer to the Erdos' problem is in the affirmative whenever {p,,: n L 1) is a dense-in-itself subset of the Stone space Y of %, then the answer for the general case is also in the affirmative. In connection with the above problem, we establish the following assertions of Maharam (1976). Let p,,, n 2 1be a sequence of 0-1 valued charges on a c+-field %. Assume that {p,, :n I1) is a dense-in-itself subset of the Stone space Y of 8. (a) If {(pl(A), p2(A), . . .): A E %} is a 7-measurable subset of (0, l}Ko, then it has .r-measure zero. (For the definition of 7 see the proof of Theorem 11.5.2.) (b) If {(pl(A), p2(A), . . .): A E has the property of Baire, then it is of first category. (1/2")p,,. If R ( p ) (a) and (b) can be interpreted as follows. Let p =Inzl is Lebesgue measurable, then it has Lebesgue measure zero. If R ( p ) has the property of Baire, then it is of first category. (a) can be established as follows. The set E = {(pl(A),pZ(A),. . .): A E %} is a subgroup of the abelian group C = {0,1}"" under coordinate addition modulo 2, and is disjoint with the set {(XI,x2, . . .) E C: xi = 0 for all but a finite number of 2s). Now, if E is 7-measurable and T(E)>0, then F = E + E = { ( x 1 + y 1 , x 2 + y 2 , . . .):(x1,x2,. . . ) E E and (y1,y2,. . .)EE} should contain a point of {(xl, x2, . . .) E C: xi = 0 for all but a finite number of i's}. See Oxtoby (1971) and Bhaskara Rao and Bhaskara Rao (1974). But F = E. This contradiction proves (a). In Theorem 11.5.2, the proof that D is not 7-measurable is essentially due to Sierpinski (1938).
CHAPTER 12 The main result of this chapter is from Maharam (1976). See also Weissacker (1982) and Talagrand (1981) for further related results.
APPENDIX 2
Selected Annotated Bibliography BOOKS To begin with, we give a list of books we have consulted at one time or the other in our study of finitely additive measures.
1. BIRKHOFF, G. “Lattice Theory” American Mathematical Society Colloquium Publications, New York, 1948. 2. DUBINS, L. E. and SAVAGE, L. J. “How to Gamble If You Must (Inequalities for Stochastic Processes)”. McGraw-Hill, London, 1965. 3. DUNFORD, N. and SCHWARTZ, J. T. “Linear Operators, Part I: General Theory”. Wiley-Interscience, London, 1954. 4. FUCHS, L. “Infinite Abelian Groups,” Vol. 1. Academic Press, London and New York, 1970. 5 . HALMOS, P. R. “Measure Theory”. Van Nostrand, London, 1950.
6. HALMOS, P. R. “Lectures on Boolean Algebras”. Van Nostrand, London, 1963. 7. KAMKE, E. “Theory of Sets”. Dover Publications, New York, 1950.
8. KELLEY, J. L. “General Topology”. Van Nostrand, London, 1955. 9. KURATOWSKI, K. “Topology”, Vol. 1. Academic Press, London and New York, 1966. 10. OXTOBY, J. C. “Measure and Category”. Springer-Verlag, New York, 1971.
11. PFANZAGL, J. and PIERLO, W. “Compact Systems of Sets”, Lecture Notes in Mathematics No. 16. Springer-Verlag, New York, 1966. 12. SCHAEFER, H. H. “Banach Lattices and Positive Operators”. SpringerVerlag, New York, 1974.
13. SIKORSKI, R. “BooIean Algebras”, Third Edition. Springer-Verlag, New York, 1969.
PAPERS We now give a list of research papers which we have come across in our quest to achieve a good understanding of the world of finitely additive measures. This list
APPENDIX 2
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is by no means exhaustive on this subject. We provide a brief description of some of the salient features of some of the papers which we think are relevant to the main theme of this book. Most of the papers contain a lot more information than the cursory annotation we provide here. ALBANO, L. (1974). Teoremi di decompozione per funzioni finitamente additive in un reticolo relativamente complementato, Ricerche Mat. 23, 63-86. Lebesgue, Jordan, Yosida-Hewitt Decomposition theorems are discussed for charges taking values in a complete vector lattice. ALEKSANDROV, I. I. (1973).The decomposition of a finitely additive set function (in Russian), Comment. Math. Univ. Carolinae 14, 87-93. Using results on vector lattices, Lebesgue Decomposition theorem for charges is proved. See Section 6.2. ALIC, M. and KRONFELD, B. (1969). A remark on finitely additive measures, Glasnik Mat., Ser. III 4(24), 197-200. The problem of embedding a charge space into a measure space is considered. See also Fefferman (1968). ANDO, T. (1961). Convergent sequences of finitely additive measures, Pacific J. Math. 11, 395-404. Vitali-Hahn-Saks theorem for sequences of charges defined on u-fields is proved. See Chapter 8. ARMSTRONG, T. and PRIKRY, K. (1978). Residual measures, Illinois J. Math.
22, 64-78. ARMSTRONG, T. and PRIKRY, K. (1981).Liapounoff’s theorem for non-atomic finitely-additive, finite-dimensional vector-valued measures, Trans. Amer. Math. SOC.266,499-514. We came across this paper at the proof-reading stage of this book. Ranges of charges defined on fields of sets is the main theme of this paper. See Chapter 11. There is some overlap of results between this paper and that of Bhaskara Rao (1981). ARMSTRONG, T. and PRIKRY, K. (1982). On the semimetric of a Boolean algebra induced by a finitely additive probability measure, Pacific J. Math. 99,
249-263.
Let p be a probability charge on a field 9 of subsets of a set n and N, the ideal of all p-null sets. On the quotient Boolean algebra 9/NW, there is a natural metric d, defined by d,([A], [B]) = p (AAB) for [A], [B] in 9/N,. The d,) is studied in detail in this paper. completion of the metric space (9/NW, See also Bhaskara Rao and Bhaskara Rao (1977). AUSTIN, D. G. (1955).An isomorphism for finitely additive measures, Proc. Amer. Math. SOC.6, 205-208. An isomorphism theorem for charge spaces analogous to the classical Halmos and von Neumann (1942) theorem for measure spaces is proved. See also Buck and Buck (1947) for a similar result.
284
THEORY OF CHARGES
BANACH, S. (1948).On measures in independent fields, StudiaMath. 10,159-177. Let (R,.9,,,EL,), a E r be a collection of probability charge spaces in which each p, is a probability measure. Let 9 be the field on R generated by is,, CY E r}.A common extension of all these probability measures to 9 as a probability measure with a special property is sought. See also Marczewski (1951). BARONE, E. (1978). Sulle misure sernplicimenten additive non continue, Atti Sem. Mat. Fix Univ. Modena 27, 39-44. An example of a nonatomic charge which is not strongly nonatomic is given. BARONE, E. and BHASKARA RAO, K. P. S. (1981). Misure di probabilita finitamente additive e continue invarianti per transforrnazioni, Boll. Un. Mat. Ital. 18, 175-184. Existence of a nonatornic probability charge invariant with respect to a transformation is discussed. BARONE, E. and BHASKARA RAO, K. P. S. (1981). PoincarC recurrence theorem for finitely additive measures, Rendiconti di Matematica 1, 521-526. The classical PoincarC recurrence theorem in Ergodic theory is discussed in the context of a charge space. BARONE, E., GIANNONE, A. and SCOZZAFAVA, R. (1980). On some aspects of the theory and applications of finitely additive probability measures, Pubbl. Istit. Mat. A p p l . Fac. Univ. Stud. Roma Quaderno 16, 43-53. Sobczyk-Hammer Decomposition theorem for charges on w-fields is proved. See Section 5.2. BAUER, H. (1955). Darstellung additiver Funktionen auf Booleschen Algebren als Mengenfunktionen, Archiv der Math. 6, 215-222. Let B* be a Boolean algebra and B a subalgebra of B*. The notion of a positive bounded charge on B being a measure relative to B* is introduced and some of the results of Yosida and Hewitt (1952) and Hewitt (1953),are generalized. BELL, W. C. (1977). A decomposition of additive set functions, Pacific J. Math. 72,305-311. Every positive bounded charge p on a field 9 of subsets of a set R can be written as a sum of positive bounded charges p l and p zon 9with the following properties. (i) p 1 and hzare mutually singular. (ii) The linear functional induced by the Lebesgue Decomposition of charges with respect to p I has a refinement integral representation. See Chapter 9. BELL, W. C. (1979).Unbounded uniformly absolutely continuous sets of measures, Proc. Amer. Math. SOC.71, 58-62. A uniformly absolutely continuous set of charges can be decomposed into bounded and finite dimensional parts. See Section 8.6. BELL, W. C. (1979). Hellinger integrals and set function derivatives, Houston J. Math. 5, 465-481.
APPENDIX 2
285
Using the concept of a refinement integral (see Chapter 9), the author introduces the notion of derivative of a bounded charge on a field 9 of sets with respect to a real valued function on 9and studies some of its properties. BELL, W. C. (1981). Approximate Hahn decompositions, uniform absolute continuity and uniform integrability, J. Math. Anal. Appl., 80, 393-405. A sequence p,,, n 2 1 of bounded charges on a field B of subsets of a set R is said to be disjoint if IpnlA (pml= 0 for all n # m. A subset G of ba(R,B) is uniformly absolutely continuous if and-only if each disjoint sequence in (&)+ is norm convergent to zero, where (G)’ is the set of positive elements in 6 = {q E baW, 9); Iq I < Ipl for some p in G}. See Theorem 8.7.7 for a related result. BELL, W. C. and KEISLER, M. (1979). A characterization of the representable Lebesgue Decomposition Projections, Pacific J. Math. 84, 185-186. Representability of the linear functional induced by the Lebesgue Decomposition of charges with respect to a fixed charge is studied. BHASKARA RAO, K. P. S. (1981). Remarks on ranges of charges, to appear in Illinois J. Math. See Chapter 11 and Armstrong and Prikry (1981). See also Notes and Comments on Chapter 11. BHASKARA RAO, K. P. S. and AVERSA, V. (1982). On Tarski’s extension theorem for group valued charges, a pre-print. See Notes and Comments on Chapter 3. See also Carlson and Prikry (1982). BHASKARA RAO, K. P. S. and AVERSA, V. (1982). A remark on E. Green’s paper “Completeness of L,-spaces over finitely additive set functions”, to appear in Coll. Math. See Notes and Comments on Chapter 7. BHASKARA RAO, K. P. S. and BHASKARA RAO, M. (1973). Charges on Boolean algebras and almost discrete spaces, Mathematika 20, 214-223. A systematic study of nonatomic, strongly continuous and strongly nonatomic charges is made. Superatomic Boolean algebras are characterized. See Chapter 5. BHASKARA RAO, K. P. S. and BHASKARA RAO, M. (1974). A category analogue of the Hewitt-Savage zero-one law, Proc. Amer. Math. SOC.44, 497-499. See Notes and Comments on Chapter 11. BHASKARA RAO, K. P. S. and BHASKARA RAO, M. (1977). Topological properties of charge algebras, Rev. Roum. Math. Pures et A p p l . 22, 363-375. Let p be a positive bounded charge on a field 9 of subsets of a set R. p induces a natural semi-metric or pseudo-metric d, on 9by d,(A, B) = p (AAB) for A, B in 9. This paper studies some topological properties of the semi-metric space (9,d,). See also Armstrong and Prikry (1982). BHASKARA RAO K. P. S. and BHASKARA RAO, M. (1978). Existence of nonatomic charges, J. Austral. Math. SOC.25 (Series A), 1-6.
286
THEORY O F CHARGES
A set of necessary and sufficient conditions for the existence of a nonatomic charge on a given Boolean algebra is provided. See Chapter 5. BHASKARA RAO, K. P. S. and BHASKARA RAO, M. (1981).On the separating number of a finite family of charges, Math. Nuchr. 101, 215-217. Given any finite number of distinct charges on a field 9 of subsets of a set R, a partition of R in 9 with minimal number of sets is sought which separates the charges. BHASKARA RAO, K. P. S., BHASKARA RAO, M. and RAO, B. V. (1982). A note on ,u-pure sub-fields, a pre-print. Let p be a probability measure on a countably generated u-field of subsets of a set 0. The following are equivalent. (i) There exists a p-pure sub-field of 9. (ii) p is perfect. (iii) p is compact. For the notions of compactness and perfectness of measures, see Ryll-Nardzewski (1953). This result was anticipated by Frolik and Pachl (1973). See also Notes and Comments on Chapter ‘ I
L.
BHASKARA RAO, M. and HALEVY, A. (1977). On Leader’s V,-spaces of finitely additive measures, J. Reine Angew. Math. 2931294, 204-216. V,-spaces (Leader (1953)) are shown to be isometrically isomorphic to L,spaces of a measure space using the Stone Representation Theorem for Boolean algebras. See Notes and Comments on Chapter 7. BOCHNER, S. (1939).Additive set functions on groups, Ann. Math. 40,769-799. V,-spaces (1s p 5 a)in the setting of charges are introduced. Radon-Nikodym theorem for charges is also proved. See Chapters 7 and 6. See also Notes and Comments on Chapters 6 and 7. BOCHNER, S. (1940). FiniteIy additive integral, Ann. Math. 41,495-504. Representation of positive linear functionals on vector lattices is provided. BOCHNER, S. (1946). Finitely additive set functions and stochastic processes, Proc. Nut. Acad. Sci., U.S.A. 32,259-261. This paper introduces a notion called stochastic phenomenon. Let P be a probability measure on a u-field ? ofIsubsets of a set S and 9 a field of subsets of a set R. A real valued function f defined on 9 x S is called a stochastic phenomenon if f(ul=, Ei, - ) = f ( E , . ) a.e. [PI for every finite number of pairwise disjoint sets El, E2,.. . , E, in 9. A stochastic phenomenon can be regarded as a general type of stochastic process and it includes many known processes. BOCHNER, S. and PHILLIPS, R. S. (1941). Additive set functions and vector lattices, Ann. Math. 42, 316-324. This is a fundamental paper on vector lattices. Riesz Decomposition Theorem in the general setting of vector lattices is proved. See Section 1.5. Lebesgue Decomposition Theorem in the setting of charges is observed. See Section 6.2. BOeDAN, V. and OBERE, R. A. (1978). Topological rings of sets and the theory of vector measures, Dissert. Math. 154. Nikodym and Vitali-Hahn-Saks type of theorems for finitely additive vector
APPENDIX 2
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measures on rings of sets are presented along the lines initiated by Drewnowski (1972a, b, c). BROOKS, J. K. (1969). On the Vitali-Hahn-Saks and Nikodym theorems, Proc. Nut. Acad. Sci., U.S.A. 64,468-471. Simplified proofs of Vitali-Hahn-Saks and Nikodym theorems for measures on cr-fields are presented. BROOKS, J. K. (1972). Weak compactness in the space of vector measures, Bull. Amer. Math. SOC., 78, 284-287. A set of necessary and sufficient conditions are given for a subset of ba(R, 9‘) to be conditionally weakly compact in a general setting. BROOKS, J. K. (1973).Equicontinuity, Absolute continuity and weak compactness in Measure Theory, a paper in “Vector and Operator valued Measures and Applications”, (D. H. Tucker and H. B. Maynard, eds). pp. 51-61. Academic Press, London and New York. Some extensions of the result in Brooks (1972) are dealt with. BROOKS, J. K. (1974).Interchange of limit theorems for finitely additive measures, Rev. Roumaine Math. Pures et A p p l . 19, 731-744. Let 9 be a field of subsets of a set R and 9*the smallest u-field on R containing 9. Let K c ba(fl, 9*). Equivalence of uniform s-boundedness of K over 9*and uniform s-boundedness of K over 9is examined. See also Brooks and Dinculeanu (1974). BROOKS, J. K. and DINCULEANU, N. (1974). Strong additivity, absolute continuity and compactness in spaces of measures, J. Math. Anal. A p p l . 45, 156-1 7 5. The notion of strong additivity of a charge studied in this paper is the same as s-boundedness we have used in this book. See Definition 2.1.4. Uniform s-boundedness of a collection of bounded charges is characterized in terms of uniform absolute continuity. See Theorem 8.7.7 for another characterization of uniform absolute continuity of a sequence of charges. BROOKS, J. K. and JEWETT, R. S. (1970). On finitely additive vector measures, Proc. Nut. Acad. Sci., U.S.A.61, 1294-1298. Vitali-Hahn-Saks and Nikodym theorems for charges on cT-fields are proved. BUCK, R. C. (1946). The measure theoretic approach to density, Arner. J. Math. 68,560-580. Density charges are constructed from simple set functions defined on a particular class of subsets of { l , 2 , 3 , . . . }. See Section 2.1. BUCK, E. F. and BUCK, R. C. (1947). A note on finitely additive measures, Amer. J. Math. 69,413-420. Isomorphism of the charge spaces (R, 9, F ) and (R’, 9:,m*), where R’= { 1 , 2 , 3 , . , . }, 9; contains all arithmetic progressions and m” is a density-like charge on 9:,is investigated.
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THEORY OF CHARGES
BUMBY, R. and ELLENTUCK, E. (1969). Finitely additive measures and the first digit problem, Fund. Math. 65, 33-42. A class S of probability charges on the power set of the set of all natural numbers is constructed such that for any p in S, p (P,) = log,, ( n + l),where P, is the set of all natural numbers whose first significant digit lies between 1 andn,lsns9. CANDELORO, D. and SACCHETTI, A. M. (1978). Su alcuni problemi relativi a misura scalari sub additive e applicazionial caso dell’additivita finita, Atti. Sem. Mat. Fis. Uniu. Modena 27,284-296. Connectedness of the range of a bounded charge is studied. CARLSON, T. and PRIKRY, K. (1982). Ranges of Signed Measures, a pre-print. Theorem A.8 is true for all abelian groups. See Notes and Comments on Chapter 3. This paper came to our notice at the proofreading stage of this book. CHENEY, C. A. and de KORVIN, A . (1976/77). The representation of linear operators on spaces of finitely additive set functions, Proc. Edinburgh Math. SOC.2(20), 233-242. An integral (Kolmogorov-Burkill type) representation of a continuous linear operator from V,(R, 9, p ) to a Banach space is provided. See also Edwards and Wayment (1974). CHERSI, F. (1978). Finitely additive invariant measures, Boll. Un. Mat. Ital. A(5) 15, 176-179. Existence of invariant probability charges is shown. See Section 2.1. CHRISTENSEN, J. P. R. (1971). Borel structures and a topological zero-one law, Math. Scand. 29, 245-255. A probability charge p on the Borel c+-field of a complete separable metric space X is a measure if p is measurable as a function on the space of all closed subsets of X equipped with a natural distance (metric) function. See Notes and Comments on Chapter 2. COBZAS, S. (1976). Hahn Decompositions of finitely additive measures, Arch. Math. 27, 620-621. Let 9 be a field of subsets of a set R. Let ba(R,9) and % ( R , F ) be as in Sections 2.2 and 4.7 respectively. ba(R, 9) is equipped with the total variation norm and %(a, 9)is equipped with the supremum norm. ba(R, 9)is the dual In this paper, it is proved that a p in b a ( R , 9 ) admits an exact of %(R, 9). Hahn decomposition if and only if p attains its norm on the unit ball of %‘(a, 9). DARST, R. B. (1961). A note on abstract integration, Trans. Amer. Math. SOC. 99,292-297. A real valued function on a set R is 9-continuous if and only iff is integrable where 9 is a field on R. See with respect to every bounded charge on 9, Section 4.7. See also Leader (1955).
APPENDIX 2
289
DARST, R. B. (1962a). A decomposition of finitely additive set functions, J. Reine Angew. Math. 210, 31-37. Lebesgue Decomposition Theorem for bounded charges is proved. See Section 6.2. DARST, R. B. (1962b). A decomposition for complete normed abelian groups with applications to spaces of additive set functions, Trans. Amer. Math. SOC. 103,549-558. A Lebesgue type decomposition theorem is proved in a general setting. Validity of Lebesgue Decomposition Theorem for unbounded charges is examined. See Section 6.2. See also Notes and Comments on Chapter 6. DARST, R. B. (1963). The Lebesgue Decomposition, Duke Math. J. 30,553-556. An extension of a result in Darst (1962b) is established. DARST, R. B. (1966). A direct proof of Porcelli’s condition for weak convergence, Proc. Amer. Math. SOC.17, 1094-1096. See Section 8.7 and Notes and Comments on Chapter 8. DARST, R. B. (1967). On a theorem of Nikodym with applications to weak convergence and von Neumann algebras, Pacific J. Math. 23, 473-477. Nikodym theorem for sequences of charges on a a-field is proved. See Section 8.4. DARST, R. B. (1970a). The Vitali-Hahn-Saks and Nikodym theorems for additive set functions, Bull. Amer. Math. SOC. 76, 1297-1298. Vitali-Hahn-Saks and Nikodym theorems are proved for charges on a-fields. See Sections 8.4 and 8.8. DARST, R. B. (1970b). The Lebesgue Decomposition, Radon-Nikodym derivative, conditional expectation and martingale convergence for lattice of sets, Pacific J. Math. 35, 581-600. The Lebesgue Decomposition Theorem and the Radon-Nikodym theorem are considered in a general setting. DARST, R. B. and GREEN, E. (1968). On a Radon-Nikodym theorem for finitely additive set functions, Pacific J. Math. 27, 255-259. Radon-Nikodym theorem for finitely additive bounded complex valued functions on a field of sets is proved. See Fefferman (1967). See also Notes and Comments on Chapter 6. DIESTEL, J. and UHL, J. J. Jr. (1977). Vector measures, American Mathematical Society Math. Surveys 15, Providence. A sharper version of Phillips’ lemma due to Rosenthal is presented. DREWNOWSKI, L. (1972a). Topological rings of sets, continuous set functions, integration I, Bull. Acad. Polon. Sci., Ser. Sci. Math. Astronom. Phys. 20, 269-276. Rings equipped with a topology such that the operations A and fl become continuous are presented. Vitali-Hahn-Saks theorem for charges taking values in a topological group is proved.
290
THEORY OF CHARGES
DREWNOWSKI, L. (1972b). Topological rings of sets, continuous set functions, integration 11, Bull. Acad. Polon. Sci., Ser. Sci. Math. Astronom. Phys. 20, 277-286. This is a continuation of the paper of Drewnowski (1972a) in which extensions of s-bounded group-valued charges on a ring of sets to the a-ring generated by the ring are sought. DREWNOWSKI, L. (1972~).Topological rings of sets, continuous set functions, integration 111, Bull. Acad. Polon. Sci., Ser. Sci. Math. Astronom. Phys. 20, 439-445. Nikodym theorem for group-valued measures is proved. DREWNOWSKI, L. (1972d). Equivalence of Brooks-Jewett, Vitali-Hahn-Saks and Nikodym Theorems, Bull. Acad. Polon. Sci., Ser. Sci. Math. Astronom. Phys. 20, 725-731. See the following paper of Drewnowski (1973). DREWNOWSKI, L. (1973). Decomposition of set functions, Studia Math., 48, 23-48. This paper and the above paper give analogues of Vitali-Hahn-Saks and Nikodym theorems for sequences of strongly bounded charges defined on o-rings of sets taking values in a commutative Hausdorff topological group. DREWNOWSKI, L. (1973a). Uniform boundedness principle for finitely additive vector measures, Bull. Acad. Polon. Sci., Ser. Sci. Math. Astronom. Phys. 21, 115-118. Nikodym theorem for s-bounded charges on a o-ring of sets taking values in a normed group is proved. DOLGUSEV, A. N. (1981). Remark on finitely additive measures, Sibirsk. Mat. 2 . 2 2 , 105-120. DUBINS, L. E. (1969). An elementary proof of Bochner’s finitely additive RadonNikodym Theorem. Amer. Math. Monthly 76, 520-523. See Notes and Comments on Chapter 6. EDWARDS, J. R. and WAYMENT, S. G. (1971). Representations for transforma154 251-265. tions continuous in the BV norm, Trans. Amer. Math. SOC. An integral representation theorem for continuous linear functionals on V,(Q 9, p ) , where a =[0,1], can be deduced using v-integrals. EDWARDS, J. R. and WAYMENT, S . G. (1974). Extensions of the v-integral, Trans. Amer. Math. SOC.191, 165-184. An integral (with respect to a charge) representation of continuous linear p ) into a Banach space can be deduced. See also Cheney operators on V,(Q 9, and de Korvin (1976/77). FAIRES, B. F. (1970). On Vitali-Hahn-Saks-Nikodym type theorems, Ann. Insti. Fourier, Grenoble 26, No. 4, 99-114. Vitali-Hahn-Saks and Nikodym type theorems are studied in the setting of
APPENDIX 2
291
Boolean algebras with interpolation property (which are same as Boolean algebras with Seever property) for Banach-space-valued s-bounded charges. See Seever (1968) and Chapter 8. FEFFERMAN, C. (1967). A Radon-Nikodym theorem for finitely additive set functions, Pacific J. Math. 23, 35-45. Radon-Nikodym theorem for bounded complex valued charges on a field of sets is proved. See also Darst and Green (1968). FEFFERMAN, C. (1968). L,-spaces over finitely additive measures, Pacific J. Math. 26, 265-271. The problem of embedding a charge space into a measure space is considered. See also AliC and Kronfeld (1969). de FINETTI, B. (1955). La Struttura delle Distribuzioni in un insieme astratto qualsiasi, Giorn. Ist. Ital. Attuari 18, 15-28. A decomposition theorem similar to the one given by Sobczyk and Hammer (1944) is proved. FROLIK, Z. and PACHL, J. (1973). Pure measures, Comment. Math. Uniu. Carolinae 14, 279-293. Properties of charges p which admit p-pure subfields of 9 are studied. Pure measures discussed here are different from pure charges studied in Chapter 10. This paper pointed out an error in M. M. Rao's (1971) paper. See Bhaskara Rao, Bhaskara Rao and Rao (1982) and also Notes and Comments on Chapter 2. GAIFMAN, H. (1964). Concerning measures on Boolean algebras, Pacific J. Math. 14,61-73. Existence of a strictly positive charge on a field 9 of subsets of a set R is related to some conditions in Set Theory. Most importantly, he exhibited a Boolean algebra satisfying countable chain condition having no strictly positive charge on it. See also Kelley (1959). GOULD, G. G. (1965). Integration over vector-valued measures, Proc. London Math. SOC.15, 193-225. Integration of scalar-valued functions with respect to vector-valued charges is developed. See Section 4.5. GRECO, G. H. (1981). The continuous measures defined on a Boolean algebra (Italian), A n n . Univ. Ferrara Ser. VII(N.S.)26, 213-218. A characterization of superatomic Boolean algebras B in terms of exact Hahn Decomposition of bounded charges on B is provided. GREEN, E. (1970/71). Completeness of L,-spaces over finitely additive set functions, Coll. Math. 22, 257-261. See Notes and Comments on Chapter 7. See also Bhaskara Rao and Aversa (1982).
292
THEORY OF CHARGES
GUY, D. L. (1961). Common extensions of finitely additive probability measures, Portugal. Math. 20, 1-5. A necessary and sufficient condition is given for the existence of a common extension of two probability charges defined on two different fields on the same set to any field containing these two fields as a probability charge. See Section 3.6. HALMOS, P. R. (1947). The set of values of a finite measure, Buff.Amer. Math. SOC. 53, 138-141. A simple proof of a result of Liapounoff on the range of a measure is given. HALMOS, P. R. (1948). The range of a vector measure, Bull. Amer. Math. Soc.
54,416-421. A simple proof of two results of Liapounoff on the range of a measure with values in a finite dimensional vector space is provided. HALMOS, P. R. and von NEUMANN, J. (1942). Operator methods in classical mechanics 11, Ann. Math. 43, 332-350. Isomorphism between two measure spaces is abstractly characterized. HATTA, L. and WAYMENT, S. G. (1973). A Radon-Nikodym theorem for the v-integral, J. Reine Angew. Math. 259, 137-146. An analogue of the classical Radon-Nikodym theorem is considered in the setting of v-integrals for charges. an der HEIDEN, U. (1978). On the representatation of linear functionals by finitely additive set functions, Arch. Math. 30, 210-214. Necessary and sufficient conditions for the existence of a charge p for a given linear functional on a Stonean lattice of functions to be expressed as an integral with respect to p are derived. HEWITT, E. (1951). A problem concerning finitely additive measures, Mat. Tidsskr. B 81-94. The structure of all bounded charges on the field 9 on Q = [0, 1) generated by all intervals of the type [a, b ) with 0 5 a I b 5 1 is determined. See Section
10.4. HEWITT, E. (1953). A note on measures on Boolean algebras, Duke Math. J. 20,
25 3-25 6. Distinction between measures on fields and measures on Boolean algebras is pointed out. See Section 10.5. HILDEBRANDT, T. H. (1934). On bounded linear functional operations, Trans. 36,868-875. Amer. Math. SOC. The dual of the Banach space of all 9-continuous functions is shown to be ba(Q, 9-),where 9is a field on Q. See Section 4.7. HILDEBRANDT, T. H. (1938). Linear operations of functions of bounded variation, Bull. Amer. Math. SOC.44, 75.
APPENDIX 2
293
Integral representation of continuous linear functionals on a subspace of is given, where R = [0, 11. ba(R, 9) HILDEBRANDT, T. H. (1940). On unconditional convergence in normed vector spaces, Bull. Amer. Math. SOC.46, 959-962. Properties of unconditional convergence in normed linear spaces are used to define some simple measures on P(R), where R = {1,2,3, . . . }. HILDEBRANDT, T. H. (1958). On a theorem in the space el of absolutely convergent sequences with applications to completely additive set functions, Math. Research Center Report No. 62 Madison, Wisconsin. HODGES, J. L. Jr. and HORN, A. (1948). On Maharam’s conditions for measure, Trans. Amer. Math. SOC.64, 594-595. One of the conditions in the set of necessary and sufficient conditions given by Maharam (1947) for a Boolean c+-algebrato admit a strictly positive bounded measure is shown to be redundant. HORN, A. and TARSKI, A. (1948). Measures on Boolean algebras, Trans. Amer. Math. SOC.64, 467497. Extension of set functions defined on a collection Y? of subsets of a set R to a field 9 on R containing %? as charges are sought. See Chapter 3. See also Notes and Comments on Chapter 3. HUFF, R. E. (1973). The Yosida-Hewitt Decomposition as an Ergodic theorem, a paper in “Vector and Operator Valued Measures And Applications”, (D. H. Tucker and H. B. Maynard, eds), pp. 133-139. Academic Press, London and New York. The Yosida-Hewitt (1952) Decomposition of a charge as a sum of a pure charge and a measure is obtained using an ergodic theorem for commutative semigroup of idempotent linear operators on a Banach space. This approach covers both the scalar valued and vector valued charges. JECH, T. and PRIKRY, K. (1979). On projections of finitely additive measures, Proc. Amer. Math. SOC.74, 161-165. There exists a translation invariant charge p on P(R), where R = {1,2,3, . . . } and a function f from R to R such that p = pf-’ and p (A)5 ; iff is one-to-one on A c R . JORSBOE, 0. G. (1966). Set transformations and Invariant measures, A Survey, Math. Inst. Aarhus Universitet Various Publications Series, No. 3, Aarhus, Denmark. Invariant charges are constructed using Banach limits. See Section 2.1. KEISLER, M. (1979). Integral representation for elements of the dual of ba(9, Z), Pacific J. Math. 83, 177-183. If 9is a superatomic Boolean algebra, then every continuous linear functional on ba(R, 5)has a refinement integral representation. See Chapter 9. See also Notes and Comments on Chapter 9.
294
THEORY OF CHARGES
KELLEY, J. L. (1959). Measures on Boolean algebras, Pacific J. Math. 9, 11651177. Necessary and suficient conditions for a Boolean algebra to admit a strictly positive charge are given. KELLEY, J. L. and SRINIVASAN, T. P. (1970/71). Pre-measures on lattices of sets, Math. Ann. 190, 233-241. Necessary and sufficient conditions are given for a positive bounded charge defined on a lattice of sets closed under countable intersections admits an extension as a measure to the cr-field generated by the lattice. KELLEY, J. L., NAYAK, M. K. and SRINIVASAN, T. P. (1973). Pre-measures on lattice of sets 11. “Proceedings of a Symposium on Vector and Operator valued measures and Applications” held at University of Utah, August 7-12, 1972, (D. H. Tucker and H. B. Maynard, eds) Academic Press, London and New York. Some improvements of the results of Kelley and Srinivasan (1970/71) are presented. KHURANA, S. S. (1978).A note on Radon-Nikodym theorem for finitely additive measures, Pacific J. Math. 74, 103-104. Radon-Nikodym theorem for charges is proved using the corresponding result for measures. The argument is essentially that of Dunford and Schwartz (1954), p. 315. KINGMAN, J. F. C. (1967). Additive set functions and the theory of probability, Proc. Camb. Phil. SOC.63, 767-775. A certain notion dense subset of a set fl in the context of a field of subsets of fl is introduced and its ramifications are studied. KISYNSKI, J. (1968). Remark on strongly additive set functions, Fund. Math. 63, 3 2 7-332. Smiley’s (1944) result on the extension of a strongly additive set function defined on a lattice of sets containing the null set to the ring generated by the lattice is reproved. See Section 3.5. LADUBA, I. (1972). Sur quelques gCnQalisations de thkorbmes de Nikodym et de Vitali-Hahn-Saks, Bull. Acad. Polon. Sci., Ser. Sci. Math. Astronom. Phys. 20,447-456. Some generalizations of Nikodym and Vitali-Hahn-Saks theorems are presented for charges on a-fields taking values in a specified space of functions. LEADER, S. (1953). The theory of L,-spaces for finitely additive set functions, Ann. Math. 58, 528-543. A systematic study of V,-spaces is presented. See Chapter 7. See also Notes and Comments on Chapter 7. LEADER, S. (1955). On universally measurable functions, Proc. Amer. Math. SOC. 6,232-234.
APPENDIX 2
295
A real valued function f on a set R is 9-continuous if and only iff is integrable with respect to every bounded charge on 9, where 9 is a field on R. See Section 4.7. See also Darst (1961). LEMBCKE, J. (1970). Konservative Abbildungen und Fortsetzung regularer Masse, 2. Wahrscheinlichkeitstheorie und Verw. Gebiete 15, 57-96. A certain order relation on the set of all real measures on a ring of sets is introduced and the maximal elements in this order are identified. LEMBCKE, J. (1972). Gemeinsame Urbilder endlich additiver Inhalte, Math. Ann. 198,239-258. LIPECKI, Z. (1971). On strongly additive set functions, Coll. Math. 22, 255-256. Another proof of a result of Smiley (1944) is presented. See Section 3.5. LIPECKI, Z. (1974). Extensions of additive set functions with values in a topological group, Bull. Acad. Polon. Sci., Ser. Sci. Math. Astronom. Phys. 22, 19-27. Extensions of group-valued charges are discussed. LIPECKI, Z. (1982). On unique extensions of positive additive set functions, a pre-print. LIPECKI, Z. (1982). Maximal-valued extensions of positive operators, a pre-print. LIPECKI, Z. (1982). Conditional and simultaneous extensions of group-valued quasi-measures, a pre-print. LIPECKI, Z., PLACHKY, D. and THOMSEN, W. (1979). Extensions of positive operators and extreme points I, Coll. Math. 42, 279-284. The result of Plachky (1976) concerning extreme points of a certain convex subsets of ba(R, F ) is generalized. Extensions of results of Jlos and Marczewski (1949) are derived in a Functional Analytic setting. LLOYD, S. P. (1963). On finitely additive set functions, Proc. Amer. Math. SOC. 14,701-704. Pure charges on Boolean algebras are characterized in terms of measures on the Stone space of the Boolean algebras. See Section 10.5. LOMNICKI, Z. and ULAM, S. (1934). Sur la thCorie de la mesure dans les espaces combinatoires et son application au calcul des probabilitks I. Variables indkpendantes, Fund. Math. 23, 237-278.
ZOS, J. and MARCZEWSKI, E. (1949). Extensions of measures, Fund. Math. 36, 267-276. The problem of extending a charge from a subfield of a field 9 of subsets of a set R to 9 as a charge is tackled. See Section 3.3. LUXEMBURG, W. A. J. (1963/64). On finitely additive measures in Boolean algebras, J. Reine Angew. Math. 213, 165-173. A special class of Boolean algebras in which every charge is a measure when restricted to some suitable ideal is studied. MAHARAM, D. (1947). An algebraic characterization of Measure algebras, Ann. Math. 48, 154-167.
296
THEORY OF CHARGES
Necessary and sufficient conditions are given for a Boolean a-algebra to admit a strictly positive bounded measure. See also Hodges and Horn (1948). MAHARAM, D. (1958). On a theorem of von Neumann, Proc. Amer. Math. SOC.
9,987-994. Lifting exists in complete measure spaces. See Chapter 12. MAHARAM, D. (1972).Consistent extensions of linear functionals and of probability measures, “Proceedings of the Sixth Berkeley Symposium on Mathematical Statistics and Probability (University of California, Berkeley, 1970/71)”, Vol. 2, Probability theory, p. 127-147, Univ. California Press, Berkeley. Let Fa,Q E r be a colIection of fields on a set R and 9 the smallest field on R containing this collection. For each a in r, let p, be a bounded charge on Fa.The existence of a charge on 9agreeing with pa on gafor every a E r is discussed. A simple case of this problem is studied in Section 3.6. MAHARAM, D. (1976). Finitely additive measures on the integers, Sankhya, Series A 38,44-59. Lifting fails to exist in the setting of charge spaces. See Chapter 12. MAHARAM, D. (1977). “Category, Boolean algebras and measures, General Topology and its relation to modern analysis and algebra”, pp. 124-135. Springer-Verlag, Berlin. MARCZEWSKI, E. (1947). Sur les mesures ti deux valeurs et les idCaux premiers dans les corps d’ensembles, Ann. SOC.Polon. Math. 19, 232-233. MARCZEWSKI, E. (1947). Two-valued measures and prime ideals in fields of sets, SOC.Sci. Lett. Varsovie C. R . Cl. ZZZ.Sci. Math. Phys. 40, 11-17. Let 9be the smallest field on [0,1] containing all sub-intervals of [0,1]. There is no non-trivial two-valued measure on 9. MARCZEWSKI, E. ~-(1947).IndCpendance d’ensembles et prolongement de mesures (Rksultats et Problkmes), Coll. Math. 1, 122-132. MARCZEWSKI, E. (1948). Ensembles d’indkpendants et leurs applications a la thkorie de la mesure, Fund.Math. 25, 13-28. MARCZEWSKI, E. (1951). Measures in almost independent fields, Fund.Math.,
38,217-229. This paper and the two papers above deal with the following problem in all its facets. Let 9-, a E r be a collection of fields on a set 0 and 9 a field on R containing all 9,’s. Let pa be a probability charge on Fa for each a in r. Is there a common extension p (with a special property) of all pa’s to 9 as a probability charge? This problem is linked with the notion of almost-independence of the fields This problem was also studied by Banach (1948) in the setting of probability measures. MARCZEWSKI, E. (1953). On Compact measures, Fund.Math. 40, 113-124. See Notes and Comments on Chapter 2. MAYNARD, H. B. (1972). A Radon-Nikodym theorem for operator valued measures, Trans. Amer. Math. SOC.173, 449-463.
APPENDIX 2
297
MAYNARD, H. B. (1979). A Radon-Nikodym theorem for finitely additive bounded measures, Pacific J. Math. 83, 401-413. Necessary and sufficient conditions for the existence of exact Radon-Nikodym derivative in the setting of charges are presented. See Section 6.3. See also Notes and Comments on Chapter 6. MOLTO, A. (1981a). On the Vitali-Hahn-Saks theorem, Proc. Roy. SOC.Edinburgh, Sec. A 90, 163-173. Boolean rings with property (f) are introduced. These include Boolean algebras with Seever property strictly. See Seever (1968) and Definition 8.4.1. Let G be a commutative Hausdorff topological group. It is proved that if wn, n 2 1 is a sequence of G-valued s-bounded charges defined on a Boolean ring with property (f), pointwise convergent and En, n 2 1 is a sequence of pairwise disjoint elements in the ring, then limp+- pn(Ep)= 0 uniformly in n. See also Faires (1976). MOLTO, A. (198 lb). On Uniform boundedness properties in exhaustive additive set function spaces, Proc. Roy. SOC.Edinburgh, Sec. A 90, 175-184. Uniform boundedness of a family of s-bounded G-valued charges defined on a Boolean ring having the property (f) is discussed. See Molto (1981a). NAYAK, M. K. and SRINIVASAN, T. P. (1975). Scalar and Vector-valued premeasures, Proc. Amer. Math. SOC.48, 391-396. Let 9 be a lattice of subsets of a set fl and 9*the smallest a-field on R containing 9. Conditions under which a charge on 9 taking values either in R or in a Banach space is extendable as a measure to 9*are presented. NAYAK, M. K. and SRINIVASAN, T. P. (1976). Vector-valued inner-measures, “Lecture Notes in Mathematics”, Vol. 541, pp. 107-1 16. Springer-Verlag, Berlin. Extension of a vector valued charge defined on a lattice of sets to the a-field generated by the lattice as a measure is discussed. See also Nayak and Srinivasan (1975). NUNKE, R. J. and SAVAGE, L. (1952). On the set of values of a nonatomic, finitely additive, finite measure, Proc. Amer. Math. SOC.3, 217-218. A nonatomic charge whose range is not convex is exhibited. See Section 11.4. OLEJeEK, V. (1977). Darboux properties of finitely additive measures on a 8-ring, Math. Slovaca 27, 195-201. An example of a nonatomic charge defined on a 6-ring which is not strongly nonatomic is given. See Definition 5.1.5, Theorem 5.1.6 and Remarks 5.1.7. OLEJCEK, V. (1981). Ultrafilters and Darboux property of finitely additive measure, Math. Slovaca 31, 263-276. The notion of an ultrafilter-atom is introduced in the setting of a charge space and some of its properties are studied. PACHL, J. (1972). An elementary proof of a Radon-Nikodym theorem for finitely additive set functions, Proc. Amer. Math. SOC.32, 225-228. See Notes and Comments on Chapter 6.
298
THEORY O F CHARGES
PACHL, J. (1972). On projective limits of probability spaces, Comment. Math. Univ. Carolinae 13, 685-691. Let p be a non-atomic probability measure on a a-field 9 of subsets of a set R. If there exists a p-pure sub-field of 9, then there is a set A in 9such that p ( A )= 0 and the cardinality of A is at least that of the continuum. See Notes and Comments on Chapter 2. PACHL, J. (1975). Every weakly compact probability is compact, Bull. Acad. Polon. Sci., Ser. Sci. Math. Astronom. Phys. 23,401-405. Let p be a probability measure on a a-field 9 of subsets of a set 0. If there is a p-pure sub-field of 9,then p is compact. See Ryll-Nardzewski (1953) for the notion of a compact measure. PETTIS, B. J. (1951). On the extension of measures, Ann. Math. 54, 186-197. Various extensions of set functions are dealt with. See Section 3.5. PHILLIPS, R. S. (1940). On linear transformations, Trans. Amer. Math. SOC.48, 5 16-54 1. Lemma 3.3 of this paper is Phillips’ lemma. See Section 8.3. PHILLIPS, R. S. (1940a). A decomposition of additive set functions, Bull. Arner. Math. SOC.46, 274-277. Let 9 be a cr-field of subsets of a set SZ and K an infinite cardinal number can be written as a not greater than the cardinal of R. Every p in ba(R, sum p l+ p z uniquely with pl,p z in ba(R, and pz vanishing on every set of cardinal %K in 9. PIERCE, R. S. (1970). Existence and uniqueness theorems for extensions of zero-dimensional compact metric spaces, Trans. Arner. Math. SOC.148, 1-21. Some comments on countable superatomic Boolean algebras are made. See Chapter 5. PLACHKY, D. (1971). Decomposition of Additive Set Functions, “Transactions of the Sixth Prague Conference on Information theory, Statistical Decision Functions, Random Processes”, pp. 715-719. Publishing House of the Czechoslovak Academy of Sciences, Prague. A general decomposition theorem is proved from which the Yosida-Hewitt Decomposition and the Lebesgue Decomposition of bounded charges follow as corollaries. PLACHKY, D. (1976).Extremal and monogenic additive set functions, Proc. Amer. Math. SOC.54, 193-196. Let 9 be a field of subsets of a set SZ and 9oa sub-field of 9. Let v be a The extreme points of the convex set of all probability probability charge on go. charges p on 9which agree with v on goare characterized. PLACHKY, D. (1980). Darboux property of measures and contents, Math. Slouaca 30, pp. 243-246. Let 9,, and g1be two cT-fields on a set such that 9,,c Let p o be a positive bounded charge on Po.Then po is strongly continuous if and only if every
APPENDIX 2
positive bounded charge strongly continuous.
p
defined on g1whose restriction to gois
299 po
is
PORCELLI, P. (1958a). On weak convergence in the space of functions of bounded variation, Math. Research Center Reports No. 39, Madison, Wisconsin. See Porcelli (1960). PORCELLI, P. (1958b). On weak convergence in the space of functions of bounded variation 11, Math. Research Center Reports, No. 68, Madison, Wisconsin. See Porcelli (1960). PORCELLI, P. (1960). Two embedding theorems with applications to weak convergence and compactness in spaces of additive type functions, J. Math. Mech. 9, 273-292. Weak convergence in ba(@ .F) is characterized using Porcelli (1958a and 1958b). See Section 8.7. See also Notes and Comments on Chapter 8. PORCELLI, P. (1966). Adjoint spaces of Abstract L,-spaces, Port. Math. 25, 105-122. V,-spaces are studied from another angle. See Chapter 7. See also Leader (1953). PTAK, V. (1969). Simultaneous extension of two functionals, Czechoslovak Math. J. 3, 553-569. The results of this paper are relevant to the problem studied by Maharam (1972). PYM, J. S. and VASUDEVA, H. L. (1975).An algebra of finitely additive measures, Studia Math. 54, 29-40. Maximal ideals in the algebra b a ( Q .F) are determined, where Cl is a discrete semigroup which is a totally ordered set with multiplication as max. RAMACHANDRAN, D. (1972). A note on finitely additive set functions, Proc. Amer. Math. SOC.31, 314-315. A counterexample is presented to a conjecture of Yosida and Hewitt (1952) concerning the correspondence between charges on a Boolean algebra and the measures on the Stone space of the Boolean algebra. RANGA RAO, R. (1958). A note on finitely additive measures, Sunkhya 19, 27-28. Another proof of the Yosida-Hewitt (1952) Decomposition of a charge as a sum of a pure charge and a measure is presented. See Chapter 10. RAO, M. M. (1971). Projective limits of Probability spaces, J. Multivariate Anal. 1, 28-57. Some conditions are given for a charge to be a measure. See Notes and Comments on Chapter 2. RICKART, C. E. (1943). Decomposition of Additive Set Functions, Duke Math. J. 10,653-665. Generalizations of a result of Phillips (1940a) are presented.
3 00
THEORY OF CHARGES
RIEFFEL, M. A. (1968). The Radon-Nikodym theorem for the Bochner integral, Trans. Amer. Math. SOC.131, 466-487. Hahn Decomposition Theorem for measures on a-fields is presented using Banach space methods. RYLL-NARDZEWSKI, C. (1953). On quasi-compact measures, Fund. Math. 40, 125-130. Perfect and compact measures are discussed. SASTRY, A. S. and SASTRY, K. P. R. (1977). Measure extensions of set functions over lattices of sets, J. Indian Math, SOC. 41, 317-330. Extension of vector-valued set functions from a lattice of sets to the ring generated by the lattice is examined. SCOZZAFAVA, R. (1978). On finitely additive probability measures, “Transactions of the Eighth Prague Conference on Information theory, Statistical Decision functions, Random Processes, (Prague 1978)”, Vol. C, pp 175-180. Reidel, Dordrecht. Let p be a strongly continuous probability charge on the power set P(n)of an infinite set a. Given 0 < a < 1, there exists a sequence F,, n 2 1of pairwise disjoint subsets of fl such that a = p(Unrl F,) = Cnrl p(F.). SCOZZAFAVA, R. (1979). Complete additivity, on suitable sequences of sets, of a simply additive and strongly nonatomic probability measure (Italian), Boll. Un. Mat. Ztal. B 5 , 16, 639-648. Sobczyk-Hammer Decomposition theorem for nonconcentrated charges p, i.e. p ( { o } )= 0 for every o in R, on the power set P(0)of R is proved. SEEVER, G. L. (1968). Measures on F-spaces, Trans. Amer. Math. SOC.133, 267-280. Nikodym and Vitali-Hahn-Saks theorems are presented for Boolean algebras having Seever property. See Sections 8.4 and 8.8. See also Notes and Comments on Chapter 8. SIERPINSKI, W. (1938). Fonctions additives non complhtement additives et fonctions non mesurables, Fund. Math. 30, 96-99. A non-Lebesgue measurable function on the unit interval [0, 11is constructed where fl ={l, 2,3,. . .}. See Notes and Comments based on a charge on P(L?), on Chapter 11. SINCLAIR, G. E. (1974). A finitely additive generalization of the FichtenholzLichtenstein theorem, Trans. Amer. Math. SOC.193,359-374. An analogue of Fubini’s theorem is established in the setting of charges. SMILEY, M. F. (1944). An extension of metric distributive lattices with an application in general analysis, Trans. Amer. Math. SOC. 56,435-447. Every strongly additive set function defined on a lattice of sets containing the empty set can be extended in a unique manner as a charge on the smallest ring containing this lattice. See Section 3.5.
APPENDIX 2
301
SOBCZYK, A. and HAMMER, P. C. (1944). A decomposition of additive set functions, Duke Math. J. 11, 839-846. Sobczyk-Hammer Decomposition Theorem is proved. See Section 5.2. See also Notes and Comments on Chapter 5. SOBCZYK, A. and HAMMER, P. C. (1944). The ranges of additive set functions, Duke Math. J. 11, 847-851. Some results on the ranges of charges are obtained. See Chapter 11. See also Notes and Comments on Chapter 11. SRINIVASAN, T. P. (1955). On extensions of measures, J. Indian Math. SOC., (N.S.)19, 31-60. Extension of measures is discussed using inner measures. STRATIGOS, P. D. (1980). Extensions of additive set functions, Serdica 6 , 197201. Extension of regular bounded charges on fields of sets generated by u-topological spaces is discussed. SUCHESTON, L. (1967). Banach limits, Amer. Math. Monthly 74, 308-311. Existence of Banach limits is shown using an old-fashioned version of the Hahn-Banach theorem. See Section 2.1. TALAGRAND, M. (1981). Non existence de relbvement pour certaines mesures finiement additives et retract& de PN, Math. Ann. 256, 63-66. Under continuum hypothesis, the author constructs a separable subset of P N - N which is not a retract of PN, where N is the set of all natural numbers with discrete topology and /3 N its Stone-Cech compactification. This example is used to show non-existence of a lifting in the setting of charges. See Maharam (1976) and Chapter 12. TARSKI, A. (1930). Une contribution 42-50.
la thkorie de la mesure, Fund. Math. 15,
TARSKI, A. (1938). Algebraische Fassung des Massproblems, Fund. Math. 31, 47-66. TARSKI, A. (1939). Ideale in Vollstandigen Mengenkorpern I, Fund. Math. 32, 45-63. Weak and strong accessibility of cardinals are discussed and existence of measures on some quotient Boolean algebras is considered. TARSKI, A. (1945). Ideale in Vollstandigen Mengenkorpern 11, Fund. Math. 33, 51-65. There exists a 0- 1 valued charge on a Boolean algebra B vanishing on all atoms of B if and only if B contains a countable set of disjoint elements. THOMSEN, W. (1978). On a Fubini-type theorem and its application in game theory, Math. Operationsforsch. Statist. Ser. Statist. 9,419-423. Sinclair's (1974) analogue of Fubini's theorem for measures in the setting of charges is generalized.
302
THEORY OF CHARGES
THOMSEN, W. (1979). The common domain of uniqueness of the products of finitely additive probability measures, “Transactions of the Eighth Prague Conference on Information Theory, Statistical Decision Functions, Random Processes, (Prague, 1978)”, Vol. C, pp. 31 1-316, Reidel, Dordrecht. Let B(X) be the Banach space of all bounded real valued functions defined on a set X, F a subset of B(X), B(X, F) the closed subspace of B(X) generated by F and B*(X) the dual of B(X). Let p be a real valued function on F such that there exists a probability charge b on P ( X ) such that p (f)= f d& for all f in F. Let U(p) = { f B(X); ~ pl(f) =p2(f) for all pi in BT(X) with pi/F = p } which is the domain uniqueness of p. It is proved that n , U ( p ) = B ( X , F ) . An extension to product spaces is also considered. TOPSOE, F. (1978). On construction of measures, “Proceedings of the Conference on Topology and Measure I (Zinnowitz 1974)”, Part 2, pp. 343-381, ErnstMaritz-Arndt Univ., Griefswald. A general result is proved in Section 8 of this paper from which the result of Smiley (1944) on the extension of strongly additive functions on a lattice of sets containing the null set to the ring generated by the lattice as a charge follows. TOPSOE, F. (1979). Approximate pavings and construction of measures, Coil. Math., 42, 377-385. A condition under which a positive bounded charge on a field of sets becomes a measure is given. See Notes and Comments on Chapter 2. TRAYNOR, T. (1972). Decomposition of Group-valued Additive Set Functions, Ann. Inst. Fourier, Grenoble 22, Part 3, 131-140. Lebesgue-type decomposition theorem is obtained for group-valued charges. TRAYNOR, T. (1972). A general Hewitt-Yosida Decomposition, Can. J. Math. 24, 1164-1169. Yosida-Hewitt (1952) Decomposition of a group-valued charge is presented using Caratheodory process. TUCKER, D. H. and WAYMENT, S. G. (1970). Absolute continuity and the Radon-Nikodym theorem, J. Reine Angew. Math. 244, 1-19. A general discussion about Radon-Nikodym Theorem in various settings is presented. TULIPANI, S. (1979). On continuous and invariant measures for a transformation (Italian), Rend. Mat. 12, 249-256. Let R be a set and T a map from R to R. Existence of a nonatomic, T-invariant probability charge on the power set P ( R ) of R is discussed.
UHL, J. J. (1967). Orlicz spaces of finitely additive set functions, Studia Math. 29, 19-58. Spaces of set functions more general than the V,-spaces (Leader (1953)) are studied. VOROB’EV, N. N. (1962). Consistent families of measures and their extensions, Theory Prob. Appl. 7 , 147-162.
303
APPENDIX 2
This paper treats the problem described in the annotation of Maharam’s (1972) paper in the setting of probability measures on cr-fields. Some combinatorial methods are used to solve the problem. WAJDA, L. (1972). Remarks on infinite products of finitely additive measures, Coll. Math. 25, 269-271. Product charge of a sequence of probability charge spaces is shown to exist. WALKER, H. D. (1975). Uniformly additive families of measures, Bull. Math. SOC.Sci. Math. R. S. Roumanie (N.S.)18,217-222. If K is uniformly Let 9 be a field of subsets of a set 0 and K c ba(0, s-bounded and pointwise bounded, then K is a bounded subset of ba(Cl, 9). For some more results in this direction, see Section 8.5. See also Brooks (1974).
a.
WEBER, H. (1982). Unabhangige Topologien, Zerlegung von Ringtopologien, Math. 2. 180, 379-393. WEBER, H. (1982). Vergleich monotoner Ringtopologien und absolute Stetigkeit von Inhalten, Comment. Math. Univ. St. Pauli 31,49-60. WEBER, H. (1982). Die atomare Struktur topologischer Boolescher Ringe und s-beschrankter Inhalte, a pre-print. WEBER, H. (1982). Der Verband der s-beschrankter monotoner Ringtopologien und Zerlegung s-beschrankter Inhalte, a preprint. WEBER, H. and VOLKMER, H. (1982). Der Wertebereich atomloser Inhalte, a pre-print. WEISSACKER, H. U. (1982). The non-existence of liftings for arithmetical density, a pre-print. The argument presented in Maharam’s (1976) paper is clearly explained. See Chapter 12. WILHELM, M. (1976). Existence of additive functionals on semi-groups and the von Neumann minimax theorem Coll. Math. 35,267-274. A general result which may be considered as a common generalization of a result on charges due to Kelley (1959) and of the von Neumann minimax theorem in Game theory is presented. WOODBURY, M. A. (1950). A decomposition theorem for finitely additive set functions, Abstract presented in Bull. Amer. Math. SOC.56, 171. A forerunner of Yosida-Hewitt (1952) Decomposition Theorem was announced. YASUMOTO, M. (1979). Finitely additive measures on N, Proc. Japan Acad. 55, Ser. A, 81-84. An improved version of a theorem of Jech and Prikry (1979) is established.
304
THEORY OF CHARGES
YOSIDA, K. (1941). Vector lattices and additive set functions, Proc. Imp. Acad. Tokyo 17,228-232. ba(R, 9)is studied from the point of view as a vector lattice. YOSIDA, K. and HEWITT, E. (1952). Finitely additive measures, Trans. Arner. Math. SOC.72,46-66. Yosida-Hewitt Decomposition of a charge into a pure charge and a measure is presented. See Chapter 10.
APPENDIX 3
Some Set Theoretic Nomenclature
1. Empty set or null set is denoted by 0. 2. The symbol R is used to denote an “abstract space” or “whole space” or “master set” which is a nonempty set of elements. The members of R are denoted generically by w . The sets in a collection of sets we consider are usually subsets of R. 3 . Membership. If w is a member of a set El we use the notation ME. If a set E is a member of a collection of sets d , we use the symbol M . 4. Inclusion. For any two sets E and F, EcF indicates that E is a subset of F, i.e. every member of E is a member of F. 5 . Union. If {Ea;a E r} is a nonempty collection of sets, we denote the E, and is defined to be the the set {w ;w E E, union of these sets by UaEr for some a in r}. 6 . Intersection. If {E, ; a E r}is a nonempty collection of sets, the intersection of these sets is denoted by neGrEa and is defined to be the set {w ; w i E, for every a in r}. 7. Difference. If E and F are any two sets, the difference of E and F is denoted by E-F and is defined to be the set {w ;w E E and w & F}. 8 . Complement. If E is any subset of R, the complement of E is denoted and is defined to be the set R - E. by 9. Symmetric difference. If E and F are any two sets, the symmetric difference of E and F is denoted by E A F and is defined to be the set (E-F)u (F-E).
Index of Symbols and Function Spaces
Function Spaces =The space of ba(C q
all bounded charges defined on the field
n. 43 space of all bounded charges defined on the u-field 9l of subsets of R vanishing on the cr-ideal9 in 9l. 140 =The space of all essentially bounded real valued functions defined on R. 90 =The space of all bounded measures defined on the field 9of subsets of R. 50 =The space of all bounded measures defined on the field 9 of subsets of R, when 9 is viewed as a Boolean algebra. 248 133 =The space of all 9-continuous functions defined on R. =The space of all real valued functions defined on R. 88 (This space is topologized in such a way that convergence in this space is precisely equivalent to hazy convergence, see p. 94.) = Thespace of all equivalence classes of B(R, 9, p ) formed under the equivalence relation f - g iff = g a.e. [ p ] . 90 =The space of all TI-measurable functions defined on R 121, 178 such that Ifl” is D-integrable, 1s p < CD. =The space of all equivalence classes of L,(R, 9, p) formed under the equivalence relation f g iff = g a.e. [P I. 178 = The space of all essentially bounded TI-measurable functions defined on 0. 122,178 =The space of all essentially bounded measurable functions defined on 0, where 9l is a u-field on R and 9 is a cr-ideal in ‘3. 137 =The space of all equivalence classes of L,(R, 9) formed under the equivalence relation f - g if f - g is a null function. 138 =The space of all bounded pure charges defined on the field 9 of subsets of a. 248 = The space of all simple functions. 101 =The space of all simple charges defined on the field 9 of subsets of R. 188 9of subsets of
ba(R, %,9)
B(R, 9, p) caW, .%
G(R,9) Y(R9 q C(R, 9, p)
= The
-
a,
INDEX OF SYMBOLS A N D FUNCT!ON SPACES
Sim(R, 9, p)
V,(n, 9, p)
(a,R P I
=The space of all D-integrable simple functions defined on R. =The space of all bounded charges A on 9 absolutely continuous with respect to p and satisfying IlA 1, < co. : Charge space, i.e. p is a charge on the field 9of subsets of a.
307 132 185 87
Operations on Boolean Algebras B Bl9
: Boolean Algebra : Quotient Boolean algebra
18 20
Operations on Charges
AB WP
=The collection of all bounded charges on 9 absolutely continuous with respect to A = a l v l + a2v2+ * ' +a,v,, where a l , a2,. . . , a , are real numbers and vl, v2, . . . , v, are 0 - 1 valued charges on 9. : A,(A)=A(AflB), A E ~BE9fixed. , = maxlsis, p (Fi), P = {Fl,F2, . . . ,F,} is a partition of R in
9. = Positive variation o f j . =Negative variation of-p.
=Total variation of p . = Outer charge induced by p. = Semi-variation
of
p.
: See Definition 3.2.6. : See Definition 3.2.6. : See Definition 2.5.2. : See Definition 2.5.2. : p is absolutely continuous with respect to u. : p is weakly absolutely continuous with respect to v. : p is strongly absolutely continuous with respect to u. : p and Y are singular. : p and v are strongly singular.
216 180 145 45 45 45 86 206 66 66 52 52 159 159 159 164 164
Operations on Functions f' fIf I f vg fAg f = g a.e. f s g a.e.
=Positive part off. =Negative part of f. = Modulus of f. = Maximum of f and g. = Minimum off and g. : See Definition 4.2.4. : See Definition 4.2.4.
11 11 11 11 11 88 88
308
INDEX OF SYMBOLS AND FUNCTION SPACES
I*
= Indicator function of the set A.
O(f, F)
12 134
: See the proof of Theorem 4.7.3.
Operations on Sets =A
=A',the complement of A. = Closure of A. =Interior of A. =Number of points in the set. =The class of all subsets of R. =The collection of all finite partitions of R in 9. = Thecollectionof all finitepartitionsof F i n F f o r F i n F . : Equivalence relation on a set. : Partial order on a set. : Relation directing a set. : The cardinality of the continuum.
6 6 15 15 41 3 15 15 14 13 13 192
Operations in Vector Lattices : Lattice supremum of x and y. : Lattice infimum of x and y. : Positive part of x. : Negative part of x. : Modulus of x. : x and y are orthogonal. : The orthogonal complement of S.
X"Y X)Y
X X
-
I
Ix xl Y
S'
24 24 24 24 24 24 29
Miscellaneous Symbols ,1
II II *
Il*IIP,
1s p
Dlf d& Slf d&
JfP a=6
{O, 1F"
39 : The space of all bounded sequences of real numbers. 33 : Norm on a linear space. : Normson2p(R,9, p)-spacesoronV,(R, 9, &)-spaces.121, 122, : See Definition 4.4.11. : See Definition 4.5.5. : Refinement Integral of f with respect to p. : a and 6 are numbers satisfying la -61 5 1. : The space of all sequences of 0's and 1's.
178,180,183 104 116 23 1 270 17
INDEX A Accumulation point, 15 of a sequence of charges, 265 Additive-class, 2 Additivity uniform, 226 uniform countable, 204 Antisymmetric relation, 13 Atom of a Boolean algebra, 22 of a charge (p-atom), 141 of a field, 7 Axiom of choice, 14 B Bake category theorem, 17, 267 property of, 17 o-field, 17 Banach lattice, 34 limit, 39 space, 33 Base topological, 16 filter, 134 Boolean algebra, 18 atomic, 22 complete, 19 nonatomic, 22 pairwise disjoint elements in a, 21 quotient, 20 Boolean algebras atomic, 22 homomorphism between, 19 isomorphic, 19 isomorphism between, 19
Stone representation theorem for, 20 Boolean a-algebra, 19 Borel-Cantelli lemma, 274 Bore1 a-field on R,12 C Cantor set, 17, 265 Caratheodory extension theorem, 81 measure, 274 Cardinal number, 14 Cartesian product space, 13 Cauchy-Schwartz inequality, 123 Cauchy sequence, 16 weak, 33 Chain, 13 Chain condition countable, 21, 211 Charge, 35 atomic, 213 bounded, 35 convex function with respect to a, 238 density, 41 finitely many valued, 249 general invariant, 41 infinitely many valued, 245 modular, 36, 60 negative variation of a, 45 nonatomic, 141 0-a valued, 35 outer, 86 positive, 35 positive bounded, 35 positive real partial, 64 positive variation of a, 45 probability, 35
310 Charge (cont.) pure, 240 range of a, 249 real, 35 real partial, 64 s-bounded, 41 shift-invariant, 39 simple, 188 strongly continuous, 142 strongly nonatomic, 142 total variation of a, 45 unbounded, 42 Charge space, 87 probability, 179 complete, 265 Chebychev’s inequality, 127 Class additive-, 2 compact, 49,245 equivalence, 14 monocompact, 273 Clopen set, 16 Closed under complementation, 5, 6 under countable intersections, 3 under countable unions, 2 under differences, 3 under finite disjoint unions, 5 under finite intersections, 3 under finite unions, 3 under proper differences, 3 under symmetric differences, 3 Closure of a set, 15 Cofinite set, 3 Compact class, 49, 245 Compact topological space, 16 Condition countable chain, 21, 211 Continuous function, 16 9-,133 Continuity absolute, 99, 159 strong absolute, 159 uniform absolute, 127, 204 weak absolute, 159 Convergence hazy, 99 in measure, 92 of a net, 15 Convergence theorem
INDEX
dominated, 88 Lebesgue dominated, 131 Convex function with respect to a charge, 238 Cover open, 16 sub-, 16
D D-integral, 96 Decomposition E-Hahn, 56 exact Hahn, 57 Decomposition theorem general Jordan, 52 Hahn, 56 Jordan, 52 Lebesgue, 168 Riesz, 29 Sobczyk-Hammer, 146 Yosida-Hewitt, 240, 241 Decomposition theorem for measures on cr- fields Hahn, 165 Jordan, 56 Dense-in-itself set, 16, 251 Dense set, 17 Density charge, 41 Derived set, 236 Determining sequence, 104 Directed set, 13 Dominated convergence theorem, 88 Lebesgue, 131 Dual space, 33 Dual of V,-space, 193
E E-Hahn decomposition, 56 Egyptian fraction theorem, 280 Equivalence class, 14 Equivalence relation, 14 Essential boundedness, 89 Exact Hahn decomposition, 57 Exhaustion, principle of, 143 Extension theorem Caratheodory, 81 Extremely disconnected topological space, 278
311
INDEX
F 9-continuous function, 133 F,-set, 15 Field, 2 atomic, 8 discrete, 3 finite-cofinite, 49 p-pure sub-, 274 nonatomic, 8 superatomic, 151 Filter base, 134 in a Boolean algebra, 19 in a field, 10 Finite-cofinite field, 49 Finite dimensional set, 216 Finite intersection property, 16 Finite partition, 8, 14 Finitely disjoint sequence of charges, 144 Finitely many valued charge, 249 First category set, 17 Function continuous, 16 %-continuous, 133 indicator, 12 @-measurable, 91 measurable, 12 modular, 61 null, 88 simple, 90 smooth, 91 strongly additive, 61 T1-measurable, 101 T2-measurable, 101 Functional induced by a real valued set function, 59 linear, 31 Functions equal almost everywhere, 88 G G8-set, 15 General invariant charge, 41 General Jordan decomposition theorem, 53 Generator, 4
H Hahn-Banach theorem, 32
Hahn decomposition E - , 56 exact, 57 Hahn decomposition theorem, 56 for measures, 165 Hamel basis, 32 HausdorfT topological space, 16 Hazy convergence, 92 Holder’s inequality, 122 Homomorphism between Boolean algebras, 19
I Ideal in a Boolean algebra, 19 in a field, 10 m-,137 Image of a set under a map, 16 Indicator function, 12 Inequality Cauchy-Schwartz, 123 Chebychev’s, 127 Holder’s, 122 Minkowski’s, 124 Infinitely disjoint sequence of charges, 145,258 Infinitely many valued charge, 249 Integral D-, 96 lower, 116 refinement, 231 S-, 116 upper, 116 Interior of a set, 15 Invariant charge general, 41 shift-, 39 Isolated point, 15 Isomorphic Boolean algebras, 19 Isomorphism between Boolean algebras, 19 J Jordan decomposition theorem, 52 for measures on (+-fields,56
K Kolmogorov’s Zero-One law, 265
312
INDEX
L L,-space, 121, 178 Lattice Banach, 34 boundedly complete vector, 29 modulus of an element in a vector, 24 negative part of an element in a vector, 24 normal sub-, 28 normed vector, 34 of sets, 1 orthogonal complement of a subset of a vector, 29 orthogonal elements in a vector, 24 positive part of an element in a vector, 24 sub-, 28 vector, 24 Lebesgue decomposition theorem, 168 dominated convergence theorem, 131 measurable set, 264 measure, 49 Lifting, 268 Limit Banach, 39 infimum, 11 supremum, 11 Linear functional, 31 Linear order, 13 Linear space complete pseudo-normed, 33 normed, 33 Linearly independent set, 32 Linearly ordered set, 13 Lower integral, 116 Lower sum, 115
M p a t o m , 141 p-measurable function, 91 p-null set, 87 @-puresub-field, 274 Maximal filter in a Boolean algebra, 19 in a field, 10 Maximal ideal in a Boolean algebra, 19 in a field, 10
Measurable function, 12 p-7 91 Ti-, 101 T2-, 101 Measurable set Lebesgue* 264 7 - 9 264 Measure, 47 bounded, 47 Caratheodory, 274 convergence in, 92 Lebesgue, 49 nonatomic, 141 positive, 47 product probability, 265 real, 41 Metric pre-compact, 273 Metric space, 16 pseudo-, 16 Minkowski’s 124 Modular charge, 36, 60 Modular function, 61 Modulus of an element in a vector latt i e , 24 Monocompact class, 273
N Negative part of an element in a vector lattice, 24 Negative variation of a charge, 45 Net, 15 convergence of a, 15 sub-, 15 weakly convergent, 33 Nikodym theorem, 204 Nonatomic charge, 141 Nonatomic Boolean algebra, 22 Nonatomic field, 8 Nonatomic measure, 141 Norm, 33 pseudo-, 33 Norm bounded set, 33 Normal sub-lattice, 28 Normed linear space, 33 Normed vector lattice, 34 Nowhere dense set, 17 Null function, 88
313
INDEX
Number cardinal, 14 ordinal, 14
0 0-a valued charge, 35
Open cover, 15 Open set, 15 Order linear, 13 partial, 13 Ordered set linearly, 13 partially, 13 well-, 13 Ordered vector space, 23 Ordinal number, 14 Orthogonal complement of a subset of a vector lattice, 29 Orthogonal elements in a vector lattice, 24 Outer charge, 87 Oxtoby’s category analogue of Kolmogorov’s zero-one law, 265
P Pairwise disjoint elements in a Boolean algebra, 21 Partial order, 13 Partially ordered set, 13 Partition finite, 8, 14 refinement of a, 15 Perfect set, 16 Phillips’ lemma, 206 Polish space, 273 Positive bounded charge, 35 Positive charge, 35 Positive measure, 47 Positive part of an element in a vector lattice, 24 Positive real partial charge, 64 Positive variation of a charge, 45 Power set, 3 Pre-compact metric, 273 Principle of exhaustion, 143 Probability charge, 35 Probability charge space, 179 Product probability measure, 265
Property of Baire, 17, 264 Pseudo-metric space, 16 complete, 16 completion of a, 17 Pseudo-norm, 33 Pure charge, 240
Q Quotient Boolean algebra, 20
R Radon-Nikodym theorem, 174,191 Range of a charge, 249 Real charge, 35 Real measure, 47 Real partial charge, 64 Refinement integral, 231 Refinement of a partition, 15 Reflexive relation, 13 Relation antisymmetric, 13 equivalence, 14 reflexive, 13 symmetric, 13 transitive, 13 Relative topology, 16 Riesz decomposition theorem, 29 Riesz representation theorem, 136 Ring, 2
S s-bounded charge, 41 S-integral, 116 Scattered set, 16,236 Seever property, 210 Semi-field, 2 Semi-ring, 1 Semi-variation, 206 Sequence Cauchy, 16 determining, 104 weak Cauchy, 33 Sequence of charges accumulation point of a, 265 discrete, 258 finitely disjoint, 144 infinitely disjoint, 145, 258 Set Cantor, 17,265 clopen, 16
314
INDEX
Set (cont.) closed, 15 closure of a, 15 cofinite, 3 dense, 17 dense-in-itself, 16, 251 derived, 236 Fu-,15 finite dimensional, 216 first category, 17 Gs-, 15 image of a, 16 interior of a, 15 Lebesgue measurable, 264 linearly independent, 32 linearly ordered, 13 p-null, 87 norm bounded, 33 nowhere dense, 17 open, 15 partially ordered, 13 perfect, 16 scattered, 16, 236 7-measurable, 264 tail, 265 weakly closed, 33 well-ordered, 13 with the property of Baire, 17, 264 a-additivity across a sequence of sets, 253 a-class, 2 a-field, 2 Baire, 17 Borel, 12, 17 discrete, 3 a-ideal, 137 a-ring, 2 Shift-invariant charge, 39 Simple charge, 188 Simple function, 90 Singularity, 164 strong, 164 Smooth function, 91 Sobczyk-Hammer decomposition theorem, 146 Space Banach, 33 Cartesian product, 13 charge, 87 compact topological, 16
complete charge, 264 complete pseudo-metric, 16 completion of a pseudo-metric, 17 dual, 33 extremely disconnected topological, 278 Hausdorff topological, 16 Lp-, 121, 178 metric, 16 normed linear, 33 ordered vector, 23 Polish, 273 probability charge, 179 pseudo-metric, 16 Stone, 21 topological, 15 totally disconnected topological, 16 VP-, 185 vector, 31 weakly complete, 33 Stone representation theorem for Boolean algebras, 20 Stone space, 21 Strong absolute continuity, 159 Strong singularity, 164 Strong topology, 33 Strongly additive function, 61 Strongly continuous charge, 142 Strongly nonatomic charge, 142 Subcover, 16 Sub-field, p-pure, 247 Sublattice, 28 normal, 28 Subnet, 15 Sum lower, 115 upper, 115 Superatomic field, 151 Symmetric relation, 13
T TI-measurable function, 101 T,-measurable function, 101 7-measurable set, 264 Tail set, 265 Theorem Baire, 17, 267 Caratheodory extension, 81
315
INDEX
Theorem (cont.) dominated convergence, 88 Egyptian fraction, 280 general Jordan decomposition, 53 Hahn-Banach, 32 Hahn decomposition, 56, 165 Jordan decomposition, 52 Lebesgue decomposition, 168 Lebesgue dominated convergence, 131 Nikodym, 204 Radon-Nikodym, 174, 191 Riesz decomposition, 29 Riesz representation, 136 Sobczyk-Hammer decomposition, 146 Stone representation, 20 Vitali-Hahn-Saks, 204 Yosida-Hewitt decomposition, 240, 24 1 Topological space, 15 compact, 16 extremely disconnected, 278 Hausdorff, 16 totally disconnected, 16 Topology relative, 16 strong, 33 weak, 33 weak*, 158 Total variation of a charge, 45 Totally disconnected topological space, 16 Transfinite induction, 14 Transitive relation, 13 Tree, 150 U Uniform absolute continuity, 127, 204 Uniform additivity, 226 Uniform countable additivity, 204 Upper integral, 116
Upper sum, 115 Urysohn’s lemma, 17
V V,-space, 185 Vector lattice, 24 boundedly complete, 29 modulus of an element in a, 24 negative part of an element in a, 24 normed, 34 orthogonal complement of a subset of a, 29 orthogonal elements in a, 24 positive part of an element in a, 24 . Vector space ordered, 23 over the real line R, 31 over the field of rational numbers, 31 Vitali-Hahn-Saks theorem, 204 W Weak absolute continuity, 159 Weak Cauchy sequence, 33 Weak topology, 33 Weak* topology, 158 Weakly closed set, 33 Weakly complete space, 34 Weakly convergent net, 33 Well-ordered set, 13 Well-ordering, 13
Y Yosida-Hewitt decomposition theorem, 240,241 Z Zero-one law Kolmogorov’s, 265 Oxtoby’s category analogue of Kolmogorov’s, 265 Zorn’s lemma, 14
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Columbia University, New Y o r k
RECENT TITLES
CARLL. DEVITO.Functional Analysis Formal Groups and Applications MICHIELHAZEWINKEL. SICURDUR HELCASON. Differential Geometry, Lie Groups, and Symmetric Spaces ROBERT B. BURCKEL. An Introduction to Classical Complex Analysis : Volume 1 JOSEPH J. ROTMAN. An Introduction to Homological Algebra C. TRUESDELL A N D R. G. MUNCASTER. Fundamentals of Maxwell’s Kinetic Theory of a Simple Monatomic Gas : Treated as a Branch of Rational Mechanics BARRY SIMON. Functional Integration and Quantum Physics A N D ARTOSALOMAA. The Mathematical Theory of L Systems. GRZEGORZ ROZENBERG DAVIDKINDERLEHRER and GUIDOSTAMPACCHIA. An Introduction t o Variational Inequalities and Their Applications. H. SEIFERT AND W. THRELFALL. A Textbook of Topology; H. SEIFERT. Topology of 3-Dimensional Fibered Spaces LOUISHALLE ROWEN. F’olynominal Identities in Ring Theory DONALD W. KAHN.Introduction to Global Analysis DRAGOS M. CVETKOVIC, MICHAELDOOB,A N D HORSTSACHS. Spectra of Graphs ROBERTM. YOUNG.An Introduction to Nonharmonic Fourier Series MICHAELC. IRWIN. Smooth Dynamical Systems JOHN B. GARNETT. Bounded Analytic Functions EDUARD PRUGOVEEKI. Quantum Mechanics in Hilbert Space, Second Edition M. SCOTT OSBORNE A N D GARTHWARNER. The Theory of Eisenstein Systems K. A. ZHEVLAKOV, A. M. SLIN’KO,I. P. SHESTAKOV, A N D A. I. SHIRSHOV. Translated by HARRY SMITH. Rings That Are Nearly Associative JEAN D I E U W N NA ~ . Panorama of Pure Mathematics ; Translated by I. Macdonald JOSEPH G. ROSENSTEIN. Linear Orderings AVRAHAM FEINTUCH AND RICHARD SAEKS. System Theory : A Hilbert Space Approach ULF GRENANDER. Mathematical Experiments on the Computer HOWARD OSBORN.Vector Bundles : Volume I, Foundations and Stiefel-Whitney Classes BHASKARA RAo A N D BHASKARA RAo. Theory of Charges IN PREPARATION
ROBERT B. BURCKEL. An Introduction to Classical Complex Analysis : Volume 2 RICHARD V. KADISONA N D J O H N R. RINGROSE. Fundamentals of the Theory of Operator Algebras, Volume 1, Volume 2 BARRETT O’NEILL.Semi-Riemannian Geometry : With Applications to Relativity EDWARD B. MANOUKIAN. Renormalization E. J. MCSHANE.Unified Integration A. P. MORSE. A Theory of Sets, Revised and Enlarged Edition
