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36
THEORY OF CHARGES
additive-classes of sets or Boolean algebras. In fact, there are some situations in the subsequent chapters where we deal with charges on these general domains. If p is a charge on a field 9 of subsets of a set a,p cannot take both the values 1-00 and -00. For, if A, B in 9 are such that p (A) = 00 and p (B) = -03, then p (a) = p (A) p (A")= 00 = p (B) p (B') = -a,a contradiction. The following proposition gives some properties of charges. In this proposition, p stands for a charge on a given field 9 of subsets of a set a.
+
2.1.2 (i). then
+
Proposition If F1, FZ, . . . ,F, are any finite number of pairwise disjoint sets in F,
(ii). If E , F E E E c F and -co
p (An+l)for every n 2 1.Take Z, = A, - A,+1, n 2 1. Now, we construct the desired sequence W,, n 2 1. Let Y1 = UnzlZ,. Write { 1 , 2 , 3 , . . .) as a disjoint union of two sets N1 and NZ such that both Zn )5 +p(Y 1 )or p(UnENz Zn)5;p(Y1). are infinite. Then either p(UnENl Let YZ= UnEN1 Z, if p (UncNI Z,) 5 +p (Yl), and Yz = UnENz Z, otherwise. Adopting the above procedure for Yz, we can obtain a set Y3 in 2l such that Y 3 c Yz, Y3 is an infinite union of sets from {Z,, n 2 1) and p ( Y 3 ) s i p (YZ). Continuing this procedure, we obtain a sequence Y1 3 YZ3 Y3 3 such that 0 < p (Y,) 5 i p (Yn-l) for every n 2 2 and each Y, is an infinite 9
< l / n (2”) for all il , iz, . . . ,in,in+lin (0, 1) and n 2 1. For each n 5 1, let A,, = Bil,i2 ....,in, where the union is taken over all il, iz, . . . , i, in (0,l). It is obvious that p (A,,) < l / n for every n z 1. Let A,,. It now follows that p(A) = 0. Since 9,is,a p-pure subfield A= of 9, it follows that Bil.iz,..., # 0 for any sequence i l , i 2 , . . . of 0’s and 1’s. (Actually, here, we use the fact that gosatisfies (i) on p. 274 and that Bil,i2,...,in’~ satisfy (iii) above.) Consequently, the cardinality of A L c.
P ( F o ) C~ PLFi). i=l
In particular,
(v). If F E F u n d p ( F ) < m , thenp(E)
C
?I21
p(E,)s~@).
(ix). p is modular, i.e. p (E)+ p (F)= p (E u F) + p (E nF) for any E and F in 9. then (x). If F1, Fz, . . . ,F, are any finite number of sets in 9,
2.
37
CHARGES
Proof. (i) follows by induction. (ii) and (iii) are easy to prove. n-1 (iv). Let El = FI, E2= F2 - F1, E3 = F3- (F1u F2), . . . , En = F, - (Ui=,F;). Then El, Ez, . . . ,En are pairwise disjoint, Ei c F i for every i = 1,2,. ,n and Fi = Ei. Consequently,
uysl u:='=,.
P ( F o ) ~ P ( G F ~ ) = P ( G Ei ~ = l )p(Ei) = I
C
i=l
pLFi).
(v), (vi) and (vii) are easy exercises. E i c E , and SO ELl p ( E i ) = @ ( U z E 1 i)I (viii). For every m 21, p(E). Hence Cizl p ( E i ) I p ( E ) (ix). Note that p(E)+ p (F)= [ p (E n F) + p (E-F)] + [ p ( E n F) + p (F-E)]. On the other hand, p ( E u F) + p (E n F) = p(E-F) + p (F-E) + p ( E n F) + p (E nF). Hence (ix) follows. (x). This is a generalization of (ix) and can be proved by induction. The result is true for n = 2 by (ix). Suppose the result is true for n = m. We prove the result for n = m + 1. Let F1, F2, . . . ,Fmcl be any m + 1 sets from 9. Then
uEl
m+l
2 i=l
m
p(Fi)= C p(Fi)+F(Frn+l) i=l
i<j
+p(c
~ i ) ]
+tL(Fm+1)
i<j
(by induction hypothesis)
(u:,
(by using the induction hypothesis for the sum p n = 2). Continuing this way, we get the desired equality.
Fi)+ p (F,+I) with
W e give some examples of charges.
2.1.3 Examples. (1). Let R be the set of all rational numbers in [0, 1) and 9= {Uy=,[ai,bi) n a;[ai, bi)n [ a , b j )= 0 for i # j , 0 I ai I bi I 1 for every i,
38
THEORY OF CHARGES
ai, bi rational for every i and n L 1). 9 is a field on R. For any set described above. let
Then p is a positive bounded charge on 9.
.
(2). Let 0 = {1,2,3, , .), 9 ={A c R; A or A" is finite}, and p on 9 be defined by if A is finite and has n elements, p(A) = n, = -p (A'), if A" is finite.
Then p is a real charge on 9 but not bounded. (3). Let 9 be a field of subsets of a set R and 4 a maximal ideal in 9. Define p on 9by p(F)=O, ifFE.9, = 1, if Fa$, F E ~ . Then p is a charge on 9. Conversely, if p is a 0-1 valued charge on 9,then 4= {F E 9; p (F)= 0) is a maximal ideal in 9and 8; = {EE 9; p (E) = 1) is a maximal filter in 9. Consequently, the Stone space of 9 can be identified as the collection of all 0-1 valued charges on 9.See Theorem 1.4.6.
(4). Let SZ={l, 2 , 3 , . . .}. There is a 0-1 valued charge p on P(R) such that p (A) = 0 if A is a finite subset of SZ. This can be shown as follows. Let = {A c SZ; A is finite}. 4l is an ideal in 9(R). There is a maximal ideal 4 in P(R) containing 4 1 . (See Section 1.2.) Define p on P(SZ)by p(A)=O, if A E ~ ,
=1, ifAa4. p
is the desired 0-1 valued charge on P(SZ).
(5). Let R = [0,1) and 9 the collection of all sets each of which is a finite disjoint union of intervals of the type [a, 6) with O s a 5 b 5 1. Let cp be any real valued function defined on [0,1]. Define pQ on 9 as follows. For O s a 5 6 ~ 1 pQ([a,b))=cp(b)-cp(a). , For any F in 9, pQ(F)is defined additively. Then pQ is a real charge on 9.
(6). A special case of the above is given by the function cp defined as follows. cp (x) = 0,
= n,
if x is irrational, x E [0, 11,
if x is rational, x = m/n, m and n are mutually prime integers, x
=0, ifx =O.
E (0,1],
2.
39
CHARGES
In this example, p+, is unbounded on every non-empty set in 9, i.e. Sup {lp (E)I; E c F, E E f l =a3 for every non-empty set F in 9?
(7). Another special case of ( 5 ) is the following. Cp(O)=O and ~ p ( x ) = l/x,
if O < x
I1.
In this case, prpis unbounded on every set in 9containing 0.
(8). B a n a c h Limits a n d Shift-inuariant Charges. Let a={1,2,3, . . ,}, 8 = the shift transformation from R to R defined by S ( n ) = n + 1 for n E a.We show that there exists an S-invariant probability charge p on 9,i.e. p is a probability charge on 9 and p (A) = p (S-'(A)) for every A c R. We use Banach Limits, which we shall now describe, to show the existence of charges of the type described above. Let embe the space of all bounded sequences of real numbers. For x = (xl,x 2 , . . .) E em,let llxllm = Sup,,, IX,,~. (em, 11 * Ilm) is a Banach space. Let
S(a)and S
L={Y
woo}.
= ( Y 1 , y 2 , . . . ) ; ( Y ~ , Y ~ + Y ~ , Y ~ + .Y. ~ + ~ ~ ,
It is clear that if y = ( y l , y 2 , . . .) E L, then y,, n 2 1 is a bounded sequence of real numbers and that L is a subspace of Em. For x E em,let p ( x ) = Sup,,,l x,,. We show that p ( .) has the following properties. (i). p ( c x ) = c p ( x ) for x and c 2 0 . (ii). p ( x + y ) s p ( x ) + p ( y ) for all x, y €ern. (iii). p ( y ) 2 0 for y E L. (i) and (ii) are obvious. We prove (iii). Suppose p ( y ) = Supnzl y n = k < O for some y = ( y l , y 2 , ...) EL. Then y l + y 2 + . . . + y n s n k for every n z l . So, y + y2 + * * + y,,, n 2 1cannot be a bounded sequence of real numbers. This contradiction shows that p ( y) 2 0 for every y in L. For y in L, let T l ( y )= 0. Then T1is a linear functional on L satisfying T l ( y ) s p ( y ) for every y in L. By Hahn-Banach Theorem 1.5.15, there exists a linear functional T on 8, such that
-
T(Y)= Tl(Y), if Y E L, and
T ( x )s p ( x ) for every x in em. This functional T has the following properties. 1". T(cx + d y ) = c T ( x )+ d T ( y ) for all x, y in emand c and d real numbers. 2". T ( x )2 0 if x = (xl, x 2 , . . .) 2 0, i.e. T is a non-negative linear functional on em.
40
THEORY OF CHARGES
1= (1, 1, 1,. . .). . . .)) = T ( ( x 2 ,x 3 . . .)) for every x = ( X I , x2, . . .) in 8m.
3". T(1)= 1, where
4". T((x1,x2, x3,
5". IT(x)I4 llxllm for every x in fa,i.e. T is a continuous linear functional on em. 6". lim inf,,, x n 5 T ( ( x 1 ,x z , . . . ) ) s l i m Supn+m x , foreveryx = ( X I , X Z , . . .) in em. 7". If x,, n z l is a convergent sequence of real numbers, then T ( ( x 1 ,xz, . . .)) = lim,,, x.,
1" is clear. We prove 2". Note that for any x in 8m, - T ( x ) = T ( - x )5 p ( - x ) = Sup,,l -xn I0 if x = ( x l , x z , . . .) 2 0. Consequently, T ( x )2 0 if x 2 0 . For 3", observe that T ( l ) s p ( l ) =1 and -T(1)= T(-l)< ~(-1) = -1, so that T(1)2 1. Hence T(1)= 1. To prove 4", we proceed as follows. For any x = ( x l , x z , . . .) in em, it can be checked that (xz, x3, . . .) - ( x l , x2, . . .) = (xz-XI, x3-xz, . . .) is in L. Consequently, T((X2,
x3, . . .))-
N X l ,
x z , . . .)) = T ( ( x 2 ,x 3 , . . .) - (x1, xz, = T ( ( X Z - X 1 , Xg-xZ,. = Tl((xZ-xl,
x3-x2,
. .>I *
. .)) * *
a))
= 0.
From this, 4"follows. For 5", we observe the following inequalities.
T ( x ) 5 p ( x ) = S u p x n <SUP IxnI=IIxllm, nzl
nz l
- T ( x ) = T ( - x ) s p ( - x ) = Sup -xn = -1nf x,, nzl
nzl
and
- 1 1 ~ llm = -SUP Ix, I 5 Inf X , tl2l
5T(X)
n z l
for any x = ( x l , x 2 , . . .) in 8,. Hence IT(x)I IIxllm. 6" is a consequence of 4"and the fact that T ( x )s p ( x ) for every x in tm.7" follows from 6". The functional obtained above is called a Banach Limit. One can define a Banach Limit, in abstract terms, as follows. A functional T on 8, is said to be a Banach Limit if T has the properties lo,2", 3" and 4"listed above. One can show that if T has properties lo,2", 3" and 4", then it has properties 5", 6" and 7". Now, we come to the problem of existence of S-invariant charges on S(R). For A c R, let X A E lm be as defined below. (xA),
= 1, if II E A,
=O,
ifn&A,nER.
2.
41
CHARGES
Let T be a Banach Limit on em.Define p on 9 by p (A) = T(xA),A c R. It is clear that p is a probability charge on 9.That, p is S-invariant follows from the observation that for any A c R, XS-'(A)
= ((xA)Z,(xA)3,
.
*)
and that T has property 4", where xA is as defined above.
(9). Banach Limits and General Invariant Charges. Let 9 be a field of subsets of a set R and S a map from R to R such that S-'(A) E 9 whenever A E 9. We show that there exists an S-invariant probability charge p on 9, i.e. p is a probability charge on 9 and p(A) = p(S-'(A)) for every A in 9.Let Y be any probability charge on 9 and T a Banach Limit on em. Define p on 9 by p(A)= T((v(A),v(S-l(A)), .@-'(A)), . . .)) for A in 9. It is clear that p is a probability charge on 9. From 4" of (8), it follows that p is S-invariant. If R = {1,2,3,.. .}, 9=P(R), S the shift transformation on R and v the charge on 9 defined by v(A)=O, if 1 & A ,
= 1 , if l E A , A c R , then the charge p defined above is precisely the S-invariant charge constructed in (8).
(10). Banach Limits and Density Charges. Let R ={l,2,3,. . .} and 9= P(R). A charge p on 9 is said to be a density charge on 9 if lim .u (A) , , = n+m
# ( A n { l , 2 , 3 , .. . ,n } ) n
whenever this limit exists for any set A c R , where for any set B, # B is the number of elements in B. It is obvious that p is positive and p(R) = 1. We show the existence of a density charge. For each n 2 1, let p n on 9be defined by #(An{1,2,3,.. . , n}), AcR. p n (A) = n Then each pn is a probability charge on 9. Let T be a Banach Limit on em.Define p on 9 by E*. (A) = T((PI(N,PZ(A),* * .)),
A = a.
It is clear that p is a probability charge on 9.From 7" of (8), it follows that p is a density charge on P(R).
Now, we introduce the concept of s-boundedness of charges.
42
THEORY OF CHARGES
2.1.4 Definition. Let p be a charge on a field 9 of subsets of a set 52. p is said to be s-bounded if for every sequence A,, n L 1 of pairwise disjoint sets in 9, p (A,) = 0. If p is a real charge, then p is s-bounded if and only if p is bounded. The following results pave the way for the validity of this assertion. 2.1.5 Lemma. Let p be a real charge on a field 9 of subsets of a set 0. If p is unbounded, there exist two sets A1and A2 in $satisfying the following conditions. (i). A l n A 2 = 0. (ii). Ip (A1)IL 1 and lp (A2)12 1. (iii). p is unbounded either on A1 or on A2, i.e. Sup {Ip(B)I; B c A 1 , B E $ } = ~ Oor Sup(Ip(C)I; C c A 2 , C E ~ } = C O .
Proof. Since p is unbounded, there exists B1 in 9 such that lp(Bl)[L Ip(fi)I+1. Then Ip(B;)I =Ip(CR)-p(Bi)ILIIp(CR)I-Ip(Bi)IILIp(Bi)IIp(CR)lL1. It is obvious that p is unbounded either on B1 or on B;. Take A1= Bl and A2= B;, 0 2.1.6 Theorem. Let p be a real but unbounded charge on a field 9 of subsets of a set CR. Then there exists a sequence A,, n L 1 of pairwise disjoint sets in 9 such that 1p (A,)I 2 1for every n 2 1.
Proof. By Lemma 2.1.5, there are two disjoint sets Al and B1 in 9 such that lp(A1)IL 1, )p(B1)l2 1 and p is unbounded on B1. Applying Lemma 2.1.5 again to B1, there are two disjoint sets A2 and B2 contained in B1 such that Ip(A2)1L 1, Ip(B2)1L 1 and p is unbounded on B2. Continuing this way, we obtain a sequence A,, n 2 1 of pairwise disjoint sets in 9such that Ip(A,)I 2 1 for every n 2 1. This completes the proof. 0 An important consequence of this theorem is the following corollary.
Corollary. Let p be a real charge on a field S o f subsets of a set 52. Then p is s-bounded if and only if p is bounded.
2.1.7
Proof. Suppose p is bounded. Let k = Sup {Ip (B)I; B E S}.Then k is finite. We show that Cnzl lp (A,)I I2k for any sequence A,, n 2 1 of pairwise disjoint sets in 9. Let m 2 1 be fixed. Let 11 = (1Ii Im ; p (Ai)2 0) and I2 = (1Ii Im ; p (Ai)< 0). Then
f
i=l
Ip(Ai)I=
c p(Ai)- c p(Ai)
i d 1
i&
2.
CHARGES
43
Hence Xi=, Ip (Ai)]I2k. From this, it follows that lim,,,oc, p (A,,)= 0. So, is s-bounded. Conversely, let p be s-bounded but not bounded. By Theorem 2.1.6, there exists a sequence A,,, n 3 1 of pairwise disjoint sets in 9 such that lp(A,)I 2 1for every n 2 1. Then p (A,) cannot be zero. This contradiction shows that p is bounded. This completes the proof. 0 p
2.1.8 Remark. The assumption that p be real valued in the statement of Corollary 2.1.7 cannot be dropped. As an example, let Q = {1,2,3,. . .} and 9 the finite-cofinite field on R. Let p on 9be defined by p (A)= 0, = CO,
if A is finite, if A is cofinite.
Then p is an s-bounded charge on 9but not bounded.
2.2 THE SPACE OF ALL BOUNDED CHARGES, b a ( i l , 9 ) The object of study in this section is the space of all bounded charges on a field 9’of subsets of a set R. We denote this space by ba(fl,@). There say p Iu if is a natural partial order Ion ba(R,9). For p, u E ba(R, 9), p (F) 5 u(F) for every F in 9. The relation Iis reflexive, antisymmetric The main result of this and transitive. So, Iis a partial order on ba(R, 9). section is that ba(R,9) is a boundedly complete vector lattice. We also introduce a norm 11.I on ba(R, 9) so that (ba(R, 9), I, II-II) becomes a Banach lattice.
2.2.1 Theorem. Let 9be a field of subsets of a set R. Then the following statements are true. (1). If p, u E ba(R, 9)and c, d are real numbers, then c p + du E ba(R, 9). (2). b a ( R , q is a vector space. (3). (ba(R,9), 5)is an ordered vector space, where Iis the partial order on ba(R, 9)introduced above. (4). Let p , u ~ b a ( R , 9 )Define . A on 9 by A(F)=Sup{p(E)+u(F-E); E c F , E E fl,F E9. Then A E ba(R, ( 5 ) . Let p, u E ba(R, 9).Then p v u exisfs irz the partial order Ion ba(fl, 9) and is equal to A defined in (4). (6). L e t p , u ~ b a ( R$, ) . D e f i n e ~ o n 9 b y ~ ( F ) = I{p(E)+u(F-E); nf EcF, EE F E9. Then T E ba(fl,F), (7). Let p , u E ba(R,9). Then p A u exists in the partial order Ion ba(R, 9) and is equal to T defined in (6). I) is a vector lattice. (8). (ba(R, 9),
a.
a,
44
THEORY OF CHARGES
(9). (ba(R, 8, I) is a boundedly complete vector lattice. (10). Forp ~ b a ( f l , 9 )let , llpII=lpl(R). (Recall I p ( = p c + p c -from Section 1.5.) Then 11.I is a norm on ba(fl, 8. (11). (ba(R, 9), I,II. 11) is a Banach lattice.
Proof. (l),( 2 ) and (3) are obvious. (4).It is obvious that A (0) = 0. Since p and v are bounded, A is a bounded function on 9. Let A1 and Az be two disjoint sets in 9. We show that A (Al u A2)= A (Al)+ A (A2).Let B E 9and B c Al u Az. Let Bl = B nA1 and B2= B nA2. Then p (B)
+ v[(Al uAz)-B]
+
= p (B1)+ p (B2) v(Ai -B1)
+ v(Az-Bz)
= [ct (Bi) + v (A1 - BdI + [CL(Bz)+ v(A2 -BdI I A (A,)+ A
(Ad,
as B1c A1,B2c A2 and B1, B2E 9. Taking supremum over all B c Al u A z with B in 9,we obtain A (Al u A2)IA (Al)+ A (A2).On the other hand, let BI, B2E 9, B1 c Al and B2c A2. Then
[II.(B~)+~(AI-B~)I+[CL(BZ)+~(A~-B~)I = p (B1 u Bz) + v [(A1 uA2) - (Bi uBz)] 5 A (A1 uA2). Taking the supremum over all B1c Al with B1E 9 and then supremum we obtain A (A1)+ A (A21IA (A1u A2). Thus over all Bz c Az with B2E 9, A (Al u A2)= A (Al)+ A (A2).Hence A E ba(R, 8. ( 5 ) . It is obvious that p 5 A and v IA. Let $ E ba(R, 9)be such that p 5 $ and v I4. Then for any F in 9,
+
A (F) = SUP{ p (E) v(F- E); E c F, E E 9)
~SU~{$(E)+$(F-E);E~F,EE~) = $03.
So, A I$. Consequently, p v v exists and is equal to A. (6) and (7) can be proved along the same lines as those of (4) and ( 5 ) . (8) is, by now, obvious. (9). Let { p a ;a E r}cba(R, 9) be bounded above, i.e. there exists A E ba(R, 9) such that pa < A for every a in r. We show that Vaorpa exists. Let d be the collection of all finite subsets of r. On d,we introduce a >*) is partial order L* as follows. For C, D ~ dsay, C z*D if C 2 D . (d, a directed set. For each C in d, let pc = VaeCpa. pc, obviously, exists in ba(R, 9)as C is only a finite set. Then the net { p c ;C E d }is an increasing net, i.e. if C, D E & and C ?* D, then p c l p D in ba(R,9). Further, p c s A
2.
45
CHARGES
for every C in d. Let T be the pointwise limit of pc, C E d, i.e. T(F)= limc.dpc(F), F E ~This . limit obviously exists for every F in 9. T is also Since p, 5 T 5 A for any fixed a in r, T is also bounded. It a charge on 9. s)is a boundedly complete is obvious that T = VaErpa. Hence (ba(n, 9), vector lattice. (10). If p = 0, then lp I = 0. So, llpII = 0. Conversely, if IlpII = 0, Ip I(n)= 0. Ipl = 0. Hence p = 0. See Theorem Since 1 ~ is1 a positive charge on 9, 1.5.4(12). If a is any real number and p E ba(R, then llap)I= IapI(C2)= IaIIpI(s2)= IaIIIpII. See Theorem 1.5.4(12). Finally, let pl, p 2 c b a ( a , F ) . Then IIcLI+cL~~I= 1~~1+1l.21(SZ)51p~~1I(R)+\1*.2)(n> =Ilpc~Il+lp211.See Theorem is a norm on ba(R, 9). 1.5.4(12). Hence ((.(( (11). If p, vEba(n, 9)and I p l 5 I v I , then ~ p ~ ( l l ) 5 ~ v Consequently, ~(fl). \\p\Isl\v\\.Hence ba(O,9) is a normed vector lattice. Now, we show that ba(n, 9)is complete under the norm 11 11. Let p n , n I1 be a Cauchy sequence in ba(O,9), i.e. limm,,,+mIIpm-pnll = 0. Since pm- p n s Ipm-pnl for all m,n 1 1 , we have for any F in 9,
m,
-~nl(n)
I ~ m ( F ) - ~ n ( F ) I ~ I ~ m - ~ n I ( F ) ( I ~ m
= llkm - ~ n l l .
This shows that p m(F), m 2 1is a uniform Cauchy sequence of real numbers Let p (F) = limm+mp m(F), F E 9. Thus p m , m 2 1 converges to p over 9. uniformly over 9. p is obviously a charge on 9. Since the convergence is uniform, p is bounded. Since II*(I is a norm, lllp,, -pII-IIpm -pIII 5 II(p,,-p)-(pm-p))II=llpn -pmll,iffollowSthatlimm,, IIpm-pll=O.Hence ( b a ( n , 9 ) , 5,11 * 11) is a Banach lattice. 0 The above theorem in conjunction with Theorem 1.5.4 brings into focus various aspects of bounded charges. Of particular interest, are the charges p c , p - , and lp I associated with p E ba(n, 9). First, we isolate these entities and write formally their computational equivalents. Let p ~ b a ( f l , 9 ) .Then p f = p v O and p - = ( - p ) v O . By Theorem 2.2.1(5),
'(F) = SUP{@(B);B C F ,B E 8, FE and
p-(F) = -1nf
{ p (B); B c F, B E 9}, F EF.
and p - ba(n, ~ and they are called the positive and negative variations of ICL respectively. The charge Ipl = p L c + p -is called the total variation of p. Combining Theorem 1.5.4 and Theorem 2.2.1, we chronicle some salient features of charges in the following theorem for future reference. pc
46
THEORY OF CHARGES
2.2.2 Theorem. Let p E ba(R, 9).Then the following are true. (1). p = p + - p - a n d p + A p - = O . (2). l p l = p + + p - = p + v p - = p v ( - p ) . A n important consequence of this is )(F)= SUP { p (B) - p (F- B); B c F, B E g}, F E 9. (3). I f p = A - v with A, vEba(R,5"), A r O and v 2 0 , then A z p + and
IF
v r p
-
.
Y E ba(R, 9)andA A v = 0 , then A = p + and v = p - . ( 5 ) . p++=$(Ip I + p 1 and p - = $(IP I - p j . + (6). If Y E ba(R, 9)and p Iv, then p 5 v and p - 2 v-. (7). I f v E b a ( R , m and ~ A V = Othen , ( p + v ) + = p + + v + and I p + v l =
(4).I f p = A - v with A,
IPI
+I4
2.2.3 Remark. Theorem 2.2.2( 1) is precisely Jordan Decomposition theorem for bounded charges. Any bounded charge p can be expressed as the difference of two positive charges with a certain minimality property in the following sense. If p = A - v also, where A, v 2 0, then A L p + and v 2 p -. We will take up the issue of writing a given charge as a difference of two positive charges in Section 2.5. In the following theorem, we give an alternative description of lpl, in addition to the one described in Theorem 2.2.2(2).
2.2.4 Theorem. Let p E ba(R, 9).Then for any F in 9, i=l
where the supremum is taken ouer all finite partitions F1, Fz, . . . ,F, of F in 9.Further, IpI(R)52 S u p { l p ( F ) I ; F ~ 9 } . Proof. Let F1,F2,. . . ,F, be any partition of F in 9. Let I = (1 5 i 5 n ; p(Fi)L 0) and J = (1Ii ~ np(Fi) ;
i
i=l
Ip(Fi)I=
c p (Fi) - Ci s J p(Fi)
is1
5 IF I(F),
xr=l
by Theorem 2.2.2(2). Consequently, Sup (p(Fi)(IIp ((F),where the supremum is taken as described above. On the other hand, let B E 9 and
2.
47
CHARGES
B c F . Then {B,F-B} is a partition of F in 9 and p ( B ) - p ( F - B ) s Ip(B)I + Ip(F-B)I 5 Sup Z;=l )p(Fi)l, where the supremum is taken as described above. Therefore, \ pI(F) = Sup { p (B) - p (F - B); B c F, B E 9) 5 SupC:=, Ip(Fi)l. This completes the proof. 0
2.3 MEASURES In this section, we introduce measures and establish some properties of measures. We also give a set of necessary and sufficient conditions under which a charge is a measure.
2.3.1 Definition. Let 9 be a field of subsets of a set a. A measure on 9is any map p from 9 to [-a, a]having the following properties. (i). p ( 0 ) = 0 . (ii). If F,, n 2 1 is a sequence of pairwise disjoint sets in 9 with UnzlF, in 9,then p(
gl 2l Fn)
=
p (Fn)*
Any measure is, obviously, a charge. If p is real valued, we call p a real measure. If p is bounded on 9,we call p a bounded measure. If p 2 0 , we call p a positive measure.
2.3.2 Proposition. Let p be a charge on a field 9of subsets of a set SZ. Then the following statements are true. (1). p is a measure on 9if and only if limn+mp (A,) = p (A) whenever A,, n 2 1 is an increasing sequence of sets in 9with A = UnzlA, in 9. (2). Let p be real. The following are equivalent. (i). p is a measure on 9. (ii). p (A,) = p(A) whenever A,,, n 2 1 is a decreasing sequence of sets in 9with A, = A in 9. (iii). limn-,m p (A,) = 0 whenever A,, n 2 1 is a decreasing sequence of sets in 9with A, = 0.
nnrl
n,,,
Proof. (1).Suppose p is a measure on 9. Let A,, n 2 1 be an increasing sequence of sets in 9 with A = U n Z 1A,, in 9.Let B l = A l , B2= A2 - Al, . . . ,B, = A, - A,-I, . . . . Then B,, n 2 1is a sequence of pairwise disjoint sets in 9 and
U An= nUr l Bn.
n z l
THEORY OF CHARGES
u An)=p( u Bn)
I*.(A)=~( n z l
n r l
m
=
C p ( B n ) =mlim C p(Bn) nzl +m n = l
The converse is simple to prove. ( 2 ) . (i)+(ii). Let A,,, n 2 1 be a decreasing sequence of sets in 9 with A, = A in 9. Then A:, n 2 1 is an increasing sequence of sets in 9 with UnZlA:=AC. Let B1=AE, B2=A;-Ai,.. . , B,=A',-A:-l, .... Then B,, n 2 1 is a sequence of pairwise disjoint sets in 9with
nnzl
u B, u A:. =
nzl
nzl
Consequently,
= lim p (A:) = lim [ p (a)- p (A,)]. n+m
n+W
Hence p (A,) = p (A). ( i i ) j (iii). This is obvious. (iii)=$ (i). Let B,, n 2 1 be a sequence of pairwise disjoint sets in 9 such that UnrlB, E 9. Let A, Bm,n 2 1. Then A,, n 2 1 is a decreasing sequence of sets in 9with A, = 0.Therefore,
=urn,, n,,,
O = lirn @(A,)= n+m
Hence
Note that Theorem 2.3.2(2)is not valid if p is not real.
m=l
2.
49
CHARGES
We now give a useful sufficient condition under which a charge is a measure. We begin with a definition.
2.3.3 Definition. A collection V of subsets of a set R is called a compact class if it has the following property: if C,, n 2 1 is a sequence of sets in V with C, = 0, then there exists m 2 1such that C, = 0.
n,,,
n:=:=,
2.3.4 Theorem. Let 9be a field of subsets of a set R and V a compact class contained in 9. Let p be a positive bounded charge on 9having the following approximation property : p (F)=Sup &(C);
C C F, C EU},F E9.
Then p is a measure on 9. Proof. In view of Proposition 2.3.2(2), it suffices to show that limn+mp (A,) = 0 for any decreasing sequence A,, n 2 1 of sets in 9 with A, = 0.Let E > 0. For each n 2 1,there exists C, in V,C, c A, such A, = 0, C, = 0. So, there that p (A,) 5 p (C,) + &/2".Since C, = 0. Consequently, exists rn 2 1 such that
n,,
n,"=,
fi
b(Am)=p(n = l An) m
5
nnrl
n,,,
CJ(An-cn))
C P (n = l
m
C p(An-Cn)= nC= l [ ~ L A n ) - ~ ( c n ) l < & . n=l
So, for every n 2 m, p (A,) < E . Hence
I.L (A,) = 0.
2.3.5 Examples (1). Let R be any infinite set and 9 the finite-cofinite field on R. Let on 9be defined by p (A) = 0, if A is finite, = 1, if A is cofinite.
p
If R is countable, then p is a charge on B but not a measure. If R is uncountable, then p is a measure on 9. In fact, in this case, every charge on 9 is a measure. (2). A standard example of a measure is the Lebesgue measure A on the Bore1 cT-field 9 of the real line R. This is the measure on 94 such that A{(a,b ) } =b - a for every -m
50
THEORY OF CHARGES
2.4 THE SPACE OF ALL BOUNDED MEASURES, c a ( i 2 , m
In this section, we take up the study of the space c a ( R , 9 ) of all bounded measures on a field 9 of subsets of a set R. The main result of is a normal vector sublattice of ba(R, 9). First, this section is that ca(R, 9) we need a proposition. + 2.4.1 Proposition. Let p, v E ca(R, 9).Then p ,p , lp I, p v v and p A v E ca(R, 9). Proof. First, we show that p + is a measure. Let A,,, n 2 1 be a sequence of pairwise disjoint sets in 9such that UnrlA,, E g. Let B c UnZl A,, and Let B, = B nA,,, n 2 1. Then B E 9.
u B n ) = C CL(B,ASC II.+(A,,).
~(B)=P(
n2l
n2l
nz1
Taking supremum over all BE^ with BcU,,,lA,,, we obtain p ~ ( ~ , , l A , ) ~ ~ , , l p ~ ( Since A , ) . p + is a positive charge on 9, Cnzl p+(A,,)%@+(Unz1 A,,). See Proposition 2.1.2(viii). Hence A,) =Inzl p+(A,,).Thus p + ca(R, ~ 9). By a similar argument, we can show that p - ~ c a ( R .F). , So, I p ) = p + + p - ~ c a ( Rq. , Now, we show that p v v and p A v are measures. By Proposition 2.3.2(2), it suffices ( p A v)(A,,) whenever A,, to show that limm-too( p v v ) ( A , ) = 0 = n 2 1is a decreasing sequence of sets in 9with A,, = 0.By Theorem 1.5.4(14),
p+(unz1
n,,,,
-IpI-IvI
Sp
A
v
I@ V v Ipl f l v l .
, E c a ( R , 9 ) , we have ( p/(A,,)= 0 = limn-toolv((A,,).From Since ( p ( (vI this, it follows that the desired limits above are each equal to zero. Hence 0 p v v and p A v are measures.
2.4.2 Theorem. ca(R, 9) is a normal vector sublattice of ba(R, 9).In particular, ca(R, 9)is a closed subspace of ba(R, 9). Proof. By Proposition 2.4.1, it follows that ca(R, 9) is a vector sublattice We now show that it is normal. Let p ~ c a ( R9) , and v E of ba(R, 9). ba(R, 9be )such that 0 5 1v1s Ip 1. Since /pi is a measure and (v(F)I5 lv((F) v is a measure by Proposition 2.3.2(2). Hence Y E ca(R,9). for every Fin 9, be such that V a o rpa exists in ba(R, 9). Let Let { p a ;a E r}c ca(R, 9) 7 = V a E pa. r We show that T E ca(R, 9). Let d be the collection of all finite subsets of r. For C, D in d,say C ?* D if C 3 D . For C in d, let pc= VaeC pa. Then pc, C E is ~an increasing net of measures whose pointwise limit is T, i.e. T(F)= lim pc(F), F E9. CE sp
2.
51
CHARGES
See the proof of Theorem 2.2.119). We show that r is a measure. Since V a s rpa = VCEd pc = r , let us assume that r itself is a directed set and pa, a E r is an increasing net of measures. Let d = SUpasr p,(R). Since p, 5 T for every a in r, d < co. Let a,,,n 2 1 be a sequence in r such that pUn(fl) = d . We can assume, without loss of generality, that pa, 5 p m 2 5* * . Let r*(F)= p,,(F), F E9. We show that r * is a measure on 9. T* is obviously a bounded charge on 9. Let A,, n 2 1be any increasing sequence of sets in 9with UnzlA, = A E 9. In view of Proposition 2.3.2(1), it suffices t o show that limk,, r*(Ak) = T*(A). A t this juncture we assume that each pa is a positive measure. Note that
-
lim T*(Ak) = SUP 7*(&) k-m
= SUP SUP p,,(Ak)
k z l
k z l nzl
= s u p s u p p,,(Ak) = s u p p,,(A) = r*(A). n z l k z l
nz1
This shows that 7* is a measure on 9. Next, we claim that r * = r . Let a in r be arbitrary. We show that pa s'r*. Note that Pa v r * = Pa v
(Ll
P%) =
v
nzl
(Pa v Fan).
If ( p , v ~*)(n) > d , then (pa v pa,)(fl)> d for some n 2 1. This contradicts the definition of d . So, (pa v r * ) ( f l )5 d = r * ( f l )5 (pa v ~*)(fl).So, (pa v ~ * ) ( f = l )r*(fl).Since r * s p a v r * and ~ * ( f l=) (pa v r*)(fl),it follows that T* = p, v r * ! (Check this.) This implies that pa ST*. Since this is true for every a in r, V a E pol r = r 5 r * . On the other hand, since pan:s r for pan= T* 5 T. Hence T = r * . This shows that r is ameasure. every n 2 1, Vn2? In the above, we have shown that r is a measure under the assumption that each p, is a positive measure. Now, we treat the general case. Choose and fix a0 in r. Let v, = polv p,, + Ipa0l,a E r. Each Y, is a positive measure since I/, 2 pa0+ Ip,,l = 2pL0 2 0. Observe also that
v
I/,
=
a E r
v
[(Pa v
+l ~ ~ , I l
11.010)
,Er
by Theorem 1.5.4(1)
By what we have proved above, r+lp,,l is a measure. Since (paOlis a is a normal sublattice measure, it follows that 7 is a measure. Hence ca(R, 9) of ba(fl, It follows from Theorem 1.5.19 that c a ( f l , 9 ) is a closed sublattice of bdfl, 0
a.
w.
52
THEORY OF CHARGES
2.4.3 Corollary. ca(R, 9 )is a boundedly complete vector lattice and also a Banach lattice in the usual norm. 2.4.4 Remark. Given any field 9 of subsets of a set R, one can find a set X and a c+-field d on X such that ba(Q, 9) and ca(X, d)are isometrically isomorphic as Banach lattices, i.e. there is a linear map T from ba(R, 9 ) onto ca(X, d)such that IIT(p)ll=llpll, p E ba(R, 9) and T ( p )1 0 whenever p 2 0 . This can be proved using the Stone Representation theorem for Boolean algebras given in Section 1.4.
2.5 JORDAN DECOMPOSITION THEOREM Jordan Decomposition theorem for bounded charges has been discussed in Remark 2.2.3. It essentially says that every bounded charge can be written as a difference of two positive charges in a minimal way. The main theorem of this section gives a simple necessary and sufficient condition for such a decomposition to prevail for charges not necessarily bounded. First, we introduce the lattice operations v and A for general charges.
2.5.1 Definitions. Let p and v be two charges on a field 9 of subsets of a set R such that either both p and v avoid the value -a or both avoid the value +a.Define the set functions A and T on 9 by
+
A (F)= SUP{ p (E) v(F-E); E c F, E E
a,
FE9,
and T(F)= Inf { p (E)+ v(F- E); E c F, E E 9}, F E9. The above set functions were defined for bounded charges in Theorem 2.2.1(4) and (6).We used the same formula for general charges in the above definition. We denote A by p v v and T by p A v. This is consistent with the notation used for bounded charges.
2.5.2 Proposition. Let p and v be two charges on a field 9of subsets of a set Q such that either both p and v avoid the value -a or both avoid the value +a.Then the set functions A ( = p v v ) and T ( = p A v ) defined above are charges on.9. Proof. The proof given for Theorem 2.2.1(4) and (6) carries through essen0 tially here. The following theorem is the main result of this section.
2. 2.5.3
of a set
53
CHARGES
General Jordan Decomposition Theorem. Let 9 b e a field of subsets n. Let p be a charge on 9.Define p + and p - by p+(F)= Sup { p (E);E c F, E E9}, F E g,
and y-(F) = -1nf {p(E);E c F, E E
w,
F E 9.
Then the following statements are true. (1). p + and p - are positive charges on $. (2). I f p does not take the value +a,then p + - p = p - . (3). If p does not take the value -00, then p + p - = p c . (4). I f p does not take the value +OO and ~ 1 p- = p 2 for some positive c h a r g e s p l , p 2 0 n 9 ,t h e n p ~ ~ p + a n d p ~ ? p - - . ( 5 ) . I f p does not take the value --CO and p + hl = A 2 for some positive charges h l , h2 on 9,then h 1 z p - and h 2 ? p C . (6). p = p + - p - if and only if p is either bounded below or bounded above. More generally, we can write p = p l - p 2 for some positive charges p l , p2 on 9 if and only i f p is either bounded below or bounded above. (7). p + A p - = 0 if and only if p is either bounded below or bounded above. ( 8 ) . I f pl and p 2 are positive charges on 9 satisfying p = p l - p 2 and pI ~ p 2 = 0 then , p~ = p+ and p2 = p - . (9). If p is a real charge, then p = p + - p - holds if and only if p is bounded. In such a case, both p + and p - are bounded. More generally, i f p is a real charge, then we can write p = p1 - p 2 for some positive charges p l and p2 on 9 if and only if p is bounded. Proof. (1)follows from Proposition 2.5.2 if we observe that p + = p v 0 and p - = ( - p ) v 0 as per Definition 2.5.1. (2). Let F E $. Suppose p(F) = -a.Then, from the definition of p - , p-(F) = co.So, p+(F)- p (F)= a = p-(F). Suppose p (F)> -a.By the given hypothesis, -a< p (F)< a.Consequently, --CO < p (E) < 00 for any E in 9 such that E c F . See Proposition 2.1.2(vii). So, CL +(F)- W
(F)= SUP{ P
(a;E c F, E
= SUP{ p (E)- p
E
SI(F)~
(F);E c F, E E $}
=SU~{-~(F-E);ECF,EEF} = -1nf {p(C);C c F, C E = p-(F).
This completes the proof. (3). This can be proved as above.
54
THEORY OF CHARGES
(4).Since p l - p = p 2 and p 1 and p2 are positive, it follows that p l z p . So, for any F in 9, p+(F)=Sup{p(E);E c F , E ~ 9 } s p ~ ( FHence ). p+5 p1. For the second part, observe that p + - p s p l - p = p 2 .By (2), p + - p = p . This completes the proof. ( 5 ) . The proof is analogous to that of (4). (6). Suppose p is bounded above. By (2), p + - p = p - . Note that pLfis a bounded charge. Consequently, - p = p - - p + or p = p + - p - . A similar argument works when p is bounded below. Conversely, if p = p + - p - , then either pLcis bounded or p - is bounded. In the former case, p is bounded above and in the later case, p is bounded below. The more general version can be established using (4). (7). Suppose p is bounded below. We show that p + A p - = 0. Since p is bounded below, p - is a positive bounded charge. So, for any F in 9, +I(
A
p-)(F) = Inf {p+(E)+p-(F-E); E c F, E E fl
+
= Inf { p (E) p-(E) +p-(F-E); E c F, E E 9},
(by (3)) =Inf{p(E)+p-(F); E c F , E € . F } = Inf { p (E); E c F, E E -+p-(F),
(since p - is bounded) = -p-(F)+p-(F)=O.
If
is bounded above, a similar argument shows that p + A p - = 0. Conversely, suppose p is neither bounded below nor bounded above. We show that p+r\ p - # O . Since p cannot take both the values +a and -a,assume, without loss of generality, that p does not take the value -a.Note that pL-(n) = a.By (3), p + p - = p + . So, p
( F +A
p - ) m = u p + p - > A p-l(n2) = Inf { ( p +p-)(F)+p-(F");
FE9}
= Inf {p(F)+p-(R); F E 5F)= a.
This completes the proof. (8). If p = p1 - p ~ where , p l and p2 are positive charges, then p is either bounded below or bounded above. See ( 6 ) .Assume p is bounded above. This implies that p l is a bounded charge. Since p = p l - p 2 = p + - p -, p1-p:-p2=-p-50. Therefore, p 1 - p + 5 p 2 . By (4),p l ? p + . So, 0 5 p1-p +5p2. Since p l ~ p ~ =and O O s p l - p c s p l , it follows that 0 s (PI - p ) 5 p1 A p2 = 0. Hence p l= p + . The other equality follows easily now.
2.
55
CHARGES
(9). If p is a real charge and p = p1- p 2 , where p 1 and p2 are positive charges, then p 1 and p2 are bounded. So, p is bounded. If p is bounded, of course, we can write p as a difference of two positive charges. This completes the proof of the theorem. 0 Some comments are in order on the above theorem.
2.5.4 Remarks. (i). The charge p described in Example 2.1.3(2) is neither bounded above nor bounded below. There is no way we can write p as a difference of two positive charges. For this charge, p + - p = p - and p f p - = p + . p + and p - work out as follows. p +(A)= n, = 03,
p-(A) = 0, = 03,
if A is finite and has n elements, if A is cofinite. if A is finite, if A is cofinite.
= 03. Observe also that (p’ A &*.-)(a) (ii). Theorem 2.5.3 goes through in toto if we replace the word “charge” by “measure”.
An interesting result emerges for measures on a-fields in contrast to the Remark 2.5.4(i), i.e. every measure on a a-field can be written as a difference of two positive measures. First, we need a lemma.
2.5.5 Lemma. Let p be a measure on a u-field % of subsets of a set R. Then p is either bounded below or bounded above. Proof. First, assume that p is real valued. In this case, we show that p is bounded. Suppose p is unbounded. By Theorem 2.1.6, there exists a sequence B,, n 2 1 of pairwise disjoint sets in % such that Ip (Bi)l r 1 for every i 2 1. Then, either there exists a sequence n1 < n2 < * * such that p (B,,) 2 1 for every i L 1 or there exists a sequence k l< k2 < such that p ( B k l ) % - l for every i r l . Assume that the former holds. Then p(UiZ1 B,,) = 03 contradicting the fact that p is real valued. If the latter holds, we do still get a contradiction. Hence, if p is a real valued measure, then it is bounded. 031 or in In the general case, observe that p takes values either in (-a, [-a, 03). Assume that p takes values in (-O3,03]. Then we show that p is bounded below. Suppose p is unbounded below. We find A1 in % such that p (A1) 5 -1. Since p is real valued on A1 n%, by what we have proved above, p is bounded on Al. Consequently, p is unbounded below on A?. So, we can find A2 in 2l such that A2 c A; and p (A2)I-1. By the same reasoning given above, we can show that p is unbounded below on Af - A2.
--
56
THEORY OF CHARGES \
Continuing this way, we obtain a sequence A,,, n 2 1 of pairwise disjoint A,,) = -a.This sets in % such that p (A,) 5 -1 for every n 2 1. So, p (UnZl is acontradiction. Hence p is bounded below. This completes the proof. 0
2.5.6 Jordan Decomposition Theorem for Measures on a-fields. Let be a measure on a u-field % of subsets of a set R. Then we can write f
F=P
p
-
-cL
with the property that p + A p - = 0. Further, the above decomposition has the following optima& property: if p = p l- p 2 , where p I and p2 are positive measures on %, then p 1 2 p + and p2 L p -. Proof. By Lemma 2.5.5, p is either bounded below or bounded above. Remark 2.5.4(ii) completes the proof. 0 2.5.7 Remark. The above theorem is not valid for charges on a-fields.
2.6 HAHN DECOMPOSITION THEOREM In this section, we prove Hahn Decomposition theorem for charges. Hahn Decomposition theorem for measures on a-fields will be proved in Section 6.1.
2.6.1 Definition. Let 9 be a field of subsets of a set R and p a charge on 5 Let E > 0. A partition {D, D? of s1 with D in 9is said to be a E-Hahn decomposition of p if the following are satisfied.
+p (B) C c D" +p (C)
B E 9, B c D C E 9,
5 E. 2 -E.
2.6.2 Hahn Decomposition Theorem for Charges. Let p be a charge on a field 9of subsets of a set R which is either bounded below or bounded above. Then for any E >0, there exists a E-Hahn decomposition for p. I f p is neither bounded below nor bounded above, there exists E > O for which there is no E-Hahn decomposition of p . Proof. We give a direct proof of this result. This can also be proved using Jordan Decomposition theorem. Assume that p is bounded below. Let d = Inf (r-L (A); A E 9). Then d is a finite number. Let E > 0. We can find D in 9 such that d 5 p (D) Id + E . This implies that -CO < p (D) < co and for any B in 9,B c D , we have -oo
2.
-
CHARGES
57
The statement that p admits E-Hahn decomposition for every E > O is equivalent t o the statement that p i h p - = 0. But this is equivalent to the statement that p is either bounded below or bounded above by Theorem 2.5.3(7). This proves the second part of the thereom. 0
2.6.3 Remark. Exact Hahn-decomposition, i.e. E-Hahn decomposition for E = 0, need not exist for every charge. For example, let R = { 1 , 2 , 3 , . . .}, 9the finite-cofinite field on R and p(A)=
1
(-1)"/2",
Ae9.
noA
There is no set D in 9 such that p ( B ) s O for every B in 9, B c D and p (C) 2 0 for every C in 9, C c D'.
CHAPTER 3
Extensions of Charges
One way of obtaining charges on a given field 9 of subsets of a set SZ is to look at subcollections % of 9and set functions v defined already on %? and see if v can be extended to 9 as a charge. For example, if v is a we can extend u from %? 0-1 valued charge on a given sub-field % of 9, to 9 as a 0-1 valued charge. The underlying theme of this chapter is to seek extensions in the spirit of the above example. In Section 3.1, we associate a natural functional T on a suitable vector space with a given set function u on a given class % of sets and study the interplay between u and T. In Section 3.2, we introduce the notion of a real partial charge on a collection %' of sets contained in 9and show that these are precisely the set functions on % for which extensions are possible. In Section 3.3, we study the extension procedure developed by Los and Marczewski. Miscellaneous useful extensions are dealt with in Section 3.5. Finally, in Section 3.6, we look for a common extension to 9 of two set functions defined respectively on two classes of sets %' and 9 contained in 9.
3.1 REAL VALUED SET FUNCTIONS AND INDUCED FUNCTIONALS Any real valued function p on a given class 9 of subsets of a set SZ, under certain conditions, gives rise to a natural linear functional on a suitable vector space. We study this correspondence in this section.
3.1.1 Definition. Let 9 be a collection of subsets of a set 0. Let
B(9)={f:SZ+R;f=
i=l
riIAiforsomeAl,A2, ..., A,
in 9, rl, r2, . . . ,r, rational numbers, It is obvious that numbers.
/
I
II 2 1
.
B(9)is a linear space over the field of all rational
3.
59
EXTENSIONS OF CHARGES
3.1.2 Definition. Let 9 be a collection of subsets of a set R and p a real valued function on 9.We set
for A1,A2, . . . , A, in 9, r l , r2, . . . ,r, rationals and n 2 1. We study the interplay between p and T. First of all, we examine the unambiguity of the definition of T. More precisely, if CF=l riIAi=, :I sJBj for some A1, Az, . . . ,A,, B1, Bz, . . . ,Bm in 9, r l , r 2 , . . . , rn, s1, SZ, . . . , Sm rationals, is Cysl rip (Ai)= C ,:, sjp(Bj)? If T is well defined, we call T the induced by p. The following proposition provides an functional on 2(9) answer to the above question.
3.1.3
Proposition. T is well defined if and only if n
m
i=l
j=l
c p ( A i ) = C p(Bj)
holds for any two finite sequences A1, A2,. . . , A,, and BI, Bz, . . . , Bm of not necessarily distinct sets from 9satisfying m
n
C i=l
C IB~.
=
j=l
Proof. In order to show that T is well defined, we have to establish k
f
C rip(Ci)= C sjp(Dj)
i=l
j=l
whenever k
C i=l
f
dci=
C
SjIDj
j=l
for C1, C 2 , . . . ,c k , D1, D 2 , .. . ,D, in 9 and r l , r2,. . . , r k , s1, SZ, . . . ,s, rationals, Since ri, r2, . .. ,rk, sl,s2,, . .,sc are rationals, we can rewrite the equality k i=l
k
I
riIci= j=l
f
miIci =
sjIDj as i=l
njIDi j=l
by cancelling the common denominator of all the given rationals leading to the situation that m l , m z , . . . , mk, n1, n2,. . . , n, are integers. This later equality can be rewritten in the form n
m
60
THEORY OF CHARGES
Retracing these steps from m
n
C p(Ai)=jC p(Bj), i=l =1 we obtain k
f
C rip(Ci) = j C= l sjp(Dj)i=l 0
The converse is trivial.
Thus, if T is well defined on the linear space 2'(.F),then it is a linear functional on 2'(9). Now, we aim to show that T is a well defined linear functional on 2(9) when 9is a field of subsets of a set and p a real charge on g.Moreover, there is a one-to-one and onto correspondence between linear functionals and ) real valued charges on 9.For this, we need on the linear space 2'(9 the following results. (For sets C1, Cz, . . . ,C, and k > n, we use the convenCii = 0.) tion that
ulsil
3.1.4 Lemma. For any two finite sequences Al, Az,. . . , A, and B1, Bz, . . . , B, of sets, the following statements are true. (a). IA, SC,:, I B ,if and onZy if
zy=l
k
k
U lsi1
n Aii c
j=l
U l s i l
n Bi,
s m j =1
for every 1 I k 5 n. IAi = II B , if, and : , only if (b).
for every 1 I k 5 max {m, n}. Moreover, one can always write
ck=lzsil
Cy=l IAi = Cyzl Ic,, where k
Proof. It is obvious that and only if
0A,, j 1
lsksn.
=
zye1IA,(w)?k, where k is a positive integer, if 6JE
u
k
n Aij.
1si 1
From this, (a) follows. (b) is a simple consequence of (a).
3.
To prove the last part, we note that C1=) C2 =) * k
1s11
U < i2<.
* =)
n A,.
1s i I < i Z < . . . < i k
s n j=1
Cn and k
u
n c,=ck=
61
EXTENSIONS OF CHARGES
0
s n j=1
Recall that a collection 9 of subsets of a set R is said to be a lattice if A u B and A n B E 9 whenever A, B E 9. If 9 is a lattice of subsets of a set R, then a real valued function p on 9 is said to be a real modular function on 9 if p (A)+ p (B) = p (A u B) + p ( A n B) for every A, B in 9.
3.1.5 Proposition. Let 9 be a lattice of subsets of a set R and p a real modular function on 9. Then for any finite number of sets A1, AZ, . . . , A, from 9, we have
In particular, if 9 is a field and p a real charge on 9,the above assertion is true. Proof. The proof given for Proposition 2.1.2(x) is a proof exactly for this assertion. 0 The following theorem shows that T is well defined in some special cases.
3.1.6 Theorem. (f). Let 9 b e a lattice of subsets of a set R with 0 in 9'. Let p be a strongly additive real function on 9, i.e. p is a real modular function on 9 satisfying the condition that p ( 0 ) = 0. Then the induced functional T o n 9(9) is well defined. (2). Let 9 be a field of subsets of a set R and p a real charge on 9. Then the induced functional T o n is defined unambiguously.
9(m
Proof. (1). In view of Proposition 3.1.3, it suffices to prove that IAi = Ci"=,I B i for A1,Az, . . . ,An, p (Ai) p ( B j ) whenever BI, Bz, . . . ,Bm in 9.By Proposition 3.1.5,
=xi"=,
f p(Ai)= f
U
P(
k=l
i=l
?IAij)
lsil
j=l
and m
C j=1 Since
p(Bj)=
xyzlIAAi=I,?,
IBj,
1 s il
f
k=l
U
P(
6Bii)*
U. . < i k s m j = l
lsil
by Lemma 3.1.4(b), k
nA, =
ik s n j = 1
U
1s i 1
k
n Bij
s m j =1
62
THEORY OF CHARGES
for every 1Ik 5 max {m,n}. Assume, without loss of generality, that m If rn < k 5 n, it is obvious that
5 n.
k
n Aij= 0.
U
l s i l
Since p is a strongly additive real function on 9, it follows that
? p(Ai)= !?P( k=l
i=l
U
lsil
AAij)+ k = m + l
j=l
~(0)
m
=
c I*(Bk)*
k=l
This proves (1).(2) is a special case of (1).
0
The following theorem looks at the converse implication of the above theorem. (
Theorem. For a given collection 9of subsets of a set 0, let T be a linear functional defined on the linear space S(9) over the field of rational numbers. For A in 9, set p(A) = T ( I A ) . Then the following statements are true. (1). If 9is a field of subsets of 0, p is a real charge on 3. (2). If 9 is a lattice on R, p is a real modular function on 9. (3). If $is an arbitrary class of subsets of SZ, the set function p on Fsatisfies p(Bj) whenever I A= ~ the following condition : I:=,p (Ai)= CJtlIB,for Al, A*, . . ,A,, B1, B2,. . . ,B, in 9.
3.1.7
c,:,
Proof. We prove (2). Let A, B E 9. Since IA+IB = I A u B +IA,B, we have /.L(A)+El.(B)= T(IA)+T(IB)=T(IA+IB)=T(IA"B +IAnB) = T(IA"B
+ T ( I A ~ B=)p (A uB) + P (A n B ) .
Hence p is a real modular function on 9. Note that if 0 ~ 3 , then p ( 0 ) = T(Iar)=T(0)=O.Now (1) is a special case of (2). (3). This is obvious. 0
3.1.8 Corollary. (1). Let 9 be a lattice of subsets of a set SZ such that 0 E 9. Then there is a one-to-one and onto correspondence between the collection of all strongly additive real functions on 9and linear functionals on LE'(9). (2). Let 9be a field of subsets of a set SZ. Then there is a one-to-one and onto correspondence between the collection of all real charges on S a n d linear functionals on Z(9). 0
3.
63
EXTENSIONS OF CHARGES
Now, we look at the relationship between positive bounded charges p on a field 9 of subsets of a set R and the induced linear functionals T on
m?.
3.1.9 Theorem. Let 9be any collection of subsets of a set R and p a real valued function on 9. Let T be defined on 2?(.9) as in Definition 3.1.2. Consider the following statements. (a). p is positive. (b). C;=l p (Ai)2 C,:, p(Bj) whenever I A i zCLl I B i for A l , Az, . . . , A,,, B1, Bz, . . . , B, in 9. (c). T is well defined and T(f ) 2 0 whenever f ~2’(9) and f 5 0. Then the following statements are true. (i). (b) and (c) are equivalent. (ii). (6) implies (a). (iii). ( a ) , ( b ) and (c) are equivalent if 9is a field on R and p a charge on 9. Proof. (i). Suppose (b) holds. By Proposition 3.1.3, T is well defined. Let a n d f r O . T h e n f = x i k _ l r i I c i f o r s o m e C 1 , C z , ..., Ck i n 9 a n d r l , r2, . . . ,rk rationals. Writing ri = mJm, i = 1 ,2 , . . . , k, where m l , m 4 . . .,mk are integers and m a positive integer, we observe that mf = Ci=lmiIci.We can rewrite this equality as mf = lai-I;=, I B i with Ai’s and Bis coming from {Cl, Cz, . . . , C,} and p, q z 1. (Why?) If f LO, I A ?I;=, ~ I B j . Since (b) holds, @(A;)?I;=,p(Bj). Conthen sequently,
fEz(9)
xy=l
xr=l
T(mf) = mT(f) = T( P
=
f
i=l
IAi
-
5
IBj)
j=l
q
C @(Ai)- C p(Bj)LO.
i=l
j=l
This proves (c). If (c) holds, it is obvious that (b) holds. (ii). Let A E 9.Then IA+IA 2 IA. Consequently, p (A) + p (A) 2 p (A) which implies that p (A) 2 0. (iii). Using Lemma 3.1.4(a), Proposition 3.1.5 and the monotonicity of the positive charge p on the field 9, one can show that (b) holds whenever (a) holds. Thus (a), (b) and (c) are equivalent when 9is a field on $2. 0
Remark. The implication (a)+ (b) is not valid if 9 is a lattice on R, 0 E 9and p a strongly additive real function on 9. Here is an example. Let R={1, 2,3} and 9 = { 0 , { 1 } , {1,2},{1,2,3}}. 9 is a lattice on R. Any function p on 9is a modular function. One can take the following function 3.1.10
o n 9 . p ( 0 ) = 0 , p ( { l } ) = l , p({1,2})=$andp({1,2,3})=$. T h e n p is a positive strongly additive real function on 9.For this p , (b) of Theorem 3.1.9 fails to hold, even though the associated functional is well defined.
p
64
THEORY OF CHARGES
3.2 REAL PARTIAL CHARGES AND THEIR EXTENSIONS In this section, we show that under some natural conditions we can extend a real valued set function on a collection V of subsets of a set R to any field 9on R containing V,as a charge.
3.2.1 Definition. Let V be a collection of subsets of a set R. A real valued p(Bj) function p on V is called a real partial charge if CrXlp (Ai)= whenever I:=,I A j=IJ:,I B for ~ A1, A 2 , .. . , A,,, B1, B 2 , . . . , B, in V,i.e. the functional T defined on 9(V)is well defined. 3.2.2 Definition. Let V be a collection of subsets of a set R and p a positive real valued function on V. p is said to be a positive real partial p(Ai)lCT=lp(Bj) whenever l a i ICY=, IB, for charge if I:='=, Al, AZ,. . . ,A,, B1,B z , . . . ,B, in V.
zrZl
If p is a real partial charge on V and p(C)?O for every C in V,it does not follow necessarily that p is a positive real partial charge on V. If V is a field or a ring of sets on R, then a real valued function on %' is a real partial charge on V if and only if it is a real charge on V.If V is a lattice on R with 0 in V, then a real valued function on V is a real partial charge on V if and only if it is strongly additive. The following proposition demonstrates that for the extension problem of charges, the given set function has to be at least a partial charge.
3.2.3 Proposition. (a). Let p be a real charge on a field 9of subsets of a set R. Let V c 9. Then the restriction fi of p to V is a real partial charge on V. (b). Let p be a positive bounded charge on a field 9of subsets of a set R. Let V c 9. Then the restriction fi of p to V is a positive real partial charge on V. Proof. (a) is a consequence of Proposition 3.1.3 and Theorem 3.1.6(2). 0 (b) is a consequence of Theorem 3.1.9(iii). Now, we come to the extension problem for charges.
3.2.4 Theorem. Let p be a real partial charge on a collection %' of subsets of a set R. Let A c R be such that A &V. Then there exists a real partial charge fi on V u{A} which is an extension of p from V to V v {A}. Proof. Let T be the linear functional on 2 ( V ) induced by p on V. If I A E ~ ( V define ), @(A)= T(IA).If I A @ Y ( V )set , fi(A)= any arbitrary but fixed real number. For C in V,set f i ( C )= p ( C ) . Now, we claim that El. is a real partial charge on V u {A}. It is obvious that fi is an extension of p from V to V u {A}.
3.
EXTENSIONS OF CHARGES
65
If I A E 2'(%'), then 2'(%u {A}) = 2'(%'). Consequently, the map T :9(% u {A})+ R is well defined and the set function fi on %' u {A} defined by fi(B)= T(IB) for B in %' u {A} is a real partial charge on %' u {A}. See Theorem 3.1.7(3). fi is the desired extension in this case. If IA&LE(%'), then2'(%'u{A})={f+rIA;f ~2'(%'), r rational}. On2'(%'u we define a functional T as follows. For f +rIA in 2'(%u{A}), T(f + T I A ) = T(f)+rd , where d is a fixed real number. We claim that is a well defined functional on 2(%' u {A}).Let f l + rlIA = f 2 + r2IA for some ') rl, rz rationals. This implies that f l -f 2 = (r2-r1)IA. Since fl, fz in 9(%and f1, f 2 ~ 2 ' ( % ' f1-f2E2'(%'). ), Since IA@2'(%'),it follows that rZ-r1= 0. Consequently, f I =f2. From this, it follows that is well defined on 2'(%' u {A}). It is indeed a linear functional on 2'(%u {A}). Consequently, the set function fi on %' u {A} defined by
{h}),
fi (B) = T(IB) for B in %' u {A} is a real partial charge on %u{A}. See Theorem 3.1.7(3).This completes 0 the proof. The following is the desired extension theorem. 3.2.5 Theorem. Let p be a real partial charge on a collection %' of subsets of a set R. Let 9be any field on R containing %'. Then there exists a real charge on 9which is an extension of p from %' to 9.
Proof. We give a proof based on Theorem 3.2.4 and transfinite induction. Let a. be the least ordinal corresponding to the cardinal number of the collection 9- V. Write 9- %' = {A, ;a ordinal, a < ao}. For each ordinal a
uBc,
u,<,,
Now, we come to the problem of extending positive real partial charges defined on a given collection %' of subsets of a set R to any field 9 on R containing %' as a positive real partial charge. This may not be possible always. Successful extension depends, to some extent, on whether R E %' or not. If R E %', we show that an extension is always possible. If fi & %', the given positive real partial charge on %' can be extended to 9as a positive partial charge but this extension might take the value co on some sets. First, we consider the case R E %'.
66
THEORY OF CHARGES
3.2.6 Definitions. Let V be any collection of subsets of a set R with R E V. Let p be a positive real partial charge on V.For any subset A of R, let
where the supremum is taken over all finite sequences A>,Az, . . . ,A, and B1, Bz, . . . , B, from %' satisfying the condition that m
kIA+
n
Iq z i=1
C
lai
j=l
for some positive integer k . Also, for any subset A of s1,let
where the infimum is taken over all finite sequences Al, Az, . . . , A, and B1, Bz, . . . ,B, from %' satisfying the condition that n
m
I B i + kIA C IA, 2 i=1 i=1
for some positive integer k . The supremum defined above always makes sense. One can take any set C in %' and note that IA+IC?Ic. The infimum defined above is also meaningful since fz E %'. The following proposition puts p i and pe into proper perspective in relation to the extension problem. 3.2.7 Proposition. Let %' be a collection of subsets of a set R with R E V. Let p be a positive real partial charge on V.Let A be any subset of R. If @ is a positive real partial charge on %' u{A} which is an extension of p, then
pi(A) 5 F (A)5 pe(A).
Proof. We prove the inequality pi(A)5 @(A). Let AI, At, . . . ,A,, B1, Bz, . . . , B, be sets from %' satisfying the condition that kIA++C,Tl I B i 2 I,"=, I*, for some positive integer k . Since @ is a positive real partial charge on %u{A}, we have k @ ( A ) + C ~ = l F ( B i ) r C ~ = , F ( A Since i ) . t;i is an extension of p from V to V v{A}, we have
Consequently, we obtain the inequality @ (A) 2 pi(A) by taking supremum
3.
EXTENSIONS OF CHARGES
67
over all finite sequences with the property specified above. The proof of the other inequality is similar. 0 From the above proposition, it is clear that when seeking an extension p from % to % u {A}, the choice of the number fi (A) should confirm to the inequalities established above. The following proposition gives some properties of the set functions pi and pe.
II. of
3.2.8 Proposition. Let % be a collection of subsets of a set R containing R. Let p be a positive real partial charge on %. Let pi and pe be the set functions on the power set B(Q)of R as defined in Definition 3.2.6. Then the following statements are true. (i). 0 5 pi(A) Ipe(A)5 p (0)for every A c R. (ii). If A E % or I A E 9(%), then pi(A) = pe(A)= T(IA),where T i s the linear functional on 9(%) induced by p. (iii). If A, B c R and A n B = 0, then pi(A)+pi(B)Ipi(AuB)Ipi(A)+pe(B) spe.(AuB)Ipe(A)+~~e(B). (iv). If A E %, B c R and A n B = 0, then
and (v). I f A , B c R , A n B = 0 a n d A u B E % , t h e n P (A u
B) = Fi(A) + pe(B) = pe(A) + pi@).
In particular, for any A contained in R, pi(A)+ pL,(AC) = pi(A‘) + pe(A)= CL
(W.
Proof. (i). For any B in %, IA +IB2 IB.Consequently,
Since R is in % and In + In 2 In + IA, we have
LetAl,AZ,.. . ,A,;B1,B2,. . . ,B,;C1,CZ,. . .,C,;andD1,D2,. . . ,D, be sets from % such that kIA + I , : I B j 2 IAj for some positive integer k , and Icj ID,+SIA for some positive integer s. From these inequalities, it follows that
xi”=,
c,!=,
68
THEORY OF CHARGES
so, P
k
C
Iq+S
j=l
m
n
4
j=l
j=l
j=l
C IB,?sC IAi+k C ID^.
Since p is a positive real partial charge on %', we obtain
Rearranging these terms, we get m
C;=I p (Cj)-Xi"= 1 P (Dj), Cjn=1 CL (Aj)- Cj=1 P (Bj) k
S
Hence pe(A)? pi(A). (ii). If I A E ~ ( % ) we , can write rA= riIci for some C1, Cz, . . . ,C, in % and r l , r2, . . . ,rm rationals. Writing ri = n i / N for i = 1,2, . . . , m, where n l , nz, . . . ,n , are integers and N is a positive integer, we can rewrite the above as
EL,
with p 2 1, q 2 1 and A1,AZ,.. . ,Ap, B1, BZ,.. . , B, {Cl, Cz, . . . , C m } .The above representation gives
coming from
Hence pi(A) = pe(A)= T(IA). (iii). Let Al,Az,. . . , A n ; B1,B2,.. . , B m ; C1, C Z ,.. . , Cp and D1, DZ, . . , , D, be sets from %' such that kIA+CLl I,, IA,for some positive integer k, and s I ~ + C ; =ID, ~ ?C:=, Ic, for some positive integer s. Then
Consequently, by observing that I A v B
=I A
+I B , we get
3.
69
EXTENSIONS OF CHARGES
The above inequality can be rewritten as
From this, it follows that pi(Au B)rpi(A)+pi(B). Now, we show that pi(Au B) Ipi(A)+p,(B). Let A1, Az, . . . ,A n ; B1, Bz, . . . , Bm;C1, Cz, . . . , Cp and D1, Dz, . . . , D, be sets in % such that k(IAUB)+C,ElI B i ZC,!=, IAi for some positive integer k, and ID, z Ici+ S I B for some positive integer s. Since I A u B = IA+ I,,
xi"=,
Multiplying throughout by ks, we obtain 4
m
k C I D ~ + SC j=l
j=l
IB,+kSIAzS
n
P
j=l
j=l
1 IAj+k C
Ici.
Consequently,
Taking supremum over A1, AZ,. . . , A,; B1, Bz, . . . , B,; k, first and then supremum over C1, Cz, . . . ,C,; D1, Dz,. . . ,D,; s, we obtain p i ( A ) z pi(Au B) - pe(B). From this, the desired inequality follows. The rest of the inequalities can be proved analogously. (iv). This follows from (ii) and (iii). (v). This follows from (iii). The second part follows if we observe that RE V. O The following theorem paves the way for the extension of positive real partial charges.
3.2.9 Theorem. Let V be any collection of subsets of a set R with R E%. Let p be a positive real partial charge on %. Let A c R be such that A & V. Then there exists a positive real partial charge on V u {A} which is an extension of p. Proof. Let T be the linear functional on Z ( V ) induced by p. Case (i). I A E Z ( % ) . Then the set function fi on Vu{A} defined by F(B) = T(IB), B in V u{A}, is a partial charge on % u{A} and it is the desired extension of p from V to V u {A}. From Proposition 3.2.8(i) and (ii), it follows that F is positive. Case (ii). I ~ & i ( g )Observe . that Z(%'u{A})= {f+rIA;f E 9(%) and r rational}. Choose and fix a real number d satisfying pi(A)~d 5 pe(A). Obviously, by Proposition 3.2.8(i), d 2 0. Define on i(% u{A}) by T ( f + r I A )= T ( f )+rd for f in A?(%) and r rational. We
70
THEORY OF CHARGES
is well defined. Suppose f l +rlIA = f2+r2IA for some f l , f i e -f 2 = (r2- rl)IA.Since la& 9 ( V ) and 2 ( V ) is a linear space, we must have r2 = rl. Consequently, f l =f 2 . Hence T is well defined on 2(% u {A}). Define fi on V u {A} by EL ( B )= F(IB)for B in % u {A}. fi is obviously an extension of p from % to V u {A}. fi (A)= d 2 0. This implies that 6 is positive. Since is well defined, fi is a partial charge. From the inequality pi(A)5 d Ipe(A), it follows that ii is a positive real partial charge on V u {A}. 0 show that
2(%) and r l , r2 rationals. Consequently, f
3.2.10 Theorem. Let V be any collection of subsets of a set R with R E %. Let p be a positive real partial charge on %. Let 9 b e any field on R containing V.Then there is a positive charge p on 9 which is an extension of p.
Proof. One can give a proof based on transfinite induction and Theorem 3.2.9. The argument is similar to the one presented in Theorem 3.2.5. 0
3.3 EXTENSION PROCEDURE OF LOS AND MARCZEWSKI The extension procedure described in the previous section consists of extending the given set function on a given collection of sets to the collection of sets obtained by adjoining just a single set to the given collection of sets. In this section, we present a procedure due to Los and Marczewski for the extension problem which can be described as follows. Let 9 be a and V a subfield of 9. Let p be a positive field of subsets of a set bounded charge on V.Let A E 9 be such that A &V. Let 9 ( V , A) denote the smallest field on R containing V and A. The procedure due to Los and Marczewski consists of extending p from V to 9 ( V , A) as a positive charge in one step. Repeating this procedure, one can extend p from V to 9 as a positive charge. First, we observe that the expressions for pi and pe introduced in Section 3.2 simplify if the domain of p is a field of sets, as the following proposition demonstrates.
3.3.1 Proposition. Let V be a field of subsets of a set R. Let p be a positive bounded charge on V. Let pi and pe be the set functions on the power set P ( R ) of R as defined in Definition 3.2.6. Then for any subset A of s1, (i). pi(A) = Sup { p ( B ) ;B c A, B E %} and (ii). pe(A)= Inf {p (C); A c C, CE V}.
Proof. We prove (i). The proof of (ii) is similar. Let A1, A2,. . . ,A,, B1, B2, . . . , B m be sets from V such that m
kLlIA+
n
1 I B i 2?i 1 IAi i=l =l
3.
71
EXTENSIONS OF CHARGES
for some positive integer ko. We show that there is a set F1 in V such that F1c A and
Since 0 E Ce, we can assume that m = n. Define for each 15 k
6Ai,
u
c k =
I n,
s n j =1
15it
and
By Lemma 3.1.4, n
n
n
i=l
k=l
n
1 I*;= k1 Ick i=l =l
C I B ~ =1 ID,.
and
Then
This leads to n
n
koIA+
n
n
i=l
i=l
c ID;-c~ + c I D ~ ~ c1, Ic;-D; +1 2
i=l
i=l
ICinDi.
This can be written in the form n
k0IA-k
c
n
2
i=l
1 Ic;-D~. i=l
-
-
Since C1 3 C2 3 ' 2 Cn and Dl 3 D23 * * 3 D,, (Di-Ci) n (Cj -Dj) for all i, j = 1,2, . . . ,n. Now, the inequality reduces to n
kora?
=0
n
IC;-Q =
i=l
c I,,
i=l
Say.
Let, for each 15 k I n , F k =
U lsiI
k
n Eii.
j=1
zy=l
--
= Note that F1 3 F22 .IF, and by Lemma 3.1.4, Let k l be the largest integer such that Fk,# 0. (If F1= 0 ,clearly F1= A and ~ ; = 1p(Ai)-C:=l
ko
p(Bi)
5 0 = p (Fi).)
72
THEORY OF CHARGES
IFi and F12 F23 * Since k J A 2 F1. Consequently,
-
*
2 F,,
it follows that A'c F; or A 3
Hence pi(A)5 Sup { p (B); B c A, B E %'}. On the other hand, for any B in %' and B c A, observe that 1, + 1, r> IB.So, gi(A) 2 [ p (B) - p (0)]/1= p (B). Therefore, pi(A) 2 SUP{ p (B); B c A, B E %'}. This proves the desired equality.
0
If %' is a field of sets, pi and pe exhibit some additional properties as the following proposition indicates.
3.3.2 Proposition. Let %' be a field of subsets of a set R and p a positive bounded charge on %. Let pi and pe be the set functions defined on P(R) as in Definition 3.2.6. If A and B are two subsets of R satisfying the conditions that A c C, B c D, C nD = 0 and C, D E %',then pi(A u B) = pi(A) + pi(B)
and pe(AuB)=pe(A)+pe(B).
Proof. We prove the first assertion. The proof of the second is similar. By Proposition 3.2.8(iii), pi(A)+pi(B)Ipi(Au B). By Proposition 3.3.1, for any given E >0, we can find E c A u B such that E E %' and p(E)2
3.
EXTENSIONS OF CHARGES
73
pi(Au B) - E . Note that pi(A uB)-E S p ( E )= p(E n(C u D ) ) = p((En C ) u (E n D ) )
+
= @ ( E nC) p(E nD) 5 pi(A) +pi@)
as E n C c A and E n D c B . Since E > O is arbitrary, it follows that 5 pi(A u B) 5 pi(A)+ pi(B). Hence pi(A u B) = pi(A) + pi(B). Now, we prove the extension theorem.
3.3.3 Theorem. Let % be a field of subsets of a set a.Let p be a positive bounded charge on %. Let A c n be such that A&%.' Let 9(%, A) be the smallest field on R containing % and A. Then there exists a positive bounded charge fi on 9 ( V , A) which is a n extension of p . Moreover, if d is any real number between pi(A) and pe(A), then there is a positive charge fi on 9(%, A) such that fi (A) = d and fi is an extension of p from % to 9(%, A). Proof. Choose and fix a number d E [pi(A),pe(A)]. Then, for C in 9(%, A), let cL(C) = a [ p i ( C ~ A ) + p e ( C ~ A ' ) l + ( l - a ) [ p e ( C n A ) + p i ( C ~ A c ) l , where a satisfies the equation d = api(A)+ (1- a)pe(A). Obviously, O s a 4 1. fi is a positive real function on 9(%, A). We show that fi is an extension of p. Let CEV. Since ( C n A ) u ( C n A ' ) = C E % and ( C n A ) n (CnA')= 0, by Proposition 3.2.8(v), p i ( C n A ) + p e ( C n A c = ) p(C)= pi(Cn A') + pe(CnA). Consequently,
fi (C) = ap (C) + (1- a l p (C) = p (C). This proves that cL is an extension of p from % to 9(%, A). It remains to A). Let p l , p2 be the set functions be shown that fi is a charge on 9(%, A) by defined on 9(%, p 1 (C) = pi(CnA)
+ pe(CnA'),
C in 9( V,A)
and
pz(C>= pi(CnA')
+ pe(Cn A), C in 9(%, A).
It suffices to show that p1 and p2 are charges on $(%, A). Now, we look at p1. Let C, D E 9(%, A) be such that C n D = 0. Then we can write C = ( C l n A ) u ( C ~ n A ' )a n d D = ( D 1 n A ) u ( D 2 n A Cfor ) some C1,C2,D1, D2 in %. See Proposition 1.1.13. By replacing C1 by CI -D1, D1 by DI -CI, C2by C2- D2 and D2by D2- C2, if necessary, we can assume that C1n D =
74
THEORY OF CHARGES
0 and CZnDZ= 0. This is possible since C nD = 0. We note that p
(C u D) = p i( (C u D) nA) + pe((C u D) nA')
+
= p i( (C nA) u (D nA)) p e( (C nA')
u (D nA'))
= pi((C1nA) u (Di nA))+ pe((Cz nA') u(Dz nA')) = Fi(C1nA) fpi(D1 n A) +pe(CznA') +pL,(Dzn A')
(See Proposition 3.3.2.) = pi(C1 nA) + p e ( G nA')
+ Fi(D1 nA) + FePZ nA")
+ pe(C nA') + pi(D nA) + pe(DnA') = Pl(C) + p 1D).
= pi(C nA)
This shows that p1 is a charge. By a similar argument, one can show that pz is a charge. Consequently, @ = a p l + (1- a ) p z is a charge. The above theorem leads to the general extension result. 3.3.4 Corollary. Let U be a field of subsets of a set 0 and p a positive bounded charge on U.Let 2F be a field on 0 containing %. Then there exists a positive bounded charge @ OIE such that @ is an extension of p from % to 9and that the range of fi is a subset of the closure of the range of p on U.
Proof. Let &)
be the closure of the range of p on U.Let
X={(B, v, 9, A ) ; U C Bc 9 c 9 , B and 9are fields on 0, v is a positive charge on B, A is a positive charge on 9,A/W = v, v/U = p and the range of A is contained in &)}. ( v / U is the charge v restricted to the subfield U.)On X, we introduce a partial order Ias follows. For (Bl,vl, g1,A l ) and (Bz,VZ, 9 2 ,Az) in X, say (281,v1, 91, A l ) 5 ( B z , v2, gZ, A2) if the following are satisfied. (i) B z c B l . (ii) v 1 / B z = vz. (iii) 9 1 c gZ. (iv) A z / 9 1 = A 1 . Let {(Bm, v,, gm, A m ) ; a E I'} be a chain in X. Let 93 =nuer Be, 9= Uacr9,, v the restriction of va0 from Boo to B for any a. fixed and A on 9be defined by A (D) = A,(D) if D E 9,, a E r. A is well defined on 9 and is a positive charge on 9. The range of A is contained in &). It follows v , ~ , A ) E Xand is indeed an upperbound of the chain that (3, {(B,, v,, 9,, A,); a E r}.By Zorn's lemma, there is a maximal element
3.
75
EXTENSIONS OF CHARGES
(930, vo, 90, Ao) in X. It obviously follows that a0= %' and v o = p. We claim that go= 9. Suppose gois properly contained in 9. Let A E 9-90 and %'*= 9(90, A). Let Aoi and hoe be as defined in Definition 3.2.6 for the charge A 0 on 90.Define A * on V* by
+
A *(C)= A oi( C nA) Aoe(CnA')
for C in %*. We have already seen in Theorem 3.3.3 that A * is a positive charge on %'*and is indeed an extension of A. from goto %*. We now show that the range of A * is contained in R ( p ) .Let B,, n z 1be a sequence in gosuch that B, c C nA for every n L 1 and Ao(B,) = Aoi(CnA). Let D,, n L 1 be a sequence in 90such that C nA' c D, for every n 2 1 and lim,,,Ao(D,)=Aoe(CnAc). Then CnA'cD,-B, for every n 21. Consequently, Ao(D, -B,) = Aoe(CnA') since Aoe(CnA') 5 Ao(D, -Bn) sAo(D,) for every n L 1. Therefore, lim Ao(Bnu D,) = lim Ao(B,) + lim Ao(D, -Bn)
n-rm
n-02
= Aoi(C nA)
n-tm
+ Aoe(CnA')
= A "(C).
Hence, the range of A * is a subset of the closure of the range of Ao. But the range of A. is contained in &). It now follows that @lo,V O , %*, A *) E X and (930, V O , 90,Ao) (30, vo, %*, A *). This is acontradiction to the maximality of (90, V O , 90, Ao). Hence go= 9, A. is the desired extension of fi from % to 9. This completes 0 the proof. This result can also be proved using Transfinite induction. As a special case of the above theorem, we can prove that every filter in a field 9 on R is contained in a maximal filter in 9.
3.3.5 Corollary. Let p be a 0-1 valued charge on a field % of subsets of a set R. Let 9 be any field on R containing V. Then there is a 0-1 valued charge fi on 9 which is an extension of p from V to 9.Equivalently, every filter in 9is contained in a maximal filter in g. Corollary 3.3.4 can also be used to extend bounded charges from % to 9.
3.3.6 Corollary. Let %' be a field of subsets of a set R and p a bounded charge on V. Let 9 be any field on R containing %'.Then there exists a bounded charge fi on 9 which is an extension of p. Proof. Apply Jordan Decomposition theorem to on %' and then apply Corollary 3.3.4 for positive and negative variations of p separately. See Theorem 2.2.2(1). 0
76
THEORY OF CHARGES
3.4 EXTENSION OF PARTIAL CHARGES IN THE GENERAL CASE In Section 3.2, we dealt with the extension of positive real partial charges defined on a given collection %' of subsets of a set R when R E U. In this section, we examine the situation when R & % and also the extension of partial charges on U taking infinite values.
3.4.1 Lemma. Let % be a ring of subsets of a set R, where R &%'. Let p be a positive real charge on U. Let 9 be the field on R generated by %.' Then there exists a positive charge fi on 9possibly taking the value infinity which is an extension of p from u to 5
Proof. Let % I = { A ~ R ; A '%}. E By Theorem 1.1.9(4), 9=% u U 1 . Let d = Sup { p ( C ) ;C E U}. Case (2). d = 00. In this case, define fi on 9 as follows.
fi (A) = p (A), if A E U, ifAEU1.
=a,
We claim that fi is a positive charge on 9. (It is obvious that fi is an extension of p from U to 9.)Let A, B E 9 and A n B = 0. Case (i). A, B E U. Obviously, fi ( A u B) = lu. ( A u B) = p (A)+ p (B) = fi (A) + fi (B). We claim that A u B E Y1. For, (AuB)'= Case (ii). A E U and B E A'n B' = B'-A E % as B'E %, A E % and %' is a ring. Consequently, fi (A u B) = co = p (A) + 00 = p (A) + fi (B) = fi (A) + fi (B). Case (iii). A E g1 Then and B E U, This is analogous to Case (ii). Case (iv). A, R = A" u B' E % which is not possible since & U. So, case (iv) does not arise. Hence fi is a charge on 9. Case (2).d < co. In this case, we define fi on 9as follows. if A E U ,
fi (A) = Er. (A), =d-p(A'),
ifAEU1.
We claim that fi is a positive bounded charge on 9.The positivity of El. is obvious. Let A, B E 9 and A n B = 0. Case (i). A, B E U. Then fi (A u B) = p ( A u B ) = p ( A ) + p ( B ) = ~ ( A ) + @ ( BCase ) . (ii). A E U and BE%'^. As proved above, A u B E U l . Therefore,
fi (A u B) = d - p ((Au B)') = d - p (B'= fi (B)
= d - p (A"nB')
A) = d - p (B")+ p (A)
+ p (A) = fi (A) + fi (B).
Case (iii). A E U1 and B E %'.This case is similar to Case (ii). Case (iv).
3.
EXTENSIONS OF CHARGES
77
A, B E As in Case (iv) under Case (1)above, this is not possible. Hence fi is a charge on 9 under Case (2) too. 0 Now, we come to the result which complements Theorem 3.2.10.
3.4.2 Theorem. Let % be a collection of subsets of a set R with R @%. Let 9be any field on R containing %. Then there exists a positive charge F on 9 (possibly taking the value co) which is an extension of p from % to 9,
p be a positive real partial charge on %. Let
u;=,
Proof. Let 9 = {AE9;A c Ci for some C1, C2,. . . ,C, in %}. It is clear that 9 is a ring on R. First, we show that + can be extended from % to 9 as a positive real partial charge on 9. Observe that pi(B) is well defined and is a real number for any B c R. pe(A)is also well defined and is a real number for any A in 9. See Definition 3.2.6 and the remarks following this definition. Using the argument given in the proofs of Theorems 3.2.9 and 3.2.10, we can find a positive real partial charge E. on 9 which is an extension of p on %. Note that every partial charge on a ring is a charge. Now, we extend E. from 9 to 9as a positive charge. This can be done as follows. Let d = Sup {$ (D); D E 9}. Case (i). d = CO. Define p on 9 as follows. p(A) = $(A), if A E9, = CO,
if A E9-9.
As in the proof of the previous lemma, one can show that is a positive charge on 9which is obviously an extension of p from % to 9. Case (ii). d
if A E ~ ,
- fi (A'), if A E
-9.
Using the argument given in the proof of Lemma 3.4.1, one can show that is a positive bounded charge on 9 1 which is obviously an extension of p from %. By Corollary 3.3.4, there exists a positive bounded charge F on 9which is an extension of p from %. This completes the proof. 0 Now, we give a condition under which a positive real partial charge on a collection % of subsets of a set R admits an extension E. on 9 which is a positive bounded charge, where 9 is any given field on R containing %. 3.4.3 Theorem. Let p be a positive real partial charge on a collection % R. Let 9 be a field on R containing %. Let 9 = {A E 9; A c Uya1Cifor some C1, Cz, . . , ,C, in %}. Let + e be the set function
of subsets of a set
78
THEORY OF CHARGES
defined on 9 as in Definition 3.2.6. Then there exists a positive bounded charge fi on 9which is an extension of p from % if Sup {pe(D);D E 9} < a.
Proof. The proof of this result is essentially contained in the proofs of 0 Theorems 3.4.2 and 3.2.9. Finally, in this section, we take up the case of charges taking infinite values.
3.4.4 Theorem. Let % be a field of subsets of a set SZ and 9be any field on R containing %. Let p be a positive charge on %possibly taking the value 00. Then there exists a positive charge fi on 9 which is an extension of p from % to g.
Proof. Let 9 ={A E %; p (A)< 00). 9 is obviously a ring on R. Let A be the restriction of p to 9. A is a partial charge on 9. Then by Theorem 3.4.2, we can find a positive charge h on 9 which is an extension of A from 9 to 9.Define fi on 9 as follows. Let A E9.
6
1-7. (A)= h(A), if A c A i for some finite number of sets i=l Al, AZ,. . . , A,, from 9, = a,
otherwise.
fi is the desired extension of is a charge on 9.
p
from % to 9.One can easily check that fi
0
Next, we deal with general charges.
3.4.5 Theorem. Let % be a field of subsets of a set R and 9 a n y field on R containing %. Let p be a charge on % not necessarily real valued. Then there exists a charge fi on 9 which is an extension of p from % to g.I f p is real valued, one can choose fi to be real valued.
Proof. One can give a proof along the lines of the ideas given in the proof 0 of Theorem 3.4.4, using Theorem 3.2.5.
3.5 MISCELLANEOUS EXTENSIONS
In this section, we give various classical extension theorems in some special cases, some of them without proofs.
3.
EXTENSIONS OF CHARGES
79
3.5.1 Theorem. (i). Let % be a semi-ring of subsets of a set 0. Let p be a real charge on V.Let 9 be the smallest ring on R containing %. Then there exists a unique real charge p on 9 which is an extension of p from % to 9. If p is positive on %, so is fi on 9. (ii). Let V be a semi-field of subsets of a set R. Let p be a real charge on V.Let 9be the smallest field on R containing %. Then there exists a unique If p is positive real charge rii on 9which is an extension of p from % to 9. on V,so is 6 on S. (iii). Let % be a lattice of subsets of a set R with 0E % and p a strongly additive real function on V.Let 9be the smallest semi-ring on R containing %. Then there exists a unique real charge p on 9 which is an extension of p from %to 9.If p is positive on %and satisfies the condition that p (A)5 p (B) whenever A, B E V and A c B, then lj; is positive on 9. (iv). Let V be a ring of subsets of a set R and p a real charge on %. Let 9 be the smallest field on R containing %. Then there exists a unique real charge p on 9 such that fi(fl) is a prescribed number. If p is a positive charge on %, there is a positive charge fi on 9 which is an extension of p. Such a p is unique if p (R) = Sup { p (B); B E %} and p is positive real valued. (v). Let p be a positive bounded charge on an additive-class % of subsets of a set R. Let 9 be the smallest field on R containing %. There need not exist a positive bounded charge on 9 which is an extension of p. Proof. Even though the results of this theorem can be proved using some of the results of the previous sections, we give direct proofs. (i). By Theorem 1.1.9(2), 9 = { A c R ; A is a finite disjoint union of sets Ci for some from V}.Define p on 9 as follows. Let A E 9.Then A = pairwise disjoint sets in V. Set p(A)=CE, p(Ci). We show that fi is unambiguously defined. Suppose A = Ci = D , where C1,Cz , . . . ,Cm are pairwise disjoint sets in % and D1,D 2 , .. . ,Dn are pairwise disjoint sets in %. Then
uy=l u?=l uyZl
u u (CinDj). m
A=
n
i=1 j=1
u;=l
Note that Ci = (Ci nDj) for every 1 5i 5 m and Ci nDj E V for all i and j . Similarly, Dj = uE1(Ci nDi) for every 15 j 5 n. Since p is a charge on %, we have m
and
m
n
80
THEORY OF CHARGES
This proves the unambiguity of the definition of fi. It is obvious that fi is a real charge on 9 and is an extension of p from V to 9.The uniqueness of fi is also clear. If p is positive on V,so is fi on 9.This completes the proof of (i). (ii). This is similar to (i). }. on9by (iii). ByTheorem 1.1.9(1),~ = { F - E ; E , F E V , E ~ F Definefi ii(F- E) = p (F)- p (E) for E, F E V and E c F. We prove the unambiguity of the definition of fi on 9.Suppose F1-El = Ft - E2 for some El, F1, El, FZ in V with E l c F 1 and E2cF2. So, F z u E 1 = F l u E 2and F z n E l = FlnE2. Since p is a modular function on %, p(F2)+p(E1)= E.L (F2 u El) + p (FznEd = CL (F1u E d + p (FI n E d = p (Fd + p 032). Hence p (Fz)- p (E2)= p (FJ - p (El).This shows that fi is unambiguously defined on 9. We now show that fi is a charge on 9. Note that f i ( 0 ) = 0 . Let F1-El and F2- E2 be two disjoint sets in 9such that (F1-El) u (F2 - Ez)= F3- E3E 9, where El, F1, EZ,F2, E3, F3E V,El c F1, Ez c Fz and E3c F3. In order to show that fi (F1- El) + fi (F2- E2)= fi (F3- E3), it suffices to show that p (FI)+ p 0%)+ p (Ed = p (F3)+ p (El)+ p W. Clearly,
F1 u F ~ E3 u = F ~ El u u Ez, (F1nF2) u (F1nEd u (Fz E3) = (El nEz)u (El F3) u (Et nF3), and
F1n F z n E3= F3nEl nEz. By Proposition 3.1.5, the desired equality follows. Since p (0)= 0, it follows that fi is an extension of p. It is obvious that fi is unique. If p is positive and monotone on V,then fi is positive on 9. (iv). Assume that R &%'.(If R E V,then 9= V.)Let Vl ={A; A'E %}. By Theorem 1.1.9(4),9=V u Vl. Let d be any prescribed number. Define fi on 9 as follows.
fi (A) = I*. (A),
if A€%':,
= d - p (A"), if A E
%I.
As in the proof of Lemma 3.4.1, one can show that fi is a charge on 9. fi is, obviously, an extension of p from V to 9.Uniqueness of fi with the property that fi (a)= d also follows easily. If p is a positive charge on %, define fi on 9by fi(A)=p(A), ifAE%,
=a,
ifAEV1.
fi is a positive charge on 9 and is an extension of
p
from V to 9.
3.
81
EXTENSIONS OF CHARGES
If p is a real positive charge on %, define r*. on 9 by if A E %,
fi (A) = p (A), =d-p(A'),
ifAEV1,
where d = Sup { p (B); B E %}. fi is the desired extension of (v). Here is an example. Let a= (1,2,3,4} and = ( 0 ,{1,21, K 3 1 , {1,41,(2,317 (2,417 (3,417
p.
a).
% is an additive-class on R. The smallest field 9on R containing % is the power set 9(a)of R. Let p on % be defined by p ( 0 ) = 0 , p{1,2})=$= p1.(3,4)),p({l,3))=1, pCL{2,41)=0, p(U,4>>=a,p({2,3))=.% p ( W = l . /Jis a positive bounded charge on %. But there is no positive bounded charge on 9 which is an extension of p. 0
Another important extension theorem is the following theorem of Caratheodory.
3.5.2 Theorem. Let p be a positive bounded measure on a field 9 of subsets of a set R. Let be the smallest a-field on R containing 9. Then there exists a unique positive measure on which is an extension of p. Proof. In any standard text book on Measure theory.
0
Related to the above theorem is the following approximation theorem.
3.5.3 Theorem. Let p be a positive bounded measure on a u-field of subsets of a set R. Let %be any field on R generating Then for any A in and E > 0, there exists F in F s u c h that p (F A A) < E.
a.
Proof. Let 9={A€ '? for I;every E > O there exists F in 9 such that p (A A F) ;< E}. Clearly, 9c 9 and 9 is closed under complementation. We shall show that 9 is closed under countable unions which will complete the proof. Let A,,, n 2 1 be a sequence in 9and E > 0. For each n L 1, let F, €9be such that p(An A F , ) < E / ~ " + ~Let . m 2 1 be such that p(UnzlF, -U:=i Fi) < ~ / 2NOW, .
+pi(
u Fn) A (6Fn))
nrl
n=l
82
THEORY OF CHARGES
We need some results about measures on topological spaces which we shall state here.
3.5.4 Definition. Let X be a compact Hausdorff space and 9 its Borel u-field. A positive bounded measure p on 9 is said to be regular if p (A) = Sup {p ( C ) ;C compact,
C c A}
for every A in 9.
3.5.5 Theorem. Let X be a compact Hausdorff space, 90its Baire a-field and 93 its Borel a-field. Then any positive bounded measure on 30can be extended uniquely as a regular measure on 9. Proof. See Theorem D on p. 239 of Halmos' Measure theory.
0
3.6 COMMON EXTENSIONS The problem considered in this section is the following one. Let 9 be a field of subsets of a set R and V and 9 be two subfields of S. Let p1 and p 2 be two positive charges on % and 9 respectively. Under what conditions can we find a positive charge p on 9 which is an extension of both p 1 and p2 to 9?The condition needed, surprisingly, turns out to be a natural one.
3.6.1 Theorem. Let % and 9 be two fields on a set R and p1 and p2 positive bounded charges on % and 9respectively. Let 9 be a field on R containing both V and 9. Then a necessary and sufficient condition for the existence of a positive bounded charge on 9which is a common extension of both p 1 and p2 is that pt(C)2 ~ . z ( D )whenever C E %, D E 9and C 2 D,
and p l ( E ) s p ~ ( F )whenever E E %, F ~ 9 a n dE c F .
Proof. The necessity of the condition is obvious. We will prove the sufficiency. From the given condition, it follows that p l(A)= p2(A) whenever A E % n9. We define a function El. on % u 9 unambiguously as follows. El. (A) = p (A), if A E %', = p2(A), if A E9.
We now show that El. is a positive real partial charge on % u9. Let At,A2,. . . , A,,, and Bt, B2, . . . ,B,,, be two finite sequences of sets from
3.
83
EXTENSIONS OF CHARGES
% v 9 such that P+q
m+n
C
C
IAi
i=l
IBi-
i=l
Assume, without loss of generality, that A l , A z , . . . , A m € % , . A m + , E 9,BI, Bz, .,Bp E %' and Bp+l, B ~ + z *,.. ,Bp+q E 9. A m + l Am+2,. , The above inequality can be written in the form m n " a C IA,+ C IA,,,2 IB,+ IB,,,.
-
9
i=l
2
i=l
2
i=l
i=l
Since 0 E % n9, we can assume, without loss of generality, that m = p and n =q. By Lemma 3.1.4, we can assume A 1 1 A z 3 . - 3 A m ;A m + l ~ A m + 2 3 . 1 A m + " ; B 13 , B ~ l* - 3 B p ; B p + l3 B p + Z3 3BP+,.Theabove inequality can be rewritten in the form
-
m
-
m
n
IAz-B, i=l
+
n
IA,+i-B,+i
IBi-Ai
i=l
+C
IB,+i-A,+,*
i=l
i=l
Let Ci=Ai-Bi, i = l , 2 , . . . , m ; Di=Bi-Ai, i = l , 2 ,..., m ; E i = Amci-Bmti, i = 1 , 2 , . . . ,n ; Fi = Bmti-Am+i, i = 1 , 2 , . . . ,n. The above inequality then becomes m
m
n
C I C , +i = l IEi i=l
n
C
ID,
+C
IFi.
i=l
i=l
Observe that Ci nDj = 0 for all i and j and Ei nFj = 0 for all i and j . By Lemma 3.1.4, we can assume that C13 C z 3 -. 3 C m ;D13 D z 3 . 3 D m ; E I =JEZ3 * * 3 E n; FI 3 FZ * 3 Fn. The above inequality leads to two inequalities:
-
-
m
n
m
n
C I c i ri C= l IF, and i-1
C I E , 2 i = l ID,. i=l
Note that C1, Cz, . . . , C,, D1, Dz, . . . , D, E %' and El, Ez, . . . , En, F1, FZ,. . . , Fn E 9. Now, observe that Fi c Ci and Di c E i for all i. Consequently, by the given condition, m
n
i=l
m
n
C p l ( C i ) r C pZ(Fi) and C i=l
pZ(Ei)>
i=l
C p1(Di).
i=l
Hence
f
i=l
fi(Ci)+
fi(Di)+
fi(Ei)z
i=l
i=l
f. fi(Fi).
i=l
From this, it follows that (using Proposition 3.1.5), m+n
P+cq
C fi(Ai)? C fi(Bi)*
i=l
i=l
84
THEORY OF CHARGES
Hence fi is a positive real partial charge on % u9.By Theorem 3.2.10, we can find a positive bounded charge p on 9 which is an extension of 6 from % u $3to 9. This completes the proof. Now, we take up the case of real charges.
3.6.2 Theorem. Let % and 9be two fields on a set R and p l and p2 two real charges on % and 9respectively. Let 9 be a field on R containing both V and 9. A necessary and sufficient condition for the existence of a real charge p on 9 which is a common extension of both p 1 and p z is that p1(A) = pz(A)for every A in % n9. Proof. The proof of this theorem is similar to the one given for the previous theorem, It may be remarked that these two theorems remain valid if p l and take infinite values.
pz
The two theorems proved above are not extendable to the case when Here are the relevant examples. there are more than two subfields of 9.
3.6.3 Example. Let R={l,2,3,4}; %1=
9=P(R);
(0, {1,21,{3,41, R1; V2 = (0, K 3 1 , (2341, n1; %3 = (0, (1,419 (2,313 n1;
pi({1,2))=& p2({2,41) =
t;
pi({3,4})=% p3({1,41) =
a,
@2({1,3))=$, p3({2,31) = .:
Each pair of p1 and p 2 , p l and p3, kz and p 3 satisfies the condition of Theorem 3.6.1. But there is n o positive charge on 9 which is a common extension of all the three charges p l , p2 and p 3 .
3.6.4 Example. Let
R = {1,2,3}; %1=
9=P(R);
(0, (11, (2,319 R1; %2 = i0, (21, {1,31, R1; V3 = (0, (31, (1,219 a);
@11({1I)=i=E.L1({2,31); p2({21)=&p2({1, 31)=f; 113({31)=
a,
p3({1,21) =
t.
Note that p l= p 2 on V l n V 2 , p l= p 3 on V 1 n V 3 and p 2 = p 3 on V2nV3. But there is no real charge p on 9 which is a common extension of pl, PZ and P3.
CHAPTER 4
Integration
In this chapter, we develop the theory of integration for real valued functions with respect to charges. Integration with respect to charges requires a good deal of tact, patience and circumspection to get around measurability problems. The treatment of this topic given here is fairly comprehensive. After presenting the preliminaries in the first three sections, we develop D-integral as presented by Dunford and Schwartz in Section 4.4. In Section 4.5, we introduce S-integrals which are of Stieltjes type and make comparisons with D-integrals. L,-spaces are introduced and studied in Section 4.6. Finally, in Section 4.7, ba(n, 9)is realized as a dual space.
4.1 TOTAL VARIATION AND OUTER CHARGES Let p be a charge defined on a field 9 of subsets of a set n. Recall the definitions of positive and negative variations, p + and p - , of p as expostulated in Section 2.5. p+(A)=Sup{p(B);BcA,B~fl,
A E ~ ,
and
@-(A)= -1nf {p(B);B c A, B ~ f l ,A E P .
As has been noted in Theorem 2.5.3, p + and p - are positive charges on g.The total variation lpl of p has been defined for bounded charges p on $. This notion can be introduced for any charge p. 4.1.1 Definition. For any charge p on a field 9 of subsets of a set 12, the total variation lp I of p is defined by
IF [(A)= p + ( ~+ )p - ( ~ ) , A E 9. Clearly, lp I is a positive charge on 2E lp I can also be described following way as in Theorem 2.2.4.
in the
86
T H E O R Y OF CHARGES
4.1.2 Theorem. For any charge A in 9,
p
on a field 9of subsets of a set R and
I~I(A)=sUP
f: Ip(Bi)l
i=l
holds true, where the supremum is taken over all partitions {Bt, B2, . . . , B,} of A in 9. Proof, If ~ ~ / ( A ) < cthen o , p is a bounded charge on the field A n 9 = { A n B; B E fl on A and the above equality follows from Theorem 2.2.4. If Ipl(A) = m, then either p+(A)= co or p-(A) = 00. From the definitions of p + and p - , it follows that SupCy=, Jp(Bi)l=a,where the supremum is taken over all partitions {B1,B2,. . . , B,} of A in 9. This proves the theorem. 0 Now, we introduce the concept of an outer charge.
4.1.3 Definition. Let 9be a field of subsets of a set SZ and p a positive charge on 9. The set function p * :P ( R ) [0, co] defined by --f
p*(A) = Inf {p(B);A c B, B E F}, is called the outer charge induced by
A cR
p.
The following proposition chronicles some of the properties of outer charges.
4.1.4 Proposition. Let F b e a field of subsets of a set SZ and p a positive charge on 9. Then the following are true. (i). p *(D)= 0. (ii). p *(A)s p*(B) if A c B c 0. (iii). p *(A) = p (A) if A E 9. (iv). p * ( A u B ) s p * ( A ) + p * ( B )if A, B c R . (v). I f 9 is a c+-field and p is a measure on 9, p*(Unat A,)S CnZlp*(A,) for any sequence A,, n 2 1 of subsets of R. Proof. Properties (i), (ii) and (iii) are obvious. To prove (iv), we proceed as follows. If either p*(A) = 00 or p*(B) = 00, the inequality obviously follows. Suppose p *(A)< co and p*(B)
4.
87
INTEGRATION
{1,2,3, . . .}). Define p on 9 by p (A) = 0, = 1,
if A is a finite subset of {1,2,3, . . .}.
,
otherwise.
is a measure on the field 9. Note that p*({l,2 , 3 , . . .)) = 1. If we let A, = { n } ,n 2 1, then ps(Un2l A,,)= 1 and p*(A,,)= 0.
p
Finally, we end this section with a result on the outer charge induced by the sum of two charges. 4.1.6 Proposition. Let subsets of a set 0. Then
p1
and p2 be two positive charges on a field 9 of
(p1+pz)*=pT+CLK
Proof. It is obvious that ( p l + pz)*(A)2 p (A)+ p ; (A) for every A c a. If eitherp?(A)=ooorp:(A)=oo, then ( p 1 + p Z ) * ( A ) s p T ( ( A ) + p $ ( A is ) true. Assume that p (A)< oo and p: (A) < CO. Let E > 0. There exist B1, BZin 9 such that A c B1, A c BZ,
T
T
and ~ nB2)5 p ? (A)+ p: (A) + E . Since E > 0 Hence ( p I + pz)*(A)5 ( p +p2)(B1 is arbitrary, the result follows. 0
4.2
NULL SETS AND NULL FUNCTIONS
In this section, we formalize the notions of a null set and a null function. We first introduce the notion of a charge space. 4.2.1 Definition. A charge space is a triple (0,9 p ), , where R is a set, 9 is a field on R and p a charge on 9. 4.2.2 Definition. Let (R, 9,p ) be a charge space. A subset A of R is said to be a p-null set, or simply, a null set if p is understood, if lp ["(A)= 0.
The following properties of null sets follow from Proposition 4.1.4. 4.2.3 Proposition. Let (R, 9, p ) be a charge space. Then the following statements are true. (i). 0 is a null set. (ii). B is a null set if B c A and A is a null set.
88
THEORY OF CHARGES
(iii). u7-1A iis a null set if Al,A2,. . . , A, are null sets, where n is any positive integer. (iv). UnzlA, is a null set if A,, n 2 1 is a sequence of null sets, p is a measure and 9 is a a-field. The concept of a null set leads to the concept of a null function. Definitions. Let (R, 9, p ) be a charge space. (i). A real valued function f on R is said to be a p-null function, or simply a null function if p is understood, if
4.2.4
IP I*GJE a; If(w>l>E N = 0 for every E >O. (ii). Two real valued functions f and g defined on R are said to be equal almost everywhere iff - g is a null function. In this case, we use the notation f = g a.e.[~]. (iii). A real valued function f on R is said to be dominated almost everywhere by a real valued function g on R, if there exists a null function h on R such that f 5 g + h. In such a case, we use the notation f 5 g a.e.[p 3.
A sufficient condition for f 5 g a.e. [ p ] to hold is that lp I*({w E R; f ( w ) > g(w)})= 0 , though not necessary. The following proposition follows easily from Proposition 4.1.4.
Proposition. Let (R, 9, p ) be a charge space. The following statements are true. (i). cf +dg, IflP(p>0) and f g are null functions whenever f and g are null functions on R and c and d are real numbers. (ii). g is a null function if l g l s If I a.e. [ p ] and f is a null function. (5).For real fimctions f , g, h on R, f = h a.e. [ p ] whenever f = g a.e. [ p ] and g = h a.e. [ p ] . (iv). For real functions f l , f 2 , g l , g2 on R and c, d real numbers with f 1 = g l a.e. [ p ] and f 2 = g2 a.e. [ p ] ] cfi+df2=cgl-t-dg2 , a.e. [ p ] and I f l [ = lgll a.e. 4.2.5
[PI.
(v). On the space C(R, 9,p ) of all real functions on R, the binary relation defined by f g i f f = g a.e. [ p ] is an equivalence relation. (vi). For real functions f , g, h on R, f 5 h a.e. [ p ] whenever f 5 g a.e. [ p ] and g Ih a.e. [ p ] . (vii). A subset A of R is a null set if and only if I, is a null function. (viii). For real functions f , g on R, f = g a.e. [ p ] i f f 5 g a.e. [ p J and g f a.e. [ p ] .
-
-
4.2.6 Remark. If f l = g l a.e. [ p ] and f 2 = g2 a.e. [ p ] , it is not true that f i f 2 = g l g 2 a.e. [ p 1. It is not even true that f 2 = g 2 a.e. [ p ] iff = g a.e. [ p ] . The following example explains this.
4. INTEGRATION
89
Example. Let R = { l , 2 , 3 , . . .), 9 the finite-cofinite field on R and p the charge on 9 defined by p (A) = 0, if A is finite, = 1, if A is cofinite.
Letf a n d g o n R b e d e f i n e d b y f ( n ) = n + ( l / n ) , n 21 a n d g ( n ) = n - ( l / n ) , n 2 1. Then f = g a.e. [ p ] . But f z = g z a.e. [ p ] does not hold.
A sufficient condition for a real function f on R to be a null function is that lp I*({w E R; f ( w ) # 0 ) )= 0, though not necessary. The following proposition amplifies this point. 4.2.1 Proposition. Let (R, 9, p ) be a charge space. Let f be a real valued function defined on R. (i). I f IpI*({o E R; f ( w )# 0))= 0, then f is a null function. (ii). The converse of (i) is not true. (iii). If 9 is a u-field and p is a measure on 9, then f is a null function if and only if lp I*({w E R; f ( w )# 0 ) )= 0. Proof. (i). Note that for any E > 0, {oE
a;1f ( w ) l >
E}
c {w E R; f ( w ) # 0).
The monotonicity of Ip I* completes the proof. CL)be as in Remark 4.2.6. Let the function f on (ii). Example. Let (R, 9, R be defined by f ( n ) = l / n , n 21. Then f is a null function. For, {w E R ; If(w)I > E } is a finite set for any E >O. On the other hand, IpI*({w € 0; f (w 1 # 0))= 1. (iii). Observe that
u
E
>O
{WEn;If(o>l>E)=
u {OEn;If(w>l>l/n>
nzl
= { w E R; f ( w ) # 0).
By Proposition 4.2.3, the result follows.
0
Now, we come to the concept of essential boundedness.
4.2.8 Definition. Let (R, 9, p ) be a charge space and f a real valued function on R. f is said to be essentially bounded if there exists a null set A contained in R such that f is bounded on A'. Iff is essentially bounded, the essential supremum of f is denoted by (1film and is defined by
llfllm
= Inf SUP{If(w)l; w E A"),
where the infimum is taken over all null sets A contained in R. The following proposition gives equivalent conditions for essential boundedness of a function.
90
THEORY OF CHARGES
4.2.9 Proposition. Let (R,9, p ) be a charge space and f a real valued function on R. Then the following statements are equivalent. (i). f is essentially bounded. (ii). There exists k > 0 such that lp I*({w E R; If(w)I> k } ) = 0. (iii). There exists k > O such that If l Ik a.e. [ p ] . (iv). There exists a bounded function g on R such that f = g a.e. [ p ] . The following proposition gives the properties of essentially bounded functions which are easily established.
4.2.10 Proposition. Let (R,9,p) be a charge space. (All functions considered below are real valued functions on R.) (i). I f f and g are essentially bounded and c and d are real numbers, then cf + dg is essentially bounded. Further, (ii). 11f Ilm = 0 if and only i f f is a null function. (iii). I f f is essentially bounded and f = g a.e. [ k ] ,then g is essentially bounded and 11film = llgIlm. (iv). If B(R, 9,p ) is the collection of all essentially bounded functions on R, then the map 11 * llm :B(R, .fF, p ) + [0,00) is apseudo-norm on B(R, 9’’ p).
-
4.2.11 Remark. Introduce an equivalence relation on B(R, g, p ) by f g for f , g in B(R, 9, p ) iff = g a.e. [ p ] .Let Cm(R, 9,p ) be the collection of all equivalence classes of B(R, 9, p ) under -. For f in B(R, 9, p ) , let [f ] denote the equivalence class containing f . The map 11 Ilm :Cm(R, 9, p )+ [O, a) defined by Il[f]ilm = l\fllm for [ f ] in Cm(R, 9, p ) is a norm on cm(ag,p ).
-
The following concept of a simple function is important for the development of D-integral.
4.2.12 Definition. Let 9be a field of subsets of a set R. A real valued function f on R is said to be a simple function if it can be written in the form
f
=
c CiIFt
i=l
for some real numbers c l , c2, . . . ,c, ; F1, F z , . . . ,F , in 9with Fi nF j = 0 for every i # j and I Fi = R.
uY=
The following properties of simple functions are clear.
4.2.13 Proposition. Let 3 be a field of subsets of a set R. I f f and g are simple functions on R and c and d are real numbers, then cf +dg, f g and If Ip(p > 0 ) are all simple functions.
4.
INTEGRATION
91
Now, we introduce smooth functions.
4.2.14 Definition. Let (R, 9, p ) be a charge space. A real valued function f on R is said to be smooth if for every E > 0 there exists k > 0 such that IPI*hJE n ; If(u)l>kl)<E.
Not every function is a smooth function as the following example demonstrates.
4.2.15 Example. Let (0,9, p ) be as in Remark 4.2.6.The function f on R defined by f ( n )= n, n E R is not smooth. However, in some cases, many natural functions are smooth. In what follows, let 93 denote the Bore1 u-field on the real line R, i.e. 93 is the smallest u-field on R containing all subintervals of R.
4.2.16 Definition. Let (R, 9, p ) be a charge space in which 9is a cT-field A real valued function f on R is said to be on R and p a measure on 9. p-measurable if there exists a null set N c R such that f-'(B) n N'E 9 for every B in 93. 4.2.17 Proposition. Let (R, 9, p ) be a charge space in which 9 i s a u-field and p a real measure on 9. Then every p-measurable function f on R is smooth. Proof. Let N be a null set such that f-'(B) nN'f 9 for every B in 93. Let B, ={u ER;If(u)l?n}, n 2 1. Note that B,, n 2 1.10 . Since p is a real measure on the u-field 9,lpl is a bounded measure. Consequently, lim,,,Ip~(B, nN')=O. For any E >0, we can find a positive integer k such that IpI(BknN') < E . Now, observe that I c L I * ( B ~bL(*((Bk = n N ) u ( B k nN")) 5 Ip I*(Bk n N) + Ip I*(Bk n N')
51pl*(N)+IpI*(BknNc)
U
The following proposition records some of the properties of smooth functions.
4.2.18 Proposition. Let ( R , 9 , p ) be a charge space. (i). I f f and g are smooth functions on R and c and d are real numbers, then cf + dg, f g and I f I"( p > 0 ) are all smooth functions. (ii). I f f is essentially bounded, then f is smooth.
92
THEORY OF CHARGES
4.3 HAZY CONVERGENCE In this section, we introduce the notion of hazy convergence in the space of all real valued functions on R of a charge space ($ p )I . This ,$ notion , is commonly known as “convergence in measure” which is a misnomer in the present context of charges.
4.3.1 Definition. Let (a,9, p ) be a charge space. A sequence f n , n 2 1 of real valued functions on R is said to converge to a real valued function f on R hazily if lim l p l * ( { ~ ~ IR f n ;( W ) - f ( u ) I > E I ) = O
n-tw
for every
E
> 0.
The following proposition establishes that the limit function in hazy convergence is essentially unique.
Proposition. Let (R, 9,p ) be a charge space. If a sequence f,, n 2 1 of real valued functions on R converges to a real valued function f on R 4.3.2
hazily and f = g a.e. [ p ] , then f n , n L 1 converges to g hazily. Conversely, if f n , n 2 1 converges to both f and g hazily, then f = g a.e. [ p ] . Proof. For the first part, observe that for any E > 0, (W
lfn(W)-g(w)l>E)C{W
lfn(~)-f(~)I>~/2)
U { W En;
If(w)-g(w)l>E/21.
For the second part, observe that for any E > 0, {W
If(w)-g(W)I>E)C{W U{W
€0;I f n ( u ) - f ( a ) l > E / 2 ) Ifn(W)-g(W)I>E/2).
The monotonicity and the sub-additivity properties of lp I* complete the proof. 0 The following theorem gives algebraic properties of hazy convergence.
4.3.3 Theorem. Let (R, 9, w ) be a charge space. Let fn, n 2 1 and g,, n L 1 be two sequences of real valued functions on R converging hazily to real valued functions f and g on R respectively. Then the following statements are true. (i). cf, +dg,, n 2 1 converges to cf +dg hazily for any two real numbers c and d. (ii). If , I, n 2 1 converges to If I hazily.
4.
INTEGRATION
93
(iii). f:, n L 1converges to f’ hazily, i f f is a null function. (iv). fnh, n 2 1 converges to f h hazily, if h is a smooth function on R. (v). I f f is smooth and $ is a real valued continuous function defined on the real line, then $ ( f n ) , n 2 1 converges to $ ( f ) hazily. In particular, If n I p , n 2 1 converges to If Ip hazily for any p > 0. (vi). f,g,,, n 2 1 converges to f g hazily i f f and g are smooth functions. (vii). f:, n 2 1 converges to f’ and f i , n 2 1 converges to f - hazily. (viii). f n v g,, n 2 1 converges to f v g and f n A g,, n 2 1 converges to f A g hazily.
Proof. (i), (ii) and (iii) are easy to prove. (iv). Let E >O. Since h is a smooth function, there exists a real number k >0 such that Ip I*({w E R; Ih ( w ) ]> k } ) < ~ 1 2Since . f,, n L 1 converges to f hazily, there exists an integer m 2 1 such that IF I*({o E R; If n ( w ) - f ( w ) l > S / k } )< ~ / whenever 2 n 2 m, where S is a given positive number. So, if nzm,
IP I*({w
If n (0) h ( w ) - f ( w ) h (w )I > 6 ) ) 5 IP I*(b E 0;Iffl ( w ) - f (w)l Ih (w)l> 8,Ih (@)I 5 k } ) + IP I * ( b E a;I f n (0)- f (w)I Ih (0)I > 8,Ih (w )I > k } ) < IP I*({w E a;If n ( w ) - f ( w )I > ~ / k )+)~ / 2 E
< & / 2 + & / 2= E . This proves (iv). (v). Let E~ > 0 and E:! > 0. Since f is smooth, there exists a real number k > 0 such that Ip I*({w E R; If ( w ) l > k } )< ~ ~ / Since 2 . $ is uniformly continuous on [-2k, 2k], there exists S > 0 such that I $ ( x ) - $(y)l< ~2 whenever 1x1, IyI 5 2 k and Ix -yI < S . Without loss of generality, assume that S < k . Since f,, n 2 1 converges to f hazily, there exists m 2 1 such that lp I*({w E Q; If, ( w )-f ( w ) l > 8)) < e1/2 whenever n 2 m. Now, if n 2 m ,
IP I*({w E I $ ( f n (w 1) - (I,( f (w ))I > ~ 2 1 ) 5 IP I*({w E a;I $ ( f n ( w ) ) - $ ( f (w))I > E Z ~ If(w)l>kI) +I~l*({w +IPI*({W
I1Cl(fn(w)>-$(f(w))I>&*, E n ; I$(fn(w))-$(f(O))I>E2,
I f ( w ) l 5 k ,I f n ( w ) - f ( w ) I > S ) ) I f ( w ) l s k ,I f n b ) - f ( W ) I ~ S H
< E1/2+&1/2+0 = E l . Thus $ ( f , ) , n 2 1 converges to $(f) hazily. (vi). Observe that for any n 2 1,
94
THEORY OF CHARGES
Since f and g are smooth, f + g and f - g are smooth. See Proposition 2 4.2.18(i). By (i) and (v), ( f n +gn)2,n 2 1and ( f n -g,) ,n 2 1converge hazily to ( f + g ) 2 and ( f - g ) 2 respectively, Again, by (i), fngn, n 2 1 converges to fg hazily. and f- = $(f The assertion now fol(vii). Observe that 'f = i(f+ lows from (i) and (ii). f, v g, = f ( f , + g, + Ifn - g,l) and f, A g n = (viii). Observe that i(fn+gn-Ifn-gnI)for a l l n r l . 0
If[).
Ifl)
4.3.4 Remark. The assertion of Theorem 4 . 3 . 3 6 ~ is ) not valid unconditionally. We need to impose some conditions on h to ensure the hazy convergence of fnh, n 2 1 to fh. The following is a relevant example. Let (0,9, p ) be as in Remark 4.2.6. For n k 1, let fn
( k )= k, = 1-l/n,
if l s k s n , if k > n ;
f ( k )= 1
for all k in R;
h ( k )= k
for all k in R.
It can be checked easily that f n , n 2 1 converges to f hazily. But fnh, n 2 1 fails to converge to fh = h hazily. Let ( R , 9 , p ) be a charge space and C(n, 9, p ) the collection of all real valued functions on R. One can introduce a pseudo-metric p on C(R, .F,p ) such that convergence in the pseudo-metric space C(R, 9, p ) coincides with hazy convergence. p ) , Let For f in C(R, 9,
$(f, c 1 = c + Ip I * ( b E 0; If(w )I > cl),
c
> 0.
$(f, c ) is a nonnegative number and could be equal to 00. If p is bounded, then $(f,c) is a real number for all c > 0. Now, define
If $(f,c ) = 00 for every c > 0, we define llfll= 1. Now, we give the properties of the function 11 * 1 .
4.3.5 Proposition. The function I\.\\ defined on C(R, g,p ) above has the fdlo wing properties. = 0 if an d only i f f is a null function. (i). (ii). IF+gII 5 llfIl+llgll. (iii). The function p defined by p(f, g) = Ilf-gll for f, g in C(n, 9, p ) is a pseudo-metric on C(R, g,p ) .
]If]
4. INTEGRATION
95
(iv). fn, n L 1 converges to f hazily if and only if p ( f n ,f),n 2 1 converges to zero.
Proof. (i). If f is a null function, then +(f, c ) = c for every c > 0. Consequently ,
MI=
fril+c C
- 0.
Conversely, let llfll= 0. Let k be any positive number. In order to show that Ip I*({w E SZ; If(w)I > k } ) = 0, it suffices to show that lgl*({w E SZ; If(w)I > k } )< E for any 0 < E < k . Let 0 < E < k . Since llfll= 0, there exists c > O such that +(f, c ) / ( l ++(f, c ) ) < E / ( ~ + E ) . This implies that +(f, c ) < E . So, c < E and lp I*({w E SZ; I f ( @ ) [ > c } ) < E . Consequently, Ip I*({w E SZ; If(w)I >k } )I:IP I*(bE a;If(w)I > E ) ) 5 lp I*({w E a; I f ( w)I > c } ) < E . This shows that f is a null function. (ii). Observe that
= Inf r>O.s>O
Inf
I :
r>o,s>o
I Inf XI
+(f+g,r+s) l++(f+g, r+s)
+(f,r) $(g,s) l++(f,r) l++(g,s) +
+(f, r ) +Inf %4 $1 1 + +(f,r ) s>o 1 + +(g, s )
Ilfll+llgll.
I
(The first of the above inequalities can be proved as follows. The function y ( x ) = x / ( l +x), 0 I:X I:co is an increasing function having the additional property y ( x + y ) Iy ( x )+ y ( y ) for all 0 S X , y ICO. We use the convention that y(00) = 1. If c 1 = ICL I * ( b E ; If (w ) + g (w )I > r + sl), c2
=
I@I * ( b E a;If(w)I > r } )
and c3 =
IcLI*(b E a;Ig(0)l >sH,
then c1 I:c2+c3. Further,
96
THEORY OF CHARGES
From this, the first of the above inequalities follows. The rest of the equalities and inequalities are obvious.) (iii). This follows from (i) and (ii). (iv). Suppose fn, n 2 1 converges to f hazily. Then for any c > 0 and E > 0, there exists m L 1 such that IPI*({W
E n ;Ifn(W)-f(W)I>CH<E
whenever n L m. Consequently, for n L m,
C & C I---+-<-+&.
I+& l + c Consequently, lim p (fn, f)I c/(l + c ) + E . Since c > 0 is arbitrary, p ( f n , f)5 E . Since E > 0 is arbitrary, it follows that f n , we have lim n L 1 converges to f in the pseudo-metric space (C(R, $, p ) , p ) . l+c
Conversely, let p ( f n , f ) , n 2 1 converge to zero. Let k be any positive number. Let 0 < E < k be arbitrary. There exists m L 1 such that p(fn, f)< ~ / ( 1E+) whenever n L m.Now, let n L m be given. Since
This shows that fn, n L 1 converges to f hazily.
0
4.4 D-INTEGRAL In this section, we develop the basic ideas concerning D-integrals. We start with simple functions.
4.4.1 Definition. Let (R, 9, p ) be a charge space and f a simple function on R with a representationf = I:=,cJFifor some real numbers c1, CZ, . . . ,c, and partition {F,, F 2 .. . ,F,} of R in $, f is said to be D-integrable if Ip I(Fi)< CX, whenever ci # 0, and the D-integral of f, denoted by D 5 f dp, is defined to be the real number I;=, cip(Fi). (We adopt the convention = 0.) that 0 (*a)
4.
97
INTEGRATION
We settle the question of unambiguity of the above definition in the following proposition. Proposition. Let (a,.IF, p ) be a charge space. (i). Let f be a simple function with a representation f =ELl C ~ Ifor E ~some real numbers cl, c 2 , . . . ,cm and partition {El, Ez, . . . ,Em}of R in $such that lpl(Ei)
xi”=, ..
m
n
(ii). Kf f and g are D-integrable simple functions, then f + g is D-integrable and D j ( f + g ) d p = D j f d p + D j g d p . (iii). I f f is a D-integrable simple function and g is a simple function on 0 such that f = g a.e. [ p ] ,then g is D-integrable and D f d p = D 1g dp. then I$is also a (iv). I f f is a D-integrable simple function and E is in 9, D-integrable simple function. Proof. (i). Note that
This implies that ci = d j whenever EinFj#O. Suppose d j # O . Let J = (1 5 i 5 m ; Ei nFi # 0). If i E J, then ci # 0 and so [ pI(Ei)< a.Further, uis~ (Ei nFj) = F+ Consequently, lp J(Fj)< co. Also,
(ii). This is simple to prove. (iii). To prove this, it suffices to show that cip(Fi)= 0 whenever h = ciIFiis a simple and null function. We claim that for every 15 i 5 m either ci = 0 or p (Fi)= 0. Suppose ci # 0. Let E > 0 be any number such that E < Icil. Then {w E R: lh(w)l>E } = Uj., Fj, where J = { j ; 1 5j 5 m and
98
THEORY OF CHARGES
Icjl>~}. Obviously, i E J. Since h is a null function, IpI*(uj.J F j ) = l p l ( u j p FJ j ) = 0. Consequently, IpI(Fi)= 0. This proves (iii). (iv). The condition of Definition 4.4.1 for I$ is easily verified from the 0 hypothesis that f is D-integrable and simple. Part (iv) of the above proposition enables us to define D D-integrable simple function f and E in 9.
sEf d p for any
4.4.3 Definition. If f is a D-integrable simple function and E E 9, then D jEf d p stands for D 5 I$ dp. The following theorem gives the properties of integrals of simple functions.
-
4.4.4 Theorem. Let (R, 9, p ) be a charge space.
(i). I f f is a simple function on R and D-integrable with respect to p , then f is D-integrable with respect to p + and p - also. Further, for any E in 9, fdp+-D
J
fdp-. E
(ii). I f f and g are D-integrable simple functions on R, c and d are real numbers and E E 9, then cf + dg is a D-integrable simple function and
D
JE
(cf+dg)dp=c
(iii). I f f is a simple function on il and D-integrable with respect to p, then If l is a simple function on R which is D-integrable with respect to Ip I and for any E in 9
(iv). I f f is a D-integrable simple function on 9, i.e. I$? 0 a.e. [ p ] , then
and f 2 0 a.e.
(v). I f f and g are D-integrable simple functions on R and f E in 9, i.e. I $ s I E g a.e. [ p ] , then
(vi). I f f and g are D-integrable simple functions on R, then
[ p ] on
5g
If l ,
E in
a.e. [ p ] on
Igl,
If
+gl,
99
4. INTEGRATION
11f I - lgll are all D-integrable simple functions and for any E in 9,
(vii). Iffis a D-integrable simple function on R and c 5 f I d a.e. [ p ] on E in 9, i.e. c I ~ Id5 5 dIE a.e. [ p ] for some real numbers c and d, then c IP 103 5 D
I
E
f dlk I5dill. KE).
(viii). Iffis a D-integrable simple function on R, then the set function A on 9defined by c
A(F)=DJfdp,
F
E
~
F
is a bounded charge on 3? Also, IA [(F)= D 5, If l dlp 1, F E97 Further, A is absolutely continuous with respect to p in the following sense. Given E > 0, there exists S >0 such that \A (E)I < E whenever E E 9and Ip l(E)<8. Proof. Note that for simple functions f and g on a, f s g a.e. [ p ] if and only if Ip l({w E R; f ( w ) > g(w)}) = 0. (The set {w E 0;f ( w ) >g(w)} does belong to 9.With ) this observation, we proceed as follows. (i). Let f = ciIEi be a representation of f. Obviously, p+(Ei)< co and p-(Ei) < 03 whenever ci # 0. Hence f is D-integrable with respect to p + as well as with respect to p - . Further, if we let J = (1s i Im ;ci# 0}, then r
D
J
m
cip(Ei) = C cip(Ei)
f dp = i=l
i=l
xi"=,
ieJ
i=l
(ii). Letf = c i E andg , = dJF, be representations off andg respectively such that lp I(Ei)< co whenever ci # 0 and f pI(Fj) < co whenever dj # 0.
100
THEORY OF CHARGES
Then
Suppose cci + ddj # 0. Then, either cci # 0 or ddj # 0. This implies that either ci# 0 or d j # 0. This means that either IpI(Ei)
If1
(iv). We can assume that E = 0 and f is a simple function having a representation ELl ciIEiwhere each ciL 0. Now, (iv) is easily proved. (v). This is a consequence of (ii). and (iv). (vi). This follows from (iii) and (v). (vii). If Ip [(E)< 00, the result is clear. If Ip I(E) = 00, observe that c 5 0 5 d. (viii). A (0) = 0 since I& = 0. Let E and F be two disjoint sets in 9. Since IEvFf=I&+IFf, it follows that A(EuF)=A(E)+A(F). So, A is a charge Further, for any F in 9, on 9.
Thus A is bounded. Next, we show that IA [(F)= D 5, If1 dlp I for any F in 9. This, we proceed to establish in stages. Clearly, the result is true for f = Ia. Next, the result is clear for cIn also, for any real number c. If f=C:, ciIEi for some real numbers cl, c 2 , .. . , cm and El, E2,. . . , E m pairwise disjoint sets in 9, then A (F)= ELl ci D jF 1~~dp. Now, considering each D jFIEid p as a charge with reference to the charge space (Ei, Ei n9, p/Ei n and applying the above argument, we obtain the desired result for f. For the last part, if we let c = max {Icil;1 5i 5 m},then for any F in ,F, \A I(F)IIc Ip I(F). Consequently, A is absolutely continuous with respect 0 to p , We introduce two important concepts before defining the integrability of a general function.
4. INTEGRATION
101
4.4.5 Definition. Let (R, 9, p ) be a charge space. A real valued function on R is said to be TI-measurable if there exists a sequence f n , n z 1 of simple functions on R converging to f hazily.
If we denote the collection of all simple functions on by S(R, 9, p), then f is T1-measurable if and only i f f ES(SZ,9, p ) , where the closure is taken in the pseudo-metric space (C(n,$, p ) , p ) . See Proposition 4.3.5. 4.4.6 Definition. Let (R, 9, p ) be a charge space. A real valued function f on R is said to be Tz-measurable if for every E > 0, there exists a partition < E and If(w)-f(w’)I < E for {Fo,F1, FZ,. . .,F,} of 0 in 9such that lp ~(FO) every w , w ’ in Fi for every i = 1 , 2 , . . . ,n.
The following theorem establishes the relation between Tl-measurability and Tp-measurability. 4.4.7 Theorem. Let (R, g, p ) be a charge space. A real valued function f on R is T1-measurableif and only if it is Tp-measurable.
Proof. Suppose f is T1-measurable. Given E > 0, there exists a simple function g on such that J pI*({w E R; I f ( @ ) - g ( w ) l > ~ / 2 }<) E . Let g = cilEi be a representation of g. Let G = {w E R; If ( w ) - g ( w ) l > ~ / 2 } . Since Ip (*(G)= Inf {lp ((F);G c F, F E9 1, there exists Fo in 9 such that G c F o a n d I p I ( F o ) < e . L e t F i = E i n F & i = 1 , 2 , ..., m. Now,if l s i l m and o,U ’ E Fi, then If ( w )- g ( w ) l s ~ / 2If,( w ’ )- g ( w ’ ) l s ~ / and 2 g ( o ) = ci= g(w‘). Therefore,
zEl
If(w ) -f(w
’11 IIf(w ) - ci I+ lci -f(o ’)I 5
I f ( w >- g(w)l+ I f (a’)- g(w ’)I 5 E .
This shows that f is Tp-measurable. Conversely, let f be T2-measurable. For each n 2 1, let {Fno, Fnlr Fnz,. . . ,Fnk,}be a partition of n in 9such that Ip I(Fno) < l / n and If ( w ) f(w’)I < l / n for every w , w ’ in Fni and for every i = 1,2, . . . ,k,. For each n 2 1 and 1 Ii 5 k,, choose and fix wni in F,i. For each n 2 1, let k, fn
=
cf
i=l
+o
(wni)lFm,
*
IFn~*
Each f n is a simple function and f n , n 2 1 converges to f hazily. For, let E > O and m 2 1be such that l / m < E . If n L m , {w €0;I f , , ( o ) - f ( w ) I > ~ ) c F,o. Consequently, Ipl*({w en; I f n ( w ) - f ( w ) I > e } ) I I p I * ( F n o ) < l / n . This 0 establishes the result. 4.4.8
Corollary. Every TI-measurablefunction is smooth.
102
THEORY OF CHARGES
Proof. From the definition of T2-measurability, observe that every T2measurable function is smooth. The result now follows from Theorem 4.4.7. 0 Relations between TI-measurable functions, Smooth functions and bounded functions can be described as follows, T1-measurable function
%\Smooth function 4.4.9 Corollary. Let (a,9,p ) be a charge space. (i). I f f and g are TI-measurable functions on SZ and c and d are real numbers, then cf + dg and f g are all TI-measurable. (ii). If I,4 is a real valued continuous function defined on the real line and f is TI-measurable,then I++( f ) is TI-measurable. In particular, I f 1’ ( p > 0 ) is TI-measurable i f f is T1-measurable. (iii). If f n , n 2 1 is a sequence of TI-measurable functions converging to f hazily, then f is T1-measurable and fcl(fn), n 2 1 converges to $ ( f ) hazily, where I,4 is as in (ii).
Proof. (i) and (ii) follow from Corollary 4.4.8 and Theorem 4.3.3(i) and (vi). (iii) follows from the Remark following Definition 4.4.5, Theorem 4.3.3(v) and Corollary 4.4.8. The following proposition is instrumental in establishing the unambiguity in the general definition of D-integral. 4.4.10 Proposition. Let (a,9,p ) be a charge space. Let f n l , n 2 1 and f n 2 , n 2 1 be two sequences of D-integrable simple functions on SZ converging to a real valued function f on R hazily. Suppose
for i = 1,2. Then D JE f n i d p converges uniformly over E in 9 for each i = 1, 2, and the limits coincide. Proof. By Proposition 4.4.4(iii), for any m , n 2 1 and i = 1, 2
for every E in 9. Consequently, D jEfni dp, n 2 1converges uniformly over E in 9for each i = 1, 2. It remains to be shown that for each E in 9, fnl
d p = lim D n-tm
f n z dp.
4.
103
INTEGRATION
This is carried out in the following steps. 1". Let g , = l f n ~ - f n 2 1 , n 2 1, and p , ( F ) = D jFgndlpl for F in 9and y1 2 1. Each pn is a positive bounded charge on 9. 2". We claim that p,, n 2 1 converges uniformly over 9. For any F in 9 and n, rn 2 1, by Theorem 4.4.4(vi),
5~
J
Ifnl-frnlI
dIFI+D
I
Ifnz-frnzl
dbI-
Thus the From this, it follows that p,, n 2 1 is uniformly Cauchy over 9. claim is established. 3". Let A(F)=lim,,+mpn(F), F E ~It .suffices to show that A = O . For, for any E in 9,
If
p,(E) = 0, then
4". Now, we claim that A is absolutely continuous with respect to p. (See Theorem 4.4.4(viii).) Let E >O. There exists N 2 1 such that ]k,(E)-A (E)(< ~ / for 2 every E in 9 and n 2N.Since p N is absolutely continuous with respect to p, there exists S > O such that ~ N ( E<)~ / 2 whenever E E 9and lp [(E)< S. Consequently, if E E 9and Ip ((E)< 6,then A(E)SIh(E)-pN(E)1+pN(E)<&/2+~/2 =E. 5". The above deliberations can be summarized as follows. Let E > 0. There exists S > 0 and N 2 1 such that A (E) < E whenever E E 9 and Ip I(E)< S, and [A (F)- pn(F)I< E for every F in 9and for every n L N. 6". We show that A ( 0 ) < 4 ~ .Let A'={w ~n;g,(o)=O}. Since g N is a simple function, A'E 9.Since gN is D-integrable, Ip,I(A)< 00. Also pN(A') = 0. From 5", we conclude that A (A') < E . If lp l(A) = 0, then A (A) = 0 and A (0)= A (A)+ A (A') < E < 4 ~ The . next step covers the case when IP KA) > 0.
104
THEORY OF CHARGES
7". Suppose Ip \(A)> 0. Since g,, n z 1 converges to 0 hazily, there exists an integer N 1> N such that lp I*({w E 0; lgn(w)I> ](A)})< S whenever n z N l . Let B" = {w E 0;IgNI(w)I> E / I ~ ~ ( A )Since } . gN1is a simple function, P B'E 3.Further, if w E B, then IgN1(w)I 5 E / ~ I(A). 8".
A (a) = A ( A n B)+A ( A n B')+ A (A")
<[pNI(A ~ B ) + E ] + +EE . The inequality A ( A n B') < E follows from the following argument. By 7", lp 1(A nB") 5 lp [(B")= Ip I*(B') < 8. So, by 5", the above inequality follows. Again, by 5", the inequality A ( A n B)
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
202
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.
204
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).
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(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
-
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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
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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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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.
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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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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.
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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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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.
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THEOREMS
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.
8.
NIKODYM AND VITALI-HAHN-SAKS
219
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 r+121. So, Ipnl(B?)ILbnl(fl)l- Ipnl(B1)ll2 Ipnl(B1)l- I C L , ~ ( R1.) ~ ~ Then either Sup,,l Ip,,I(Bd = 00 or Sup,,l lpnI(B;)= 00. If Sup,,l k&%) = co, take A1 = B;. Using the same argument given above for B1,we can find n z > n l and BZ in 9 such that Bz c B1, lp,(Bz)l 2 1 and Ipnz(B1 -B2)1 2 1 and so on. If Sup,,l Ip,,l(BT) = co,take Al = B1 and then work with B ; as above. Thus we obtain two sequences n l < nz < * * and A,,, n 2 1 of painvise disjoint sets in 9such that Ip,,(Ak)lL 1 for every k L 1.Since $ is bounded, limn,, $(A,) = 0. Hence p,, n 2 1 cannot be uniformly absolutely con0 tinuous with respect to $. This contradiction proves the result.
-
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),
+
222
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).
224
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
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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
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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,
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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
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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,,,
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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.
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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.
278
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.
APPENDIX
1
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
APPENDIX
1
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
283
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.
288
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
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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
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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
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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
Pure and Applied Mathematics A Series of Monographs and Textbooks Editors
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Columbia University, New Y o r k
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Consequently, A (R) < 4.5. Since E > 0 is arbitrary, it follows that A (0)= 0. SinceA isapositivecharge,wehavethatA = 0.Thiscompletestheproof. 0 Now, we are in a position to define D-integrability of a general function.
4.4.11 Definition. Let (R, 9, p ) be a charge space. A real valued function f on R is said to be D-integrable if there exists a sequence f n , n Z 1 of D-integrable simple functions on such that (i). f n , n 2 1 converges to f hazily. 69. limm.,+m D J I f n -f m l dJpI= 0. Iff is D-integrable, the D-integral off is denoted by D J f d p and is defined to be the number limn+mD J f n d p . The sequence f n , n 2 1 is called a determining sequence of D-integrable simple functions for f , or simply a determining sequence for f . From Proposition 4.4.10, it is clear that limn,m D J f n d p exists and is finite, and is independent of the choice of the determining sequence fn, n 2 1. Also, observe that every D-integrable function is TI-measurable. Iff is D-integrable and E E 3,then the function I& is also D-integrable. If f n , n 2 1 is a determining sequence of D-integrable simple functions for f, it is easily checked that Id,,, n 2 1 is a determining sequence of Dintegrable simple functions for I&,. Consequently, we define D 5, f d p = D JkfdCL. We need a lemma which will be useful in the proof of Theorem 4.4.13(xi). 4.4.12 Lemma. Let ( a , F , p )be a charge space and f a D-integrable function. Let f n , n 2 1 be a determining sequence for f. Then f , -f is Dintegrable for every n L 1 and
lim D n-m
J
Ifn-fldlpl=O.
105
4. INTEGRATION
Proof. Fix n 2 1. Then fn -fm, m 2 1 converges to f n -f hazily. Further, each f,, -fm is a D-integrable simple function. Also, limm,p+m D I( fn -f m ) (fn -fp)l dJpI= 0. Hence f n -f is D-integrable, and Ifn -fml, m L 1 is a determining sequence for If,,-f I. Thus lim D
n-m
J
lfn-fJdlp/=lim n-tm =
[ Iim D J Ifn-fmIdl~I] m-m
lim D I Ifn-fmldlpl=O.
m,n+m
This proves the lemma. Now, we give a comprehensive list of properties of D-integrable functions. 4.4.13 Theorem. Let (R, 9, p ) be a charge space. (i). If a real valued function f on R is D-integrable with respect to p, then f is D-integrable with respect to p + as well as with respect to p - . Further, for every E in 9,
D [ E f d p = D /E f d p + - D (ii). Iffand g are D-integrable and c and d are real numbers, then cf+dg is D-integrable and for every E in .F,
(iii). Iff is D-integrable with respect to p, then If1 is D-integrable with and for any E in 9, respect to p as weEl as with respect to
(iv). f is D-integrable if and only iff' and f- are D-integrable. Further, if f is D-integrable, then D / E f d p = D IE f ' d p - D j
E
f-dp
for every E in 9. More generally, iff and g are D-integrable, then f v g and f A g are D-integrable and g d p = D [ E ( f v g ) d p + + DJE(fAg)dlr for every E in 9.
106
THEORY OF CHARGES
(v). I f f is D-integrable and f
20
a.e. [ p ] on E in g, i.e. I j r 0 a.e. [ p ] ]then ,
(vi). I f f and g are D-integrable and f I g a.e. [ p ] on E in 9, i.e. I J s I E g a.e. [ p ] , then
(vii). I f f is D-integrable and c ~f numbers c and d, then
Id
ClFI(E)SD E
a.e. [ p ] on E in 9for some real
f dlCLIsdlll.I(E).
In fact, if c > 0 , then Ip I(E)< 00. (viii). I f f and g are D-integrable and E E 9, then
(ix). f is a null function if and only i f f is D-integrable and I f I dlp I = 0. (x). If f is D-integrable and g = f a.e. [ p ] ] ,then g is D-integrable and D jE f d p = D 5, g dp for all E in 9. (xi). I f f is D-integrable, then the set function A on $defined by
A(F)=D
[ f dp,
F
E
~
F
is a bounded charge on 9satisfying IAl(F)=D[F I f l d l p l ,
FE$.
Further, A is absolutely continuous with respect to p in the following sense. Given E > O , there exists 6 > O such that IA(E)I < E whenever E E and~ IF I(E)<6.
4.
107
INTEGRATION
(xii). If p is positive, f is D-integrable and A is the charge on $defined by IFf dp, F E 9, then
A (F)= D
f'dp
and A-(F)=D
f-dp IF
for all F in 9. (xiii). For a D-integrable function f, D I 1f 1 dlp 1 = 0 if and only if D 5, f d p = 0 for every E in $. Iffand g are D-integrable, then f = g a.e. [ p ] if and only if
for every E in $. Proof. (i). If fn, n 2 1 is a determining sequence for f with respect to p, then it has the same property with respect to Jpl,p + and p - . This follows from the inequalities that pfslpI and p - s l p l . Consequently, f is Dintegrable with respect to p + as well as with respect to p - . By Theorem 4.4.4(i), we have for any E in g,
D
f d p = n-m lim D = lim n+m
D
JE
fn
JE
fn
d p = n+cC lim [D
JEfn
d p + - lim D
I,
n+m
dp+-D
J
fn
dp-1
E
fn
dpc-
(ii). This follows from Theorem 4.3.3(i) and Theorem 4.4.4(ii). (iii). If fn, n 2 1 is a determining sequence €or f with respect to p, then 1fnl, n 2 1 is a determining sequence for 1f l with respect to p as well as withsrespect to ( p ( .This follows from Theorem 4.3.3(ii) and Theorem 4.4.4(vi). Further, from Theorem 4.4.4(iii),
= D lE IfldlPl for any E in 9. (iv). If f n , n 2 1 is a determining sequence for f, g , n 1 1 is a determining sequence for g, then fn v g,, n 5 1 converges to f v g hazily, by Theorem
108
THEORY OF CHARGES
4.3.3 (viii). Further,
(Note that \(a v 6) -(c v d ) l s (a- c l + (6- d ( and ( ( aA 6) - (C ~ d 5 ) ((a-c(+ Ib - dl for any numbers a , 6, c and d.) Hence f, v g,, n z 1 is a determining sequence forf v g. Similarly, one can show that f, A g,, n I1is a determining sequence for f A g. Since f + g = (f v g ) + (f A g), by (ii), we have D ,j (f+ g ) d p = D j E ( f v g ) d p + D j E ( f A g ) d pf o r a n y E i n P . T a k i n g g = O , w e get the first part of (iv). 0I a.e. [ p ] ,there exists a null function h such that h +I&z 0. (v). Since I& If fa, n L 1 is a determining sequence for f,then Id,, n z 1 is a determining sequence for h +I&. See Proposition 4.3.2. Hence h +I& is D-integrable, and (I&,,)+, n L 1 is a determining sequence for ( h +I&)+= h +I& with respect to lp I. Since f: 2 0 for every n L 1, by Theorem 4.4.4(iv), and (ii) and (iii) above, we have
= lim n+w
D J (I&,)+ dlp 1 2 0.
This proves (v). (vi). This is a consequence of (v) if one looks at the function IE(g -f). (vii). If lp I(E)
for all m, n 2 1. Hence f,,A c, n L 1 is a determining sequence for the constant function c. Hence the constant function c is D-integrable. But = m. This shows that c s 0. Hence, it follows this is not possible since JpJ(f2) dip]. If Ipl(E)= co, we can show that d L 0 by an that cIpl(E) ID analogous argument. Thus the desired inequalities are established for all cases.
If
4.
109
INTEGRATION
(viii). This is a consequence of Theorem 4.4.4(vi). (ix). Iff is a null function, it is easily verified that 0, 0, . . . is a determining sequence for f and hence D j If1 dlp I = 0. Conversely, let f be D-integrable and D 5 dlp I = 0. Let f,, n z 1 be a determining sequence for f. Then limn+mD 5 lfnl dlpLJ=0. For any S>O and n z l , let E , ( S ) = ( w ~ f l ; lfn(w)1>6) and F,(S)= {w E a; ( w ) -f(o)I> 8). Since f n is a simple function, E,(S) E 9.Since f,, is D-integrable, lp I(E,(S)) < 00. Further, for any n 2 1, by (vi) and (vii),
If1
If,
D
J
lfnl
J'
dlp12~
lfnl
dlpl2SlpI(En(S))*
E"(6)
Consequently, limn+mIp J(E,(S))= 0. Since f,, n 2 1 converges to f hazily, Ipl*(F,(S)) = 0. Thus, for any n 2 1,
IF I*({w E a;If(w )I > 26)) 5 I@I*({w E a;Ifn (w )I >6)) + IP I*({w E a;Ifn(w> -f(w>l>
6))
which converges to zero as n tends to co.Consequently, Ip I*({w E R; If(w)I > 28)) = 0 for any 6 > 0. Hence f is a null function. (x). This follows from (ix) and (ii). (xi). It is now obvious that A is a charge. The boundedness of A follows from IA (F)I= ID jFf d p J5 D If1 dlpl s D I If1 dlp 1 for any F in 9. Now, we show that \Al(F)= D SF I f ] d1p1 for every F in 9. Let f,, n 2 1 be a determining sequence for f. Let A,, n z 1 be defined by A,(E) = D IEfn dp, E E 9. We show that limn.+mIA,l(E) = IA [(E)for every E in 9. Then, by Theorem 4.4.4(viii), and (ii) above, we would have
IF
Let F E 9be fixed and E > 0. Since fn, n 2 1is a determining for every E in 9. sequence for f, by Lemma 4.4.12, there exists N 2 1 such that D SF [fdlp 1 < E whenever n 2 N. Let {Fl,F2, . . . , Fk} be any partition of F in 9. Then for any n 2 N ,
I,f
110
THEORY OF CHARGES
We show that for any n 2 N , I lA,l(F) - [A I(F)I < 3.5. Let n 2 N be fixed. From the definition of IA,l(F) and IA I(F), we can find a partition {Al, AZ,. . . ,Am} of F in 9and a partition {B1,B2,. . . ,Bk}of F in 9such that
and k
k
lAnl(F)-
i=l
lAn(Bi)l=lAnl(F)-
i=l
ID /
Bi
f n dpl <&.
LetFij=AinBj,I s i s m a n d l s j s k . T h e n { F i j ; l s i s r n a n d l s j s k } Further, is a partition of F in 9.
and
Therefore, for n 2 N,
<&
+ E +&
=3E.
This completes the proof of the desired assertion. Now, we prove that A is absolutely continuous with respect to p. Let E > 0. By Lemma 4.4.12, there exists a D-integrable simple function g such that D If-gl dlpl < E . Let M be a positive number such that lg(u)I s M for every u in SZ. Take 6 = E/M.Let E E 9and Ipl(E)< 6. Then
J
E
s~ If-gIdlpI+MIpl(E)<&+M-=2~. M
4.
INTEGRATION
111
This shows that A is absolutely continuous with respect to p. (xii). From Theorem 2.2.2(5), A + =$(A + IA I) and A - = +(A - IA I) for the bounded charge A. Then,
= D j f'dy. F
Similarly, the result about A - is proved. if D J F f d p = 0 (xiii). Let A (F)= D l, f dp. Then A (F)= 0 for all F in 9, for all F in 9.Hence [A I(R) = 0. So, by (xii) above, we have D 5 I f l dlp I = 0. The converse is clear. Now, the second part follows from the first part and (ix). This completes the proof of the theorem. Now, we enumerate all the D-integrable functions for the simple example given below.
Example. Let (R, 9, p ) be a charge space in which p is a 0-1 valued charge on 9.A real valued function f on R is D-integrable if and only i f f = c a.e. [ p ] , where c is a constant, and i f f = c a.e. [ p ] ] ,then Djfdp=c. 4.4.14
Proof. Since p is bounded, every constant function c is D-integrable. By Theorem 4.4.13(x), any function f which satisfies f = c a.e. [ p ] for some constant c is D-integrable and D 5 f d p = c. Conversely, suppose f is D-integrable. Let f,, n 2 1 be a determining sequence for f . Since each f , is a simple function and p is a 0-1 valued charge, f , = c, a.e. [ p ] for some constant c,. Without loss of generality, we can assume that f, =c, for every n 2 1. Note that c,, n 2 1 is a Cauchy sequence of real numbers since D If , -f , I d p = Jc, - c, I which converges c,. Then to zero as n, m + co. Let c =
We show that f = c a.e. [ p ] .Let E > 0. Let N 1L 1 be such that Ic, - c I < ~ / 2 for every n r N 1 .Since f,, n 2 1 converges to f hazily, there exists N2r1 such that p *({w E R; I f , ( w ) -f(w)I > ~ / 2 } = ) 0 for every n LN2. Let N = max { N I ,Nz}. If n LN, then p * ((0E R; If (0)- c I > E }) 5 p * ( { w E R; If (w ) f,(w)I > .5/2}+p*({w E R; If,(w) - c J > ~ / 2 } 5 ) O + 0 = 0. Hence f = c a.e. [PI.
The following three lemmas are needed to prove Theorem 4.4.18.
0
112
THEORY OF CHARGES
4.4.15 Lemma. Let (R, 3,p ) be a charge space and f a D-integrable function on R. Let A on 9 be defined by A (F)= D jF f d+, F E 9.Then given E > 0, there exists A in 9 s u c h that I+ [(A) < co and IA [(A')< E .
Proof. By Lemma 4.4.12, there exists a D-integrable simple function f on R such that D J lg -f I dlp I < E . Let A' = {w E R; g ( w ) = 0). Since g is a simple function, A' E 9. Since g is D-integrable lp I(A)< 00. Further,
CD
J
If-gldlpI+O<E.
0
This completes the proof.
4.4.16 Lemma. Let (R, 9, p ) be a charge space and f a D-integrable function. Suppose g is a simple function and I g l s l f i a.e. [ p ] . Then g is D- integrable.
0
Proof. This follows from Theorem 4.4.13(vii).
4.4.17 Lemma. Let (R, 9, p ) be a charge space. Let g,, n 2 1 be a sequence of simple functions converging to g hazily. Then there exists a sequence h,, n 2 1 of simple functions converging to g hazily such that lhnis 2 ( g ]for every n21.
Proof. From the definition of hazy convergence, we can find a subsequence g,,, k 2 1of g,, n 2 1 such that ]pl*({w E R; lgnk(w)1> l / k ) )< l / k for every k r 1 . Consequently, we can find Ak in 9 such that I p I ( A k ) < l / k and { w E S 1 ; I g n , ( w ) - g ( w ) l > l / k } c A k for every k r l . So, if ~ E A Cthen ~, ]g,,(w) -g(w)l< l / k for every k 2 1. Now, for each k 2 1 , define hk(W)=gnk(w), if w E A C k and lgnk(w)I> 2 / k , = 0, otherwise. Since each g,, is a simple function, each hk is a simple function. Let w E ACk and satisfy the inequality Ig,,(w)l > 2 / k . Then Ihk(w)I= lgnk(U)l 5 Ignk(w)-g(w)I+Ig(w)I 5(1/k)+Ig(w)I* But I l k rlgnk(w)I-Ig(w)I> 2 / k -Ig(w)l. Therefore, Ig(w)l> l / k . Consequently, ]hk(w)I5 l / k +lg(w)l<2lg(w)I.Thus, for any w in S1, Ihk(w)ls2]g(w)Ifor all k 2 1 . Now, we show that hk, k r 1 converges to g hazily. Let E > 0. Then there exists N 2 1 such that 1 / N < E . If k r N, then
I*
((0 E
; Ihk (w - g (w )I > E 1) IF I* ({w E A
k
; Ihk (w
- g (w )I > E 1)
Ihk(W)-g(w)l>E)) S IpI(Ak)+o < l / k . 0 Hence hk, k 2 1 converges to g hazily. +Ip]*({w
113
4. INTEGRATION
The following theorem gives a necessary and sufficient condition under which a given function dominated by a D-integrable function is Dintegrable. 4.4.18 Theorem. Let (R, 9, p ) be a charge space. Let f and g be two real valuedfunctions on R such that l g l s a.e. [ p ] and f is D-integrable. Then g is D-integrable if and only if g is TI-measurable.
If1
Proof. “Only if” part follows from the definition of D-integrability. “If” part. Suppose that g is TI-measurable. By Lemma 4.4.17, there exists a sequence h,, n 2 1 of simple functions converging to g hazily and such that lhn1s21gl for every n r l . Since each lhn1s21f1 and f is Dintegrable, by Lemma 4.4.16, we have that h, is D-integrable. Now, we show that limm,n+oO D J Ih, - hml dlp I = 0. Given E > 0, we exhibit N L 1 such that D Ih, - h, Idlp I < 3.5 for every n, m 2 N . Observe that for every n, m 2 1 and E in 9, D JE Ih, -hml d l p l 5 4 D 1, dlpl. L e t s > O . ByLemma4.4.15, thereexistsaset A i n 9 s u c h that IpI(A)
If1
D
I
Ihn -hml dlpl = D
I
Ihn -hml dlpl + D
A
I,.
Ihn -hmI dIpI
If1
This shows that h,, n 2 1 is a determining sequence for g. Hence g is D-integrable. 0 p ) be a charge space and f a T1-nteasurable 4.4.19 Corollary. Let (R, 9, function on R. Then f is D-integrable if and only i f I is D-integrable.
fl
114
THEORY OF CHARGES
The following theorem is an extension of Lemma 4.4.12.
(a,
4.4.20 Theorem. Let 9, p ) be a charge space and f,, n 2 1 a sequence of D-integrable functions such that
lim D m,n-co
J Ifn-fmldlpl=O.
Let f be a real valued function on R such that fn, n z 1converges to f hazily. Then f is D-integrable an d lim D n-a,
J Ifn-fldlpl=O.
Proof. If fn's are simple functions, the assertion follows from the definition of D-integrability off and by Lemma 4.4.12. Since f n is D-integrable, by Lemma 4.4.12, there exists a D-integrable simple function h, on R such that D
Ifn-hnIdIpI
and
IP I*({w E a;Ifn(w)-hn(w)I>
l/n>)
for every n L 1. We claim that h,, n z 1 converges to f hazily. Let E >O. Since f,, n z 1 converges to f hazily, there exists N1L 1 such that IE.L I*({w E R; If n ( w ) -f(w)l> ~ / 2 } < ) ~ / whenever 2 n 2 N1. Let N L 1 be such that 1/N < ~ / 2and N L N ~If. n rN, then
( ~ € Ihn(o)-f(U)l>E)C{U 0;
Ihn(w)-fn(w)l>~/2)
4.J
Ifnb)-f(W)I>E/2)
€0;Ihn(w)-fn(w)l>l/n) u{w
€02If"(W)--f(dI>EI2). ;
From these set inclusions, it follows that IpI*({w ER; I h , ( w ) - f ( w ) l > ~ } ) < : ~ / 2 + ~ / 2 = ~
whenever n LN. Next, observe that
+D
J Ifm-hmldl/l.l
for all m, n z 1. From this, it follows that limn.m+a,D 5 )h, -hml dip( = 0.
4.
INTEGRATION
115
Hence h,, n 2 1 is a determining sequence for f . So, f is D-integrable. Further, for n 2 1,
The second term on the right above converges to zero as n + CO, by Lemma 4.4.12. Hence lim D n+m
I Ifn
-f
I dlp I = 0.
This completes the proof.
4.5
S-INTEGRAL
In this section, we introduce S-integrals which are of Stieltjes type in the framework of charge spaces. We also show that D-integrals and Sintegrals coincide in the case of positive bounded charges and bounded functions. I n what follows, we assume that all the charges are positive bounded unless otherwise specified. For a given field 9 on a set R, let 8 denote the collection of all finite partitions of R in 9. On 8, we define a partial order by PI 2 P2 for PI, PZ in 8 if PI is a refinement of P2, i.e. every set in P2 is a union of sets in PI. Indeed, ( 8 , ~ is a)directed set. 4.5.1 Definition. Let ( R , 9 , p ) be a charge space. Let f be a bounded real valued function on R. For P = {Fl, F2, . . ,F,} in 8, let
.
and
L(P) is called the lower sum associated with P and U(P) is called the upper sum associated with P. (Since f and p are bounded, L(P) and U(P) are real numbers.) The following proposition gives some inequalities between these sums.
Proposition. Let (R, 9,p ) be a charge space and f a bounded real valued function on R. Then for any PI L P in ~ 8) 4.5.2
116
THEORY OF CHARGES
Proof. Since P 1 z P Z , every set in Pz is a union of sets in PI. Hence the above inequalities easily follow. Of course, we use the fact that p is positive in proving the above inequalities. Thus, we observe that the net {U(P);P E P} defined on the directed set (9,L) is a decreasing net of real numbers bounded below and hence has a limit. The net {L(P); P E 9)defined over the directed set ( 9 ,is~an) increasing net of real numbers bounded above and therefore, has a limit.
4.5.3 Definitions. Let (fl, 9, p ) be a charge space and f a bounded real valued function on R. Let
J
f d p = Inf U(P) = lim U(P) PEP
PEP
and
-
f d p = Sup L(P) = lim L(P). PE B
-
j f d p is called the upper integral of f with respect to
p
and
5f
d p the
lower integral off with respect to p. The following proposition is obvious in view of Proposition 4.5.2.
4.5.4 Proposition. Let (fl, 9, p ) be a charge space and f a bounded real valued function on fl. Then
Now, we define the S-integral.
4.5.5 Definition. Let (O,?F, p ) be a charge space and f a bounded real valued function on fl. f is said to be S-integrable if
If f is S-integrable, the S-integral of f is denoted by S I f d p and is
-
defined to be the common number 5 f d p = f d p . -
4.5.6 Remark. I f f is a simple function, then S j f d p = D f d p .
117
4. INTEGRATION
We link S-integrability and D-integrability of a function in the following result. 4.5.7 Theorem. Let (a, 9, t ~ be ) a charge space and f a bounded real valued function on Q. Then the following statements are equivalent. (i). f is TI-measurable. (ii). f is T2-measurable. (iii). f is S-integrable. (iv). There exists a real number a with the following property. For every E >0, there exists a partition Po in B such that for every partition P in B with P = {Fl, F2, . . . , F,} 2 POand for every wi in Fi, i = 1 , 2 , . . . , n,
holds. (v). For every E > 0, there exists a partition Po in B such that for every partition P in 9 with P ={F1, F 2 , . . . , F,}?Po and for every w i l , wi2 in Fi, i = l , 2, .. . , n ,
IC
I
I n
i = l ( f ( w i i ) - f ( W i Z ) ) ~ ( F i ) l< E
holds. (vi). For every E >0 , there exists a partition Po in B such that for every partition P in 9 with P = { F I , F2, . . . ,F,) 2 PO,
i[
i=l
SUP
If(wi1)-f(Wi2)11tL
(Fi) < E
w,l,w,zeFi
holds. (vii). For every E > 0 , there exists a partition PO= { E l , E2. . . . ,E m } in $9’ such that for any partition { E l l , E12,. . . , Elkl, E21, E22,.. ., E2k2,. .., Eml, Em2,. . ., Emk,} in 9 with E i = U f ~ , E i i , i = 1, 2, . . . ,m and for every choice Aii, j = 1 , 2 , . . . , ki, i = 1 , 2 , . . . ,m, of real numbers satisfying
holds. (viii). f is D-integrable.
118
THEORY OF CHARGES
Proof. (i)+(ii). This follows from Theorem 4.4.7. (ii) (iii). Let E > 0. We show that
+
This then would prove that f is S-integrable. Let M = SupwEnlf(w)l. Since f is Tz-measurable, there exists a partition P = {Fo,F1, Fz, . . . ,F,} in 9 such that p(Fo)<&/4M and
If(wil)-f(wiz)I
<~/2~((n>
for every oil,wi2 in Fi for i = 1 , 2 , . . . , m.Then
< &/2+ &/2= e. This shows that f is S-integrable. -
(iii)J(iv). Let a = S Jfd p = f dk
=
J f dp. Let
-
E
> 0. There exists a
partition POin P such that for every P z Po, U(P)- a < E and a - L(P) < 8. Let P = {FI,Fz, . . . ,F,} be any partition in 9 such that P z PO.Let wi in Fi, i = 1, 2, . . . ,YE be arbitrary. Then n
n
C f(Wi)P(Fi)-a i=l
5
C (Supf(w))P(Fi)-a
i=l
w€Fi
=U(P)-a
<&,
and "
n
= a -L(P)
<&.
119
4. INTEGRATION
\xy=l
Consequently, f(wi)p(Fi) - a\ < E . This proves (iv). (iv).$(v). This is obvious. ( v) j ( vi ) . This is obvious. The modulus sign can be taken inside the summation in (v) and the inequality still remains valid. (vi) (vii). This is obvious. (vii)j(viii). For each k L 1, let P k = {Ekl,Ek2,.. . , E k p k } be a partition in B satisfying (vii) for E = l/k. Assume, without loss of generality, that P I I P ~ S P* ~. ~For. each k 2 1 and i = 1,2,. . . , p k , choose and fix Wki in E k i . Let
+
pk fk
=
k
f(wki)IEk,, i=l
2
1.
We claim that limm,k+ao D Ifm -fkl d p = 0. To begin with, we look at the function Ifk -fml. Suppose m 2 k. Since P k < P m , every set in P k is a union of sets in Pm.Assume, without loss of generality, that the sets in P k are of the following form. E k l = E m l U E m Z U * * *uEmrl, E k 2 = Emr1+1 U E m r l + 2 u * * u Emr2,
.............................. Ekpk
= EmrPk-l+1 u Emrpk-l+2
U
* * *
u Emp,,,.
Write ro = 0 and rpk= pm. Then
IC pk
Ifk
-fml
p, f(Wki)IEki
i=l
pk
1c Pk
f(wmi)IE,,,iI i=l
P,
i‘
C 1 = (i-1 j=ri-1+1 = i-1
-
f(Uki)IE,,
-
1f ( w m i ) I E , i l
i=l
‘i
j = ri - l + l
(f(wki)-f(Omj))lE,jI.
Therefore,
D
I
pk Ifk-fml
dp =
‘8
C C i =1j = ri
I+
If(wki)-f(wmj)lp(Emj). 1
Obviously, If(Wki) -f(wrnj)l = Aii 5 S U ~ I~f ( w, )-f(w’)I ~ ’ ~ for~ ~every ~ = + 1,ri-l + 2, . . . ,ri and i = 1,2, ... ,P k . Hence D 1( f k - f m ( dp < l / k . This shows that limm,k-rm D 5 Ifk -fml d p = 0. Now, we claim that fk, k L lconverges to f hazily. Observe that for any kzl,
120
THEORY OF CHARGES
Hence limk,m p * ( { w E R; If k ( w )-f ( w ) l > integrable.
E})
= 0.
This shows that f is D-
o
The definition of S-integral can be extended to general charge spaces and real functions. See Gould (1965). We link S-integrals and D-integrals in the following theorem. 4.5.8 Theorem. Let ( R ,9, p ) be a charge space and f a bounded real valued function on R. If f is S-integrable, then f is D-integrable and S S f dP = D S f dP.
Proof. The proof of this theorem can be recovered from the proof of the above theorem as follows. For any partition P = {Fl, Fz, . . . , F,} in B and fixed points mi E Fi, i = 1 ,2 , . . ,n, define n
Then, by Theorem 4.5.7 (iii)+ (iv), limPEBu p = a = S f dp. Next, observe that the proof of Theorem 4.5.7 (vii)3 (viii) can be improved to show that if P 2 P k , Q 1P , and k 5 m, then lap- aQ(5 l/k.These two observations
4. INTEGRATION
121
together show that
=
I
lim a p , = a = S f dp.
n-m
As a consequence of Theorem 4.5.7, we obtain the following result. 4.5.9 Corollary. Let 9 be a u-field of subsets of a set R and p a bounded charge on 9. Let f be a bounded measurable function on fl, i.e. f-'(B) E 9 for every Bore1 set B c R.Then f is D-integrable.
Proof. We show that f is the uniform limit of a sequence of simple functions. Let M = SupmEnI f(w)I and d = 2M. For n L 1, let for w in R, fn (0)=
-M
i-1
+7 d, 2
if -M
i-1 + 71 f ( w ) < -M 2
1
+-2"'
f o r i = 1 , 2 , . . . , 2"-1,
=M-- d 2"'
if M
d 2" -
---= f ( w )s M .
It can be checked that fn, n 2 1 converges to f uniformly. Since f is measurable, each fn is a simple function. Hence f is TI-measurable. An application of Theorem 4.5.7completes the proof. 0 Note that i f f is not bounded in the above proof, there is no sequence of simple functions converging to f uniformly.
4.6
L,-SPACES
Let (R, 9, p ) be a charge space. In this section, we introduce some p ) and study function spaces, namely L,-spaces, associated with (R, 9, some of the properties of these spaces. We establish Holder's inequality and Minkowski's inequality to aid the study of these spaces. We also prove Lebesgue dominated convergence theorem. First, we introduce L,-spaces. 4.6.1
Definitions. Let (R, 9, p ) be a charge space. For 1~p
L,(R, 9, p ) = {f ; f is a TI-measurable real valued function on R such that
Ifl"
is D-integrable},
122
THEORY OF CHARGES
Lm(R,9, p ) = {f; f is essentially bounded and T1-measurable}, and
Ilfllm
If1
= Essential supremum of = Inf
{k > 0 ; Ip I*({@E R; If(w)I > k}) = 0).
(We use the convention that the infimum over an empty set is co.) If f is a null function, obviously, l f 1 1, = 0 for any 1~p I03. If f and g are such that f = g a.e. [ p ] , f c L P ( f l , 9 , p ) for some 1 s p S c 0 , then g E L,(R, 9, p ) and Il f l , = llgllp We want to show that the nonnegative function II.IIp defined on L,(fl, 9 , p ) for 11p1co is a pseudo-norm on L,(fl, 9,p ) . We need the following inequalities for this purpose. The first of these is Holder’s inequality. 4.6.2 Theorem. Let (R, 9,p ) be a charge space and p and q be two positive numbers satisfying l / p + l / q = l . 1 f f € L p ( f l , 9 , p )and g E L q ( R , 9 , p ) , then fg E Ll(fl, 9, p ) and
llfgll1 5 llfllPllg11q. Proof. Assume, first, that p > 1 and q > 1. The function $(t) =
t P t-+-,
P
9
t >o
has a global minimum at t = 1. Therefore, for every t > 0, $ ( t ) 2 $(1)= l / p + l / q = 1. Let a and b be any two positive numbers and t = (al’q)/(bl’p). Then
This implies that ab s a P / pfbq/q. This inequality is valid even if a = 0 or b = 0. Now, we turn to the proof of the theorem. Iff or g is a null function, then fg is a null function. This can be proved as follows. Suppose f is a null function. Since any T1-measurable function is smooth, g is smooth. So, for a given E > O , there exists k > O such that Ip I*({@ E R; Ig(w)l> k}) I E . Consequently, for any s > 0 , IPI*({@Efln; If(@)g(w)l>S})~ICLI*({@ E n ;I f ( ~ ) l > S l W
+ IPl*({@E 0; Ig(w)l > kl) SO+&
=&.
This shows that fg is a null function. In this case, the theorem is evidently true.
4.
123
INTEGRATION
Since f and g are T1-measurable, fg is TI-measurable. See Corollary 4.4.9(i). By Theorem 4.4.18, fg is D-integrable. It is now obvious that llfgll1 5 ( l / P + ~/q~llfllPll~ll~ = IlfllPllgllq. If p = 1, then q =a.Therefore, lfgl s k l f l a.e. [ p ] for any number k > llgllm. Since fg is TI-measurable, by Theorem 4.4.18, it follows that fg is D-integrable and lFgll1s kllflll for any k >llgllm. Consequently, llfgllls Ilflllllgllm. The case p = 00 and q = 1 can be disposed of in a similar vein. 0
A more general version of the above theorem is the following result.
4.6.3 Corollary. Let (a,9, p ) be a charge space and p , q, r be numbers satisfying 1s p , q, r 5 00 and l / r = l / p + l/q. If f E L p ( R , 9 ,p ) and g E Lq (a,9, P ), then f g E Lr (a,9, CL 1 and llfgllr 5 Ilfllpllgllq* Proof. There are only three possibilities involving 00. Case (i). p = 1,q = 00, r = 1. Case (ii). p = 00, q = 1, r = 1. Case (iii). p = 00, q = 00, r = 00. In Cases (i) and (ii), the result follows from Theorem 4.6.2. For the case (iii), we proceed as follows. For any k > 0 and t > 0 satisfying k > llfl1m and t > llgllm, we have Ir.Ll"({w E n ; I f ( w ) g ( w ) l > k t } ) ~ I l l l * ( { w E n ;
If(w)l>kH
+lPl*(b ER; Ig(4l>tN = 0.
This shows that fg is essentially bounded. Further, llfgllmskt for any k > IIfIIm and t > IIgIIm. Hence IIfgIIm 5 IIfIImIkII.o. Let us look into the case 1< p
If('
Consequently, llfgllr 5 ~ ~ f ~ ~Since p ~ fg ~ is g TI-measurable, ~ ~ q . fg E Lr(R, 9,p ) .
0 A special case of Holder's inequality is Cauchy-Schwartz inequality.
124
THEORY OF CHARGES
Corollary. Let (a,9,p ) be a charge space and f , g E L2(Q 9, p). Then f g E L l ( f l ,9, p ) and
4.6.4
llfgll1 5 llfl1211g112.
A consequence of Corollary 4.6.3 is the inclusion relations among the L,-spaces. 4.6.5 Corollary. Let (a,9, p ) be a charge space in which p is bounded. Let r and s be any numbers satisfying 15 r 5 s 5 03. Then
Ll(a,~,)~Lr(R,9,p)=,L,(a,9,pCL)LLm(a,~,~CL).
Proof. Since p is bounded, every constant function (which is obviously TImeasurable) is D-integrable. Consequently, If(” is D-integrable whenever f E L,(n, 9, p ) and 15 p < 03. So, the last inclusion relation in the corollary follows. Now, let 1Ir < s
(a,9, p)
be a charge space and 1‘ p + g E L,(n, 9, p ) and
503. If
f,
Ilf+gll, 5 llfll, fIIgl1,. Proof. The case p = 1 follows from Theorem 4.4.13(viii). If p = 03, it is easy to check that [ I f + gll.. 5 llfll.. + llg1lrn by using an argument similar to the one used in the proof of Case (iii) of Corollary 4.6.3. Assume 1< p COO.Let q be the positive number satisfying l / p + l / q = 1.For any four real numbers al, b l , a2,b2, we have lalbl +u2b215 (/allP+ /az/P)1’P(~bl14 + 1b21q)1/q. This can be seen as follows. Let a ={ol, w z } , 9= P(n)and p ({ol}) = 1= p ({wz)). Let f h = )0 1 , f ( 4 = a2, g ( w d = bl and g ( w ) = b2. Applying Holder’s inequality, we have precisely the above inequality. Note that
lf+glP~ I f l l f + ~ l P - l + I ~ l l f + ~ l P - l 5
( ( f p+I g l p ) l / p ( 2 ( f + g ) q ( p - - l ) ) l / q
5
(IflP
+ l,lp~1~p(21~q>(lf+gop~q.
so, I f + g ) p - ( p / q ) = If+g1521’q(lf(p+ ( g l p ) l / pThus, . we obtain the inequality If+gl” 52p/q(1flp+ Igl”). By Corollary 4.4.9 and Theorem 4.4.18,
4.
125
INTEGRATION
This follows from Holder's inequality. Therefore,
From the above inequality, it follows that (D
[ If+gl" dlul)
'-(l/q)
=Ilf+gllp ~llfllP+llgllP.
0
This completes the proof.
Theorem. Let (fl, 9, p ) be a charge space. Then,for each 15 p 5 00, (Lp(fl,9, p ) , 1) [Ip) is a linear space with a pseudo-norm 11 IIp.
4.6.7
-
Proof. It is now obvious that each Lp(fl,9, p ) is a linear space. Further, if f = 0, then Ilfl , = 0. For any real number c and f in L,(fl, 5 .F, p ) , it is ~ 9,p ) and that Ilcfll, = Ic I l fl1,. The inequality obvious that c f L,(fl, Ilf+gllp 5 l f11, +llgllp for f,g in L,(R, 9, E L ) follows from Theorem 4.6.6.0
Remark. If p is a 0-1 valued charge on a field 9 of subsets of a p ) / - is isometrically isomorphic to the real line R set fl, then Lp(fl,9, for any 15 p 5 CO, where Lp(fl,9,p ) / - is the collection of all equivalence induced by the classes of L,(fl, 9,p ) under the equivalence relation notion of a null function. (See Example 4.4.14.) Consequently, L,(fl, 9,p ) is complete. In general, (Lp(fl,9,p ) , 11 )1, need not be complete. Let fl = {1,2, . .}, 9the finite-cofinite field on fl and p the charge on 9defined by 4.6.8
-
.
p(A) =
1
12"'
if A is finite,
noA
= 2-
1 ".AC
1 2"'
-
if A is cofinite.
126
THEORY OF CHARGES
.
Let A,, = { 1 , 2 , 3 , . . ,n } , n 2 1. We claim that lim D m,n+m
J lIA,,-
This claim is established if we observe that D J IIA, -IA,/ dp = p(A,AA,) which converges to zero as m, n + co.Suppose la,, n L 1 converges to some function f on R hazily. We show that f- 1. For every k 2 1, there exists nk 1 1 such that p*({w E a; IIA,,(W)-~(WI > l/2k}) < 1/2k. Let Bk in 9be ~ ER;l I ~ , ~ ( ~ ) - f ( w )1l/>2 k } ~ B kfor any set such that ~ ( B k ) < l / 2and{w k 2 1. Assume, without loss of generality, that nl < 112 < n3 < * . We now give the properties of the sets Bk, k L 1. (i). Each Bk is a finite set. For, for any infinite set A, p*(A) 2 1. (ii). B k ~ { k + l , k + 2 , .. .}, k 2 l . (iii). I I ~ , , ( w ) - f ( w ) l 5 1/2&,if w&Bk,for k 2 1. (iv). k &Bk,for each k 2 1. Now, let ko in R be fixed. Let E > 0. We show that lf(ko)- 11 < E . This then would imply that f ( k o )= 1. Let N 2 1 be such that 1/2N < E . Let p 2 max{N, ko}. Since B, c { p + 1, p +2, . . .} and p 2 ko, ko & B,. Further, ko E A h c Anko c Anp. Therefore, If(ko)- I A , ~ ( ~=oI f)( ~ k 0 )- 11 5 1/2' 5 1/2N< B . This shows t h a t f s 1. Next, we show that IAn,n 2 1 does not converge to the constant function identically equal to 1 hazily. Let E = i. Then the set {w E 0;[IA,(W) - 11 >b} is a cofinite set and consequently, p ({w E R; IIA,(w)- 11 >b}) 2 1. So, IA., n 2 1 fails to converge to hazily. Thus, we have a Cauchy sequence in L1(R, 9 , p ) not convergent in L1(R, 9, p ) . Hence L l ( R , 9 , p ) is not complete.
4.6.9 Remark. In L,(R, $, p ) , if we introduce the equivalence relation by f - g for f, g in L,(R, 9, p ) if f = g a.e. [ p ] ] ,then the collection of p ) / - of L,(R, 9, p ) equipped with the all equivalence classes L,(R, 9, norm
-
Il[fIIIP = IlfIIP for f in L,(R, 9, p ) is a normed linear space, where [f]is the equivalence p ) containing f. class in L,(R, 9, Next, we aim at proving Lebesgue dominated convergence theorem. For this, we need the following theorem on convergence in L,-spaces.
4.6.10 Theorem. Let (a,9, p ) be a charge space and 1 S p
4.
127
INTEGRATION
(i). f,, n z 1 converges to f hazily. (ii). The charges A, on 9defined by A, (F) = D 5,l f , 1’ dlp 1, F in 9,n z 1 are uniformly absolutely continuous with respect to w, i.e. given E >0, there exists S > 0 such that A, (E)< E for every n 2 1 whenever E E 9 and IP I(E)< 6.
(iii). For each E > 0, there exists a set E, A,(Ez) < E for every n z 1.
E 9 such
that
I(E,) < 00 and
Proof. The proof is carried out in the following steps. 1”. “Only if” part. If h is a nonnegative simple D-integrable function on 0, the following inequality known as Chebychev’s inequality is easy to establish. D IPl({wEa;h(W)>r))s
J h dlPl r
for any r > 0. 2”. In order to show that f,, II 2 1 converges to f hazily, it suffices to show that for any given E > 0 and s 2 > 0, there exist N z 1 and sets A, in 9 for n 2 N such that \pI(A,)< E~ and If , ( w ) -f(w)I < E~ for w in A: whenever n 2N. 3”.Let r and E be two positive numbers satisfying (2r)”’ < s2 and 3 ~ /
and
IP
I*({@E a;Ih, (W f - g, (W >I2 TI)< &IT.
See Lemma 4.4.12. Let B , in 9 be any set such that Ipl(B,)<E/r and {W E a; Ih,(w) -g,(w)l 2 r ) c B,, n 2 1. Let C , = {W E a; h,(w) >r}, n z 1. Then, by lo,lpl(C,) 5 (D 5 h , d(FI)/r for every n z 1. Let A, = B , u C,, n z 1. Then
J
I P I ( A . ) ~ l P I ( B f l ) + I P l ~ c n ~ ~ ~ /hr,+dlPl)/r (D
<E/~+(D
J g, d ~ w ~ + E ) / r
< E / I + ~ E / ~ < E ~ ,
if n rN.
.
128
THEORY OF CHARGES
If w EAE=BznC;, Ifn(w)-f(w)IP= g n ( w ) i h , ( w ) f r 1 2 r which implies that If,(w) -f(w)I i (2r)ljp< E:! for every n 2 1. This shows that f,, n 5 1 converges to f hazily and thus (i) is established. 6”. Since is D-integrable, for E >0, there exists 61 > O such that d l p l < ~ / 2wheneverFE9andIpI(F)<61.Sincelim,+m ~ l f, -flip = D ,5 0, there exists N 2 1 such that D dlp I < ~ / 2 ’for all n >N. Since A l , AZ, . . . ,AN are all absolutely continuous with respect to p , by Theorem 2 F E 9 and 4.4.13(xi), there exists SZ> 0 such that A, (F) < ~ / whenever Ipl(F) < & for n = 1 , 2 , . . . ,N . Let S =min (61,Sz). If F E and~ IpI(F)<S, then for 1i n S N , and ArI(F) < E D ,
Ifl”
Ifl”
If, -fl”
A, (F) = D
\ If, F
-f+fI”
dlp I
< [(&/2”l/P + (&/2”)’/”]”= E ,
for n >N .
This proves (ii). 7”. We now prove (iii).Let E > 0. There exists Eoin 9such that lp ~(EO) < 00 and D jE; dlpl< ~/2’. See Lemma 4.4.15. Further, there exists N 2 1 such that D d l p l < ~ / 2for ~ every n > N . Also, for each n = 1,2, . . . ,N, there exists E, in 9 such that ( pI(E,) < 00 and A,(E‘,) < E . See Lemma 4.4.15. Let E, = EOu El u uEN. Obviously, lp I(EE) < 00. If 15 n s N , then A,(ES)IA,(E;)<E. If n > N , then also,
Ifl”
If, -fl”
-
< [(&/2”)’/”+ (&/2”)’/”1”= E . This completes the proof of “only if” part. 8”. Now, we prove “if” part. First, we show that g = is D-integrable. Let g , = Ifnlp, n L 1. In view of Theorem 4.4.20, it suffices to show that g,, n 1 1 converges to g hazily and
Ifl”
lim D
m,n-co
J
lg, -gml dlpl= 0.
Hazy convergence can be proved as follows, Since each f , is TI-measurable and f,, n L 1 converges to f hazily (by (i)),f is T1-measurable and so, n 2 1 converges to hazily. See Corollary 4.4.9(iii).
Ifl”
lf,lp,
4.
129
INTEGRATION
9". It remains to be shown that the above limit is indeed equal to zero. For this, first, we show that if E E 9 and I(E)< 00, then
Let E > 0. By (ii), there exists 6 > 0 such that A,(F) < ~ / ( + 2 ( p((E))for every n z 1 whenever F E 9 and lp [(F)< 6. Since g,, n 2 1 converges to g hazily, we have m,n+m lim l~l*(b l ~ n ( ~ ) - g , ( ~ ) J ~ ~ / ( ~ + l ~ l ( E ) ) ~ ) = ~ . Consequently, there exist N L 1 and sets En, in 9for n, m r N such that lp I(E,,) < 6 and lg, (0)- gm(w)l< ~ / ( + 2 1~ I(E)) for w in Ei,,, whenever n, m 2 N. Now, if n, m 2 N,
D / lgn-gm/dlPI=D/ E
Ign-grnIdlPI+D E,,nE
J
Ez,nE
Ign-grnIdlPI
This establishes the desired assertion. 10". Next, we show that Iimm,n+w D lg,, -g,l d(pI= 0. Let E >O. By (iii), there exists a set E in 9 such that l(E)< co and A,(E') < E for all n 2 1. Consequently,
sD/EIgn-gmIdIPI+E+E l . < 00, for all n, m 2 1. Taking limits as n, m -f CO, we obtain, by 9" as ( ~ I(E) that limm,,+wD (g,, - g, 1 d ( pI 5 2.5. Since E > 0 is arbitrary, the above limit is indeed equal to zero. Hence g is D-integrable. = 0. Let A. on 9be defined by 11". Finally, we show that limn-tml f, - f [ l , Ao(F)= D jFIflp dip[, F in 9.By Theorem 4.4.13(xi) and Lemma 4.4.15, (ii) and (iii) hold for the sequence Ao, AI, AZ, . . . also.
130
THEORY OF CHARGES
Let E >O. By (iii), there exists E, in such that Ipl(E,)
+(D
JA“IflP dliLI)l/p
<E+E+~[J~~(E,-A,)]~/~+E+E
< E + E + r[lp I(EE)I1”+ E
+E <5
~ .
Hence limn,oo D If, - f l p dlpl= 0. This completes the proof of the theorem. 4.6.11 Remark. If p is bounded, condition (iii) is not required in Theorem 4.6.10. We need the following notion of hazy convergence for nets and the attendant result in the proof of Lebesgue dominated convergence theorem. 4.6.12 Definition. Let (a, 3,p ) be a charge space and f a , a E D be a net of real valued functions on 0. fa, a EDis said to converge to f hazily if for every E >0,
4.
131
INTEGRATION
The following result says that the limit function f is T1-measurable if each f, is T1-measurable.
4.6.13 Proposition. Let (Q, 9,p ) be a charge space and f a , a E D a net of T1-measurable functions on Q converging to a function f hazily. Then f is T1-measurable. Proof. In view of Theorem 4.4.7, it suffices to show that f is T2-measurable. Let E > 0. There exists a. in D and a set A in 9 such that 1p I(A) < ~ / 2 and If,,(w) -f ( w ) l < ~ / for 3 all w in A‘. Since f,, is T2-measurable, there exists a partition {Fo,F1,F2, . . . , F,} of Q in 9such that Ip ~(FO) < ~ / and 2 If,,(w) -fa0(w’)I< ~ / for 2 all w , w ’ in Fi for i = 1,2, . . . , n. Let Eo = A u Fo, and Ei = Fi nA‘, i = 1,2, . . . , n. {Eo,El, E2, . . . , En}is obviously a partition & .w , w ‘ E E ~ for any i = of Q in 9.Further, I p 1 ( E o ) < ~ / 2 + ~ / 2 = If 1 , 2 , . . . , n, then If(d-f(w’)l~If(w)-fa,(~)l+lfao(~)-f~,(w’)l
+Ifa,(W’)-f(4
< &/3-k &/3-k &/3= E . 0
This completes the proof.
Now, we are ready to prove Lebesgue dominated convergence theorem.
4.6.14 Theorem Let (Q, 9, p ) be a charge space and g E L,(Q, 9,p ) for some p 2 1. Let f,, a E D be a net of T1-measurable functions on Q such that Ifal 5 lgl a.e. [ p ] .Let f be a real valued function on Q. Then the following statements are equivalent. (a). f a , a E D converges to f hazily. (b). ~ E L ~9, (Q p ),and lim,.D Ilfa -flip = 0. If p = 1, the following statements are equivalent. (a’). f a , a E D converges to f hazily. (b‘). f E L1(R, 9,p ) and lim,,D D 1, ( f a -f ) dlpl= 0 uniformZy over F in
9. (c’). f
E L1(fl, 9, p)
and lim,.,,
D1 , If a -f l dlp I = 0 uniformly over F in
9. (d’). f ELI(Q,g, p ) and lim,.D D
Ifa - f l
d b l = 0.
Proof. We prove the first part. By Theorem 4.4.18, each f a E L,(R, 9,p ) . Assume that the net f a , a E D is really a sequence f n , n 2 1. Suppose (a) holds. We verify (i), (ii) and (iii) of Theorem 4.6.10. (i) holds by virtue of the validity of (a). We show that (ii) holds. Let E >O. By Theorem 4.4.13(xi), there exists 8 > O such that D jF lglp dlp I < E whenever FE9 and Ip [(F)< 8. Since Ifnl c Igl a.e. [ p ] for every n 1 1,it follows that D jFIf n 1’ dlp I < E whenever F E 9and lp I(F) < 8. Thus (ii) holds. Next, we show that (iii) of Theorem 4.6.10 holds. By Lemma
132
THEORY OF CHARGES
4.4.15,thereexistsasetFEi n 9 s u c h t h a t IpI(F,)
+
We end this section with a result on the denseness of D-integrable simple p ) for 1 5 p < a. functions in L, (R, 9, 4.6.15 Theorem. Let ( R , 9 , p ) be a charge space. Let Sim(R,9, p ) be the space of all D-integrable simple functions on R. Then Sim(R, 9, p ) is dense in L, (R, 9,p ) for every 1 5 p < CO.
Proof. Let 1 s p < a be fixed and f~ L,(R, C ,F , p ) . For a given E >O, by Lemma 4.4.15, there exists a set A in 9 such that IpI(A)
<0+EP.
From this, it follows that Ilf-I~fll,< E . Now, we look at IAf.Let f n , n 2 1 be a sequence of simple functions converging to IAf hazily. By Lemma 4.4.17, we can assume that Iffl I 5 2IA(fl for every n 2 1. By Lemma 4.4.16, Iffl/’ is D-integrable for every n 2 1. By Lebesgue dominated convergence theorem, limn+mllfn - I ~ f l l = , 0. Consequently, there exists N ? 1 such that llfN -IAfllp < E . From this, it follows
4.
133
INTEGRATION
that
0
4.7 ba(fk,9) AS A DUAL SPACE In this section, we introduce a Banach space whose dual is the space of on the field 9 of subsets of a set R. We also all bounded charges ba(R, 9) consider some extensions of this result.
4.7.1 Definition. Let 9 be a field of subsets of a set 0. A real valued function f on R is said to be 9-continuous if given E >0, there exists a partition {F1, FZ,. . . , F,} of R in 9 such that 1 f ( w ) -f (w’)I < E for all w , W ’ in Fi and for every i = 1 , 2 , . . . , n. Let %‘(a,9)denote the collection of all 9-continuous functions on 0. It is obvious that every 9-continuous function is bounded. Further, the space %‘(a, is a linear space. All simple functions are available in % ( R , 9 ) . On U ( 0 , %‘), we introduce a norm by 11fll= Sup {I f(w)l; w E R} for f in %(a, It is obvious that 11 11 is indeed a norm on %(a, 9)If.A E9 and A # 0, then IIIAll = 1.
a a.
4.7.2 Proposition. Let 9 b e a field of subsets of a set R. Then the following statements are true. (i). (%(a, 9), 11 * 11) is a Banach space. (ii). The collection of all simple functions is a dense subset of %(a, 9). (iii). f is $-continuous if and only iff is the uniform limit of a sequence of simple functions on 0. n s l be a Cauchy sequence in %(R,S). Let f ( o ) = Then fn, n 2 1 converges to f uniformly over R. We show that f is 9-continuous. Let E >O. There exists N 2 1 such that Sup {I f N ( 0 ) -f (o)l;w E R}< ~ / 3For . fN, there is a partition {Fl, F2, . . . ,Fk} for all o,w ’ in Fi and for every of R in $ such that I f N ( W ) - f N ( o ’ ) I < E / 3 i = 1,2, . . . , k . Let i E {1,2, . . . ,k} and w , W ’ E Fi. Then
Proof. (i). Let
fn,
f,(w), w
E R.
If(d-f(w’>I 5 If(4 -f”
+lfN(W)
-fdw’)I
+IfN(W’)
-f(w’>I
< &/3+ &/3+&I3= E. This shows that f is 9-continuous. It is obvious that limn+mllfn -f l l = 0. II.II) is a Banach space. Hence (%‘(R,
134
THEORY OF CHARGES
(ii). Let f be $-continuous. For n rl, there is a partition {F,I, Fn2, . . . , Fnk,} of R in 9 such that for every i E {1,2, . . . , k,}, I f ( w ) - f ( w ' ) l < l / n for all w , w' in Fni.For each n 2 1 and i = 1,2, . . . , k,, choose and fix wni in Fni.Let
If(@)
-f,(w)I = We claim that f,, n 2 1 converges to f uniformly. If w E F,i, If(w)-f(wni)< l l / n for all n 2 1 and i = 1,2, . . . ,k,. Consequently, Ilf-f,ll< l / n for all n 2 1. Hence Ilf-f,ll = 0. This shows that the class of all simple functions on R is a dense subset of %(a,8. (iii). This is now obvious. 0
We now give a characterization of $-continuous functions based on D-integrability.
Theorem. Let 9be a field of subsets of a set R. Then f E %'(a,fl if and only i f f is D-integrable with respect to every bounded charge A on $.
4.7.3
Proof. Let f~ %(R, 9). Note that if A is a bounded charge on 9, then f is Tz-measurable with respect to A and hence is D-integrable. See Theorem 4.5.7. Conversely, suppose f is D-integrable with respect to every bounded Suppose f is not $-continuous. There exists E > O such that charge on 9. there exists 1Ii In satisfying given any partition {F1,F2, . . . ,F,} of R in 9, O(f,Fi) = Sup{lf(w)-f(w')l; w , O'E Fi}> E . For each partition P = {F1,F2, . . . ,F,} of R in 9, let A(P) = U{Fi; 1 si I n , O(f, Fi)> E } . Let B denote the collection of all finite partitions of R in 9. Then {A(P); P E 9 ) is a filter base in 9. To see this, let P1, Pz E 8 and P any partition in B finer than both P1 and Pz. Suppose for some F in P, O ( f ,F) > E . Then F is contained in some G1in PI and in some GZin P2. Consequently, O ( f ,GI)> E and O ( f ,Gz)> E . Thus F c G1nG2 and from this, it follows that A(P) c A(P1)nA(P2). Further, note that A(P) # 0 for every P in 8.Thus we have proved that {A(P); P E P } is a filter base in $. Hence there is a maximal Define A on 9by filter 8 in 9containing {A(P); P E 8}. A(E)=l, ifEE'iY, =0, ifEg'iYandEE9. A is a 0-1 valued charge on 9.
We claim that f is not D-integrable with respect to A. Suppose f is D-integrable with respect to A. Then f = c a.e. [A] for some constant c. See Example 4.4.14. Then A*({w E R; if(w)-cl > s / 2 } ) = 0. Consequently, there exists B in 9 such that {w E R; I f ( w ) - c I > ~ / 2 c} B and A(B)=O. Now, we look at the partition P={B,B"} in B. Then
4.
135
INTEGRATION
B ' c ( w ~ 0 ;I f ( w ) - c l 5 & / 2 } . For any w , w' in B', I f ( w ) - f ( w ' ) l I If(w)-cl+)f(w')-cI S E / ~ + E / ~ = E . Therefore, O ( f ,B')IE. Hence A(P) = B E 8. By the definition of A, A (B) = 1.This contradiction proves the 0 desired assertion. Now, we characterize the continuous linear functionals on (%(a, 9), II-II). 4.7.4 Theorem. Let 9 b e a field of subsets of a set 0.Let T b e a continuous linear functional on %(a, Then there exists a unique bounded charge p on 9 s u c h that
e.
for every f in %(a, 9).Further, llTll= Sup {IT(f)l; llflls1}=Ipl(a). Conversely, for any given bounded charge A on 9, the functional T' on %'(a, 9) defined by T ' ( f )= D J f dA, f in %(a, 9) is a continuous linear functional on %'(a, 9-with ) IlT'll= IA /(a). If T is a nonnegative linear functional on %?(a, 9), i.e. T (f ) 2 0 i f f 2 0 , then p is a positive charge. I AE %?(a, 9)and S O , define p (A)= T(IA).It is obvious Proof. For A in 9, that p is a charge on 9.Also, IpLA)I= IT(IA)I 511TllllrAl1511Tll
for every A in 9. This shows that p is a bounded charge on 9.Now, we 9), T ( f )= D J f d p . Iff ciIAi is a simple claim that for any f in %?(a, function in %(a, then
=xi"=,
a,
T ( f )=
i=l
ciT(IAi)=
f
i=l
c i p ( A i )= D
J f dp.
For any f in %(a, 9), by Theorem 4.7.2, there exists a sequence f n , n 5 1 of simple functions in %(a, 9) converging to f uniformly. It is obvious that f,,, n 2 1 converges to f hazily. Further, since p is bounded,
Thus, f n , n 2 1 is a determining sequence for f with respect to p. Consequently,
D
J'
f dp
= lim n+m
D
as T is continuous on %(a, 9).The claim is thus established.
136
THEORY OF CHARGES
Now, if fE%‘(R,.F) and llfilsl,then ~ T ( f ) ~ = ~ D ~ f d p ~ ~ D ~ Ipl(R). See Theorem 4.4.13 (iii). Hence ~ ~ T \ ~ I ~ Since ~ ~ Ipl(R) ( R ) =. Sup Ip (Ai)/,where the supremum is taken over all finite partitions {Al, A2, . . . ,A,} of R in 9, for any given E >0, there exists a partition (FI,F2, . . . ,F,,,} of R in 9such that m
Since E > 0 is arbitrary, we have Ip I(R) IIlTll. This shows that llTll= Ip [(a). The rest of the theorem is obvious. 0
4.7.5 Corollary. The dual of %‘(a,9)= %‘*(a, 9) = ba(R, F). 4.7.6 Corollary. Let X be a compact Hausdorff totally disconnected space, 9 the field of all clopen subsets of X, 93 the Bore1 u-field on X, %(X)the space of all real continuous functions on X and ca(X, 93) the space of all bounded regular measures on 93. (%(X) is a Banach space under supremum norm and ca(X, 93) is a Banach space under total variation norm.) Then the following statements are true. (9. %(X, 9) = The space of all .%continuous real functions on X = %(X).
(ii). (Riest Representation Theorem.) If T is a continuous linear functional on %(X), there exists a unique bounded regular measure p on 93 such that
T(f)=D
If
dp
for f in %(X) having the property that llTll= lpl(X). (iii). The dual of %(X)= %*(X)= ca(X, 93).
Proof. (i). Suppose f is a continuous real valued function on X. For E > O and for each x in X, there exists a clopen set C, containing x such that
4.
137
INTEGRATION
I f ( y ) - f ( x ) l < ~ / for 2 all y in C,. {C,; x EX} is an open cover for X. Since X is compact, there exists a finite subcover {C,,, C,,, . . . ,CXn}of X. For each 11i%n and for every x , y in CXi, we have I f ( x ) - f ( y ) l I I f ( ~ ) - f ( ~ i ) l + I f ( ~ i ) - f ( y ) ( < ~ / 2 + ~ / 2 = ~ . Let Di=Cxl, D2= C,, - C,,, . . . ,D, = CXn-(GIu C,, u . * u CXn-,). Then {Dl,D*,. . . ,D,} is a partition of X in 9and for this partition, we still have I f ( x ) -f(y)I < E for all x , y in Di and for all i = 1,2, . . . ,n. Hence f is $-continuous. . Conversely, if f is $-continuous, it follows easily that ~ E % ( X )This proves (i). (ii). Let T be a continuous linear functional on %(X).By Theorem 4.7.5, there exists a bounded charge C; on 9 such that T ( f )= D J f dC; for every f in %(X). C; can be extended as a measure on the Baire a-field Boof X. can be extended as a regular measure p on B. See Theorem 3.5.5. This completes the proof of (ii). (iii). This is obvious now. 0
-
4.7.7 Remark If 9 is a field of subsets of a set R, then %(a, 9) can be realized as %(X)for some compact Hausdorff totally disconnected space X.
Now, we obtain some natural subspaces of ba(Cl, 9) as dual spaces when a-field.
$ is a
We consider the following set-up. Let R be a set, 8 a a-field of subsets (ii) I. C, E $ if C c A, C E %?I of R and 9 a proper cr-ideal in 8, i.e. (i) 4 c ? and A E 4 , (iii) R @ 4 and (iv) Unal A,, E 9 if A,, n 2 1 is a sequence in 9. A real valued function f on R is said to be measurable if f-'(B) E 8 for every Bore1 set B c R. A measurable function f on R is said to be essentially bounded if there exists k > O such that {w E R; \ f ( w ) \> k } E 9 . If f is an essentially bounded measurable function on R, define
Ilfllm
= Inf
{ k ;k > 0 and
{w E R; If(w)I > k} E 9).
Obviously, 0 Illfllm < a. Let Lm(RZ, a, 9)denote the collection of all essentially bounded measurable functions on R. If f E Lm(Q%, 9), then A={w =
u
nzl
If(w)l>IIfllmI @En;
If(w>l>llfllm-tl/n}
€9.
Further, we show that A has the following properties. (i). SUP{lf(w)t; 0 E A'} = IFll==. ; E A'- B} = llfllm for any B in 4. (ii). Sup { l f ( w ) / w Obviously, Sup { I f ( w ) l ;w E A'} 5 llfllm. If Sup {If(o)(; w E A'} < llfllm, then {w'ER; If(w')l>Sup{If(w)l; w E A ' } } ~ A and hence {w'ER;If(w')I> Sup {If(w)l; w E A'}} E 9.But this contradicts the definition of llflloo. (ii) can be proved analogously.
138
THEORY OF CHARGES
-
4.7.8 Proposition. The function 11 Jlmdefined on L,(R, M, 9)above has the following properties. (i). llcfllm = IcI llfllm for any real number c and f in L,(R, a, 9 ) . 6).Ilf + gllm 5 llfllco + llgllm for allf, g in Lm(R, M, 9). Proof. (i). This is obvious. (ii). Note that {w €0;l f < ~ ~ + ~ ( ~ ) l > I l f l l m + + l ~ I I ~ ~ C{W
€0;If(~)l>llfil~I
u l w En; Id~I>llSll~}E.a.
Consequently, Ilf + gllm 5 llfllm + llgllm-
0
4.7.9 Proposition. (L,(R, M, 9 ) ,1)- llm) is a h e a r space and pseudo-norm under which L,(R, M,9) is complete.
11.1)m
is a
Proof. It is clear the L m ( R , M,9) is a linear space and that Il*llm is a pseudo-norm on Lm(R, a,9).We show that Lm(R, %,9)is complete. Let fn, n 2 1 be a Cauchy sequence in Lm(R,Vl, 9). For every n, m 2 1, let Anrn={w Ea;I f n ( ~ ) - f r n ( ~ ) l > l l f n - - f r n I l m } . Then for any B in 9, SUP{Ifn ( w )-frn(w>I; w E A;rn--B}
=
llfn -frnll,*
LetA=Un,,Urn,,Anrn. T h e n A E 9 and lim Sup(~fn(w)-frn(w)~;o €A'}=
rn,n-m
lim
llfn-frnllm=O.
rn,n-m
Let f ( w ) = limn,m f n ( w ) for w in A'. Then f n , n L 1converges to f uniformly on A". For w in A, define f ( w ) = 0. We claim that l f n -film, n L 1converges to zero.
llfn -fIL
= IKfn - f ) l A c + Il(fn -f)lAcllm
(fn
+ Il(fn
-f)lAllm
0 This completes the proof.
5 SUP {Ifn(w)-f(w)I;
which converges to zero as n + 00.
-fV*lIm w E A?+
In the present context, we define a null function as follows. A real valued measurable function f on R is said to be a null function if {w E R; If(o)I> k } E 9 for every k > 0. This is equivalent to the condition that )lfllrn = 0. The space of all null functions is a linear subspace of Lm(R, M, 9).We introduce an equivalence relation on Lm(R, M, 9)by f-g if f - g is a denote the collection of all equivalence null function. Let 2?m(R,8,9) classes of Lm(R, M, 9). The pseudo-norm 11. llm defines unambiguously a norm on Z,,(R, a, 9) which is again denoted by 11 llm. Now, it is apparent that Ym(R,a,9)is a Banach space. We work out its dual.
-
6
4.
139
INTEGRATION
4.7.10 Theorem. Let T be a continuous linear functional on (Lm(R,%, 9), 11 llm). Then there exists a unique bounded charge p on % with the following properties. (i) T ( f ) = D J f d p f o r e u e r y f i n Lm(R,%,9). (ii). llTll= IpI(W. (iii). p (A) = 0 i f A E 4;.
Proof. For A in 8,let @(A)= T(IA). p is obviously a charge on a. Note that T(f) = 0 whenever f is a null function. If A E 9,then IAis a null function and so, p (A)= 0. This proves (iii). The boundedness of p follows from the inequalities IpL(A)I= IT(IA)l
5
~
~
~
~
~
~
~
~
A
~
for any A in a. If f is a simple function, it is obvious that T(f)= D jf dp. Let f be any function in Lm(R, %, 9) with llfllm>O. We show that T(f)= D Jfdp. Let B = {w E R; If(w)I 5 Ilfllm}. Note that B“E 4. Let d = 211fllm. For each n z 1, define (similar to the construction given in the proof of Corollary 4.5.9)fn as follows. For w in R,
for i = 1 , 2 , . . . , 2 ” -1,
It is easy to check that f,,,n z 1 converges to f uniformly on B. We claim that f,,,n z 1 is a determining sequence for f with respect to p. It is obvioils that each fn is a simple function. For any E > 0, Ip
I({@
E
a;Ifn
(@)-f(@)l>
Elf =
l({w
E
B; Ifn (@> -ff@)l>
+ IPKb E B‘;
Ifn(4
I))
-f(d > &I)
= 0,
if n is sufficiently large. This assertion follows from lp [(B‘) = 0 and fn, n 2 1 converges to f uniformly on B. This shows that fn, n 2 1 converges to f
~
m
5
~
\
140
THEORY OF CHARGES
hazily. Also,
= 0.
Hence f n , n L 1 is a determining sequence for f. Since IIfn - fllm IIl(fn - f)I~ll~ + Il(fn - f ) I ~ c l l ~ , it follows that limn+m\Ifn - film = 0. Since T is a continuous linear functional on Lm(fl, %, $1,
T ( f )= lim T ( f n = ) lim D n+m
n-m
I
fn
d p =D
I
f dp.
Next, we show that ~ ~ T ~ ~ = ~For p ~any ( f lf )in. Lm(R,%,9),IT
If
If1
For the following corollary, let ba(R, %, 9) stand for the space of all bounded charges on % vanishing on 9. We equip ba(fl, %, 9)with the total variation norm.
4.7.11 Corollary. The dual of Ym(fl,%, 9 ) is isometrically isomorphic to ba(fl, %, 9).
CHAPTER 5
Nonatomic Charges
Classification of charges is an obvious pursuit one embarks on for a good understanding of charges. We have already come across 0-1 valued charges in Chapter 2. In this chapter, we examine what could be an antithesis of the notion of a 0-1 valued charge. Section 5.1 develops the relevant classification of charges. The Sobczyk-Hammer decomposition theorem is presented in Section 5.2. We prove some existence theorems for nonatomic charges in Section 5.3. Finally, in Section 5.4, we consider the plentitude of nonatomic charges.
5.1 BASIC CONCEPTS The notion corresponding to nonatomicity of measures on cr-fields can be introduced for charges in three different ways. We show that these three ways are actually distinct for charges.
5.1.1 Definition. Let 9 be a field of subsets of a set R. A set F in 9 is said to be a p-atom if the following conditions are satisfied. (i). p( F ) # 0. (ii). If E E 9 and E c F, then either p (E)= 0 or p (F- E) = 0.
If F is a p-atom, then the restriction of p to F n9 is a 0-p (F) valued charge on the field F n9of F. However, if p restricted to the field F n9 for an F in 9 is two valued, F need not be a p-atom. Also, it is not difficult to check that an F in 9 is a p-atom if and only if F is a Ipl-atom. For, if F is a p-atom, E E 9, E c F and p (E)= 0, then Ip I(E) = 0. Note that if F E 9, F is an atom of 9 and p (F)# 0, then F is a p-atom. However, a p-atom need not be an atom of 9. Non-existence of p-atoms is one way of defining a class of charges. More formally, we give the following definition. 5.1.2 Definition. Let 9 be a field of subsets of a set R and p a charge on 9. p is said to be nonatomic on 9 if there are no p-atoms in 9. Equivalently, if F E 9 and p (F) # 0, then there exists E in 9 such that EcF,p(E)#Oandp(F-E)#0.
142
THEORY OF CHARGES
In view of the remarks made after Definition 5.1.1, it follows that p is nonatomic if and only if Ip I is nonatomic. If p is a positive bounded charge on 9, then p is nonatomic on 9, if for every F in 9 with p (F)> 0, there exists E in 9such that E c F and 0 < p (E)< p (F).Now, we prove a simple property of nonatomic charges.
5.1.3 Proposition. Let 9 be a field of subsets of a set R and p a positive bounded nonatomic charge on 9. Then given F in 9 with p(F)>O and E > 0 , there exists E in 9 such that E c F and 0 < p (E)< E .
Proof. Since p is nonatomic, there exists El in 9 such that El c F and 0