GENERAL VECTOR MEASURE THEORY 9
(i) F is strongly additive.
(ii) {x*F : x* e X*, \\x*\\ :g 1} is uniformly strongly additive.
(iii) F is strongly bounded, i.e., if (En) is a sequence of pairwise disjoint members
of '&, then lim„F(£„) = 0.
(iv) || F|| is strongly bounded, i.e., if(E„) is a sequence of pairwise disjoint members
of&,thenhmn\\F\\(En) = 0.
(v) {|x*F| : x* e X*, \\x*\\ ^ 1} is uniformly strongly additive.
(vi) limnF(En) exists for every nondecreasing monotone sequence (En) of members
of &.
(vii) limMF(F„) exists for every nonincreasing monotone sequence (En) of members
of &.
PROOF.
The equivalence of statements (i) through (v) is clear from Proposition
17. The equivalence of (vi) and (vii) follows from the identity F{E) + F{Q\E) =
F(Q). To see that (i) implies (vi), let (En) be a nondecreasing sequence of members
of &. Then
lim F(En) = F(EX) + lim £F(EJ+1\EJ)
n n j=2
exists since the sequence(EJ+X\EJ)consists of disjoint members of J*. This proves
that (i) implies (vi).
On the other hand, if(Fn) is a sequence of pairwise disjoint members of F, then
limnF((JjJ=1 Ek) exists by (vi). Thus
lim F(En) = lim
0.
This completes the proof.
Another basic fact about strongly additive vector measures is contained in
COROLLARY
19. A strongly additive vector measure on afield is bounded.
PROOF.
LetFbe a field of sets and F: & -• X be a strongly additive measure. If
||F||(0) = + oo, choose Hx e & such that
\\F(HX)\\ ^ 1 + 2||F(£)||.
Then since F(HX) = F{Q) - F(Q\HX), it follows that
\\F{HX)\\ - \\F(Q)\\ ^ \\F(Q\HX)\\.
Thus \\F(Q\HX)\\ ^ 1. Now \\F\\ is subadditive on disjoint sets so either ||F||(#i)
or IIFIKO^) is infinite. If ||F||(#i) = oo, let Ex = Hx; otherwise, let Ex =
Q\HX. In either case,
\\F\\(EX) = oo and \\F(EX)\\ £ 1.
Replacing 0 by Ex in the above line of reasoning produces a member F2 of fF
contained in Ex such that
|F|(£
2
) = oo and |F(£2)|| £ 2.
Iterating this procedure yields a nonincreasing sequence (En) of members of SF such
Previous Page Next Page