8 J. DIESTEL AND J. J. UHL, JR.

subsets Eh E2, -, En of E such that /l(£;) = X{E)jn for all « = 1, •••, n. Note that

t

\\F(Ei)\\

= t

(KE)l")Wp

=

"a~1/p) KE)l/*.

«=i 1=1

Plainly this means that \F\(E) = oo.

There are many alternative and useful formulations of strong additivity. Most

of them are consequences of the next result.

PROPOSITION

17. Any one of the following statements about a collection {FT :zeT}

ofX-valued measures defined on afield 3F implies all the others.

(i) The set {FT : z e T) is uniformly strongly additive.

(ii) The set {x*FT: z e T, x* e X*, \\x*\\ g 1} is uniformly strongly additive.

(iii) If(E„) is a sequence of pairwise disjoint members of IF, then \\mn\\Fx(E^)\\ = 0,

uniformly in % e T.

(iv) If(En) is a sequence of pairwise disjoint members of $F then limn||Fr||(£n)

= 0, uniformly in z e T.

(v) The set {\x*Fx\ : z e T, x* e X*, \\x*\\ ^ 1} is uniformly strongly additive.

PROOF.

That (i) implies (ii) and (ii) implies (iii) are obvious. To prove that (iii)

implies (iv), suppose (iv) fails. Then there exists a d 0 and a sequence (En) of

pairwise disjoint members of !F for which suprer||Fr||(£n) ^ 45 0 holds for

all n. By Proposition 11(b), for each n there is Hn e F such that Hn £ En and

suprerj|Fr||(£n) ^ 4 suprer||Fr(//n)||. The sequence (Hn) consists of pairwise dis-

j oint members ofFwhich satisfy

sup||Fr(//„)|| ^ 8 0

for each n. Thus (iii) fails to hold. This shows that (iii) implies (iv).

To prove that (iv) implies (v), suppose that {|x*Fr|: z e F, x*eX*, \\x*\\ ^ 1} is

not uniformly strongly additive. Then there exists a disjoint sequence (En) in fF

and a 5 0 such that for all m one has

sup{ 2 \x*Ft\(EH) :zeT, x* e X*, \\x*\\ l\ =t 25 0.

Thus there is an increasing sequence (mj) of positive integers such that for ally

sup 2 I x*FT \(En) :reT,x*e X*, \\x* \\ g 1

U=my+1

= sup{|x*FT|( l j En):ze T, x* e X*, \\x*\\ £ l} ^ d 0.

^ \»=my+l / /

Therefore setting H; = (J«=£*+! ^n produces a sequence (//,) of pairwise disjoint

members of & such that

sup{||Fr||(//;-):rer} = sup{\x*FT\(Hj):zeT,x*eX*,\\x*\\ ^ 1} ^ 5 0.

This denies (iv) and proves that (iv) implies (v). That (v) implies (i) is obvious.

The following corollary is the principal result of this section.

COROLLARY

18. Any one of the following statements about a vector measure F

defined on afield J* implies all the others.