Chapter Vector Measures In this chapter we present a survey of known results about vector measures. For the convenience of the readers some of the results are given with proofs, but neither results nor proofs pretend to be ours. The only exception is the material of Section.5 that covers the results of our joint paper with V. M. Kadets [8].. Elementary properties The results of this section may be found in [7]. Definition.. A function µ from a field F of subsets of a set Ω to a Banach space X is called a finitely additive vector measure, or simply a vector measure, if whenever A and A 2 are disjoint members of F then µ (A A 2 )=µ(a )+µ(a 2 ). ( ) If in addition µ A n = µ (A n ) in the norm topology of X n= n= for all sequences (A n ) of pairwise disjoint members of F such that A n F,thenµis termed a countably additive vector measure, or n= simply µ is countably additive. Example..2 Finitely additive and countably additive vector measures. Let Σ be a σ-field of subsets of a set Ω, λ be a nonnegative countably additive measure on Σ. Consider p. Define 3
4 CHAPTER. VECTOR MEASURES µ p :Σ L p (Ω, Σ,λ)bytheruleµ p (A)=χ A for each set A Σ (χ A denotes the characteristic function of A). It is easy to see that µ p is a vector measure, which is countably additive in the case p< and fails to be countably additive in the case p =. Definition..3 A vector measure µ : F X is said to be nonatomic if for every set A F with µ (A) 0there exist A,A 2 F \ Ø such that A A 2 = A and µ (A i ) 0(,2). Definition..4 Let µ : F X be a vector measure. The variation of µ is the nonnegative function µ whose value on a set A F is given by µ (A) =sup µ(b), π B π where the supremum is taken over all partitions π of A into a finite number of pairwise disjoint members of F.If µ (Ω) <, then µ will be called a measure of bounded variation. The semivariation of µ is the extended nonnegative function µ whose value on a set A F is given by µ (A) =sup{ x µ (A) :x X, x }, where x µ is the variation of the real-valued measure x µ. Example..5 A measure of bounded variation. Let µ be the measure discussed in Example..2. Since µ (A) λ(a), it is plain that µ (A) λ (A), so that µ is of bounded variation. The next proposition presents two basic facts about the semivariation of a vector measure. Proposition..6 Let µ : F X be a vector measure. A F one has (a) µ (A) = sup {ε n},π B n π Then for ε n µ (B n ), where the supremum is taken over all partitions π of A into a finite number of pairwise disjoint members of F and over all finite collections {ε n } satisfying ε n ;
.. ELEMENTARY PROPERTIES 5 (b) sup{ µ(b) : A B F } µ (A) 2sup{ µ(b) : A B F }. Consequently a vector measure is of bounded semivariation on Ω if and only if its range is bounded in X. It is easy to define the integral of a bounded measurable function with respect to a bounded vector measure. To this end, let F be a field of subsets of a set Ω and µ : F X be a bounded vector measure. If f is a scalar-valued simple function, say f = n α i χ Ai,whereα i are nonzero scalars and A,..., A n are pairwise disjoint members of F, define T µ (f) = n α i µ (A i ). It is easy to show that this formula defines a linear operator T µ from the space of simple functions of the above form into X. Moreover,iffis as above and β =sup{ f (ω) : ω Ω}, then n n T µ (f) = α i µ (A i ) = β α i β µ (A i) β µ (Ω). by Proposition..6 (a). Thus T µ is a continuous linear operator, which acts on the space of simple functions on F, equipped with the supremum norm. Another look at the Proposition..6 (a) and the above calculations shows that T µ = µ (Ω) (.) Then T µ has a unique continuous linear extension, still denoted by T µ, to B (µ), the space of all scalar-valued functions on Ω that are uniform limits of simple functions modeled on F. (Note that in the case F is a σ-field, B (µ) is precisely the familiar space of bounded F -measurable scalar-valued functions defined on Ω). This discussion allows us to make Definition..7 Let F be a field of subsets of a set Ω and µ : F X be a bounded vector measure. For each f B (µ), the integral fdµ is defined by fdµ = T µ (f), where T µ is as above.
6 CHAPTER. VECTOR MEASURES It is easy to see that this integral is linear in f (and also in µ) and satisfies fdµ f µ (Ω). Moreover, if x X,thenx fdµ = fdx µ; indeed, for simple functions this equality is trivial and density of simple functions in B (µ) proves the identity for all f B (µ). Definition..8 A family {µ τ : F X τ T } of countably additive vector measures is said to be uniformly countably additive whenever for any sequence (A n ) of pairwise disjoint members of F, the series µ τ (A n ) converges in norm uniformly in τ T. n= There are many alternative and useful formulations of countable additivity. We present two of them. Proposition..9 Any one of the following statements about a vector measure µ defined on a field F implies all the others. (i) µ is countably additive. (ii)theset{x µ:x X, x }is uniformly countably additive. (iii) Theset{ x µ : x X, x }is uniformly countably additive..2 Bartle-Dunford-Schwartz theorem The results of this section may be found in [7]. Definition.2. Let F be a field of subsets of a set Ω, µ : F X be a vector measure, and λ be a finite nonnegative real-valued measure on F. If µ (A) =0, then µ is called λ-continuous and this is lim λ(a) 0 denoted by µ λ. Sometimes we will say µ is absolutely continuous with respect to λ.
.2. BARTLE-DUNFORD-SCHWARTZ THEOREM 7 It should be noted that writing µ λ is not the same as saying µ vanishes on λ-null sets unless both λ and µ are countably additive and defined on a σ-field. Theorem.2.2 (Pettis) Let Σ be a σ-field of subsets of a set Ω, µ :Σ Xbe a countably additive vector measure, and λ be a finite nonnegative real-valued measure on Σ. Then µ is λ-continuous if and only if µ vanishes on sets of λ-measure zero. Proof. To prove the sufficiency, suppose µ vanishes on sets of λ- measure zero, but lim µ (A) > 0. Then there exist an ε>0anda λ(a) 0 sequence (A n ) in Σ, such that µ (A n ) εand λ (A n ) 2 n for all n. Foreachnselect an x n X such that x n and x nµ(a n ) ε 2. As µ is countably additive on Σ, then in accordance with Proposition..9 the set { x µ : x X, x }is uniformly countably additive. Now set B n = ( ) A i.evidentlyλ B n = 0. Consequently µ i=n n= vanishes on every set D Σ that is contained in B n = B. It follows n= that x µ (B) =0. Let C = Ω \ B and C n+ = B n \ B n+ for all n. Then (C n ) constitutes a sequence of pairwise disjoint members of Σ for which B n \ B = C i. i=n Also, since x n µ (B) = 0 for all n, one has ( ) lim x m n µ (B m) = lim x m n µ C i = lim x m n µ (C i)=0 i=m i=m uniformly in n, by the uniform countable additivity of the family { x n µ }. But now x n µ (B n ) x n µ (A n ) x n µ (A n ) ε 2.
8 CHAPTER. VECTOR MEASURES This contradicts the last calculation and proves the sufficiency; the converse is transparent. The following theorem is central to the theory of vector measures. Theorem.2.3 (Bartle-Dunford-Schwartz) Let µ be a bounded countably additive vector measure defined on a σ-field Σ. Then there exists a nonnegative real-valued countably additive measure λ on Σ such that µ is absolutely continuous with respect to λ. Ifµis nonatomic, then λ can be chosen nonatomic. (λ can be chosen such that 0 λ(a) µ (A) for all A Σ.) Corollary.2.4 Let Σ be a σ-field, µ :Σ Xbe a bounded countably additive vector measure, λ be a measure from Theorem.2.3. Then there exist an operator T : L (λ) X, which is continuous for the weak topology on L (λ) and weak topology on X such that Tχ A = µ(a) for any A Σ. Proof. Define T : L (λ) X by Tf = fdµ. Then for each Ω x X, one has x Tf = fdx µ = f dx µ dλ dλ, where dx µ/dλ = g x L (λ) is the Radon-Nikodym derivative of x µ with respect to λ. If (f α )isanetinl (λ)convergingweak to f 0, then for each x X, lim α x Tf α = lim α f α g x dλ = f 0 g x dλ = x Tf 0, i.e., (Tf α ) converges weakly to Tf 0. Hence T is a weak to weak continuous linear operator. Corollary.2.5 (Bartle-Dunford-Schwartz) Let µ be a bounded countably additive vector measure on a σ-field Σ. Then µ has a relatively weakly compact range. Proof. Let T be the operator from Corollary.2.4. It follows that T maps the weak compact set {f L (λ) : f }ontoaweakly compact set R X. But now {µ (A) :A Σ}={T(x A ):A Σ} {T(f): f } R, and the proof is complete.
.3. LYAPUNOV S CONVEXITY THEOREM 9 Corollary.2.6 Let µ and λ be as in Corollary.2.4. Then co (µ (Σ)) = fdµ :0 f,f L (λ). Ω Proof. Let U = {f L (λ) :0 f } and V = Ω fdµ : f U. Note that U is weak compact and { A : A Σ} =exu(where ex U is the set of the extreme points of U). By the Krein-Milman theorem U = co w (ex U). Then a glance at Corollary.2.4 proves that V = co (µ (Σ)). The following theorem shows that the measure λ from the Bartle- Dunford-Schwartz theorem may be taken of the form x µ for a certain x X. Theorem.2.7 (Rybakov) Letµ:Σ Xbe a countably additive vector measure. Then there is x X such that µ x µ..3 Lyapunov s convexity theorem The results of this section may be found in [7]. One of the most important results in the theory of vector measures is the Lyapunov convexity theorem which states that the range of a nonatomic vector measures with values in a finite dimensional space is compact and convex. As it was mentioned above, this theorem fails in every infinite dimensional Banach space. Example.3. (Uhl) A nonatomic vector measure of bounded variation whose range is closed but nonconvex and noncompact. Let Σ be the Borel sets in [0, ] and λ be Lebesgue measure. Define µ :Σ L (λ)byµ(a)=χ A.IfΠ Σ is a partition of [0, ], it is evident that A Π µ (A) = A Π λ (A) =. Since every L -convergent sequence contains an almost everywhere convergent subsequence, µ (Σ)
0 CHAPTER. VECTOR MEASURES is closed in L (λ). To see that µ (Σ) is not a convex set, note that χ 2 [0,] co µ (Σ) while if A Σ µ (A) 2 χ [0,] =[µ([0, ] \E)+µ(E)] /2 = 2. To see that µ (Σ) is not compact let A n = {t [0, ] : sin (2 n πt) > 0} for each positive integer n. A brief computation shows that µ (A n ) µ (A m ) = 2 for m n. Hence µ (Σ) is not compact. Example.3. suggests that nonatomicity may not be a very strong property of vector measures from the point of view of the Lyapunov theorem in the infinite dimensional context. Let us attempt to understand what nonatomicity means. Let µ :Σ R n have the form µ (A) =(λ (A),...,λ n (A)),A Σ, where each λ i is a countably additive finite signed scalar measure on Σ. Set λ (A) = n λ k (A). Then µ (A) 0 if and only if λ (A) 0. k= Now if µ is nonatomic, then λ is nonatomic. Consequently, if A Σandλ(A)>0 the mapping on the infinite dimensional subspace {fχ A : f L (λ)} that takes f into fdµis never one-to-one. We will A see that in the infinite dimensional case, this latter condition is precisely what is needed to make the Lyapunov theorem work. Throughout, Σ is a σ-field of subsets of Ω and X is a Banach space. Theorem.3.2 (Knowles; Lyapunov convexity theorem in the weak topology) Letµ:Σ Xbe a countably additive vector measure and λ be a finite nonnegative countably additive measure on Σ from Theorem.2.3. The following statements are equivalent. (i) If A Σ and λ (A) > 0, then the operator f fdµ on L (λ) is A not one-to-one on the subspace of functions in L (λ) vanishing off A.
.3. LYAPUNOV S CONVEXITY THEOREM (ii) For each A Σ, {µ (B A) :B Σ}is a weakly compact convex set in X. (iii) If 0 f L (λ), there exists a function g L (λ) such that fg > 0 but fgdµ =0. Ω Proof. Note that (iii) implies (i). To show that (i) implies (iii) let f L (λ), f > 0. Then there is an ε>0anda Σ such that fχ A >εand µ (A) > 0. According to (i), there is an h L (λ) such that hχ A > 0and Ahdµ =0. Setg=h/f on A and g =0offA. Then fg = h on A and so fgχ A > 0. Also fgdµ = hdµ =0. Ω A Hence (iii) holds. To check that (ii) implies (i), suppose (i) is false. Without loss of generality, we can assume that A = Ω, so that the operator f fdµ (f L (λ)) is one-to-one. It is not difficult to see that this A { } means that µ (Σ) = χ A dµ : A Σ is a proper subset of the set { Ω } V = fdµ : f L (λ),0 f. But by Corollary.2.6, V = Ω co (µ (Σ)); thus µ (Σ) cannot be both closed and convex. Hence (ii) is false. To complete the proof, we shall verify that (iii) implies (ii) again. It is enough to show that µ (Σ) is convex and weakly compact. Let f L (λ) be such that 0 f. Since by Corollary.2.4 the operator ( ) dµ is continuous for the weak topology on L (λ) and Ω the weak topology on X, theset H= g L (λ):0 g, fdµ = gdµ is a weak compact convex set in L (λ) and therefore has extreme points. If we can show that the extreme points of H, denoted by ext (H), are all in {χ A : A Σ} it will follow that there exists A Σ such that µ (A) = fdµ. Then an appeal to Corollary.2.6 will yield Ω Ω Ω
2 CHAPTER. VECTOR MEASURES the equalities co (µ (Σ)) = Ω fdµ : f L (λ),0 f = µ(σ) and prove that µ (Σ) is weakly compact and convex. To this end, suppose f 0 ext (H) but f 0 χ A > 0 for each A Σ. A simple calculation shows that there exists f L (λ) with f > 0 such that 0 f 0 ± f. An appeal to (iii) gives us a g L (λ), that may be selected with g such that f g > 0 but f g dµ =0. Thenf 0 ±f g H;thusf 0 is not an Ω extreme point of H. This completes the proof. Note that in light of the remarks before Theorem.3.2 the classical Lyapunov theorem is contained in this theorem; in fact, the above proof of (iii) (ii) inthecasex=r n is Lindenstrauss s argument [23] to prove the Lyapunov theorem. Theorem.3.3 (Lyapunov) Let Σbe a σ-field of subsets of Ω, X be a finite dimensional Banach space and µ :Σ Xbe a countably additive vector measure. If µ is nonatomic, then the range of µ is a compact convex subset of X. Corollary.3.4 Let Σ be the σ-field of Borel subsets of [0, ]. IfXis an infinite dimensional Banach space, then there is a countably additive vector measure of bounded variation µ :Σ X and a set A Σ such that {µ (B A) :B Σ}is not a weakly compact convex set in X. Proof. Let λ be Lebesgue measure on Σ. Select a sequence (f n )in L (λ) such that f n = and the only g L (λ) with f n gdλ =0 for all n is g = 0. Choose a sequence of pairs (x n,x n) such that x n X, x n X, x m (x n ) = 0 if m n and x n (x n ) = for all n. Define T : L (λ) X by ( T (g) = x n 2 n x n ) f n gdλ. n= [0,] [0,]
.4. l p -VALUED MEASURES 3 If T (g) =0,thenx n T(g)=( 2 n x n ) [0,] f n gdλ = 0 for all n. Hence T is one-to-one on L (λ). To produce the advertised measure, define µ (A) =T(χ A ) for A Σ. It is not hard to see that µ is countably additive and T (.) = (.) dµ. Hence µ violates (i) of Theorem.3.2. [0,] By Theorem.3.2, µ is as advertised..4 l p -valued measures The main results of this section are contained in [7]. Another way of generalizing the Lyapunov convexity theorem is considering the closure of the range of the vector measure in the norm topology. Further by an X-valued vector measure we mean a countably additive vector measure µ defined on a σ-field Σ of subsets of a set Ω. Definition.4. A nonatomic X-valued vector measure µ is said to be Lyapunov if the closure of its range is convex. In 969 Uhl [7] proved the following theorem. Theorem.4.2 Let Σ be a σ-field of subsets of Ω and suppose X has the Radon-Nikodym property. If µ :Σ X is a nonatomic and countably additive measure of bounded variation, then µ is a Lyapunov measure. If we waive the requirement that the variation be bounded, then Uhl s theorem no longer holds, in particular there exists a nonatomic measure with values in a Hilbert space such that the closure of its range is not convex. (It is enough to consider the measure µ 2 from Example..2.) In 99 V. M. Kadets and M. M. Popov [6] obtained the following strengthening of this result. Theorem.4.3 The following conditions are equivalent for the space X:
4 CHAPTER. VECTOR MEASURES. Each nonatomic X-valued measure of bounded variation is a Lyapunov measure; 2. There does not exist a sign-embedding of the space L [0, ] in X. Definition.4.4 A Banach space X is said to have the Lyapunov property if every nonatomic countably additive vector measure with values in X is a Lyapunov measure. As already mentioned above, an infinite dimensional Hilbert space does not have the Lyapunov property. Consequently, all spaces containing isomorphic copies of the space l 2 do not have the Lyapunov property; in particular, L p [0, ], C [0, ], l do not have the Lyapunov property. Nevertheless, spaces having the Lyapunov property exist. In 992 V. M. Kadets and G. Schechtman proved that l p ( p<, p 2)andc 0 have the Lyapunov property [7]. Before proving this result we need some lemmas. Lemma.4.5 If the measure µ is not a Lyapunov measure, then there exist a number ε>0and a set V Σ such that µ (U) µ (V 2 ) ε for any subset U Σ V (here and below, Σ V denotes the family of elements of the σ-algebra Σ that are subsets of the set V : the restriction of Σ to V ). Proof. Assume the contrary, i.e., assume that for any ε>0andany V Σ there exists a U V such that µ (U) µ (V ) 2 <ε. Let x and y be arbitrary elements of µ (Σ), x = µ (A), y = µ (B). We prove that (x + y) /2 µ (Σ). This means that µ is a Lyapunov measure, contrary to assumption. We choose U ε A \ B and U 2 ε B \ A such that 2 µ (A \ B) µ (U ) ε <ε, 2 µ(b\a) µ(uε 2) <ε. Then (x + y) µ (W ε ) 2ε for the set W ε = U ε 2 U2 ε (A B), i.e., (x + y) /2 can be approximated by values of the measure to any degree of accuracy. The lemma is proved.
.4. l p -VALUED MEASURES 5 Lemma.4.6 Suppose that λ is a nonatomic positive numerical measure on Σ, and µ :Σ Xis a vector valued measure that is absolutely continuous with respect to λ but is not a Lyapunov measure. Then there exist a number a>0and a subset Ω Σ with λ (Ω ) > 0 having the following property : for any subsets A,A Σ Ω,A A, µ (A ) 2 µ (A) aλ (A). (.2) Proof. Suppose that ε>0andv Σ are as in the statement of Lemma.4.5. Take a = ε/λ (V ) as the required number a. A set A Σ V is said to be a narrow if there exists a subset A A such that the inequality converse to (.2) holds. By construction, V is not a narrow set. It suffices for us to prove the existence of a set Ω Σ V with λ (Ω ) > 0 that does not have narrow subsets. Assume the contrary, i.e., assume that the quantity narr (A) =sup{λ(a ):A Σ A, A is narrow} is positive for any A Σ V with λ (A) 0. We use induction to construct a chain V = V 0 V V 2... of sets such that at each step the set W n = V n \ V n+ is narrow and large : λ (W n ) 2 narr (V n). Then W = n=0 V n is a narrow set. Consequently, since V is not narrow, λ (Ω ) 0 holds for the set Ω = V \ W = n=0 V n. At the same time, the sets W n are disjoint and n= λ (W n )=λ(w)< ; therefore, λ (W n ) 0. We get that narr (Ω ) narr (V n ) 2λ (W n ) 0; i.e., narr (Ω ) = 0, although λ (Ω ) > 0. This contradiction proves the lemma. Lemma.4.7 The following two conditions are equivalent for a Banach space X: (i) X does not have the Lyapunov property. (ii) There exist a triple (Ω, Σ,λ) with a nonatomic numerical measure λ on Σ and an operator T : L (Ω, Σ,λ) X such that (a) T is continuous for the weak topology on L (Ω, Σ,λ) and the weak topology on X, and
6 CHAPTER. VECTOR MEASURES (b) there exists an ε>0such that Tf ελ (supp f) for any sign f L (Ω, Σ,λ), i.e., any function taking only the values 0 and ±. Proof. We first show that (i) implies (ii). Let (Ω, Σ,λ) be a triple (Ω, Σ Ω,λ) as in Lemma.4.6, we get the existence of a measure µ :Σ Xand a number a>0 such that the inequality (.2) holds for any A A Σ. According to Corollary.2.4, there exists a weak - weak continuous operator T : L (Ω, Σ,λ) X connected with the measure µ by the relation T (χ A )=µ(a), A Σ. (.3) It remains to verify the condition (b). Suppose that f is a sign ; then f = χ A χ A2,whereA,A 2 ΣandA A 2 =Ø. LetA=A A 2 and ε =2a.Weget Tf = µ(a ) µ(a 2 ) =2µ(A ) µ(a) 2 2aλ (A) =ε λ(supp f). (ii) implies (i). Suppose that the operator T satisfies the conditions (a) and(b). The measure µ is given by (.3). It is easy to verify that µ is nonatomic and not a Lyapunov measure. The lemma is proved. Remark.4.8 Since µ : Σ X, µ (A) = Tχ A is absolutely continuous with respect to λ, the operator T will also have the following property: (c) for every ε>0there is δ>0such that Tf ε f for any f L (Ω, Σ,λ) with λ (suppf) δ. Let us introduce some additional symbols for the sequel. Let r i = r i (ω) be a sequence of independent random variables taking values + and with probability.fixanumbern Nand denote the 2 random variable T (ω) =inf j: j r i (ω) N.
.4. l p -VALUED MEASURES 7 Further define the stopped martingale k s i by the rule s i (ω) = { ri (ω) if i T (ω) 0 if i>t(ω) The constructed martingale is evidently symmetric and its absolute value is bounded by N +. Lemma.4.9 Let k beafunctionofn.ifn=o(k), then ( k lim P s N i < ) N =0. If k = o (N), then Proof. We have, ( k P s i < ) N lim P N ( k s i < ) N =. ( j = P max r j k i < ) N ( k P r i < ) ( N = P k k r i < N k By the central limit theorem, the last expression tends to zero if N = o (k). On the other hand, if k = o (N), then ( k P s i < ) k N = P( s i ) N e N/2k. N The lemma is proved. Lemma.4.0 Suppose that x, x n l p, p< and x n Then for any ε>0there exists an index n N such that x p + x n p ε< x+x n p < x p + x n p +ε. ). w 0. But if x, x n c 0, then under the same assumptions it is possible to ensure that max { x, x n } ε< x+x n <max { x, x n } + ε.
8 CHAPTER. VECTOR MEASURES The method in the following theorem due to V. M. Kadets and G. Schechtman will often be used in the main results of the thesis. Theorem.4. The spaces l p with p 2, p<, and the space c 0 have the Lyapunov property. Proof. The case p<2. Fix N N for the present, and let k =[N/ ln N]. We assume that X = l p fails the Lyapunov property, and we use Lemma.4.7. Let Ω, Σ,λ,ε, and T : L X be as in that lemma, with λ (Ω) =. We prove by induction on j that there exist functions {t j } L (Ω, Σ,λ) such that for each j the functions {t i } j are jointly equidistributed with the {s i} j in Lemma.4.9 and p j T j t i > T (t i ) p. (.4) Indeed, everything is obvious for j =. Suppose that the assertion is already proved for j = m, and that {t i } m have been constructed { that satisfy (.4) with j = m. We consider the set A = ω Ω: m t i < } N and a sequence {z n } n= of functions independent on A, equal to zero off A, and taking the values ± with probabil- ity λ (A) /2. For any n the system {{t i } m,z n} is equidistributed with {s i } m+. The sequence z n tends to zero in the weak topology; hence w Tz n 0, and by Lemma.4.0 it is possible to choose n such that ( m ) T p m p t i + T (z n ) > T (t i ) + T (z n ) p δ, where δ is arbitrarily small. Choosing δ sufficiently small and letting t m+ = z n, we get the required inequality (.4) with j = m +. We now use the fact just proved. Suppose that {t i } k is equidistributed with {s i } k and subject to the condition (.4) with j = k. Then T p k p ( k ) t i T p k t i T (t i ) p. In view of the condition (ii) of Lemma.4.7, which T satisfies by construction, we have T (t i ) ε λ(supp t i ) ε λ (supp t k ) for i k.
.4. l p -VALUED MEASURES 9 Since {t i } k and {s i} k are equidistributed, this gives us by the choice of k that ( ) p T N + p + k T (t i ) p kε p (P (s k 0)) p (N/ ln N ) ε p ( P ( k )) p s i < N. By Lemma.4.9, the last factor tends to ; hence this inequality cannot hold for large N. This contradiction completes the analysis of the first case. Thecase2<p<. Assume that X = l p fails the Lyapunov property. By analogy with the first case, we fix an N N, let k = [N ln N], and construct functions {t i } k equidistributed with {s i} k and satisfying the requirement ( k ) T p k t i < T (t i ) p + T p k+. (.5) We define the auxiliary function f N by k if t i N f N = k if t i. N 0 for the rest k and denote supp f N by A. Since t i < N +, f N k t i on A. N N Consequently, ( χ A f N ) k t i 0. (.6) N N Further, according to Lemma.4.9, λ (Ω\A) 0, and by Remark.4.8 ( T χ Ω/A N k t i ) N 0. (.7)
20 CHAPTER. VECTOR MEASURES Comparing (.6) and (.7), we get that ( ) k T t i T(f N ) +o(). (.8) N By (.5) and the property (.8) of T, [ε λ(a)] p T p k ( N ) p +o() = o (). This contradicts Lemma.4.9, which asserts that lim λ (A) =. N The case X = c 0 is analysed just like the preceding case. The theorem is proved. There are simpler proofs for c 0 and l, due to V. Kadets, which were never published before. We use the chance to present (with the kind permission of V. Kadets) these proofs too. Proposition.4.2 c 0 has the Lyapunov property. Proof. Let µ : Σ c 0 be a nonatomic countably additive vector measure. According to Corollary.2.4 there is an operator T : L (Ω, Σ,λ) c 0 such that µ (A) =Tχ A for all A Σandwhichis continuous for the weak topology on L (Ω, Σ,λ) and the weak topology on c 0,whereλis a nonnegative nonatomic real-valued countably additive measure on Σ such that µ is absolutely continuous with respect to λ. By Lemma.4.5, it is sufficient to show that for every A and every ε>0 there exists a set B Σ A such that µ (B) 2 µ (A) ε. (.9) Let {e i } be the unit vector basis of c 0 and {f i } be the coordinate functionals on c 0. Then Tx = f i (Tx)e i. Denote T i x = f i (Tx). Take an arbitrary A Σandanε>0. Fix N N for the present. As µ is a nonatomic measure we can decompose A into the union of mutually disjoint sets A k Σ A (k =,...,N)withµ(A k ) 0 for all k, λ (A k )=λ(ω) /N. Let rbe a Rademacher function on A. Put n =andm 0 = 0. Select m N such that Tr m T i r
.4. l p -VALUED MEASURES 2. Consider a sequence {r 2 2 N+ n } n= of Rademacher functions on A 2.By Corollary.2.5, Trn 2 n 0 in the weak topology. Hence there are a m number n 2 such that T i rn 2 2 and a number m 2 N+3 2 such that T i rn 2 2. Consequently 2 N+3 i=m 2 + m 2 Tr2 n 2 T i r 2 n 2 i=m + 2. N+2 Continuing this process we obtain the sequences of numbers {n k } N k= and {m k } N k=0 and the set of functions { } N rn k such that k k= rk n k is a Rademacher function on A k and m k Trk n k T i r k n k i=m k + 2 N+k for k =,...,N. Consequently ( N ) T rn k N m k k T i rn k k + 2 = max N k= k= i=m k + m k k N i=m k + T i rn k k + 2. N If the maximum is reached at k 0 then applying the equation (.) we have ( N ) T m rn k k0 k T i r k 0 n k0 + 2 T N Ak0 + 2 µ (A N k 0 )+ 2. N k= i=m k0 + As µ is absolutely continuous with respect to λ and λ (A k )=λ(ω) /N for k =,..., N and by Proposition..6, the last sum tends to zero as N. Thus choosing N sufficiently large and taking B = { Nk= ω Ω:r k nk = } we obtain the required inequality (.9). The proposition is proved. Before we prove that l has the Lyapunov property, let us recall that Schur has proved that in l the norm-convergence and the weakconvergence coincide. We will show that all Banach spaces, which have this property, have the Lyapunov property.
22 CHAPTER. VECTOR MEASURES Definition.4.3 A Banach space X is said to have the Schur property if a sequence (x n ) of elements of X converges in the weak topology if and only if it converges in the norm topology. Proposition.4.4 If a Banach space X has the Schur property, then it has the Lyapunov property. Proof. Let µ be a nonatomic countably additive vector measure on a σ-field Σ and T : L (λ) X be the operator from Corollary.2.4. Take an arbitrary A Σ and consider on A the sequence of functions f n = +rn,where(r 2 n ) are Rademacher functions on A. Asf n χ n 2 A in the weak topology, then by Corollary.2.5 we have Tf n µ (A) 2 in the weak topology. It follows that Tf n µ(a) 2 f n = χ An,whereA n Σ. Thus we have µ (A n) 2 µ (A) 0. n By Lemma.4.5, the proposition is proved. n n 0. But.5 The three-space problem The results of this section have been published in [8], [5]. The three-space problem arises for every Banach space property. Namely, do a subspace Y and the quotient space X/Y have the property if X does and does X have the property if Y and X/Y do? Some of these problems can be solved easily in the case of the Lyapunov property. Indeed, if X has the Lyapunov property and Y X,then Y has the Lyapunov property, but X/Y needs not have the Lyapunov property (for instance, X = l because the set of its quotient spaces contains all separable Banach spaces [24, p.08]). At the same time the answer to the last problem runs into difficulties. The purpose of this section is the positive solution of the above mentioned problem : let Y be a subspace of a Banach space X. If Y and X/Y have the Lyapunov property, then X has the Lyapunov property. The following three lemmas are of technical nature.
.5. THE THREE-SPACE PROBLEM 23 Lemma.5. Let X have the Lyapunov property and µ :Σ Xbe a nonatomic measure. Then there is a nonatomic nonnegative measure λ :Σ Rsuch that for every A Σ and n N there exists B n Σ A satisfying the following inequalities: µ (B n) 2 µ (A) 2 and λ (B n n ) 2 λ (A) 2. (.0) n Proof. By Theorem.2.7 there is a functional x X with x = for which µ x µ. Put λ = x µ. Let A Σ. In accordance with the Hahn decomposition theorem let us denote by Ω + and Ω the positivity and negativity sets for x µ respectively. Then λ (C) = x µ(c Ω + ) x µ(c Ω ) for any C Σ. Put A + = A Ω + and A = A Ω. Since X has the Lyapunov property, we can choose B n + Σ A + and B n Σ A such that µ ( B n + Then x µ ( B n + Define B n = B n + (.0) are true. ) 2 µ ( A +) 2 and µ ( ) B n+ n 2 µ ( A ) 2. n+ ) 2 x µ ( A +) 2 and x µ ( ) B n+ n 2 x µ ( A ) 2. n+ B n. It is easy to check that for B n the inequalities Lemma.5.2 Let X have the Lyapunov property, µ :Σ X be a nonatomic measure, λ be the measure from Lemma.5.. Then for every A Σ with λ (A) 0and ε>0there exist G Σ A,G = A\G such that (i) λ ( G ) = λ ( G ) = λ (A) /2, (ii) µ ( G ) 2 µ (A) <ε. Proof. Let us choose B n as in Lemma.5.. We can select C n Σ Bn if λ (B n ) λ (A) orc 2 n Σ A\Bn if λ (B n ) < λ (A) for which 2 λ (C n )= λ(a) λ(b 2 n) (because λ is a nonatomic real-valued measure). By (.0), λ (C n ) n 0. PutG n =B n C n,g n = A\G n.then
24 CHAPTER. VECTOR MEASURES we obtain λ ( ) ( ) G n = λ G n = λ (A), and λ ( G 2 n \B ) n 0, and n λ ( B n \G n) 0. Since µ is absolutely continuous with respect to n λ the last condition implies that µ ( B n \G n) 0andµ ( G n n \B ) n n 0. Together with inequality (.0) this gives us µ ( ) G n 2 µ (A) 0. n So for sufficiently large n the sets G = G n and G = G n the conditions (i) and(ii). will satisfy Lemma.5.3 Under the conditions of Lemma.5.2 for every A Σ with λ (A) 0and ε>0there exists a σ-algebra Σ Σ A such that for every B Σ we have µ (A) µ (B) λ (B) ελ (B) (.) λ (A) and the measure λ is nonatomic on Σ. Proof. Take A Σandε > 0. Employing Lemma.5.2, we choose sets A Σ A, A 2 = A\A with λ (A ) = λ(a 2 ) = λ(a) and 2 µ(a ) µ(a) ε (note that 2 4 µ (A ) 2 µ (A) = 2 µ (A ) µ (A 2 ) = µ (A 2) 2 µ (A) 4 ε). Employing Lemma.5.2 twice (for A = A and A = A 2 )weobtain A, Σ A,A,2 =A \A, ;A 2, Σ A2,A 2,2 =A 2 \A 2, with λ (A, )= λ(a,2 )=λ(a 2, )=λ(a 2,2 )= 4 λ(a)and µ(a,) 2 µ(a ) ɛ 6, µ (A 2, ) 2 µ (A 2) ɛ 6. When we continue this process, we obtain a tree of sets A i,i 2,...,i n,i k {,2},n Nwith A i,i 2,...,i n+ A i,i 2,...,i n,a i,i 2,...,i n,2 = A i,i 2,...,i n \A i,i 2,...,i n,, µ (A i,i 2,...,i n ) µ ( ) A 2 i,i 2,...,i n ε, λ (A 4 n i,i 2,...,i n )= λ(a) 2 n
.5. THE THREE-SPACE PROBLEM 25 Let Σ be a σ-algebra generated by the sets A i,i 2,...,i n. We are going to show that the algebra Σ has the required property. Let B = A i,i 2,...,i n.then µ(b) λ(b) λ(a) µ(a) µ(ai =,i 2,...,i n ) µ (A) 2 n µ (A i,i 2,...,i n ) µ ( ) A 2 i,i 2,...,i n + µ ( ) A 2 i,i 2,...,i n µ ( ) A 2 i,i 2,...,i n 2 +... + µ 2 n (Ai ) µ (A) 2 ε + ε +... + ε 4 n 2 4 n 2 n 4 = ε ( ) + +... + 2 2n 2 2(n ) 2 ε n+ 2 n = ελ (A i,i 2,...,i n ). Hence by the triangle inequality and the σ-additivity of µ and λ we get (.) for B = B n, where the B n are disjoint sets of the form n= A i,i 2,...,i k. Then we obtain (.) for every B Σ using approximation of λ (B) by bigger sets of the form B = B n. Lemma.5.4 The statements of Lemmas.5.2 and.5.3 are valid for an arbitrary nonatomic measure λ. Proof. Let µ :Σ X, λ :Σ R + be nonatomic measures. Let ν be the measure which played the role of λ in Lemma.5.. Consider two cases. Case : λ ν.take A Σ, n N. In view of Lemma.5.3, there is Σ Σ A such that for any B Σ n= ν (B) µ (B) ν (A) µ (A) ν (B), (.2) 2n ν is a nonatomic measure with respect to Σ. Applying the Lyapunov theorem to the measure σ : Σ R 2 : σ(a) = (ν(a),λ(a)), we obtain sets G,G Σ such that λ ( G ) = λ ( G ) = λ (A) and 2 ν ( G ) =ν ( G ) = ν (A). Then (.2) implies inequality (ii) which 2 we need.
26 CHAPTER. VECTOR MEASURES Case 2: λ ν. We decompose λ into the sum of absolutely continuous and strictly singular measures with respect to ν : λ = λ + λ 2. Then λ 2 is concentrated on a ν-negligible set S. Now we consider A\S and S A separately. By the case chose G ( ) A\S so that λ G = λ 2 (A), µ ( ) G µ 2 (A) and G 2 n 2 S A with ( ) λ 2 G 2 = λ (S ( ) ( A). Because λ 2 G 2 = 0, µ G 2) = 0 and ( ) λ 2 G = 0 it is clear that G = G G 2 satisfies (ii). The proof of the Lemma.5.3 shows that if Lemma.5.2 is valid for an arbitrary λ then Lemma.5.3 is valid too for an arbitrary nonatomic measure λ. The following statement is evident. Lemma.5.5 If X is a Banach space, Y isasubspaceofx, µ :Σ X is a nonatomic measure, then µ :Σ X/Y ( µ (A)-the equivalence class of µ (A)) is a nonatomic measure too. Theorem.5.6 Let Y be a subspace of a Banach space X. If Y and X/Y have the Lyapunov property, then X has the Lyapunov property too. Proof. Let µ :Σ X, λ :Σ R + be nonatomic measures, λ (Ω) =. Fix A Σ, λ(a) 0,andε>0. By Lemmas.5.4 and.5.5 there is a σ-algebra Σ Σ A such that µ (B) λ (B) µ (A) λ (A) < ελ (B) (.3) 2 for all B Σ and λ is nonatomic on Σ. Define σ :Σ Xby the rule σ (B) =µ(b) λ(b) µ(a).now we show that there exist a σ-algebra λ(a) Σ Σ and a nonatomic measure β : Σ Y such that σ (B) β (B) < 2ε for all B Σ. Indeed, by inequality (.3) and nonatomicity of λ we can choose sets A, A 2 Σ such that A = A A2,λ(A )=λ(a 2 )= 2 λ(a),and σ (A ) <ε 2 λ(a). Clearly, there is x σ (A ) such that x ε 2 λ(a).denote α (A) =x, α (A2) = x,
.5. THE THREE-SPACE PROBLEM 27 Applying step by step Lemma.5.2 and Lemma.5.4, we get sets A i,...,i k Σ, k = 2,3,...; i,..., i k =,2, such that A i,...,i k = ) A i,...,i k, Ai,...,i k,2, λ(a i,...,i k ) = 2 λ( A i,...,i k = λ (A), and 2 k σ ( ) A i,...,i k, σ ( ) A 2 i,...,i k <ε λ(a).it is readily seen that 2 2k in every equivalence class σ ( ) A i,...,i k, σ ( ) A 2 i,...,i k X there exists an element x i,...,i k such that xi,...,i k ε λ (A). Put 2 2k α ( ) A i,...,i k, = α ( ) A 2 i,...,i k + xi,...,i k, α ( ) A i,...,i k,2 = α ( ) A 2 i,...,i k xi,...,i k. Evidently, α ( ( ( A i,...,i k,) σ Ai,...,i k,) and α Ai,...,i k,2) σ ( A i,...,i k,2). By Σ denoteσ-algebra generated by the sets A i,...,i k. Iterating the inequality α (A i,...,i k ) α ( ) A 2 i,...,i k + ε λ (A), 2 2k we obtain α (A i,...,i k ) ε λ(a)=2ελ (A 2 k i,...,i k ). Let us extend α to Σ andshowthat α(b) 2ελ (B) (.4) for any B Σ. For this purpose we take B = A i,...,i k,wherei I is a finite set of indices and A i,...,i k are mutually disjoint sets. Put α (B) = α(a i,...,i k ). Clearly, I α (B) I α(a i,...,i k ) 2ε I λ(a i,...,i k )=2ελ (B). This proves that inequality (.4) is valid for elements of the algebra S, generated by the sets A i,...,i k. Now applying the Kluvanek-Uhl extension theorem [6, Chapter 6, Section 8, Addition 4] we obtain an extension of α to Σ. Thus, we have constructed a measure α : Σ X such that α (B) σ (B) for any B Σ and α(b) 2ελ (B).αis a nonatomic measure because λ is nonatomic by construction. It is clear that β = α σ has the required property. The statement is proved. Let us complete the proof of the theorem. Since β : Σ Y is a nonatomic measure and Y has the Lyapunov property, we see that by Lemma.5.2 there is B Σ such that λ (B) = λ(a)and 2 β(b) 2 β(a) ε.
28 CHAPTER. VECTOR MEASURES Thus we have σ (B) σ (A) 2 σ(b) β(b) + 2 σ(a) β(a) 2 + β(b) 2 β(a) 3ε. Since σ (B) σ (A) 2 =µ(b) µ(a),the proof is completed. 2 Let us stress an important particular case of this theorem. Corollary.5.7 Let X = X X 2 where X,X 2 have the Lyapunov property. Then X has the Lyapunov property too. It is clear that by induction we can extend this property to any finite sum of Banach spaces. Corollary.5.8 The spaces X p from the work [9] ( twisted sums of l p spaces) have the Lyapunov property for p 2.