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1 This article was downloaded by: [Polska Akademia Nauk Instytut Matematyczny] On: 0 April 204, At: 05:08 Publisher: Taylor & Francis Informa Ltd Registered in ngland and Wales Registered Number: Registered office: Mortimer House, 37-4 Mortimer Street, London WT 3JH, UK Quaestiones Mathematicae Publication details, including instructions for authors and subscription information: TH COMPLTNSS PROBLM IN SPACS OF PTTIS INTGRABL FUNCTIONS Kazimierz Musial a a Instytut Matematyczny, Uniwersytet Wroclawski, Pl.Grunwaldzki 2/4, , Wroclaw, Poland -mail: Published online: 8 Oct 200. To cite this article: Kazimierz Musial (200) TH COMPLTNSS PROBLM IN SPACS OF PTTIS INTGRABL FUNCTIONS, Quaestiones Mathematicae, 24:4, , DOI: 0.080/ To link to this article: PLAS SCROLL DOWN FOR ARTICL Taylor & Francis makes every effort to ensure the accuracy of all the information (the Content ) contained in the publications on our platform. However, Taylor & Francis, our agents, and our licensors make no representations or warranties whatsoever as to the accuracy, completeness, or suitability for any purpose of the Content. Any opinions and views expressed in this publication are the opinions and views of the authors, and are not the views of or endorsed by Taylor & Francis. The accuracy of the Content should not be relied upon and should be independently verified with primary sources of information. Taylor and Francis shall not be liable for any losses, actions, claims, proceedings, demands, costs, expenses, damages, and other liabilities whatsoever or howsoever caused arising directly or indirectly in connection with, in relation to or arising out of the use of the Content. This article may be used for research, teaching, and private study purposes. Any substantial or systematic reproduction, redistribution, reselling, loan, sub-licensing, systematic supply, or distribution in any form to anyone is expressly forbidden. Terms & Conditions of access and use can be found at

2 Quaestiones Mathematicae 24(200), c 200 NISC Pty Ltd, TH COMPLTNSS PROBLM IN SPACS OF PTTIS INTGRABL FUNCTIONS Kazimierz Musia l Instytut Matematyczny, Uniwersytet Wroc lawski, Pl.Grunwaldzki 2/4, Wroc law, Poland. -Mail musial@math.uni.wroc.pl Abstract. Two subspaces of the space of Banach space valued Pettis integrable functions are considered: the space P (µ, X, var) of Pettis integrable functions with integrals of finite variation in a Banach space X and LLN(µ, X, var), the space of functions satisfying the law of large numbers. It is proved that LLN(µ, X, var) is always complete and P (µ, X, var) is complete if Martin s axiom and the perfectness of µ are assumed. Moreover, a non-trivial example of a non-conjugate Banach space X with non-complete P (µ, X, var) is presented. Mathematics Subject Classification (2000): Primary: 46G0; Secondary: 28B05, 28A5. Key words: Pettis integral, lifting, completeness, vector measures.. Preliminaries. It has been proven already by Pettis [2] that the space P (µ, X) of Pettis integrable functions may be non-complete when endowed with the semivariation norm of the integrals. Then Thomas [6] proved that the space is almost always non-complete. In view of the Open Mapping Theorem in such a case no complete equivalent norm can be defined on P (µ, X). The question is now whether there are interesting linear subspaces of P (µ, X) where a complete norm does exist. In this paper we consider two such subspaces, both endowed with the norm given by the variation of the indefinite Pettis integral: the space P (µ, X, var) of Pettis integrable functions having integrals of finite variation and the space LLN(µ, X, var) of functions satisfying the strong law of large numbers. Throughout the paper (Ω, Σ, µ) stands for a complete probability space, ρ is a lifting on L (µ) and X is a Banach space. λ is the Lebesgue measure on the unit interval [0, ] and L denotes the corresponding σ-algebra of Lebesgue measurable sets. We say that Axiom L (cf. [6]) is satisfied if [0, ] cannot be covered by less than the continuum closed sets of the Lebesgue measure zero. It is known (cf. [3]) that Axiom L is a consequence of Martin s Axiom. A function f : Ω X is said to be Pettis integrable with respect to µ if it is weakly measurable (i.e. x f is measurable for every x X ) and for each Σ there exists ν f () X satisfying for each functional x X the equality The research of the author was supported by KBN Grant No. 2 P03A

3 442 K. Musia l x ν f () = x f dµ. We will write ν f () = P X fdµ or ν f () = P fdµ if no confusion is possible. Identifying weakly equivalent Pettis integrable functions we get a linear space which we denote by P (µ, X). It is well known that the space can be normed by setting f P : = sup x f dµ. x Ω Rybakov proved (cf. [9]) that the measure ν f : Σ X is always of σ-finite variation. We denote by P (µ, X, var) the set of those f P (µ, X) for which ν f (= the variation of ν f ) is finite. P (µ, X, var) can be considered as a subspace of cabv(µ, X) - the space of µ-continuous X-valued measures of finite variation, endowed with the variation norm. We set f V : = ν f = ν f (Ω). Clearly f P f V for each f P (µ, X). ba[0, ] is the Banach space of all bounded additive real-valued set functions defined on the σ-algebra of all subsets of [0, ] and equipped with the variation norm. L (µ) is the space of all bounded µ measurable real-valued functions. If f : Ω X is a weak -measurable (i.e. xf is measurable for every x X) and weak -bounded function (i.e. there exists M > 0 such that for each x X the inequality xf M x holds µ-a.e.) and ρ : L (µ) L (µ) is a lifting, then ρ 0 (f) : Ω X is the unique function satisfying for each x X the equality x, ρ 0 (f) = ρ( x, f ) (see [8]). It is a consequence of Theorem III.3.3 from [8] that ρ 0 (f) : = sup{ ρ( x, f ) : x } is a measurable function. It is known (cf. [9]) that given a weak -measurable function f : Ω X there exists a non-negative measurable function ψ f on Ω satisfying the following properties: (i) For every x X x, f ψ f x µ-a.e. (ii) ψ f f µ-a.e. (iii) If ϕ : Ω [0, ) is measurable and satisfies (i) and (ii) (with ϕ rather than ψ f ), then ψ f ϕ µ-a.e. It has been also proved in [9] that if f is weak -integrable and ν is its weak -integral, then () ν () = ψ f dµ for every Σ. Moreover, if f is weak -bounded, then ψ f = ρ 0 (f) µ-a.e.

4 Completeness in spaces of Pettis integrable functions 443 Following [7] and [5] we are going to introduce now the space LLN(µ, X) of X valued functions satisfying the law of large numbers. It is defined in the following way: LLN(µ, X) = { f : Ω X : a f X lim n a f n } n f(ω i ) = 0 for µ -a.e. (ω i ) Ω where µ is the countable direct product of µ on Ω the countable product of Ω. The space LLN(µ, X) will be considered with the Glivenko-Cantelli seminorm, defined in [5] for an arbitrary function f : Ω X by setting i= where Moreover, let One can easily see that f GC = lim sup n h n (ω) = sup x n h n dµ, x (f(ω i )). i n f := f dµ. (2) f GC f. According to [5] each function satisfying the law of large numbers is Pettis integrable. Moreover, the GC-seminorm and the Pettis seminorm are equivalent on LLN(µ, X). In particular functions in LLN(µ, X) that are weakly equivalent are not distinguishable by the GC-norm. This permits us to identify weakly equivalent elements of LLN(µ, X) and investigate the quotient space (denoted also by LLN(µ, X)). 2. Completeness of P (µ, X, var). In many cases P (µ, X, var) = L (µ, X) the space of Bochner integrable functions with respect to µ. In particular, it is so in case of separable X, weakly compactly generated X and for X possessing RNP. We are going to show now that there are several situations when the above equality does not hold but in spite of this P (µ, X, var) is complete. It has been proven in [4] that P (µ, X) is always barrelled. xactly in the same way we get Theorem 2.. P (µ, X, var) is barrelled. The first non-trivial completeness of P (µ, X, var) is obtained in the case of X possessing the weak Radon-Nikodym property. The completeness of P (µ, X, var) is a direct consequence of the completeness of the space of X-valued measures equipped with the variation norm.

5 444 K. Musia l Proposition 2.2. If X has the weak Radon-Nikodym property, then P (µ, X, var) is a Banach space. In particular if Y is a Banach space not containing any isomorphic copy of l, then P (µ, Y, var) is complete (see [9]). We do not know whether the space P (µ, X, var) is always complete. Assuming the validity of Axiom L, we can prove however the completeness of P (µ, X, var) for an arbitrary perfect measure µ. Theorem 2.3. (Axiom L) If µ is perfect then P (µ, X, var) is complete. Proof. Let (f n ) be a Cauchy sequence in P (µ, X, var)) and let ν be the limit of (ν fn ) in cabv(µ, X ). The classical Radon Nikodym theorem guarantees the existence for each n N of a function h n L (µ) such that (3) ν ν fn () = h n dµ, for each Σ. Moreover, since X has the W RNP (see [], p. 238), there exists a weak measurable function f : Ω X such that for each x X and each Σ xν() = xf dµ. The theorem will be proved if we show that there exists a Pettis integrable function f that is weak -equivalent to f. To do it let us notice first that for each x from the closed unit ball of X and for each n N the equality (3) yields the relation (4) x, f f n h n µ a.e. Then, let B be the union of all weak -closures in X of countable subsets of the unit ball of X. It follows that for each n N and each x B we have x, f f n h n µ a.e.. Since the sequence h n is convergent to zero in the norm of L (µ), we obtain the measurability of all functions x f, with x B. If f is weak -bounded, then the Pettis integrability of ρ 0 (f) follows from [4], Theorem 6-2- and we may put f = ρ 0 (f). If f is arbitrary, then for each k N let Ω k Σ be defined by Ω k = ρ{ω Ω : ψ f (ω) k}. Then xfχ Ωk k x µ a.e. for all x X. It is a consequence of the first part of the proof that x ν( Ω k ) = x, ρ 0 (fχ Ωk ) dµ

6 Completeness in spaces of Pettis integrable functions 445 for all Σ, k N and x X. Since ν is of finite variation, one can easily show that f : Ω X defined for all ω k= Ω k by f(ω) = lim k ρ 0 (fχ Ωk ) (ω) is Pettis integrable and for all Σ ν() = P f dµ. This completes the proof. The next result shows that sometimes the assumptions concerning Axiom L and the measure space are superfluous. Proposition 2.4. Let X be a separable Banach space and let Y be a closed linear subspace of X. Then P (µ, Y, var) is complete for an arbitrary µ. Proof. Let (f n ) be a Cauchy sequence in P (µ, Y, var) and let ν be the limit of (ν fn ) in cabv(µ, Y ). Moreover, let f : Ω X be a weak -density of ν with respect to µ. It follows from the separability of X that for each n N and Σ (5) ν ν fn () = Hence there exists a subsequence (n k ) such that (6) lim k f n k f = 0 f f n dµ. µ a.e. In particular f(ω) Y for µ-almost all ω Ω. Since (6) yields the weak measurability of f and (5) implies the convergence of (x f nk ) to (x f) in L (µ) for every x X it follows that f P (µ, Y, var) and lim n f n f V = 0. This proves the completeness of P (µ, Y, var). Corollary 2.5. For an arbitrary µ the space P (µ, l, var) is complete. We do not know whether the space P (µ, X, var) is always complete. We have however an example of a non-conjugate X, such that P (λ, X, var) is non-complete. Since λ is perfect, it follows that in general the structure of the Banach space X is more important, than the properties of µ.

7 446 K. Musia l 3. xample. A normed space X is said to have property (K) if every sequence (x n ) converging in X to zero has a subsequence (x n(k) ) such that x n(k) is convergent. This notion, introduced by S. Mazur and W. Orlicz has been used to give non-categorical proofs of several classical theorems, as well as many new results (see []). Clearly each Banach space has the property (K). Answering my question, Fremlin constructed an example of a non-complete space P (λ, X, var), with X being a closed linear subspace of L [0, ] l [0, ] (unpublished). I am going to present now Fremlin s example (with his kind permission) with a modification of Fremlin s proof showing that the space P (λ, X, var) is not only non-complete but does not even possess the (K) property. For each t (0, ] let e t l [0, ] be the unit vector at t: e t (s) = 0 if s t and e t (s) = if s = t. Consider χ [0,t] as an element of L [0, ] and set for each t (0, ] x(t) = χ [0,t] + t 2 e t. Then, let X be the closed linear subspace of L [0, ] l [0, ] generated by all x(t), t (0, ] and all (w, 0) with w C[0, ]. Lemma 3.. If (ϕ, ψ) X, then for each t (0, ) and exist, and ϕ(t ) := lim δ 0 δ ϕ(t + ) := lim δ 0 δ t t δ t+δ t ϕ(s) ds ϕ(s) ds (7) ϕ(t + ) ϕ(t ) t 2 ψ(t) for each t (0, ). Proof. The inequality (7) holds true for the elements (χ [0,t], t e 2 t ) and for (w, 0) with w C[0, ]. It easily follows from this that (7) holds true for an arbitrary (ϕ, ψ) X. Theorem 3.2. The space P (λ, X, var) does not have the (K) property. Proof. If x X, then there are y L [0, ] and z l [0, ] such that x (f, g) = y (f) + z (g) for every f L [0, ] and g l [0, ]. We may assume that z = z + z2 with zi ba[0, ] such that supp z is at most countable and z2 vanishes on countable sets. It is clear that z (e t ) 0 for at most countable number of t (0, ] and so x, x(t) = y χ [0,t] for λ-almost all t (0, ]. Hence, we have x, x(t) dt = y, χ [0,t] dt = y, ν = x, (ν, 0),

8 Completeness in spaces of Pettis integrable functions 447 where ν (t) = λ ( [t, ]). Notice that ν C[0, ] L [0, ] and so we get x P (λ, X) with P x(t) dt = (ν, 0) X and (8) ν L [0,] λ(). Now, if a (0, ] and L, then x(at) dt = a a x(s) ds = a (ν a, 0) X. For each i 0,, 2,... and each t (0, ] define now a function g i setting g i (t) = 2 i x(t2 i ) We have g i P (λ, X) and P g i (t) dt = 2 i x(t2 i ) dt = (ν 2 i, 0) X. It follows that lim g i V = 0, i since ν 2 i L [0,] 2 i by (8). We shall prove that for no subsequence ( ) g n(i) the series g n(i) is convergent. To do it, suppose a series g n(i) is convergent in P (λ, X, var) to a function f. Let n f n (t) = 2 n(i) x(t2 n(i) ) for all t (0, ] and κ n () = f n (t) dt for all L. By our conjecture (f n ) is convergent in P (λ, X, var) to f, what is equivalent to the convergence of (κ n ) in cabv(l, X) to a measure κ such that (9) κ() = P X f dλ for each L. Consider now the series 2 n(i) x(t2 n(i) ). Since for each t (0, ] and each s [0, ] we have e t2 i(s) = if s = t2 i and e t2 i(s) = 0 if s t2 i we get for each t (0, ] the equality 2 n(i) x(t2 n(i) )(s) = 2 n(i) χ [0,t2 n(i) ](s), for all but countably many s [0, ].

9 448 K. Musia l Since the right hand side series is uniformly convergent on [0, ] we can define a function f 0 : [0, ] L [0, ] by setting (0) [f 0 (t)] (s) = 2 n(i) χ [0,t2 n(i) ](s) for every s [0, ]. Clearly (f 0 (t), 0) L [0, ] l [0, ] for each t (0, ]. If x X, then and so x, f n (t) = y, for λ-almost all t (0, ]. In particular n 2 n(i) χ [0,t2 n(i) ] + z, x, f n (t) = y, n n 2 n(i) χ [0,t2 n(i) ] 2 n(i) t 2 e t2 n(i), lim n x f n (t) = x (f 0 (t), 0) for λ- a.a. t (0, ]. Since at the same time we have also x f n (t) x for every n, x and λ-a.e. t (0, ] we have by the Lebesgue Convergence Theorem x, κ() = lim n x, κ n () = x, (f 0 (t), 0) dt. for every L. Thus () κ() = P L [0,] l [0,] (f 0, 0) dλ. Since f : [0, ] X, there are functions ϕ : [0, ] L [0, ] and ψ : [0, ] l [0, ] satisfying for each t (0, ) the relation f(t) = (ϕ(t), ψ(t)). Hence we get from (0) and () that for every L P L [0,] ϕ(t) dt = P L [0,] f 0 (t) dt. Since L [0, ] is weak -separable, this yields (2) ϕ(t) = f 0 (t) for λ a.e. t (0, ]. Let us fix some t > 0 satisfying (2). Then (f 0 (t), ψ(t)) X and so in virtue of (7) f 0 (t)(s + ) f 0 (t)(s ) s 2 ψ(t)(s) s 2 ψ(t),

10 Completeness in spaces of Pettis integrable functions 449 for every s [0, ]. But f0 (t)(2 n(i) t + ) f 0 (t)(2 n(i) t ) = 2 n(i) and so we have for each i the inequality 2 n(i) t 2 ψ(t) what is clearly impossible. 4. Completeness of LLN(µ, X, var). It has been proven in [5] that the Pettis integrals of functions from LLN(µ, X) are measures of finite variation. Thus, it makes sense to equip the space LLN(µ, X) with the variation norm of the integrals. It will be denoted by LLN(µ, X, var). Contrary to P (µ, X, var), the space LLN(µ, X, var) is always complete. Theorem 4.. LLN(µ, X, var) is a Banach space. Proof. Let ρ be a consistent lifting (see [4]) on (Ω, Σ, µ) and, let (f n ) be a Cauchy sequence in LLN(µ, X, var). Moreover, let ν be the limit of (ν fn ) in cabv(µ, X ) and, let f : Ω X be a weak -density of ν with respect to µ. We shall split the proof into two parts. Assume first that f is weak -scalarly bounded. Without loss of generality, we may assume that f = ρ 0 (f). Moreover, let for each n a function ψ n L (µ) be the RN-density of the measure ν fn and let Ω n,m : = ρ({ω Ω : ψ n (ω) m}). Since ψ n L (µ), there exists for each n N a number m n such that µ(ω c n,m n ) < /n. Since f n χ Ωn,m is scalarly bounded and properly measurable (see [4] for the definition), it follows from the consistency of ρ that g n : = ρ 0 ( fn χ Ωn,m ) is bounded and properly measurable, for each n, m N (see [4]). In particular g n LLN(µ, X, var). Since f g n = ρ 0 (f g n ), applying (2) and (), we see that for each n N f g n GC for some positive M and so f g n dµ = ν ν fn (Ω n,mn ) + ν (Ω c n,m n ) ν ν fn (Ω) + Mµ(Ω c n,m n ) (3) lim n f g n GC = 0. That is f is approximated in the Glivenko-Cantelli norm by elements of LLN(µ, X ) what means - in virtue of Theorem 26 from [5] - that f satisfies the law of large numbers.

11 450 K. Musia l Assume now that f is an arbitrary weak -density of ν with respect to µ and let ψ f be the RN-derivative of ν f with respect to µ. Then, for each k N let Ω k : = ρ({ω Ω : ψ f (ω) k}). It is a consequence of the first part of the proof that h k : = ρ 0 (fχ Ωk ) LLN(µ, X ) for all k N. Since ν is of finite variation, one can easily show that g : Ω X defined for all ω k= Ω k by g(ω) = lim k ρ 0 (fχ Ωk ) (ω) is a Pettis integrable density of ν with respect to µ and lim n g f n V = 0. One has to prove yet that g LLN(µ, X ). Notice however that for every x X we have xf χ Ωk ψ f χ Ωk x µ-a.e. and so xh k ρ 0 (ψ f χ Ωk ) x everywhere. Consequently, h k ψ f µ-a.e. This yields the inequality g ψ f µ-a.e. Since ψ f L (µ) we get g h k GC g h k ψ f dµ k 0. This completes the proof. Dobric posed in [3] a question about completeness of LLN(µ, X). We are going to prove that the space is almost never complete and in general it is even not barrelled. Theorem 4.2. If X is infinite dimensional and µ is not purely atomic, then LLN(µ, X) is non-complete. Proof. If P c (µ, X) : = {f P (µ, X) : ν f (Σ) is norm relatively compact } then P c (µ, X) endowed with the Pettis norm is non-complete (see [6]). Let (f n ) be a Cauchy sequence in P c (µ, X) that is not convergent in P c (µ, X). Since by the assumption each set ν fn (Σ) is norm relatively compact, it follows from [0] Proposition 3 (or [] Lemma 9.4), that for each n N there exists a simple function h n : Ω X with f n h n P < /n. Obviously, the sequence (h n ) is Cauchy in P c (µ, X). Since simple functions are properly measurable and LLN(µ, X) P c (µ, X), the sequence (h n ) is also Cauchy in LLN(µ, X). Clearly, the sequence (h n ) is divergent in LLN(µ, X). This completes the proof. Now we are in a good position to consider the problem of barrelledness of LLN(µ, X). It has been proven in [4] that the space P (µ, X) is always barrelled. Then, it has been proven in [2] that the spaces P (µ, X) and P c (µ, X) are even ultrabornological. It turns out that in general the space LLN(µ, X) does not behave so well. Theorem 4.3. If LLN(µ, X, var) is complete, X is infinite dimensional and µ is not purely atomic, then LLN(µ, X) is not barrelled. Proof. One can easily check that the identity map from LLN(µ, X) to LLN(µ, X, var) has closed graph and so, if LLN(µ, X) were barrelled it would be continuous in virtue of the closed graph theorem. Hence there would exist a positive number a such that f V a f P for each f LLN(µ, X). This is however impossible if X is infinite dimensional and µ is not purely atomic. Ω c k

12 Completeness in spaces of Pettis integrable functions 45 Corollary 4.4. Let µ be non-atomic. If X is infinite dimensional and each X- valued Pettis integrable function is weakly equivalent to a strongly measurable function, then LLN(µ, X) is not barrelled. In particular, if X has the RNP, is reflexive or measure compact (cf. [5] for the definition), then LLN(µ, X) is not barrelled. Proof. Since it is well known that L (µ, X) LLN(µ, X, var) (see [7]), we get at once the equality LLN(µ, X, var) = L (µ, X). Applying Theorem 4.3 we get the required result. Corollary 4.5. If X is infinite dimensional and µ is not purely atomic, then the ultrabornological space P c (µ, X) contains a dense non-barrelled subspace. Proof. According to [0] simple functions are dense in P c (µ, X). This completes the proof. Corollary 4.6. Let µ be non-atomic. If X is infinite dimensional then L (µ, X) endowed with the Pettis norm is not barrelled. References. P. Antosik and J. Burzyk, Sequential conditions for barrelledness and bornology, Bull. Acad. Polon. Sci. Sér. Sci. Math. 35 (987), S. Diaz, A. Fernandez, M. Florencio and P.J. Paul, A wide class of ultrabornological spaces of measurable functions, J. Math. Anal. Appl. 90 (995), V. Dobric, The decomposition theorem for functions satisfying the law of large numbers, J. of Theoretical Probability 3 (990), L. Drewnowski, M. Florencio and P.J. Paul, The space of Pettis integrable functions is barrelled, Proc. Amer. Math. Soc. 4 (992), G. dgar, Measurability in a Banach space, Indiana Univ. Math. J. 24 (977), D.H. Fremlin and M. Talagrand, A decomposition theorem for additive set functions, with applications to Pettis integrals and ergodic means, Math. Z. 68 (979), J. Hoffmann-Jørgensen, The law of large numbers for non-measurable and nonseparable random elements, Astérisque 3 (985), A. and C. Ionescu-Tulcea, Topics in the theory of lifting, rgebnisse Math. Grenzgebiete, Band 48, Springer-Verlag, K. Musia l, The weak Radon-Nikodym property in Banach spaces, Studia Math. 4 (979), , Martingales of Pettis integrable functions, Measure Theory (Oberwolfach 979), Lecture Notes in Math. Vol. 794 (980), , Topics in the theory of Pettis integration, Rendiconti Ist. di Matematica dell Universita di Trieste 23 (99),

13 452 K. Musia l 2. B.J. Pettis, On integration in vector spaces, Trans. Amer. Math. Soc. 44 (938), M.. Rudin, Martin s axiom, In: Handbook of Mathematical Logic, (J. Barwise, ed.), North Holland, M. Talagrand, Pettis integral and measure theory, Memoirs of the Amer. Math. Soc. No. 307, , The Glivenko-Cantelli problem, Annals of Probability 5 (987), G..F. Thomas, Totally Summable Functions with Values in Locally Convex Spaces, Measure Theory (Oberwolfach 975), Lecture Notes in Math. Vol. 54 (976), 7 3. Received 4 January, 2000 and in revised form 20 August, 200.