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On the integration of vector-valued functions

arXiv:math/9202202v1 [math.FA] 21 Feb 1992amsppt.styFMp91.texamsppt.styFMp91a.texVersion of 20.1.92On the integration of vector-valued functionsD.H.Fremlin & J.MendozaUniversity of Essex, Colchester, EnglandUniversidad Complutense de Madrid, Madrid, SpainWe discuss relationships between the McShane, Pettis, Talagrand and Bochner integrals.Introduction A large number of different methods of integration of Banach-space-valued functions havebeen introduced, based on the various possible constructions of the Lebesgue integral. They commonly runfairly closely together when the range space is separable (or has w∗-separable dual) and diverge more orless sharply for general range spaces.

The McShane integral as described by [Go90] is derived from the‘gauge-limit’ integral of [McS83]. Here we give both positive and negative results concerning it and the otherthree integrals listed above.1A Definitions We recall the following definitions.

Let (S, Σ, µ) be a probability space and X a Banachspace, with dual X∗. (a) A function φ : S →X is Pettis integrable if for every E ∈Σ there is a wE ∈X such thatRE f(φ(x))µ(dx) exists and is equal to f(wE) for every f ∈X∗; in this case wS is the Pettis integral of φ,and the map E 7→wE : Σ →X is the indefinite Pettis integral of φ.

(b)Afunctionφ:S→XisTalagrandintegrable,withTalagrandintegralw,if w = limn→∞1nPi

)(c) A function φ : S →X is Bochner integrable, with Bochner integral w, if for every ǫ > 0 we canfind a partition E0, . .

. , En of S into measurable sets and vectors x0, .

. .

, xn ∈X and an integrable functionh : S →R such thatRh ≤ǫ, ∥φ(t) −xi∥≤h(t) for t ∈Ei, i ≤n and ∥w −Pi≤n µEi.xi∥≤ǫ.1B Now we come to an integral which has been defined for functions with domains which are intervalsin R. In fact it can be satisfactorily generalized to very much wider contexts; but as the extension involvesideas from topological measure theory unnecessary for the chief results of this paper, we confine ourselveshere to the original special case.Definitions A McShane partition of [0, 1] is a finite sequence ⟨([ai, bi], ti)⟩i≤n such that ⟨[ai, bi]⟩i≤n is anon-overlapping family of intervals covering [0, 1] and ti ∈[0, 1] for each i. A gauge on [0, 1] is a functionδ : [0, 1] →]0, ∞[.

A McShane partition ⟨([ai, bi], ti)⟩i≤n is subordinate to a gauge δ if ti −δ(ti) ≤ai ≤bi ≤ti + δ(ti) for every i ≤n.Now let X be a Banach space. Following [Go90], we say that a function φ : [0, 1] →X is McShaneintegrable, with McShane integral w, if for every ǫ > 0 there is a gauge δ : [0, 1] →]0, ∞[ such that∥w −Pi≤n(bi −ai)φ(ti)∥≤ǫ for every McShane partition ⟨([ai, bi], ti)⟩i≤n of [0, 1] subordinate to δ.1C Summary of results With four integrals to play with, a good many questions can be asked; and thesituation is complicated by the fact that certain natural restrictions which may be put on the space X andthe function φ change the answers.

We therefore set out the facts in a semi-tabular form. We give referencesto the literature for those which are already known, and references to paragraphs below to those which webelieve to be new.

(a) Consider first the situation in which no restriction is placed on the Banach space X nor on thefunction φ : [0, 1] →X. In this context, a Bochner integrable function is Talagrand integrable, a Talagrandintegrable function is Pettis integrable, and the integrals coincide whenever defined ([Ta87], Theorem 8).A Bochner integrable function is McShane integrable ([Go90], Theorem 16) (in fact, a measurable Pettisintegrable function is McShane integrable – see [Go90], Theorem 17); a McShane integrable function isPettis integrable (2C below).1

2None of the implications here can be reversed. To see this, it is enough to find a McShane integrablefunction which is not Talagrand integrable (3A, 3G) and a Talagrand integrable function which is notMcShane integrable (3E).

(b) Now suppose that the unit ball B1(X∗) of the dual X∗of X is separable and that φ is bounded. In thiscase, a McShane integrable function must be Talagrand integrable (2M).

We ought to observe at this pointthat in these circumstances the continuum hypothesis, for instance, is enough to make any Pettis integrablefunction Talagrand integrable ([Ta84], 6-1-3), and that it remains conceivable that this is a theorem of ZFC(see [SFp90]). But our result in 2M does not depend on any special axiom.In this context it is still true that a McShane integrable function need not be Bochner integrable (3F)and that a Talagrand integrable function need not be McShane integrable (3E).

(c) If we take X to be separable, but allow φ to be unbounded, then the Bochner and Talagrand integralscoincide (see 2K below), and the McShane and Pettis integrals coincide (2D).There is still a McShane integrable function which is not Talagrand integrable (3G). (d) For separable X and bounded φ, the Bochner and Pettis integrals coincide (2K), so all four integralshere coincide.

(e) Finally, the same is true, for unbounded φ, if X is finite-dimensional.amsppt.styFMp91b.texVersion of 20.1.922. Positive results In this section we give our principal positive results.

A McShane integrable function isPettis integrable (2C); using this we are able to prove a convergence theorem for McShane integrable functions(2I) with some corollaries (2J). We conclude by showing that a bounded McShane integrable function from[0, 1] to a space with w∗-separable dual unit ball is Talagrand integrable (2M).2A We approach the first result by means of the ‘Dunford integral’.

Recall that a function φ : [0, 1] →Xis Dunford integrable if hφ : [0, 1] →R is integrable for every h ∈X∗; in this case we have an indefiniteDunford integral ν : Σ →X∗∗, where Σ is the algebra of Lebesgue measurable subsets of [0, 1], given bythe formula (νE)(h) =RE hφ for every h ∈X∗, E ∈Σ ([Ta84], 4-4-1 or [DU77], p. 52, Lemma 1). Thus aPettis integrable function is just a Dunford integrable function whose indefinite integral takes values in X(identified, of course, with its canonical image in X∗∗).

Now we have the following general characterizationof Pettis integrable functions on [0, 1] (which in part, at least, is already known; cf. [Dr86], 4.2).2B Proposition Let X be a Banach space and φ : [0, 1] →X a Dunford integrable function withindefinite integral ν : Σ →X∗∗.

Suppose that ν([a, b]) ∈X for every subinterval [a, b] of [0, 1]. Then thefollowing are equivalent:(i) φ is Pettis integrable;(ii) for every sequence ⟨[ai, bi]⟩i∈N of non-overlapping subintervals of [0, 1], Pi∈N ν([ai, bi]) exists in X(for the norm of X);(iii) for every ǫ > 0 there is an η > 0 such that ∥νE∥≤ǫ whenever µE ≤η;(iv) ν is countably additive.proof (i)⇒(ii) is a theorem of Pettis (see [DU77], II.3.5).

(ii)⇒(iii) The point is that {hφ : h ∈B1(X∗)} is uniformly integrable. To see this, it is enough to showthatlimn→∞sup{RGn |hφ| : h ∈B1(X∗)} = 0for every disjoint sequence ⟨Gn⟩n∈N of open sets in [0, 1] (see [Di84], VII.14).

But given such a sequence, anda sequence ⟨hn⟩n∈N in B1(X∗), set αn =RGn |hnφ| for each n. Then we can find for each n a set Fn ⊆Gn, afinite union of closed intervals, such thatRFn |hnφ| ≥αn −2−n. Now we can find a sequence ⟨[ai, bi]⟩i∈N ofnon-overlapping intervals, and an increasing sequence ⟨kn⟩n∈N of integers, such that Fn = Skn≤i

3Now it follows that for every ǫ > 0 there is an η > 0 such thatRE |hφ| ≤ǫ whenever h ∈B1(X∗), E ∈Σand µE ≤η, writing µ for Lebesgue measure; so that ∥νE∥≤ǫ whenever µE ≤η. (iii)⇒(iv) is elementary.

(iv)⇒(i) Our original hypothesis was that νE ∈X for intervals E; it follows that νE ∈X whenever Eis a finite union of intervals. Because ν is countably additive, νE ∈X whenever E is open, and thereforewhenever E is Gδ; but also of course νE = 0 ∈X whenever µE = 0, so νE ∈X for every E ∈Σ, and φ isPettis integrable.2C Theorem Let X be a Banach space and φ : [0, 1] →X a McShane integrable function.

Then φ isPettis integrable.proof As remarked in [Go90], Theorem 8, φ is Dunford integrable; let ν : Σ →X∗∗be its indefinite Dunfordintegral. We know also that ν([a, b]) ∈X for every subinterval [a, b] of [0, 1] ([Go90], Theorem 4).

So weseek to show that (ii) of 2B above holds true.Let ǫ > 0.Let δ : [0, 1] →]0, ∞[ be a gauge such that ∥ν([0, 1]) −Pi≤n(bi −ai)φ(ti)∥≤ǫ when-ever ⟨([ai, bi], ti)⟩i≤n is a McShane partition of [0, 1] subordinate to δ. Fix a particular McShane partition⟨([ai, bi], ti)⟩i≤n of [0, 1] subordinate to δ, and set M = supi≤n ∥φ(ti)∥.

We claim that if E ⊆[0, 1] is a finiteunion of closed intervals then ∥νE∥≤MµE + 2ǫ. To see this, express E as Sj≤m[cj, dj] where the [cj, dj]are non-overlapping, and let η > 0.

For each i ≤n, we can express [ai, bi] \ int E as a (possibly empty) finiteunion of non-overlapping intervals [aik, bik] for k < r(i); write tik = ti for i ≤n, k < r(i). Then we see that∥Pi≤n(bi −ai)φ(ti) −Pi≤nPk

Then a function φ : [0, 1] →X is McShane integrableiffit is Pettis integrable.proof If φ is Pettis integrable, it is measurable, because X is separable; so it is McShane integrable byTheorem 17 of [Go90]. Now 2C gives the reverse implication.2E As a further consequence of 2C we have the following.Theorem Let X be a Banach space and φ : [0, 1] →X a McShane integrable function.

Then for anymeasurable E ⊆[0, 1] the function φE = φ × χ(E) : [0, 1] →X, defined by writing φE(t) = φ(t) if t ∈E and0 otherwise, is McShane integrable.proof Again write ν for the indefinite integral of φ; we now know that ν takes all its values in X. Let ǫ > 0.By 2B, we can find an η > 0 such that ∥νH∥≤ǫ whenever µH ≤η.

Let F1 ⊆E, F2 ⊆[0, 1]\E be closed setssuch that µF1 + µF2 ≥1 −η. Let δ : [0, 1] →]0, ∞[ be a gauge such that ∥ν([0, 1])−Pi≤n(bi −ai)φ(ti)∥≤ǫwhenever ⟨([ai, bi], ti)⟩i≤n is a McShane partition subordinate to δ.

Let δ1 : [0, 1] →]0, ∞[ be such thatδ1(t) ≤δ(t) for every t and [t −δ1(t), t + δ1(t)] ∩Fj = ∅whenever t /∈Fj, for both j ∈{1, 2}.Suppose that ⟨([ai, bi], ti)⟩i≤n is a McShane partition of [0, 1] subordinate to δ1. We seek to estimate∥νE −Pi≤n(bi −ai)φE(ti)∥.

Set I = {i : i ≤n, ti ∈E}, H = Si∈I[ai, bi], Then we must have F1 ⊆H ⊆

4[0, 1] \ F2. Consequently µ(H△E) ≤η and ∥νE −νH∥≤2ǫ.

But now recall that by Theorem 5 of [Go90]we know that ∥νH −Pi∈I(bi −ai)φ(ti)∥≤ǫ. So∥νE −Pi≤n(bi −ai)φE(ti)∥= ∥νE −Pi∈I(bi −ai)φ(ti)∥≤2ǫ.As ǫ is arbitrary, φE is McShane integrable.2F For the next theorem of this section, we need to recall some well-known facts concerning vectormeasures.

Suppose that Σ is a σ-algebra of sets and X a Banach space. (a) Let us say that a function ν : Σ →X is ‘weakly countably additive’ if f(ν(Si∈N Ei)) = Pi∈N f(νEi)for every disjoint sequence ⟨Ei⟩i∈N in Σ and every f ∈X∗.

The first fact is that in this case ν is countablyadditive, that is, Pi∈N νEi is unconditionally summable to ν(Si∈N Ei) for the norm topology whenever⟨Ei⟩i∈N is a disjoint sequence of measurable sets with union E (‘Orlicz-Pettis theorem’, [Ta84], 2-6-1 or[DU77], p. 22, Cor. 4).

(b) If now µ is a measure with domain Σ such that νE = 0 whenever µE = 0, then for every ǫ > 0 thereis a δ > 0 such that ∥νE∥≤ǫ whenever µE ≤δ. ([DU77], p. 10, Theorem 1.

)(c) Thirdly, suppose that ⟨νn⟩n∈N is a sequence of countably additive functions from Σ to X such thatνE = limn→∞νnE exists in X, for the weak topology of X, for every E ∈Σ; then ν is countably additive. (Use Nikod´ym’s theorem ([Di84], p. 90) to see that ν is weakly countably additive.

)2G Lemma Let X be a Banach space. If φ : [0, 1] →X is McShane integrable with McShane integral w,then∥w∥≤R∥φ(t)∥µ(dt).proof Take any f in the unit ball of X∗.

By [Go90], Theorem 8, f(w) is the McShane integral of fφ :[0, 1] →R, and by 6-4 and 6-5 of [McS83] this is the ordinary integral of fφ. So we have|f(w)| = |Rfφ| ≤R|fφ| ≤R∥φ∥.As f is arbitrary, ∥w∥≤R∥φ∥.2H Lemma Let X be a Banach space and φ : [0, 1] →X a McShane integrable function; let ǫ > 0.

Thenthere is a gauge δ : [0, 1] →]0, ∞[ such that ∥RE φ −Pi≤n µEiφ(ti)∥≤ǫ whenever E0, . .

. , En are disjointmeasurable subsets of [0, 1], t0, .

. .

, tn ∈[0, 1] and Ei ⊆[ti −δ(ti), ti + δ(ti)] for every i.proof Let δ be a gauge such that ∥Rφ−Pi≤n(bi−ai)φ(ti)∥≤ǫ whenever ⟨([ai, bi], ti)⟩i≤n is a McShane par-tition of [0, 1] subordinate to δ. Let E0, .

. .

, tn be as in the statement of the lemma; set M = maxi≤n ∥φ(ti)∥.Take η > 0; let η′ > 0 be such that (n + 1)Mη′ ≤η and ∥RH φ∥≤η whenever µH ≤(n + 1)η′(see 2B(iii)). Then we can find a family ⟨[aij, bij]⟩i≤n,j≤r(i) of non-overlapping closed intervals such thatµ(Ei△Sj≤r(i)[ai, bi]) ≤η′ and ti −δ(ti) ≤aij ≤bij ≤ti + δ(ti) for each i ≤n, j ≤r(i).

Write tij = ti fori ≤n, j ≤r(i). Then ⟨([aij, bij], tij)⟩i≤n,j≤r(i) can be extended to a McShane partition of [0, 1] subordinateto δ.

So writing Fi = Sj≤r(i)[aij, bij] for each i, F = Si≤n Fi, we have∥RF φ −Pi≤n,j≤r(i)(bij −aij)φ(tij)∥≤ǫby [Go90], Theorem 5; that is,∥RF φ −Pi≤n µFiφ(ti)∥≤ǫ.Next,∥RE φ −RF φ∥≤ǫbecause µ(E△F) ≤(n + 1)η. Also∥Pi≤n µFiφ(ti) −Pi≤n µEiφ(ti)∥≤M(n + 1)η′ ≤η.Putting these together,∥RE φ −Pi≤n µEiφ(ti)∥≤ǫ + 2η;as η is arbitrary we have the result.2I Theorem Let X be a Banach space. Let ⟨φn⟩n∈N be a sequence of McShane integrable functions from[0, 1] to X, and suppose that φ(t) = limn→∞φn(t) exists in X for every t ∈[0, 1].

If moreover the limitνE = limn→∞RE φnexists in X, for the weak topology, for every measurable E ⊆[0, 1], φ is McShane integrable andRφ =ν([0, 1]).

5proof Fix ǫ > 0. Write µ for Lebesgue measure, Σ for the algebra of Lebesgue measurable subsets of [0, 1].

(a) For t ∈[0, 1], n ∈N set qn(t) = supj≥i≥n ∥φj(t) −φi(t)∥. For each t, write r(t) = min{n : qn(t) ≤ǫ, ∥φ(t)∥≤n}; set Ak = {t : r(t) = k} for each k. For each k ∈N, let Wk ⊇Ak be a measurable set withµ∗(Wk \Ak) = 0; set Vk = Wk \Sj 0 such that ∥νE∥≤2−kǫ whenever µE ≤ηk (see (b) and (c) of 2F above); let Gk ⊇V ∗k be an open setsuch that µ(Gk \ V ∗k ) ≤min(ηk, 2−kǫ).

(b) If k ∈N and E ⊆V ∗k is measurable, then ∥νE −RE φk∥≤ǫµE. To see this, it is enough to considerthe case E ⊆Vj where j ≤k.

In this case, observe that∥νE −RE φk∥≤lim supn→∞∥RE φn −RE φk∥≤supn≥kRE∥φn(t) −φk(t)∥µ(dt)by Lemma 2G. Now µ∗(E \ Aj) = 0 and for t ∈Aj we have ∥φn(t) −φk(t)∥≤qj(t) ≤ǫ for every n ≥k, soRE∥φn(t) −φk(t)∥µ(dt) ≤ǫµEfor every n ≥k, giving the result.

(c) For each k ∈N let δk : [0, 1] →]0, ∞[ be a gauge such that∥RE φk −Pi≤n µEiφk(ti)∥≤2−kǫwhenever E0, . .

. , En are disjoint measurable sets with union E and t0, .

. .

, tn ∈[0, 1] are such that Ei ⊆[ti −δk(ti), ti + δk(ti)] for each i; such a gauge exists by Lemma 2H. Choose δ : [0, 1] →]0, ∞[ such thatδ(t) ≤min(ǫ, δk(t)) and [0, 1] ∩[t −δ(t), t + δ(t)] ⊆Gk for t ∈Ak.

(d) Let ⟨([ai, bi], ti)⟩i∈N be a McShane partition of [0, 1] subordinate to δ. We seek to estimate ∥ν([0, 1])−w∥, where w = Pi≤n(bi −ai)φ(ti).Set Ik = {i : i ≤n, ti ∈Ak} for each k; of course all but finitely many of the Ik are empty.

For i ∈Ik,set Ei = [ai, bi] ∩V ∗k . We have [ai, bi] ⊆[ti −δ(ti), ti + δ(ti)] ⊆Gk, so Pi∈Ik µ([ai, bi] \ Ei) ≤2−kǫ, andPi∈Ik µ([ai, bi] \ Ei)∥φ(ti)∥≤2−kkǫ, because ∥φ(t)∥≤k for t ∈Ak.

Consequently, if we writew1 = Pi≤n µEiφ(ti),we shall have ∥w −w1∥≤Pk∈N 2−kkǫ = 2ǫ.For each i ≤n, let k(i) be such that ti ∈Ak(i). Then we have ∥φ(ti) −φk(i)(ti)∥≤ǫ for each i. SoPi≤n µEi∥φ(ti) −φk(i)(ti)∥≤Pi≤n(bi −ai)ǫ ≤ǫ,because ⟨[ai, bi]⟩i≤n is non-overlapping.

Accordingly, writingw2 = Pi≤n µEiφk(i)(ti),we have ∥w −w2∥≤3ǫ.Set Hk = S{Ei : i ∈Ik} for each k. Because Ei ⊆[ti −δk(ti), ti + δk(ti)] for each i ∈Ik, we have∥Pi∈Ik µEiφk(ti) −RHk φk∥≤2−kǫ.Consequently, writingw3 = Pk∈NRHk φk,we have ∥w −w3∥≤5ǫ.Next, for any k, Hk ⊆V ∗k , so we have∥νHk −RHk φk∥≤ǫµHk,by (b) above. So writing w4 = Pk∈N νHk we have ∥w3 −w4∥≤ǫ and ∥w −w4∥≤6ǫ.If we set H′k = S{[ai, bi] : i ∈Ik}, then µ(H′k \ Hk) ≤ηk, so that ∥νH′k −νHk∥≤2−kǫ, for each k.Accordingly ∥w −w5∥≤8ǫ, wherew5 = Pk∈N νH′k = ν(Sk∈N H′k) = ν(Si≤n[ai, bi]) = ν([0, 1]).As ǫ is arbitrary,Rφ exists and is equal to ν([0, 1]).Problem In this theorem we are supposing that φ(t) = limn→∞φn(t) in the norm topology for every t. Isit enough if φ(t) is the weak limit of ⟨φn(t)⟩n∈N for every t?2J Corollary Let X be a Banach space.

(a) Let ⟨φn⟩n∈N be a sequence of McShane integrable functions from [0, 1] to X such that φ(t) =limn→∞φn(t) exists in X for every t ∈[0, 1]. IfC = {fφn : f ∈X∗, ∥f∥≤1, n ∈N}is uniformly integrable, then φ is McShane integrable.

In particular, if {∥φn∥: n ∈N} is dominated by anintegrable function, then φ is McShane integrable.

6(b) Let φ : [0, 1] →X be a Pettis integrable function and ⟨Ei⟩i∈N a cover of [0, 1] by measurable sets.Suppose that φ × χ(Ei) is McShane integrable for each i. Then φ is McShane integrable.proof (a) The point is that φn, φ satisfy the conditions of Theorem 2I.

To see this, take E ∈Σ and ǫ > 0.Because C is uniformly integrable, there is an η > 0 such thatRH |g| ≤ǫ whenever g ∈C and µH ≤η;consequently ∥RH φn∥≤ǫ for all n ∈N whenever H ∈Σ and µH ≤η. Now setAn = {t : ∥φi(t) −φj(t)∥≤ǫ ∀i, j ≥n};then ⟨An⟩n∈N is an increasing sequence with union [0, 1], so there is an n such that µ∗An ≥1−η.

Let G ∈Σbe such that An ⊆G and µG = µ∗An. Then whenever i, j ≥n we have∥RE∩G φi −RE∩G φj∥≤RE∩G∥φi(t) −φj(t)∥µ(dt) ≤µG supt∈An ∥φi(t) −φj(t)∥≤ǫ.Also ∥RE\G φi∥and ∥RE\G φj∥are both less than or equal to ǫ, so ∥RE φi −RE φj∥≤3ǫ.

This shows that⟨RE φi⟩i∈N is a Cauchy sequence and therefore convergent, for every E ∈Σ. Accordingly the conditions of2I are satisfied and φ is McShane integrable.

(b) We apply 2I with φn(t) = φ(t) for t ∈Si≤n Ei, 0 elsewhere.Remark Part (a) is a version of Vitali’s lemma. Part (b) is a generalization of [Go90], Theorem 15.2K We now give a result connecting the McShane and Talagrand integrals.

Recall that if (S, Σ, µ) is aprobability space, a set A of real-valued functions is stable (in Talagrand’s terminology) if for every E ∈Σ,with µE > 0, and all real numbers α < β, there are m, n ≥1 such that µ∗m+nZ(A, E, m, n, α, β) < (µE)m+n,where throughout the rest of paper we write Z(A, E, I, J, α, β) for{(t, u) : t ∈EI, u ∈EJ, ∃f ∈A, f(t(i)) ≤α ∀i ∈I, f(u(j)) ≥β ∀j ∈J},and µ∗m+n is the ordinary product outer measure on Sm × Sn. Now if X is a Banach space, a functionφ : S →X is properly measurable if {hφ : h ∈X∗, ∥h∥≤1} is stable.

Talagrand proved ([Ta87],Theorem 8) that φ is Talagrand integrable iffit is properly measurable and the upper integralR∥φ(t)∥µ(dt)is finite.In particular, a Talagrand integrable function φ : S →X must be scalarly measurable for the completionof µ ([Ta84], 6-1-1). So if X is separable, φ must be measurable for the completion of µ ([DU77], II.1.2 or[Ta84], 3-1-3); now asR∥φ∥dµ < ∞, φ is Bochner integrable.2L Proposition Let X be a Banach space such that the unit ball of X∗is w∗-separable.

If φ : [0, 1] →Xis a McShane integrable function then it is properly measurable.proof Let w be the McShane integral of φ. Set A = {hφ : h ∈X∗, ∥h∥≤1} ⊆R[0,1]; we have to show thatA is stable.

Note that because the unit ball of X∗is separable for the w∗-topology on X∗, and the maph 7→hφ : X∗→R[0,1] is continuous for the w∗-topology on X∗and the topology of pointwise convergenceon R[0,1], A has a countable dense subset A0.Take a non-negligible measurable E ⊆[0, 1] and α < α′ < β′ < β in R. For m, n ≥1 set Hmn =Z(A, E, m, n, α, β), H′mn = Z(A0, E, m, n, α′, β′); then Hmn ⊆H′mn and H′mn is measurable for the usual(completed) product measure on Em × En. We seek an m with µ2mH′mm < (µE)2m, writing µ for Lebesguemeasure on [0, 1] and µ2m for Lebesgue measure on [0, 1]m × [0, 1]m.Set ǫ = 16(β′ −α′)µE, and choose a function δ : [a, b] →]0, ∞[ such that ∥w −Pi≤n(bi −ai)φ(ti)∥≤ǫfor every McShane partition ⟨([ai, bi], ti)⟩i≤n of [a, b] subordinate to δ.Take k ≥1 such that µ∗D ≥12µE, where D = {s : s ∈E, δ(s) ≥1k}.

Let ⟨[ai, bi]⟩i

Write ∆(s) = {s′ : |s′ −s|

Because [a, b]\G is itself a finite union of closed intervals, we can find a family ⟨[ai, bi]⟩m≤i≤nof non-overlapping closed intervals such that [0, 1] \ G = Sm≤i≤n[ai, bi] and t(i) ∈[ai, bi] ⊆∆(ti) for eachi. Now if we set u(i) = t(i) for m ≤i ≤n, we see that ⟨([ai, bi], t(i))⟩i≤n and ⟨([ai, bi], u(i))⟩i≤n are bothMcShane partitions of [a, b] subordinate to δ.

So we must have∥Pi≤n(bi −ai)(φ(t(i)) −φ(u(i)))∥≤2ǫ.

7Now (t, u) ∈H′mm, so there is an f ∈A such that f(t(i)) ≤α′ and f(u(i)) ≥β′ for every i < m. f is ofthe form hφ for some h of norm at most 1, so| Pi

Examples In this section we give examples to show that the results above are more or less complete intheir own terms. In particular, a McShane integrable function need not be Talagrand integrable (3A, 3G)and a Talagrand integrable function need not be McShane integrable (3E).3A Example There is a bounded McShane integrable function φ : [0, 1] →ℓ∞(c) which is not properlymeasurable and therefore not Talagrand integrable.proof Enumerate as ⟨Hξ⟩ξ

Then we can choose inductively a disjointfamily ⟨Dξ⟩ξ 0. For each ξ < c let Gξ ⊇Dξ be a relativelyopen subset of [0, 1] of measure at most ǫ; let δ(s) be the distance from s to [0, 1] \ Gξ if s ∈Dξ, 1 ifs ∈[0, 1] \ Sξ

If ⟨([ai, bi], ti)⟩i∈I is any McShane partition of [0, 1] subordinate to δ, then|(Pi∈J(bi −ai)φ(ti))(ξ)| = Pi∈J,ti∈Dξ(bi −ai) ≤µGξ ≤ǫfor any J ⊆I, ξ < c, so that∥Pi∈J(bi −ai)φ(ti)∥≤ǫfor every finite J ⊆I. As ǫ is arbitrary, φ is McShane integrable, with integral 0.To see that φ is not properly measurable, setA = {hφ : h ∈(ℓ∞(c))∗, ∥h∥≤1},and consider, for m, n ≥1, the set Hmn = Z(A, [0, 1], m, n, 0, 1).Suppose, if possible, that there are m, n ≥1 such that (µ)∗m+nHmn < 1.

In this case there is a non-negligible measurable H ⊆([0, 1]m × [0, 1]n) \ Hmn. SetH′ = {u : u ∈[0, 1]n, (µ)n{t : (t, u) ∈H} > 0};then µnH′ > 0, so there is a ξ < c such that Hξ ⊆H′, and a u ∈Dnξ ∩H′.

Now H−1[{u}] = {t : (t, u) ∈H}is non-negligible, and Dξ is finite, so there is a t ∈H−1[{u}] such that no coordinate of t belongs to Dξ. Inthis case, taking hξ(x) = x(ξ) for x ∈ℓ∞(c), and f = hξφ ∈A, we see that f(u(j)) = 1 for each j < n, butthat f(t(i)) = 0 for each i < m; so that (t, u) ∈Hmn and (t, u) /∈H, which is absurd.Thus φ is not properly measurable and therefore not Talagrand integrable.Remark Observe that φ is not measurable, and either for this reason, or because it is not Talagrandintegrable, cannot be Bochner integrable.3B The next example will be of a Talagrand integrable function which is not McShane integrable.

Itrelies on a couple of lemmas.Lemma Let (S, Σ, µ) be a probability space and B ⊆RS a stable set of functions. Suppose that A ⊆RS isa countable set of functions such that A \ U is stable whenever U is an open subset of RS (for the pointwisetopology of RS) including B.

Then A is stable.

8proof Let E ∈Σ be such that µE > 0, and take α < β in R.Let m, n ≥1 be such that( µ∗m+nZ(B, E, m, n, α, β) < µE)m+n. Let G ⊆(Em × En) \ Z(B, E, m, n, α, β) be a measurable set ofnon-zero measure.

For any (t, u) ∈G, we see thatUtu = {f : f ∈RS, ∃i < n, f(t(i)) > α} ∪{f : f ∈RS, ∃j < m, f(u(j)) < β}is an open set including B, so A \ Utu is stable, and there are mtu≥1, ntu≥1 such thatµmtu+ntuZ(A, Utu, E, mtu, ntu, α, β) < (µE)mtu+ntu. Let p, q be so large that µ∗m+nG1 > 0, whereG1 = {(t, u) : (t, u) ∈G, m + mtu ≤p, n + ntu ≤q}.Now observe that if (v, w) ∈Z(A, E, p, q, α, β) and (t, u) = (v↾m, w↾n) ∈G1 then we must have an f ∈A such that f(v(i)) ≤α for every i and f(w(j)) ≥β for every j; but in this case f /∈Utu, so that(v↾p \ m, w↾q \ n) ∈Z(A \ Utu, E, p \ m, q \ n, α, β).

Because p −m ≥mtu, q −n ≥ntu we must have themeasure of Z(A \ Utu, E, p \ m, q \ n, α, β) strictly less than (µE)p−m+q−n. But this means that for every(t, u) ∈G1,µp−m+q−n{(v, w) : (t⌢v, u⌢w) ∈Z(A, E, p, q, α, β)}is less than (µE)p−m+q−n.Because Z(A, E, p, q, α, β) is measurable and µ∗m+nG1 > 0, we see thatµp+qZ(A, E, p, q, α, β) must be less than (µE)p+q.As α, β, E are arbitrary, A is stable.3C Lemma Let H ⊆R be a measurable set such that µ(H ∩[a, b]) > 0 whenever a < b in R, writing µfor Lebesgue measure.

(a) If D is a non-negligible subset of R, then µ∗(H ∩(t + D)) > 0 for almost all t ∈R. (b) If ⟨Dn⟩n∈N is a sequence of non-negligible subsets of R, then there is a sequence ⟨tn⟩n∈N such thattm ∈Dm and tm + tn ∈H whenever m < n ∈N.proof (a) Let E be a Lebesgue measurable set such that D ⊆E and (µ)∗(E \ D) = 0.

The functiont 7→µ∗(H ∩(t + D)) = µ∗((H −t) ∩D) = µ((H −t) ∩E)is continuous, so F = {t : µ(H ∩(t+D)) = 0} is closed. Let θ be a translation-invariant multiplicative liftingof Lebesgue measure ([IT67]).

Then(θ(H) −t) ∩θ(E) = θ(H −t) ∩θ(E) = θ((H −t) ∩E) = ∅for every t ∈F, so θ(H) ∩(θ(E) + F) = ∅. But µ(θ(H)△H) = 0 so µ(θ(H) ∩[a, b]) > 0 whenever a < band θ(E) + F cannot include any interval.

Consequently one of θ(E), F must be negligible ([Ru74], chap.8, exercise 12). But µθ(E) = µE = µ∗LD > 0, so µF = 0, as claimed.

(b) Choose tm, Dnm inductively, as follows. Start with Dn0 = Dn for every n. Given that µ∗Dnm > 0for all n ≥m and that ti + t ∈H whenever i < m, t ∈Sn≥m Dnm, observe that by (a) we haveµ{t : ∃n ≥m, µ(Dnm ∩(H −t)) = 0} = 0.So we can find tm ∈Dmm such that µ∗(Dnm∩(H −tm)) > 0 for every n > m. Set Dn,m+1 = Dnm∩(H −tm)for n > m, and continue.3D Lemma Let H ⊆R be a measurable set such that µ(H ∩[a, b]) > 0 whenever a < b ∈R.

Let C bethe set of subsets C of [0, 1] such that s + t /∈H whenever s, t are distinct members of C. Then the set Bof characteristic functions of members of C is stable.proof Let E be a non-negligible measurable subset of [0, 1] and α < β.If α < 0 or β > 1 thenZ(B, E, 1, 1, α, β) = ∅. If 0 ≤α < β ≤1 thenZ(B, E, 1, 2, α, β) ⊆{(t, (u0, u1)) : u0 = u1 or u0 + u1 /∈H}.But we know from Lemma 3Ca that µ{u : u ∈E, u0+u ∈H} > 0 for almost all u0, so that γ = µ2{(u0, u1) :u0, u1 ∈E, u0 + u1 ∈H} > 0, and now µ3Z(B, E, 1, 2, α, β) ≤µE((µE)2 −γ) < (µE)3.3E Example There is a bounded Talagrand integrable function φ : [0, 1] →ℓ∞(N) which is not McShaneintegrable.proof (a) Let H be an Fσ subset of R such that 0 < µ(H ∩[a, b]) < b −a whenever a < b in R; let ⟨Hn⟩n∈Nbe an increasing sequence of closed sets with union H. Let C, B ⊆{0, 1}[0,1] be defined from H as in Lemma3D, so that B is stable.

For each n ∈N let An be the countable set of functions f : [0, 1] →{0, 1} whichhave total variation at most n, jump only at rational points, and are such that min(f(s), f(t)) = 0 whenevers < t and s + t ∈Hn; set A = Sn∈N An. Then we see that in the compact space {0, 1}[0,1]Tn∈NSm≥n Am ⊆B.

9Accordingly every neighbourhood U of B in R[0,1] must include all but finitely many of the An. On theother hand, each An is stable, being comprised of functions of variation at most n. So Lemma 3B tells usthat A is stable.Enumerate A as ⟨fn⟩n∈N and define φ : [0, 1] →ℓ∞(N) by setting φ(t) = ⟨fn(t)⟩n∈N for each n ∈N.

Then{hφ : h ∈(ℓ∞)∗, ∥h∥≤1} is precisely the balanced closed convex hull of A, which by [Ta84], 11-1-1, isstable. So φ is Talagrand integrable (since of course ∥φ(t)∥≤1 for every t).

(b) But suppose, if possible, that φ is McShane integrable.Let w be its McShane integral, and letδ : [0, 1] →]0, ∞[ be such that ∥w −Pi≤n(bi −ai)φ(ti)∥≤15 for every McShane partition ⟨([ai, bi], ti)⟩i≤nof [0, 1] subordinate to δ. Let k ≥5 be such that D = {t : δ(t) ≥1k} has outer measure at least 45.

Let⟨[ai, bi]⟩i

So we must have25 ≥∥Pi≤n(bi −ai)φ(ti) −Pi≤n(bi −ai)φ(ui)∥= 1k∥Pi

It is easy to see that such a function must be both McShaneintegrable and Talagrand integrable.Because the unit ball of L∞([0, 1]) is w∗-separable, there is an isometric embedding of L∞[0, 1] in ℓ∞(N)(indeed, L∞([0, 1]) is isomorphic to ℓ∞(N) – see [LT77], p. 111), so there is a bounded McShane integrable,Talagrand integrable function from [0, 1] to ℓ∞(N) which is not Bochner integrable.3G Example There is a McShane integrable function φ : [0, 1] →ℓ2(N) which is not Talagrand integrable.proof Let en be the nth unit vector of ℓ2(N) and set φ(t) = 2n(n + 1)−1en for 2−n−1 ≤t < 2−n. ThenR∥φ∥= ∞so φ is not Talagrand integrable, but by [Go90], Theorem 15, it is McShane integrable.amsppt.styFMp91d.texVersion of 20.1.92Acknowledgement Part of the work of this paper was done while the first author was visiting the Uni-versidad Nacional de Educaci´on a Distancia in Madrid.The second author was partially supported byD.G.I.C.Y.T.

grant PB88-0141. We are grateful to L. Drewnowski for useful comments.References

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