LIMITATIONS on the EXTENDIBILITY of the RADON-NIKODYM THEOREM 0. Notation and Basic Definitions Throughout the Paper X and Y

Home , Dirac measure

PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 131, Number 8, Pages 2491–2500 S 0002-9939(03)07046-1 Article electronically published on March 11, 2003

LIMITATIONS ON THE EXTENDIBILITY OF THE RADON-NIKODYM THEOREM

GERD ZEIBIG

(Communicated by N. Tomczak-Jaegermann)

Abstract. Given two locally compact spaces X, Y and a continuous map r : Y X the Banach lattice 0(Y ) is naturally a 0(X)-module. Following → the Bourbaki approach to integrationC we deﬁne generalizedC measures as 0(X)- 1 linear functionals µ : 0(Y ) 0(X). The construction of an L (µ)-spaceC and → the concepts of absoluteC continuityC and density still make sense. However we exhibit a counter-example to the natural generalization of the Radon-Nikodym Theorem in this context.

0. Notation and basic definitions Throughout the paper X and Y will denote locally compact spaces and r : Y X will be a ﬁxed continuous function. → We use standard notation as in [3, 5]. In particular (X) stands for the space of C0 all complex-valued continuous functions on X which vanish at inﬁnity and b(X) is the space of all complex-valued bounded continuous functions on X. The injectiveC and the projective tensor products are respectively denoted ˇ and ˆ .Themost relevant feature of the injective tensor product for the present⊗ work is⊗ that we can identify (X Y )with (X) ˇ (Y )[2]. C0 × C0 ⊗ C0 Recall that a Banach module M over the Banach algebra 0(X) (under the pointwise operations) is a Banach space together with a contractiveC bilinear action- . map 0(X) M M.IfX is compact this action is subject to the usual axioms: α.(β.mC)=(×αβ).m−→ and 1.m = m for all ‘scalars’ α, β in (X)andallelements C0 m of M.IfX is not compact, in which case the algebra 0(X)doesnothavea unit element, the second axiom is replaced by the non-degeneracyC requirement that (ϕj.m)j converges to m,where(ϕj)j is any contractive approximate identity for

0(X). Clearly 0(X)isa 0(X)-module under the pointwise operations. C Given two BanachC modulesC M and N over (X), a bounded linear map T : C0 M N is said to be 0(X)-linear if T (α.m)=α.T (m) for every scalar α 0(X) and→ every m M. TheC Banach space of all (X)-linear maps from M∈toCN is ∈ C0 denoted L 0(X)(M,N). C

Received by the editors March 20, 2002. 2000 Mathematics Subject Classiﬁcation. Primary 46B22; Secondary 46J10, 46E30. Key words and phrases. Banach module, Radon-Nikodym Theorem, Riesz Theorem.

c 2003 American Mathematical Society

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1. Generalized measures µ and their L1(µ)-spaces The starting point for our work is the fact that the continuous function r : Y X → yields a natural (X)-module structure on (Y ), the action being α.f := (α r) f C0 C0 ◦ · for α 0(X)andf 0(Y ). Consider∈ C the special∈ caseC where X is the one-point space . Now the function {∗} r : Y X is necessarily the constant function r , the Banach algebra 0(X) → ≡∗ . C identifies with the field C and the action 0(X) 0(Y ) 0(Y ) ‘is’ the vector C × C −→ C space action of C on 0(Y ). Hence any 0(X)-linear map µ : 0(Y ) 0(X) iden- C C C → C tifies with a usual bounded linear functional on 0(Y ). The Riesz Representation Theorem [4] in turn identifies µ with a finite RadonC measure on Y. This special case motivates

Deﬁnition 1.1. A generalized measure µ on Y is a bounded 0(X)-linear map µ : (Y ) (X). We say that µ is positive if µ(f) 0 for everyC f 0. C0 → C0 ≥ ≥ Example 1.1. If X is the one-point space, the generalized measures on Y simply identify with the ﬁnite Radon measures on Y . For more examples assume that r : Y X is the canonical projection pr : X Z → Z × X for some locally compact space Z. Every Radon measure µsc : 0(Z) C on Z induces a generalized measure µ : (X Z) (X)asthecompositeC → C0 × → C0 id (X) ˇ µsc C0 ⊗ µ : 0(X Z) 0(X) ˇ 0(Z) 0(X) ˇ C 0(X). C × ≡ C ⊗ C −−−−−−−→ C ⊗ ≡ C We have

µ(f)(x)= f(x, .) dµsc ZZ for f 0(X Z)andx X.Inparticularµ is positive if µsc is positive. To∈ getC more× generalized∈ measures on X Z we can start with any bounded × linear map Tµ : 0(Z) 0(X): we obtain a generalized measure µ by taking the composite C → C

ˇ id (X) Tµ multiplication µ : (X Z) (X) ˇ (Z) C0 ⊗ (X) ˇ (X) (X). C0 × ≡ C0 ⊗ C0 −−−−−−−→ C0 ⊗ C0 −−−−−−−−−→ C0 We then have µ(f)(x)=Tµ(f(x, .))(x)forf 0(X Z)andx X. Again µ is ∈ C × ∈ positive if Tµ is positive. It can be shown that every generalized measure µ : (X Z) (X)comes C0 × → C0 from a unique map Tµ : 0(Z) b(X): if X is compact we simply have Tµ(g)= µ( g pr )forg (ZC). If X→ isC merely locally compact, we use a contractive ◦ Z ∈ C0 approximate identity of 0(X)toobtainTµ. We have in fact an isometric identi- ﬁcation of Banach latticesC

L 0(X)( 0(X Z), 0(X)) L( 0(Z), b(X)). C C × C ≡ C C We now come to the definition of the space L1(µ) for some fixed positive generalized measure µ. To begin we observe that we can define a semi-norm on (Y ) C0 by setting f 1 := µ( f ) 0(X),where . 0(X) denotes the usual sup-norm. The subadditivityk k followsk from| | k theC classicalk trianglekC inequality. Indeed, let f,g be in

0(Y ).Sinceµ( f + g ) is a positive element of 0(X) there is some x0 in X such C | | C that µ( f + g ) 0(X) = µ( f + g )(x0)=δx0 µ ( f + g ), where δx0 denotes the k | | kC | | ◦ | | Dirac measure at x0.Asδx0 µ : 0(Y ) C ‘is’ a positive measure in the usual ◦ C →

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sense, the triangle inequality assures us that

1 f + g 1 = δx0 µ ( f + g )= f + g L (δx µ) k k ◦ | | k k 0◦ 1 1 f L (δx µ) + g L (δx µ) = δx0 µ ( f )+δx0 µ ( g ) ≤kk 0◦ k k 0◦ ◦ | | ◦ | | µ( f ) 0(X) + µ( g ) 0(X) ≤k | | kC k | | kC = f 1 + g 1 . k k k k We denote by L1(µ) the completion of the normed space obtained from ( (Y ), C0 . 1) by quotienting out the null-space of the semi-norm . 1. k k 1 k k By the construction of our space L (µ) we have a canonical map i : 0(Y ) 1 C → L (µ). This map is injective precisely when the semi-norm . 1 on 0(Y ) is actually a norm, and this occurs when µ has “full support” in a sensek k whichC will be made precise in a subsequent paper [7]. Example 1.2. a) If X is the one-point space our positive generalized measure µ identifies with a positive (finite) Radon measure on Y .ThespaceL1(µ)inthe sense we just defined now identifies with the usual space L1(µ). b) It can be shown that in the case where Y r X is the identity X id X, 1 −→ −→ the space L (µ) identifies with a certain 0(U). Specifically U is the complement 1 C in X of the subset m− ( 0 ), where m b(X) denotes the function such that µ(f)=f m for every f { } (X). ∈ C · ∈ C0 Proposition 1.1. The space L1(µ) is naturally a (non-degenerate) Banach (X)- C0 module and a Banach lattice, in such a way that the canonical map i : 0(Y ) 1 C → L (µ) is a morphism of 0(X)-modules and of Banach lattices. Furthermore this mapC satisfies the universal property that for every Banach

0(X)-module M and every contractive morphism T :( 0(Y ), . 1) M of semi- k k → normedC (X)-modules there exists a unique contractiveC morphism H : L1(µ) M 0 → of (X)C-modules such that the following diagram commutes: C0 T 0(Y ) /M C < i H L1(µ)

Proof. L1(µ)isa (X)-module. • C0 Recall that the 0(X)-module structure of the semi-normed space ( 0(Y ), . 1) is given by α.f :=C (α r) f for α (X)andf (Y ). As C k k ◦ · ∈ C0 ∈ C0 α.f 1= µ( α r f ) 0(X)= µ( α r f ) 0(X)= µ( α . f ) 0(X) k k k | ◦ · | kC k | |◦ ·| | kC k | | | | kC = α µ( f ) 0(X) α 0(X) µ( f ) 0(X) = α 0(X) f 1 , k| | | | kC ≤k kC k | | kC k kC k k 1 the nullspace K := . − ( 0 )ofoursemi-norm . 1 is algebraically an ideal k k1 { } k k in (Y ), so the module action of (X) on (Y ) quotients to a (X)-action on C0 C0 C0 C0 0(Y )/K. Furthermore we still have α.f α 0(X) f in 0(Y )/K. It follows C 1 k k≤k kC k k C that the completion L (µ) of this quotient space is still a 0(X)-module (with 1 C contractive action). The map i : 0(Y ) L (µ) is obviously a morphism of 0(X)- modules and satisﬁes the assertedC property.→ C 1 We now show that L (µ) is in fact a non-degenerate 0(X)-module in the sense 1 C that (ϕj .f)j converges to f for every f L (µ), where (ϕj)j is any contractive ap- ∈ proximate identity of 0(X). As the bounded canonical map i :( 0(Y ), . 0(Y )) C C k kC →

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1 (L (µ), . 1) has dense range it suﬃces to consider elements f = i(g)forsome k k compactly supported g in (Y ). By taking an appropriate contractive approxi- C0 mate identity we can assume that for all j greater than or equal to some j0 the function ϕj is identically equal to 1 on r(supp(g)). We then have ϕj.i(g)=i(g), so ϕj.i(g) i(g). −→j L1(µ) is a Banach lattice. • 1 The nullspace K := . 1− ( 0 )ofthesemi-norm . 1 is a lattice ideal in ( (Y ), ), since if 0 k fk g{and} g belongs to K,thenk k f also belongs to K. C0 ≤ ≤ ≤ Therefore the quotient 0(Y )/K is naturally a normed lattice, and so is its completion L1(µ). The mapCi is obviously a morphism of lattices.

1 Remark. For future reference (Deﬁnition 2.2) we observe that L (µ)isevena 0(Y )- 1 C module in the natural way: given α 0(Y )andg L (µ)oftheformi(f)for some f (Y ), we have α.g = i(α ∈fC). Using the∈ projective tensor product to ∈ 0 · linearize theC action, we get a contraction (Y ) ˆ L1(µ) L1(µ)whichis (X)- 0 ⊗ → 0 linear when we consider on (Y ) ˆ L1(µ)theC (X)-module structure it inheritsC 0 ⊗ 0 from (Y ) . C C C0 We can now state the following

Proposition 1.2. The positive generalized measure µ : 0(Y ) 0(X) induces 1 C → C a unique positive, contractive, 0(X)-linear map µ˜ : L (µ) 0(X) such that the diagram C → C µ 0(Y ) /0(X) C vC; vv i vv vv µ˜ vv L1(µ) commutes. 1 1 Furthermore we have f L (µ)= µ˜( f ) 0(X) for every f in L (µ),where f k k k | | kC | | denotes the absolute value of f for the lattice structure of L1(µ). We stress thatµ ˜ is contractive, whatever the norm of µ. This is possible because µ already intervenes when we compute the norm of an element of L1(µ). In practice we might use the letter µ to denote both the positive generalized measure µ and the mapµ ˜ of the proposition.

Proof. To prove the ﬁrst part of the proposition it is enough to show that µ • is contractive when we consider it as a map ( 0(Y ), . 1) ( 0(X), . 0(X)) C k k → C k kC between semi-normed spaces. Let f 0(Y ). We need to show that µ(f) 0(X) ∈ C k kC ≤ µ( f ) 0(X).Wechoosex0 in X and c C of modulus one such that µ(c.f)(x0)= k | | kC ∈ µ(f) 0(X). k kC Since δx0 µ : 0(Y ) C is a positive Radon measure we have ◦ C →

µ(f) 0(X)= µ(c.f)(x0)=δx0 µ(c.f)= c.f d(δx0 µ) c.f d(δx0 µ) k kC ◦ ◦ ≤ | | ◦ ZY ZY = δx0 (µ( c.f )) µ( c.f ) 0(X)= µ( f ) 0(X) . | | ≤k | | kC k | | kC 1 1 Let us prove that f L (µ)= µ˜( f ) 0(X) for every f L (µ). As the • k k 1 k | | kC ∈ isometry i :( 0(Y ), . 1) L (µ) has dense range and since f f is continuous C k k → 7→ | |

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it is suﬃcient to verify the assertion for f = i(g),g (Y ). But in this case ∈ C0 µ˜( f ) 0(X) = µ˜( i(g) ) 0(X) = µ˜(i( g )) 0(X) k | | kC k | | kC k | | kC 1 1 = µ( g ) 0(X) = g 1= i(g) L (µ) = f L (µ) . k | | kC k k k k k k We should remark that if µ : 0(Y ) 0(X) is merely a positive bounded map the construction of the BanachC lattice→L1C(µ) still makes sense and we still get a 1 contractive positive mapµ ˜ : L (µ) 0(X) as in the proposition. We have already seen what our space→ C L1(µ) looks like in two special cases. The following proposition shows that these are actually extreme cases, with X and Z respectively being the one-point space.

Proposition 1.3. Suppose r : Y X is the canonical projection X Z X with Z being some locally compact→ space. Additionally assume that the× positive generalized measure µ : (X Z) (X) arises from some positive Radon C0 × → C0 measure µsc : 0(Z) C. Then we have the natural isometric 0(X)-module and C →1 1 C lattice identiﬁcation L (µ) 0(X, L (Z, µsc)). ≡ C 1 1 By L (Z, µsc) we mean the usual space L coming from the measure µsc on Z. The way µ arises from µsc was illustrated when we saw examples of generalized

measures. Recall in particular that µ(f)(x)= Z f(x, .) dµsc. 1 Proof. Let j : 0(Z) L (Z, µsc) be the canonicalR map. We deﬁne a morphism of C → 1 0(X)-modules T : 0(X Z) 0(X, L (Z, µsc)) as the composite C C × → C 0(X, j) 1 T : 0(X Z) 0(X, 0(Z)) C 0(X, L (Z, µsc)), C × ≡ C C −−−−−→ C where 0(X, j) denotes the map g j g. C 7→ ◦ 1 This morphism T is isometric as a map ( 0(X Z), . 1) 0(X, L (Z, µsc)). C × k k → C 1 Indeed, ﬁx f 0(X Z)andx X:wehaveT (f)(x)=j(f(x, .)) in L (Z, µsc). Hence ∈ C × ∈

T (f)(x) 1 = T (f)(x) dµsc = j(f(x, .)) dµsc k kL (Z,µsc) | | | | ZZ ZZ = f (x, .) dµsc = µ( f )(x). | | | | ZZ Therefore T (f) = µ( f ) 0(X) = f 1 and T is isometric. k k k | | kC k k As T is a contractive morphism of 0(X)-modules there is a unique contractive 1 C 1 morphism of 0(X)-modules H : L (µ) 0(X, L (Z, µsc)) such that the following diagram commutes:C → C

T 1 0(X Z) /0(X, L (Z, µsc)) C × C 6 i H L1(µ) Since T is an isometry and the isometry i has dense range it follows that H is also an isometry. H is clearly a morphism of lattices. It remains to show that H is onto, and for this it is suﬃcient to show that T has 1 dense range. But this follows from the facts that j : 0(Z) L (Z, µsc) has dense range and that the map T can also be described as C →

id (X) ˇ j C0 ⊗ 1 1 T : 0(X Z) 0(X) ˇ 0(Z) 0(X) ˇL (Z, µsc) 0(X, L (Z, µsc)). C × ≡ C ⊗ C −−−−−−→ C ⊗ ≡ C

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2. Absolute continuity and densities We continue to work with some ﬁxed continuous map r : Y X between locally compact spaces. Also µ and ν denote two positive generalized→ measures

0(Y ) 0(X). C As we→ cannotC use Borel sets to define what it would mean for ν to be absolutely continuous with respect to µ, we adopt the following Definition 2.1. We say that ν is absolutely continuous with respect to µ if for every >0thereisanη>0 such that for every continuous function f in (Y ) C0 with 0 f 1, µ(f) 0(X)<ηimplies ν(f) 0(X)<. ≤ ≤ k kC k kC This definition is justified by Proposition 2.1. Suppose that X is the one-point space, so that the positive generalized measures µ, ν : 0(Y ) 0(X) C identify with usual positive Radon measures on Y . The generalizedC → measureC ν≡is absolutely continuous with respect to the generalized measure µ in the sense of the above definition if and only if the Radon measure ν is absolutely continuous with respect to the Radon measure µ in the usual sense. Proof. Absolute continuity in the traditional sense implies absolute continuity in the above• sense. We can assume that ν(Y ) > 0. Let >0 be given. There exists aη> ˜ 0such that for every Borel set A Y, µ(A) η˜ implies ν(A) </2. We show that η := η/˜ 2ν(Y ) works in the⊆ above definition.≤ Let f (Y ) , 0 f 1, with µ(f) <η.Wethenhave ∈ C0 ≤ ≤ 2ν(Y ) 2ν(Y ) µ([f ]) = 1 dµ fdµ η =˜η. ≥ 2ν(Y ) [f /2ν(Y )] ≤ [f /2ν(Y )] ≤ Z ≥ Z ≥ Hence ν([f ]) < ,andso ≥ 2ν(Y ) 2

ν(f)= fdν+ fdν [1 f /2ν(Y )] [f</2ν(Y )] Z ≥ ≥ Z ν([f ]) + dν < + ν(Y )=. ≤ ≥ 2ν(Y ) 2ν(Y ) 2 2ν(Y ) ZY Absolute continuity in the above sense implies absolute continuity in the traditional• sense. Fix >0. Since ν and µ are inner regular it is enough to show that there exists an η>0 such that µ(K) <ηimplies ν(K) <for all compact subsets K of Y .

There isη> ˜ 0 such that µ(f) < η˜ implies ν(f) <for every f in 0(Y ), 0 f 1. Then η :=η/ ˜ 2works.ToseethisletK Y be compact,C with µ(≤K) <η≤.Asµ is outer regular there exists an open subset⊆ U of Y containing K such that µ(U) < η˜. Next, there is a continuous function f on Y ,0 f 1, such that f 1onK and f 0onX U. ≤ ≤ ≡ ≡ \ Since f 1U it follows that µ(f) µ(U) < η˜, and hence ν(f) <.But1K f, so ν(K) ≤ν(f) <. ≤ ≤ ≤ Remark. The deﬁnition of absolute continuity for generalized measures has been chosen in a way such that the proposition holds. One might be inclined to deﬁne ν as absolutely continuous with respect to µ if and only if for every continuous function

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f on Y ,with0 f 1, µ(f) = 0 implies ν(f) = 0. To see that this is not a good ≤ ≤ deﬁnition take for X the one-point space and for Y the real line. Let (qn)n be an enumeration of the rational numbers. Then the Lebesgue measure and the measure 1 n 2n δqn would be mutually absolutely continuous (the only continuous positive function they annihilate is the zero function), although neither one is absolutely Pcontinuous with respect to the other in the usual sense. We now generalize the concept of density in our setting. Recall from the Remark 1 before Proposition 1.2 that L (µ)isa 0(Y )-module and that the contractive action- 1 1 C map 0(Y ) ˆ L (µ) L (µ)is 0(X)-linear. Also recall the 0(X)-linear contraction µ˜ : L1C(µ) ⊗ (X)→ of PropositionC 1.2. Any d L1(µ) yieldsC a composite morphism → 0 ∈ of (X)-modulesC C0 f f d action µ˜ (Y ) 7→ ⊗ (Y ) ˆL1(µ) L1(µ) (X) C0 −−−−−→ C0 ⊗ −−−−→ −→ C0 which is the subject of the following Deﬁnition 2.2. Given d in L1(µ), the (not necessarily positive) generalized mea-

sure d.µ : 0(Y ) 0(X)withdensity d with respect to µ is defined as the above compositeC map.→ HenceC d.µ is given by d.µ(f)=˜µ( f d ). · Clearly d.µ has norm at most d 1 . The following proposition asserts that k kL (µ) equality holds. In analogy with the classical case let us denote the “dual” of (Y ), C0 the space L 0(X)( 0(X Z), 0(X) ) of all 0(X)-linear maps 0(Y ) 0(X), by C C × C C C → C 0(Y )∗. Since the algebra 0(X) is commutative, 0(Y )∗ is still a 0(X)-module. AsC in the classical situationC we have C C Proposition 2.2. The map d d.µ is an isometric embedding of Banach lattices 7→ and 0(X)-modules C 1 L (µ) , (Y )∗. → C0 Proof. The map is isometric. As we• noted before, it is contractive. We can assume that µ =0.Letd in L1(µ) 6 and fix >0. We need to show that d.µ d L1(µ) . Now, since the isometry 1 k k≥k k − ˜ i :( 0(Y ), . 1) L (µ) has dense range, there exists some d in 0(Y ) such that C k k → C d i(d˜) 1 </3. For simplicity we will write d˜ instead of i(d˜). Next, as the k − kL (µ) subset [ d˜ /6 µ ]ofY is compact, there exists some ϕ in (Y ) ,0 ϕ 1, | |≥ k k 0 ≤| |≤ such that (ϕ d˜)(y)= d˜(y) for every y for which d (y) /6Cµ . Therefore we · ˜ ˜ | | | | ≥ k k have that ϕ d d 0(Y ) /3 µ . It follows that k · −| |kC ≤ k k d.µ d.µ(ϕ) = µ˜(ϕ d) k k≥k k k · k µ˜( d˜) µ(ϕ d˜ d˜) µ˜(ϕ (d d˜)) ≥k| | k−k · −| | k−k · − k ˜ ˜ ˜ ˜ d L1(µ) µ ϕ d d sup µ˜ ϕ sup d d L1(µ) ≥kk −k k·k · −| |k −k k·k k ·k − k d˜ 1 µ 1 1 ( d 1 ) ≥kkL (µ) −k k·3 µ − · · 3 ≥ k kL (µ) −3 − 3 − 3 k k = d 1 . k kL (µ) − The map is an embedding of lattices. It• is obvious from the definition of the lattice structure of L1(µ)thatapositived in L1(µ) yields a positive generalized measure d.µ.Nowfixd L1(µ) and suppose ∈

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that d.µ is positive. We need to show that d is positive in L1(µ). For this we show that d = d in L1(µ). | | 1 When the product space x X L (δx µ) is endowed with the sup-norm the 1 ∈ 1 ◦ natural embedding L (µ) , x X L (δx µ) is injective. Hence it is enough to →1 Q ∈ ◦ show that d equals d in L (δx µ)forﬁxedx X. But this follows from the fact | | Q◦ ∈ that the measure δx (d.µ) is positive on 0(Y ):since fd d(δx µ) 0 for every ◦ C Y ◦ ≥ positive f in 0(Y ), d is positive δx µ-a.e. C ◦ R As we might expect we still have the following Proposition 2.3. If a positive generalized measure ν = d.µ has a density d L1(µ) with respect to µ,thenν is absolutely continuous with respect to µ. ∈ Proof. Let >0 be ﬁxed. Since ν is positive we know from Proposition 2.2 that d ˜ ˜ is also positive. Hence there is a positive d in 0(Y ) such that d d L1(µ)</2. ˜ C k − k Take η>0 such that d 0(Y ) η</2. Then, for every f in 0(Y ) ,0 f 1, with µ(f) <ηwe havek kC · C ≤ ≤ k k ν(f) = µ˜(f d) µ˜( f (d d˜)) + µ(f d˜) k k k · k≤k · − k k · k ˜ ˜ + d 0(Y ) µ(f) 0(X) + d η ≤ 2 k kC ·k kC ≤ 2 k k· < + = . 2 2 3. Radon-Nikodym fails We now arrive at the main result of this paper. In the classical situation, which corresponds to the case where X is the one-point space, the Radon-Nikodym The- orem [5] asserts the converse to Proposition 2.3: if ν is absolutely continuous with respect to µ,thenithasadensityinL1(µ). The following theorem shows that this converse is no longer true in our setting. Theorem 3.1. There are compact spaces X, Z and positive generalized measures

µ, ν : 0(X Z) 0(X) such that ν is absolutely continuous with respect to µ, but hasC no density× →d C L1(µ). ∈ In order to give such a counter-example we need the following 1 Lemma 3.1. There exists a family (gn)n of positive functions in L (0, 1), ∈N∪{∞} taking values 1, such that (gn)n N converges to g in the weak-star topology, but not in norm.≤ By the weak-star topology∈ we mean the∞ subspace topology on L1(0, 1) induced by the weak star-topology on ([0, 1]) . C0 ∗ 1 Proof of Lemma 3.1. The functions gn(t):= (1 + cos(2nπt)) for n N and 2 ∈ g (t):= 1 satisfy the conditions: they are clearly positive and take values 1. ∞ 2 ≤ Since t cos(2πnt) is the real part of t ei2πnt it follows from the Riemann- 7→ 7→ Lebesgue lemma that the sequence ( t cos(2πnt))n converges to 0 in the weak- 7→ star topology. Hence (gn)n converges weak-star to g . But it is easy to compute ∈N ∞ that gn g L1 =1/π for all n N, so that there is no convergence in norm. k − ∞ k ∈ Proof of Theorem 3.1. To prove the theorem we consider the following: We take for X the one-point-compactiﬁcation X := N of the integers. • ∪{∞} Hence the space 0(X) can be identiﬁed with the space c of all convergent sequences. We choose ZC to be the closed interval [0, 1]. •

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We now deﬁne ν : (X Z) (X) by setting • C0 × → C0

ν(f)(n):= f(n, .)gn dλ Z[0,1]

for f 0(X Z)andn X,whereλ denotes Lebesgue measure. This map ν is well-deﬁned:∈ C × as is the∈ only accumulation point of X,toverifythatν(f)is ∞ continuous on X it is enough to show that (ν(f)(n))n converges to ν(f)( ). ∈N ∞ But, as f belongs to 0(X Z) 0(X, 0(Z)), for given >0thereisann1 N C × ≡ C C ∈ such that f(n, .) f( ,.) 0(Z)</2 for all n n1. k − ∞ kC ≥ Next, as (gn)n converges weak-star to g ,thereisn2 N such that ∞ ∈

f( ,.)(gn g ) dλ </2 | ∞ − ∞ | Z[0,1] for all n n2. Hence for all n N :=max n1,n2 ,wehave ≥ ≥ { }

ν(f)(n) ν(f)( ) = f(n, .)gn dλ f( ,.)g dλ | − ∞ | | − ∞ ∞ | Z[0,1] Z[0,1] [f(n, .) f( ,.)]gn dλ + f( ,.)[gn g ] dλ + = . ≤| − ∞ | | ∞ − ∞ |≤2 2 Z[0,1] Z[0,1]

It is now obvious that ν is bounded (in fact contractive), positive and 0(X)- linear, so ν is a positive generalized measure. C For the positive generalized measure µ we simply take the one which comes • from µsc := λ, the Lebesgue measure on Z := [0, 1]. Hence we have µ(f)(n)= f(n, .) dλ for f (X Z)andn X. [0,1] ∈ 0 × ∈ Having chosen the spaces X, Z and the positiveC generalized measures ν and µ R we now show that ν is absolutely continuous with respect to µ, but does not have adensity. ν is absolutely continuous with respect to µ. • Given >0, we can take η := to ensure that for every f 0(X Z), 0 f 1, µ(f) <ηimplies ν(f) <. Indeed, evaluating ν(f)atany∈ C n ×X,wehave≤ ≤ k k k k ∈

0 ν(f)(n)= f(n, .)gn dλ f(n, .) dλ = µ(f)(n) µ(f) <η= . ≤ ≤ ≤k k Z[0,1] Z[0,1] ν does not have a density d L1(µ) with respect to µ. • ∈ 1 1 By Proposition 1.3 we know that L (X Z, µ)identiﬁeswith 0(X, L (0, 1)). × C 1 Arguing by contradiction, we assume that ν does have a density d 0(X, L (0, 1)) with respect to µ. ∈ C

For every n X and every g 0(Z) we then have, using the canonical projection pr : X Z∈ Z, ∈ C Z × g d(n) dλ = g pr (n, .) d(n) dλ =˜µ (g pr d)(n) · ◦ Z · ◦ Z · Z[0,1] Z[0,1]

= ν(g pr )(n)= g pr (n, .) gn dλ = g gn dλ. ◦ Z ◦ Z · · Z[0,1] Z[0,1] 1 Hence d(n)=gn λ-a.e., for all n X.Asd : X L (0, 1) is continuous, the ∈ → sequence (d(n))n converges in norm to d( ). But this contradicts the fact that ∞ (gn)n does not converge in norm. ∈N

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Acknowledgment The author thanks his advisor, Andrew M. Tonge, for providing encouragement as well as numerous helpful suggestions and improvements. References

[1] David P. Blecher, Paul S. Muhly, Vern I. Paulsen, Categories of Operator Modules (Morita Equivalence and Projective Modules), Memoirs of the American Mathematical Society 143 (2000). MR 2000j:46132 [2] Johann Cigler, Viktor Losert, Peter Michor, Banach Modules and Functors on Categories of Banach Spaces, Lecture Notes in Pure and Applied Mathematics 46 (1979). MR 80j:46112 [3] Joseph Diestel, John J. Uhl, Vector Measures, American Mathematical Society, Mathematical Surveys 15 (1977). MR 56:2216 [4] Gerald B. Folland, Real Analysis, John Wiley & Sons, Inc. (1999). MR 2000c:00001 [5] Paul Malliavin, Integration and Probability, Graduate Texts in Mathematics 157,Springer Verlag (1995). MR 97f:28001a [6] Helmut H. Schaefer, Banach Lattices and Positive Operators, Springer Verlag (1974). MR 54:11023 [7] Gerd Zeibig, Generalized Lp(µ)-spaces, to appear.

Department of Mathematical Sciences, Kent State University, Kent, Ohio 44240 E-mail address: [email protected]

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