Notes on Tensor Product Measures 7 1 YAMAGAMI Shigeru 0 2 Nagoya University Graduate School of Mathematics n a J January 13, 2017 2 1 ] Introduction A F ConsideraspectraldecompositionU(t)= ReitτE(dτ)ofaone-parameteruni- h. tarygroupU(t)onaHilbertspaceH,whereE()isaprojection-valuedmeasure t on R. In quantum mechanics, the dynamRical b·ehavior of a physical system is a m described by the associated automorphic action of R on the algebra L(H) of bounded linear operators. The relatedtransitionprobabilities are associatedto [ (ξ U(t)TU(t)∗η) = (U(t)∗ξ TU(t)∗η) (ξ,η H, T L(H)), which takes the | | ∈ ∈ 1 form v ( e−itτE(dτ)ξ T e−itτ′E(dτ′)η) 5 R | R 1 Z Z 2 in terms of the spectral measure. At first glance, it seems quite natural to 3 rewrite this to the product measure form like 0 1. eit(τ−τ′)(E(dτ)ξ TE(dτ′)η), R R | 0 Z × 7 whichmeansthatweexpectacomplex-valuedmeasure(E(dτ)ξ TE(dτ′)η)tobe :1 well-defined on R2. This anticipation is reasonably generalized|to the following v question: LetT L(H)andξ(),η()beH-valuedmeasuresonaσ-algebraBin Xi a setS. Itis imm∈ediate to chec·k tha·tthe mapB B A B (ξ(A)Tη(B)) r is extended to a finitely additive function µ on×the∋Boo×lean7→algebra| B B a generated by B B. Is it then possible to extend µ to a complex measure⊗on theσ-algebragen×eratedbyB B? WhenT isafiniterankoperator,µcertainly × admitssuchanextensionasalinearcombinationofproductmeasuresandwith a litte more effort we can show that the question is answered affirmatively for a trace class operator. The general answer, however, turns out to be negative: A bounded linear operator T has the measure extension property if and only if T is in the Hilbert-Schmidt class ([Swartz1976,Theorem 8]1 ). Our main purpose here is to collect relevant results together and combine them to give a self-contained proof of it. 1Wewouldliketopointout,however,thattheproofthereisbasedonatheoreminanother paper,whichseemsdifficulttobeidentified. 1 Notation: ForaBanachspaceV,itsunitballisdenotedbyV anditsdualspace 1 by V∗. Given a set T, ℓ∞(T) denotes the Banach space of bounded complex- valued functions on T with the sup-norm, which is the dual Banach space of ℓ1(T) of summable functions. We then have a canonical isometric embedding V ℓ∞(V∗) for a Banach space V as a restriction of the canonical pairing V →V∗ C1: For v V, v ℓ∞(V∗) is defined by v(v∗)= v,v∗ (v∗ V∗). × → ∈ ∈ 1 h i ∈ 1 More generally, if T V∗ satisfies v = sup v,v∗ ;v∗ T , then the ⊂ 1 k k {|h i| ∈ } restrictionmapℓ∞(V1∗)→bℓ∞(T)is isometric onV b={v;v ∈V} andwe getan embedding V ℓ∞(T). → The semi-variation of a finitely additive measubre φbis denoted by φ, while | | φ is set aside to designate the total semi-variation of φ. k k 1 Vector Valued Measures WeshallmainlydealwithBanachspacesasvectorspacesandnominate[Diestel- Uhl] as a basic reference. See also [Ricker, Chap.1] for a friendly survey on the subject. In a (Hausdorff) topological vector space V, a family of vectors v is i i∈I { } said to be summable if we can find a vector v V fulfilling the following ∈ condition: Given any neighbourhood N of v, we can find a finite subset F I ⊂ sothat v N foranyfinite subsetF′ I F. Thevectorv isunique j∈F∪F′ j ∈ ⊂ \ if it exists and denoted as v = v . P i∈I i When V is a Fréchet space, the condition is equivalent to the following: P Given any neighborhood N of 0, we can find a finite subset F of I so that v N for any finite subset F′ I F. j∈F′ j ∈ ⊂ \ When I is countable, any counting labeling i ;n 1 gives n P { ≥ } n v = lim v . n→∞ ik k=1 X Conversely, in a Banach space V, if any counting labeling satisfies the above convergence relation, then v is summable and v = v . In fact, if not, { i} i∈I i P ǫ>0, F ⋐I, F′ ⋐I F, v ǫ j ∃ ∀ ∃ \ ≥ (cid:13)jX∈F′ (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) andwe canfind apartition F ofI byfinite subsetssatisfying v n k j∈Fn jk≥ ǫ. Let i be a counting labeling adapted to the increasing sequence F k 1 { } F P ⊂ F F . Then v cannot be a Cauchy sequence by looking at 1 ∪ 2 ⊂ ··· { ik}k≥1 k = F + + F (n=1,2, ). 1 n | | ··· | | ··· WhenV isfinite-dimensionalwith anycompatiblenorm,thesummability k·k of v is equivalent to v < , the so-called absolute convergence. { i}i∈I i∈Ik ik ∞ In fact, for a basis v∗, ,v∗ , the summability implies absolute convergence of v∗(v ) for 1{ 1 j···Pn,n}which is equivalent to n v∗(v ) < . i∈I j i ≤ ≤ j=1 i∈I| j i | ∞ Note that n v∗(v) (v V) defines a norm on V. P j=1| j | ∈ P P P 2 Example 1.1. Let δ be an ONB in a separable Hilbert space H. Then, n n≥1 for ξ H, (δ ξ)δ{ } is summable and ξ = (δ ξ)δ , whereas its ∈ { n| n}n≥1 n≥1 n| n absoluteconvergenceisequivalenttothestrongercondition (δ ξ) < . P n≥1| n| | ∞ LetV beaBanachspaceandBbeaBooleanalgebrainaPsetS. AV-valued semi-measure is an additive map φ : B V. We say that φ is countably ∞ → additive if A = A is a countable partition in B, then φ(A) = φ(A ). n n n=1 n≥1 G X Bythe correspondancebetweenB = A =A ( A )andA = B B , countable additivity ismequivanl≥emnt tno the c\ond1it≤ionn<:mIf Bn innB, n n+1 n \ F S ↓ ∅ then lim φ(B )=0. n→∞ n A semi-measure φ:B V is called a measure if B is a σ-algebra and φ is → countably additive. Clearly countable additivity implies lim φ(A ) = 0. We say that a n→∞ n semi-measureφissqueezing2 iflim φ(A ) =0foranydisjointsequence n→∞ n A in B ( A B being not assumked). Rkemark that a squeezing semi- n n≥1 n n m{ eas}ure φ is co∪ntinuou∈s, i.e., A in B implies lim φ(A ) = 0, and, if B is n n n ↓ ∅ a σ-algebra, a continuous semi-measure is squeezing. This squeezing property together with finite additivity of φ in turn assuresthe summability of φ(A ) . n { } In fact, non-summability n ǫ>0, N, n m N, φ(A ) ǫ k ∃ ∀ ∃ ≥ ≥ ≥ (cid:13)kX=m (cid:13) (cid:13) (cid:13) allowsustofindasubsequence1=l 0,thereexistsanN 1satisfying ≥ v ǫ for any finite subset F of N+1,N+2, . Converselygiven k j∈F jk≤ { ···} a sequence v satisfying the Cauchy condition, v is summable for any n n n∈A suPbset A {N a}nd a countably additive map φ:B{ }V is defined by ⊂ → φ(A)= v . n n∈A X Example 1.3. Let B be the Boolean algebra generated by finite subsets of N: A B if and only if either A or N A is finite. A semi-measure φ : B Z is ∈ \ → then defined by A if A < , φ(A)= | | | | ∞ ( N A otherwise. −| \ | 2Acommonterminologyforthisisstrongadditivity,whichis,however,aboutsummability ratherthanadditivity. 3 Example 1.4. Let (S,B,µ) be a probability space. Then B A 1 A ∋ 7→ ∈ Lp(S,µ) defines a measure for 1 p< and a semi-measure for p= . ≤ ∞ ∞ LetT :V W beaboundedlinearoperatorbetweenBanachspaces. Given a semi-measu→re (resp. measure) φ : B V, the composite map Tφ : B W → → is a semi-measure (resp. measure). As a special case of this, we have a semi- measure (resp. measure) φ : B ℓ∞(V∗) as a composition of φ with the → 1 canonical embedding V ℓ∞(V∗). → 1 b Definition 1.5. Given a semi-measure λ : B C, the variation of λ is a function λ :B [0, ] defined by → | | → ∞ n λ(A)=sup λ(A ); A is a finite partition of A in B . j j | | { | | { } } j=1 X The value λ(S) is called the total variation of λ and denoted by λ . A semi- | | k k measure φ is said to be of bounded variation when λ < . k k ∞ The following are standard facts on complex (semi-)measures. Proposition 1.6. (i) Thevariation λ ofacomplexsemi-measureλisadditiveandsatisfiesthe | | inequality sup λ(A);A B,A B λ(B) πsup λ(A);A B,A B for B B. {| | ⊂ ∈ }≤| | ≤ {| | ⊂ ∈ } ∈ (ii) The variation of a complex measure λ is countably additive and satisfies ∞ λ(A)=sup λ(A ); A is a countable partition of A in B . j j | | { | | { } } j=1 X (iii) Anycomplexmeasuredefinedonaσ-algebraBhasafinitetotalvariation and the vector space L1(B) of all complex measures on B is a Banach space with the norm of total variation. Lemma 1.7 (Half Average Inequality). For each positive d N, there exists ∈ C > 0 (C = 1, C = 1/π, C = 1/4 and so on) with the following property: d 1 2 3 Given a finite family v Rd of euclidean vectors,we can find a finite subset j J 1, ,n so tha{t ∈n }v v /C . ⊂{ ··· } j=1| j|≤| j∈J j| d Proof. WemaysupposePthatv =0. FPoraunitvectore,set(v ,e) =(v ,e) 0, j j + j 6 ∨ which is a continuous function of e. In view of the inequality n v (v ,e)= (v ,e) , j j j + ≥ (cid:12)(vjX,e)>0 (cid:12) (vjX,e)>0 Xj=1 (cid:12) (cid:12) (cid:12) (cid:12) 4 let e be a unit vector which maximizes the function n (v ,e) of e and set 0 j=1 j J = j;(v ,e )>0 . Then v n (v ,e ) n (v ,e) for any { j 0 } | j∈J j|≥ j=1 j 0 +P≥ j=1 j + e and have P P P n n v (v ,e) de=C v with (v,e) de=C v for any v. j j + d j + d ≥ | | | | (cid:12)Xj∈J (cid:12) Xj=1Z|e|=1 Xj=1 Z|e|=1 (cid:12) (cid:12) (cid:12) (cid:12) Definition 1.8. Let φ be a V-valued semi-measure on a Boolean algebra B with V a Banach space. The semi-variation (variation)3 of φ is a function φ :B [0, ] ( φ :B [0, ]) defined by | | → ∞ ||| ||| → ∞ φ(A)=sup v∗,φ (A); v∗ 1,v∗ V∗ , | | {|h i| k k≤ ∈ } n φ (A)=sup φ(A ) ; A is a finite partition of A with A B . j j j ||| ||| { k k { } ∈ } j=1 X Here v∗,φ denotes a complex semi-measure v∗(φ(A)) (A B). h i ∈ A semi-measure is said to be bounded (resp. strongly bounded) if φ | | (resp. φ ) is bounded. We say that φ is squeezing if lim φ(A ) = 0 n→∞ n for any|||di|||sjoint sequence A in B|. | | | n n≥1 { } Proposition 1.9. (i) The variation of a V-valued measure is a positive measure. (ii) A V-valued strongly bounded semi-measure φ defined on a σ-algebra is countably additive if φ is countably additive. ||| ||| (iii) The semi-variationof a V-valued semi-measure is monotone,subadditive; φ(A) φ(A B) φ(A) + φ(B) for A,B B, and satisfies the | | ≤ | | ∪ ≤ | | | | ∈ inequality sup φ(A) ;A B,A B φ(B) πsup φ(A) ;A B,A B , B B. {k k ⊂ ∈ }≤| | ≤ {k k ⊂ ∈ } ∈ Consequently the range of a semi-measure φ is a bounded subset of V if and only if φ is bounded, i.e., φ(S)< . | | ∞ (iv) A semi-measure is bounded if it is squeezing. In particular, measures are bounded. (v) A semi-measure φ is squeezing if and only if so is φ. | | Proof. (i) (iii) are consequences of definitions by standard arguments. (iv): A∼ssume that φ(B) = for some B B. Then, given any r > 0, we can find A B in B s|uc|h that∞φ(A) r and∈ φ(B A) r. In fact, if we ⊂ k k ≥ k \ k ≥ 3Warning: In literatures, semi-variation is denoted by k k, whereas | | is used to indicate variation. 5 chooseAsothat φ(A) r+ φ(B) , φ(A) = φ(B) φ(B A) φ(B) + k k≥ k k k k k − \ k ≤k k φ(B A) . Byasqueezingargument,weobtainadecreasingsequence A n n≥1 ikn B s\atiskfying φ(A ) = and φ(A ) 1+ φ(A ) for n {1. T}hus, n n+1 n | | ∞ k k ≥ k k ≥ the squeezing property is violated for the disjoint sequence A A . n n+1 n≥1 { \ } (v): This is a consequence of (iii). The if part is trivial, whereas the only if partis checkedas follows: If φ is notsqueezing,there existadisjoint sequence | | B and δ > 0 such that φ(B ) δ for n 1. Then, thanks to the π- n n i{neq}uality,we canfind A |B| in B≥so that φ(≥B ) π φ(A ) +1/n,which n n n n ⊂ | | ≤ k k denies lim φ(A ) =0. n→∞ n k k Lemma 1.10. φ(A)=sup α φ(A ) ; A is a finite partition of A with A B and α 1 with α C . j j j j j j | | { { } ∈ | |≤ ∈ } (cid:13)X (cid:13) Proof. (cid:13) (cid:13) v∗( α φ(A )) v∗(φ(A )) = eiθjv∗(φ(A ))= v∗( eiθjφ(A )) . j j j j j ≤ | | (cid:12) X (cid:12) X X (cid:12) X (cid:12) F(cid:12)(cid:12)romthefirstineq(cid:12)(cid:12)uality, v∗( αjφ(Aj)) v∗φ(A)andth(cid:12)(cid:12)en αjφ(Aj) (cid:12)(cid:12) | |≤| | k k≤ φ(A). Fromtheequalities, v∗(φ(Aj)) eiθjφ(Aj) sup αjφ(Aj) | | P| |≤k k≤ Pk k and then φ(A) sup α φ(A ) . j j | | ≤ k P k P P Corollary 1.11. Semi-Pvariation remains invariant under taking composition with an isometric embedding. In particular, φ = φ. Here φ denotes the | | | | composition of φ with the canonical embedding V ℓ∞(V∗). → 1 b b Let φ : B V be a semi-measure. For a simple function f : S C, → → i.e., a function with f(S) a finite set, we note that f = 1 and z∈f(S) [f=z] set φ(f) = φ([f = z]). Here [f = z] = s S;f(s) = z . By z∈f(S) { ∈ P } subpartitioning and regrouping,the correspondence f φ(f) is linear and the P 7→ above lemma means φ = φ(S). Therefore, if φ(S) < , φ is continuously extended to the unifokrmkclos|u|re C(B) of the set o|f|simple∞functions. Note that C(B) is a commutative C*-algebra. When B is a σ-algebra, C(B) is the set of boundedmeasurablefunctionsandtheobviouspairingC(B) L1(B) Cgives rise to inclusions C(B) L1(B)∗, L1(B) C(B)∗. × → Conversely, any bou⊂nded linear map⊂φ : C(B) V arises in this way. → Thus bounded semi-measures form a Banach space with respect to the norm φ = φ(S). k k | | Asequence f ofcomplex-valuedfunctionsonS issaidtoσ-converge n n≥1 { } to a function f onS if f is uniformly bounded andlim f (s)=f(s) for n n→∞ n every s S. Let Bσ be{the}σ-algebra generated by B. Then C(Bσ) is minimal among s∈ets which contain C(B) and have the property of being closed under σ-convergence. A V-valued measure φ on a σ-algebra B is σ-continuous in the sense that, if f C(B) σ-converges to f C(B), then φ(f ) φ(f) in the n n ∈ ∈ → weak topology. Consider a set Λ of complex semi-measures on a Boolean algebra B and assume that it is bounded in the sense that sup λ(S);λ Λ < . We {| | ∈ } ∞ 6 introduce then a bounded linear map φ : C(B) ℓ∞(Λ) by φ (f) : λ Λ Λ λ(f) C. From mutual estimates → 7→ ∈ α φ (A ) = sup α λ(A ) sup λ(A)= sup δ ,φ (A) φ (A), j Λ j j j λ Λ Λ k k | |≤ | | |h i| ≤| | λ∈Λ λ∈Λ λ∈Λ X X we see that φ (A)=sup λ(A);λ Λ for A B. In particular, Λ | | {| | ∈ } ∈ φ =sup λ(f);λ Λ,f C(B) =sup λ ;λ Λ < Λ 1 k k {| | ∈ ∈ } {k k ∈ } ∞ and φ = φ . Here we set Λ = λ;λ Λ , which is again a bounded set Λ |Λ| | | | | | | {| | ∈ } of semi-measures in view of the π-inequality. Proposition 1.12. Consider the following conditions on a bounded set Λ of complex semi-measures on a Boolean algebra B. (i) φ is squeezing. Λ (ii) φ is a measure. Λ (iii) φ is squeezing. |Λ| (iv) φ is a measure. |Λ| (i) and (iii) are equivalent. If B is a σ-algebra and Λ consists of complex measures, all the conditions (i) (iv) are equivalent. ∼ Proof. (ii) = (i) and (iv) = (iii) are trivial, whereas (i) follows from (iii) in ⇒ ⇒ view of λ(A) λ(A) and (iii) = (iv) is a special case of (i) = (ii). | |≤| | ⇒ ⇒ (i) = (ii): If φ is not countably additive, we have a disjoint sequence Λ A in⇒B andδ >0 suchthat φ ( A ) δ for all m 1. Thenwe can n Λ n≥m n { } k ⊔ k≥ ≥ inductively find a sequence λ Λ and a subsequence n < n < so that m 1 2 ∈ ··· λ ( A ) δ/2. Now the disjoint sequenceB = A | m ⊔nm≤n 0 such that φ (B ) δ. We can therefore find a sequence λ Λ |Λ| n n sothat λ (B ) δ/k2andthekn,≥bythe π-inequality,a sequenceA B i∈nB n n n n | | ≥ ⊂ fulfilling λ (B ) π λ (A ) +δ/3. Now the disjoint sequence A satisfies n n n n n | | ≤ | | { } φ (A ) λ (A ) δ/6 and denies the squeezing property of φ . Λ n n n Λ k k≥| |≥ Definition 1.13. Let µ be a finite positivesemi-measureonaBoolean-algebra B. A vector semi-measure φ on B is said to be µ-continuous if ǫ > 0, δ > 0, A B,µ(A) δ = φ(A) ǫ. ∀ ∃ ∀ ∈ ≤ ⇒k k≤ Theorem 1.14 (Pettis1938). Suppose that both of φ and µ are measures (B being a σ-algebra necessarily). Then φ is µ-continuous if and only if µ(A) = 0 (A B) implies φ(A)=0. ∈ 7 Proof. We follow [DU] I.2. By taking the composition with the canonical em- § beddingV ℓ∞(V∗),wemaysupposethatφ=φ ,whereΛ= v∗φ;v∗ V∗ is a bounde→d subset1of L1(B). Λ { ∈ 1} If φ is not µ-continuous, there exists δ > 0 and a sequence A B such n that φ(A ) δ for n 1 and ∞ µ(A ) < . Let B =∈ A be a kdecrenasking≥sequence ≥in B. Fromn=th1e latnter ine∞quality, µ(mB ) ∪n0≥,mi.e.n, m P ↓ µ( B )=0. From the former inequality, we have m ∩ φ (B ) =sup v∗φ(B );v∗ V∗ sup v∗φ(A );v∗ V∗ k |Λ| m k {| | m ∈ 1}≥ {| | m ∈ 1} sup v∗,φ(A ) ;v∗ V∗ = φ(A ) δ. ≥ {|h m i| ∈ 1} k m k≥ Since φ is a measure, φ is also a measure by Proposition 1.12 and the limit Λ |Λ| m is applied to get φ ( B ) δ. Therefore we can find a functional |Λ| m v∗→∞V∗ such that v∗φk( B ∩) > δk/2≥and then A B in B such that ∈ 1 | | ∩ m ⊂ ∩ m π v∗,φ(A) >δ/2. Thus φ(A) =0, whereas µ(A) µ( B )=0. m |h i| k k6 ≤ ∩ Theorem 1.15 (Doubrovsky1947). Let Λ be a bounded set of complex measures on a σ-algebra B. If φ :B ℓ∞(Λ) is a measure, there exists a positive Λ measure µ L1(B) for which φ→is µ-continuous with a reverse inequality Λ ∈ µ(A) φ (A)=sup λ(A);λ Λ . Λ ≤| | {| | ∈ } Proof. This is [DH], Theorem I.2.4. We first establish a kind of compactness of a bounded Λ: Given any ǫ > 0, we can find a finite subset F Λ such that if A B satisfies λ(A)=0 for λ F, then λ(A) ǫ for any λ⊂ Λ. ∈Ifnot,there|ex|istsδ >0such∈thatforan|y|F ⋐≤Λ,wecanfind∈A Bsatisfy- ∈ ing λ(A)=0foranyλ F but ν (A) δ forsomeν Λ. Thenwecaninduc- tive|ly|choose a sequence∈λ and|A| B≥so that λ (A∈)= = λ (A )=0 n n 1 n n n ∈ | | ··· | | and λ (A ) δ for n 1. Let B = A be a decreasing sequence n+1 n m n≥m n | | ≥ ≥ ∪ and set B = B . Since λ (B ) λ (B ) λ (A ) = 0 for ∞ ∩n n | m| ∞ ≤ | m| m ≤ n≥m| m| n m 1 and lim sup λ(B B )=0, we have ≥ m→∞ λ∈Λ| | m\ ∞ P 0= lim λ (B B )= lim λ (B ), m+1 m ∞ m+1 m m→∞| | \ m→∞| | which contradicts with λ (B ) λ (A ) δ. m+1 m m+1 m | | ≥| | ≥ Now we use the boundedness of Λ again to construct a control measure µ over Λ. For each n 1, choose F ⋑ Λ so that λ(A) = 0 with A B ≥ n λ∈Fn| | ∈ implies λ(A) 1/nforanyλ Λ. Thenthe positivemeasureµ = λ satisfies|µ| (A)≤ F φ (A) fo∈r A B and, if wPe define n λ∈Fn| | n n Λ ≤| || | ∈ P ∞ 1 µ= µ , F 2n n n n=1| | X it is a positive finite measure on B with the property µ(A) φ (A). Assume Λ ≤| | that µ(A) = 0. Then from µ (A) = 0, λ(A) 1/n for any λ Λ and n | | ≤ ∈ any n 1, i.e., λ(A) = 0. Thanks to the Pettis theorem, this means the ≥ | | µ-continuity of φ . Λ 8 Corollary 1.16 (Bartle-Dunford-Schwartz1955). For a vector measure φ on a σ-algebra,we can find a finite positive measure µ so that φ is µ-continuousand µ is majorized by φ. | | Proof. Apply the theorem to Λ= v∗φ;v∗ V∗ , which is bounded by Propo- { ∈ 1} sition 1.9 (iv). Recall that countable additivity of a semi-measure µ: B [0, ) is equivalent to the condition that, if A in B, then µ(A ) 0→. Le∞t Bσ be the n n σ-algebrageneratedbyB. Theclass↓ic∅alextensiontheorem4↓saysthat,ifafinite positive semi-measure on B is countably additive, it is uniquely extended to a positive measure on Bσ. In the framework of Daniell integral (see [12] for example), this can be ex- plainedinthefollowingfashion: LetLbethevectorlatticeofreal-valuedsimple functions on the base set S. Then a semi-measure µ can be interpreted as a positive linear functional L R, which is also denoted by µ. Let f 0 in L. n Then, given ǫ > 0, [f ǫ]→ in B and µ(f ) f µ([f ǫ])↓+ǫµ(S), n n 1 ∞ n ≥ ↓ ∅ ≤ k k ≥ together with the continuity of µ imply lim µ(f ) ǫµ(S). Thus, µ is contin- n n ≤ uous as a linear functional and we can apply the whole construction of Daniell integral to get a measure extension to Bσ. Lemma 1.17. Let µ be a finite positive measure on Bσ and embed Bσ into L1(S,µ). Then Bσ is closed in L1(S,µ) and B is dense in Bσ. Proof. Sinceanysequentialconvergenceinmeanimpliesalmostallconvergence by passing to a subsequence, Bσ (more precisely 1 ;B Bσ ) is closed in B { ∈ } L1(S,µ)(pointwiseconvergenceof 0,1 -valuedfunctionsproduce 0,1 -valued { } { } functions). In view of 1 = 1 1 and the dominated convergence theorem, A∩B A B onseesthattheclosureofB(moreprecisely 1 ;B sB )inL1(S,µ)provides B a σ-algebra and hence coincides with Bσ. { ∈ } Theorem 1.18 (Kluvánek1961). Let φ : B V be a semi-measure on a Boolean algebra B. If φ is µ-continuous for so→me countably additive positive semi-measure µ on B, then φ is uniquely extended to a measure Bσ V on the σ-algebra Bσ generated by B. → Proof. The uniqueness is as usual: Given two extensions, sets of their coinci- dency form a σ-algebra containing B, whence extensions coincide on the whole Bσ. For the existence, first note that µ is extended to a measureby the classical extension theorem, which is again denoted by µ. From the previous lemma, a complete (pseudo)metric on Bσ is defined by d(A,B) = 1 1 = A B 1 µ(A B) = µ(A B)+µ(B A) so that (Bσ,d) is the complketion−of (Bk,d). △ \ \ In view of the inequality φ(A) φ(B) = φ(A B) φ(B A) φ(A B) + φ(B A) , k − k k \ − \ k≤k \ k k \ k 4This can be attributed to many researchers: Fréchet, Carathéodory, Kolmogorov, Hahn andHopf(VladimirI.Bogachev,MeasureTheoryI).ItseemsfairtoaddthenameofDaniell tothelistbecausethetheorem itselfisjustanexampleofDaniellintegral. 9 φisuniformlycontinuouswithrespecttod,whichadmitsthereforeacontinuous extension φ to (Bσ,d). Now φ is countably additive thanks to the d-continuity: finite additivity of φgoesovertothatofφandmonotoneconvergenceassurestheσ-additivity. 2 Cross Norms This is a veryoldbut still developing subjectandthere arelots of referencesto be mentioned. We nominate, however, just [Raymond 1973] and [Diesel 1985] here to follow them. GivenBanachspacesX andY,letB(X,Y)betheBanachspaceofbounded bilinear forms on X Y and L(X,Y) be the Banach space of bounded linear maps of X into Y. ×There are natural identifications L(X,Y∗) = B(X,Y) = L(Y,X∗). Recall that a (semi)norm on the algebraic tensor product X Y is ⊗ called a cross (semi)norm if it satisfies x y = x y . The obvious bilinear map X Y k B⊗(Xk∗,Yk∗)kgkiveks rise to a linear map X Y B(X∗,Y∗) by (x y)×(x∗,y→∗) = x∗(x)y∗(y), which is injective. In ⊗ → ⊗ fact, let z = x y X Y and express z = z e f l l ⊗ l ∈ ⊗ 1≤j≤m,1≤k≤n j,k j ⊗ k with e X and f Y linearly independent. The dual bases e∗ and f∗ {arje}c⊂ontPinuous{onk}fi⊂nite-dimensional subspacesPand can be ext{enjd}ed to { k} bounded linear functionals. We then have z = z(e∗,f∗). Thus the norm on j,k j k B(X∗,Y∗) induces a cross norm on X Y, which is denoted by B(X∗,Y∗). In the embedding X Y B(X∗,Y∗)⊗= L(X∗,Y∗∗), x yk·kX Y is ⊗ → l l ⊗ l ∈ ⊗ realized by the operator x∗ x∗(x )y and the image of X Y is included in the subspace L(X∗,Y) 7→L(Xl∗,Y∗∗l). lThus we have P ⊗ ⊂ P k xl⊗ylkB(X∗,Y∗) =sup{k x∗(xl)ylk;x∗ ∈X1∗}=sup{| x∗(xl)y∗(yl)|;x∗ ∈X1∗,y∗ ∈Y1∗} l l l X X X and a similar expression holds with the role of X and Y exchanged. There is another natural way to get a cross norm on X Y. Each f B(X,Y)definesalinearfunctionalonX Y byf(z)= n f(⊗x ,y )satisfyin∈g f(z) n f x y , whence it ind⊗uces a linear mapl=X1 Yl l B(X,Y)∗ | |≤ l=1k kk lkk lk P ⊗ → and the associated seminorm B(X,Y)∗ satisfies P k·k n n z B(X,Y)∗ =inf xl yl ;z = xl yl . k k { k kk k ⊗ } l=1 l=1 X X The inequality is clear. To get the reverse inequality, we first notice that the right hand side≤defines a seminorm on X Y. Let ϕ:X Y C be a inf k·k ⊗ ⊗ → -bounded linear functional with its dual norm denoted by ϕ . Then the inf k·k k k associated bilinear functional f(x,y)=ϕ(x y) satisfies f ϕ , whence ⊗ k k≤k k z =sup ϕ(z); ϕ 1 sup f(z);f B(X,Y), f 1 = z B(X,Y)∗. k k {| | k k≤ }≤ {| | ∈ k k≤ } k k The bilinear map Φ:X∗ Y∗ B(X,Y) defined by × → Φ(x∗,y∗):(x,y) x∗(x)y∗(y) 7→ 10

Notes on Tensor Product Measures

Similar Notes on Tensor Product Measures