CORRECTIONS AND ADDITION TO THE PAPER KELLERER-STRASSEN TYPE MARGINAL MEASURE PROBLEM. Rataka TAHATA. Received March 9, 2005

Transcription

1 Scientiae Mathematicae Japonicae Online, e-5, CORRECTIONS AND ADDITION TO THE PAPER KELLERER-STRASSEN TPE MARGINAL MEASURE PROBLEM Rataka TAHATA Received March 9, 5 Abstract. A report on an error in the paper [5] is given. The error is in Theorem.4 of [5]. The condition imposed on the dominant measure ρ in the theorem of [5] is insufficient in order to deduce the conclusion of the theorem. In this note a corrected theorem is given. See Theorem A in of this note. In this note a stronger condition on ρ is imposed. Corrections for the two corollaries of the theorem in [5] are also given. See Corollaries A1 and A in this note. 1. Introduction. The fact that Theorem.4 in [5] is not true is seen by giving a counter example for a corollary to the theorem. See Example below in this note. Since Theorem.4 in [5] is not true, there must be one or more errors in the proof in [5] of the theorem. In fact the author found an error in the Step 4 of the proof of the theorem. The statement For any f C(T ) there is a monotone increasing sequence {h n } of step functions such that lim n h n (t) =f(t) for all t T and T [h n ]={t T ; h n (t) } T n for any n in Step 4 of the proof of the theorem in [5] is not true. It follows that the statement η defined in Step belongs to Λ(T ). Therefore one has η(t )=1. in Step 5 cannot be proved. Our aim in this note is to give a stronger condition for the dominant measure ρ in order for the conclusions of the theorem and its corollaries in [5] to remain true. In the case where the component spaces are Polish spaces the new condition for the dominant measure ρ is the condition that it is σ-finite on the algebra consisting of all finite unions of measurable rectangles of the product space. In the cases where the component spaces are separable metric spaces or separable measurable spaces, the condition that ρ is rectangle-normal is imposed.. Results. Let be a Hausdorff space. We denote by β() the set of all Borel subsets of. Let B b () denote the set of all bounded real valued Borel measurable functions on. In what follows, unless the contrary is explicitly stated, (, A) and (, B) denote separable measurable spaces. We denote by (, A B) the product space of these measurable spaces. Let π and π be the canonical projections from onto and, respectively. We denote by R() the algebra consisting of all finite unions of measurable rectangles of. P () denotes the set of all probability measures on. When ρ is a σ-finite measure on, we put Mathematics Subject Classification. 6A1. Key words and phrases. Maiginal measures. Λ() ={λ P (); λ ρ}.

2 19 RATAKA TAHATA For any set A we denote by χ A the characteristic function of A. First we shall give a counterexample for Corollary.7 in [5]. Example. Put = =(, 1] and =(, 1] (, 1]. Let λ be the Lebesgue measure on (, 1]. Let λ λ be the product measure of λ. Let <ɛ< 3. Put S ɛ = {(x, y) ;<x<y min{x + ɛ, 1}}. Put ϕ ɛ (x, y) = ɛ( ɛ) χ S ɛ (x, y) We denote by θ ɛ the measure on having ϕ ɛ as the density function concerning λ λ. Put 1 h(x, y) = (y x) if <x<y 1 otherwise We denote by ρ the σ-finite measure on having h as the density function concerning λ λ. Lemma.1. If <ɛ< 3, then θ ɛ Λ() ={θ P (); θ ρ}. ɛ( ɛ) Proof. Since the area of S ɛ is, we have θ ɛ P (). If x<y x + ɛ and <x, then we have <y x<ɛ.in this case we have 1 ɛ < 1 (y x). If <x<y 1 and 1 ɛ<x 1, then we have <y x<ɛ.in this case we have 1 ɛ < 1. Therefore for any (x, y) in with <x<y 1 we have (y x) ϕ ɛ (x, y) = Hence we have θ ɛ Λ(). ɛ( ɛ) χ S ɛ (x, y) 1 ɛ < 1 = h(x, y). (y x) For any k B b () put P (k) = sup{ kdθ; θ Λ()}. Lemma.. For any f B b () and g B b ( ) one has 1 f(x) dλ(x)+ 1 g(y) dλ(y) P (f π + g π ). Proof. Since the measures treated in this lemma are probability measures, it is sufficient for us to prove the inequality for non-negative functions f and g. (1) 1 1 χ Sɛ (x, y)f(x) dλ λ(x, y)

3 MARGINAL MEASURE PROBLEM 191 and () ɛ( ɛ) Hence we have = =ɛ f(x) dλ(x) =ɛ( ɛ)( x+ɛ f(x) dλ(x) x f(x) dλ(x)+ 1 dλ(y)+ 1 1 f(x) dλ(x) x (1 x)f(x) dλ(y) 1 f(x) dλ(x)+ f(x) dλ(x)). χ Sɛ (x, y)f(x) dλ λ(x, y) ɛ( ɛ) f(x) dλ(x) 1 ɛ f(x) dλ(x) ɛ( ɛ) f(x) dλ(x). Accordingly we have 1 (3) f π dθ ɛ f(x) dλ(x) = ɛ( ɛ) ɛ ( ɛ) 1 1 ɛ ( ( ɛ) Letting ɛ, then we have 1 χ Sɛ (x, y)f(x) dλ λ(x, y) f(x) dλ(x) ɛ 1 f(x) dλ(x). 1 f(x) dλ(x) By a similar argument we have 1 ɛ 1 ɛ (4) g π dθ ɛ g(y) dλ(y) g(y) dλ(y) g(y) dλ(y). ( ɛ) ɛ Since θ ɛ Λ(), we have ( 1 1 ) P (f π + g π ) f(x) dλ(x)+ g(y) dλ(y) 1 ) ( 1 ɛ f(x) dλ(x)+ g(y) dλ(y) f(x) dλ(x)+ ( 1 P (f π + g π ) f(x) dλ(x)+ This shows that Lemma. holds. 1 ) g(y) dλ(y). dλ(y) ) g(y) dλ(y). In this example there does not exist a θ in Λ() having µ = λ and ν = λ as marginals (see Kellerer [1], page 196). This is a counterexample for Corollary.7 in [5]. It follows that Theorem.4 in [5] is not true. We correct Theorem.4 in [5] as follows. Theorem A Let and be Polish spaces and let =. Let ρ be a measure on β() such that the restriction ρ of ρ to R() is σ-finite on R() and ρ() 1. Given

4 19 RATAKA TAHATA µ P () and ν P ( ), then the following two conditions are equivalent to each other: (1) There exists a θ Λ() having µ and ν as marginals. () For any functions f β() and g β( ) one has fdµ+ gdν sup{ (f π + g π ) dλ; λ Λ()}. Proof. The implication (1) () is almost obvious. We shall show the converse. Put B b () ={f π + g π ; f B b () and g B b ( )}. We define a linear functional W on B b () by the equation W (f π + g π )= fdµ+ gdν. For any function h B b () put P (h) = sup{ hdλ; λ Λ()}. W is a linear functional on B b () satisfying W P on B b (). Since P is subadditive and positively homogeneous on B b (), W can be extended, by the Hahn Banach theorem, to a linear functional W on B b () such that W P on B b (). For any set E R() put θ (E) =W (χ E ). Clearly θ is a finitely additive measure on R() having µ and ν as marginals. Since and are Polish spaces, µ and ν are Radon measures on and, respectively. Accordingly θ is a finite Radon measure on R(). By Theorem 16 in [3] (page 51) the measure θ can be extended to a probability measure θ on β(). Since ρ is a σ-finite measure on R(), ρ can be uniquely extended to the measure ρ on β(). By the uniqueness of extension we have ρ = ρ on β(). Since θ ρ on R(), we have θ ρ = ρ on β(). Clearly θ has µ and ν as marginals. Thus the theorem has been proved. We correct Corollary.7 in [5] as follows. Corollary A1 Let and be separable metric spaces and let =. Let ρ be a rectangle-normal measure on β() with ρ() 1. Given µ P () and ν P ( ), then the following conditions are equivalent to each other: (1) There exists a measure θ Λ() having µ and ν as marginals. () For any functions f B b () and g B b ( ) one has fdµ+ gdν sup{ (f π + g π ) dλ; λ Λ()}. Proof. The implication (1) () is almost obvious. We shall show the converse. Let and Ỹ be the completions of and, respectively. Put = Ỹ. Since ρ is rectanglenormal measure on, there exist disjoint sequences { n } n=1 and { m} m=1 of sets in β() and β( ) respectively such that = n and = m and ρ( n m ) <,n,m= n=1 m=1

5 MARGINAL MEASURE PROBLEM 193 1,,.... For any positive integer n there exists a set A n β( ) such that A n = n. Put n 1 n = A n A i, n =1,,.... i=1 { n } n=1 is a disjoint sequence of sets in β( ) such that n = n. Put = n. { n } n= is a disjoint sequence of sets in β( ) such that = a disjoint sequence {Ỹm} m= of sets in β(ỹ ) such that Ỹ = n= m= n=1 n. Similarly there exists Ỹ m and for any positive integer m m = Ỹm and Ỹ =. For any Ẽ β( ) put ρ(ẽ) =ρ(ẽ ). ρ is a rectangle-normal measure on. In fact, we have ρ( Ỹm) =ρ(( ) (Ỹm )) = ρ( m )=, m =, 1,... and, for any positive integer n, ρ( n Ỹm) =ρ( n m ) <, m =, 1,.... Hence ρ is a rectangle-normal measure on. i, etc., denote the canonical injection from into. For every µ P () we denote by i (µ) the image measure of µ by i. Put i (µ) = µ and i (ν) = ν. We have i (Λ()) = { θ P ( ); θ ρ} =Λ( ). In fact, it is obvious that i (Λ()) is contained in Λ( ). We shall show that the converse. Let θ Λ( ). For any set E β() there exists an Ẽ β( ) such that E = Ẽ. Put θ(e) = θ(ẽ). θ is well defined, that is, if E = Ẽ = F (Ẽ, F β( )), then we have θ( Ẽ)= θ( F ). In fact, since we have (Ẽ F ) = E E =, θ(ẽ F ) ρ(ẽ F )=ρ((ẽ F ) ) =ρ( ) =, where Ẽ F denotes the symmetric difference of Ẽ and F. Hence we have θ(ẽ) = θ( F ). Since, for any disjoint sequence {E n } n=1 of sets in β(), there exists a disjoint sequence {Ẽn} n=1 of sets in β( ) such that Ẽn = E n,n=1,,...,θis a measure on. Hence we have i (θ) = θ. For any functions f B b ( ) and g B b (Ỹ ) put f = f i and g = g i. Then we have f d µ + g d ν = e e = sup{ f i dµ + g i dν fdµ+ gdν (f π + g π ) dλ; λ Λ()} = sup{ ( f i π + g i π ) dλ; λ Λ()}. Since π e i = i π and π e i = i π and i (Λ()) = Λ( ), we have

6 194 RATAKA TAHATA e f d µ + g d ν sup{ ( f π e + g π e ) d λ; λ Λ( )}. e e Since and Ỹ are Polish spaces, by Theorem A there exists a θ Λ( ) having µ and ν as marginals. Since i (Λ())=Λ( ), there exists a θ Λ() such that i (θ) = θ. Then we have i (µ) = µ = π e ( θ) =π e i (θ) =i π (θ). Since i is an injective map from P () intop ( ), we have µ = π (θ). Similary we have ν = π (θ). Since θ Λ( ), we have θ ρ. Accordingly θ ρ holds. Thus Corollary A1 has been proved. In the case where component spaces are separable measurable spaces Corollary.8 in [5] is corrected as follows. Corollary A Let (, A) and (,B) be separable measurable spaces and (, A B) the cartesian product of these spaces. Let ρ be a rectangle-normal measure on with ρ() 1. Given probability measures µ and ν on and, respectively, then the following two conditions are equivalent to each other: (1) There exists a probability measure θ on such that θ ρ and π (θ) =µ and π (θ) =ν. () For any real valued bounded measurable functions f and g on and respectively one has fdµ+ gdν sup{ (f π + g π ) dλ; λ Λ()}. Proof. Any separable mesurable space is isomorphic to a measurable space consisting of a subset of the closed interval [, 1] and its Borel σ-algebra (see [], page 5, 4 ). Therefore this corollary can be proved by transforming to the problem into one for the case of separable metric spaces Acknowledgement. The author would like to express his hearty thanks to Prof. T. amanaka for carefully reading the manuscript. References [1] H. G. Kellerer, Maßtheoretische Marginalprobleme, Math Ann., 153(1964) [] D. Ramachandran, Perfect measures, Part 1, The Macmillan Company of India Limited, [3] L. Schwartz, Radon measures on arbitrary topological spaces and cylindrical measures, Tata. Inst. Fund. Res., Oxford Univ., [4] V. Strassen, The existence of probability measures with given marginals, Ann. Math. Statistics, 36(1965), [5] R. Tahata, On Kellerer-Strassen type marginal measure problem, Math. Japonica, 45, No. 3(1997), Department of Mathematics College of Science and Technology Nihon University, 1-8, Kanda Surugadai, Chiyoda-ku, Tokyo, Japan. tahata@math.cst.nihon-u.ac.jp