Raphaël CERF CNRS, Université Paris Sud Mathématique, Bâtiment Orsay Cedex France.

Transcription

1 ORIENTED MEASURES Rphël CERF CNRS, Université Pris Sud Mthémtique, Bâtiment Orsy Cedex Frnce. Crlo MARICONDA Università di Pdov Diprtimento di Mtemtic pur e pplict vi Belzoni 7, Pdov, Itly. Revised version, My 1995 Abstrct. A vector mesure µ = (µ 1,, µ n ) defined on [, b] is oriented if for ech k tuple of disjoint mesurble sets (A 1,, A k ) such tht A 1 < < A k the determinnt µ 1 (A 1 ) µ 1 (A k ) µ k (A 1 ) µ k (A k ) is positive. We study the rnge R of n oriented mesure: R = { µ(e) : χ E hs n discontinuity points }, R = { µ(e) : χ E hs less thn n 1 discontinuity points } Mthemtics Subject Clssifiction. 15A15, 28B05. Key words nd phrses. oriented mesure, rnge of vector mesure, Lypunov, extreme points. We re deeply grteful to Professor Arrigo Cellin for suggesting the initil problem, for his useful dvices nd for his wrm encourgement; we lso wish to thnk Professors Jen Pierre Aubin nd Helen Frnkowsk who gve the uthors the opportunity to meet together. We wrmly thnk the Referee for sking us precisions bout n uncler point concerning theorem 2.2. We thnk the D.M.I. of the E.N.S. for the technicl support during the preprtion of this pper. The second uthor (C.M.) ws prtilly supported by Grnt of the Consiglio Nzionle delle Ricerche (Grnt ). 1 Typeset by AMS-TEX

2 Title: ORIENTED MEASURES Nme nd miling ddress of the uthor (for proofs, etc...): Crlo Mricond Universitá di Pdov Diprtimento di Mtemtic Pur e Applict Vi Belzoni, Pdov, Itly. emil ddress: MARICONDA@PDMAT1.MATH.UNIPD.IT 2

3 1. Introduction A theorem of Lypunov sttes tht the rnge R of non tomic vector mesure µ on [, b] R = { µ(a) : A mesurble subset of [, b] } coincides with the convex set { } b ρ dµ : 0 ρ 1. However for given ρ, 0 ρ 1, the usul proofs bsed on convexity extreme points rguments [4,5] do not give ny informtion bout the existence of nice set E such tht µ(e) = ρ dµ. Consider for instnce the two dimensionl vector mesure µ(a) = ( A, A + 2 A B ) where B is borelin subset of [, b] nd denotes the Lebesgue mesure. For ech set E, the equlity µ(e) = µ(b) implies B = E. When the mesure µ dmits density f, Hlkin [3] showed tht if for ech vector p R n the set { } t [, b] : p f(t) > 0 (where is the usul sclr product) is finite (respectively countble) union of intervls then there exists set E which is finite (resp. countble) union of intervls. In our pper [2] we introduced the stronger orienttion condition : we sy tht n rel functions f 1,, f n verify condition on n intervl [, b] if for ech k in {1,, n}, the determinnt f 1 (x 1 ) f 1 (x 2 ) f 1 (x k ) f 2 (x 1 ) f 2 (x 2 ) f 2 (x k ) f k (x 1 ) f k (x 2 ) f k (x k ) is not equl to zero whenever the x i s in [, b] re distinct nd its sign is constnt on the k tuples (x 1,, x k ) such tht x 1 < x 2 < < x k b. We showed tht if mesure µ dmits density function whose components re continuous nd stisfy the orienttion condition then the set E my be built in such wy tht its chrcteristic function hs t most n discontinuity points. Moreover, if 0 < ρ < 1 there exist two such sets E 1 nd E 2 whose chrcteristic functions χ E1 nd χ E2 hve exctly n 3

4 discontinuity points (one set is neighbourhood of wheres the other is not). Our proofs relied upon the fct tht the mp (α 1,, α n ) α2 α 1 f(x) dx + α4 α 3 f(x) dx + is differentible nd hs n invertible Jcobin whenever < α 1 < < α n < b. We lso showed tht whenever function x stisfies x(0) = = x (n 2) (0) = 0 nd x (n 1) (0) = 1 then the n functions (x (n 1),, x, x) verify on neighbourhood of 0. We pplied these results to the study of rechble sets of constrined bng bng solutions nd to non convex problems of the clculus of vritions. In this pper we del with mesures which re not necessrily bsolutely continuous with respect to the Lebesgue mesure. Oriented mesure. If A 1,, A k re k mesurble sets of [, b], by A 1 < < A k we men tht for ll k tuple (x 1,, x k ) of A 1 A k we hve x 1 < < x k. A mesure µ = (µ 1,, µ n ) is sid to be oriented if for ech k tuple of mesurble sets A 1,, A k such tht A 1 < < A k the determinnt µ 1 (A 1 ) µ 1 (A k )..... µ k (A 1 ) µ k (A k ) is positive. In this more generl frmework we give new proof of the results stted in [2]. We crry out deep study of the rnge R of the mesure: for ech point q of its interior R there exist exctly two distinct dul sets E 1, E 2 whose chrcteristic functions hve n discontinuity points such tht µ(e 1 ) = q = µ(e 2 ); the set R coincides with { } b ρ dµ : 0 < ρ < 1 so tht the bove set is open; the set R is strictly convex; point µ(e) belongs to the boundry R of R if nd only if the chrcteristic function of E hs less thn n 1 discontinuity points; finlly we give recursive decomposition of the boundry R. 2. Oriented mesures Throughout the pper we will work with n intervl [, b] equipped with the Lebesgue σ field L. Mesurble will men mesurble with respect to this σ field. A negligible set 4

5 is mesurble set of Lebesgue mesure zero. A vector mesure on [, b] is countbly dditive set function defined on the Lebesgue σ field with vlues in R n for some integer n. Nottion. If A 1,, A k re k mesurble sets of [, b], by A 1 < < A k we men tht A 1,, A k hve non zero Lebesgue mesure nd for ll k tuple (x 1,, x k ) of A 1 A k we hve x 1 < < x k. Let µ = (µ 1,, µ k ) be vector mesure. If ρ belongs to L 1 µ([, b]), we note µ i (ρ) = ρ dµ i, µ(ρ) = ρ dµ = ( µ 1 (ρ),, µ k (ρ) ). Definition 2.1. A vector mesure µ = (µ 1,, µ n ) on [, b] is sid to be oriented on [, b] if it is non tomic nd if for ech k in {1,, n} nd for ech k tuple of mesurble sets A 1,, A k such tht A 1 < < A k the determinnt is positive. µ 1 (A 1 ) µ 1 (A k )..... µ k (A 1 ) µ k (A k ) Remrk. If µ is oriented then µ 1 is positive mesure which ssigns positive vlues to sets of positive Lebesgue mesure. In prticulr, if I is non trivil intervl, then µ(i) is non zero. Remrk. If µ is oriented nd I 1,, I n re n disjoint non trivil intervls, then the vectors µ(i 1 ),, µ(i n ) form bsis of R n. A very importnt fct concerning oriented mesures is tht their chrcteristic property crries on from sets to positive functions. Nottion. If ρ is function its support is the set supp ρ = { x : ρ(x) 0 }. Theorem 2.2. Let µ = (µ 1,, µ n ) be n oriented mesure. If ρ 1,, ρ n re n µ integrble non negtive functions such tht supp ρ 1 < < supp ρ n then the determinnt µ 1 (ρ 1 ) µ 1 (ρ n )..... µ n (ρ 1 ) µ n (ρ n ) is positive. Let us first stte preprtory lemm. 5

6 Lemm 2.3. Let µ = (µ 1,, µ n ) be vector mesure nd ρ 1,, ρ n be n mesurble µ integrble functions. The determinnt ρ 1 dµ 1 ρ n dµ ρ 1 dµ n ρ n dµ n is equl to ( ) ρ 1 (s 1 ) ρ n (s n ) d ɛ(σ)µ σ(1) µ σ(n) (s 1,, s n ). σ S n Proof of the lemm. The identity is obviously true whenever ρ 1,, ρ n re chrcteristic functions. The monotone clss theorem yields the result. Proof of theorem 2.2. We pply the lemm. The domin of integrtion of the n fold integrl is reduced to supp ρ 1 supp ρ n. We first prove tht the mesure ˆµ = σ Σ n ɛ(σ)µ σ(1) µ σ(n) is positive on the product spce (supp ρ 1, L) (supp ρ n, L) equipped with the product σ field (where L denotes the one dimensionl Lebesgue σ field). Notice tht the product σ field L n does not coincide in generl with the n dimensionl Lebesgue σ field (i.e. the completion of the n dimensionl Borel σ field). Consider first the cse of subset of supp ρ 1 supp ρ n which is product set A 1 A n (where the A i s re mesurble). Necessrily, ech A i is subset of supp ρ i. If none of the A i s is negligible, then we hve A 1 < < A n nd ˆµ(A 1 A n ) = det[µ(a 1 ),, µ(a n )] is positive by definition. Suppose now some of the A i s re negligible. For ech index i, 1 i n, there exists decresing sequence (Bm) i m N of non negligible mesurble subsets of supp ρ i hving n empty intersection (this is consequence of the fct tht supp ρ i is not negligible). Now for ech m we hve A 1 Bm 1 < < A n Bm n whence ˆµ(A 1 Bm 1 A n Bm) n is positive. By the continuity of the mesure µ we hve ˆµ(A 1 A n ) = lim ˆµ(A 1 Bm 1 A n Bm) n m so tht ˆµ(A 1 A n ) is non negtive. It follows tht ˆµ is non negtive on the boolen lgebr of the finite (disjoint) union of product sets: its unique extension to the σ field L n generted by these products is lso non negtive. 6

7 The function (s 1,, s n ) ρ 1 (s 1 ) ρ n (s n ) is positive everywhere on this set nd is mesurble with respect to the σ field L n : thus the integrl ρ 1 (s 1 ) ρ n (s n ) dˆµ(s 1,, s n ) is positive. Remrk. If µ is bsolutely continuous with respect to the Lebesgue mesure then Lypunov theorem yields n lterntive proof of theorem 2.2. In fct k {1,, n} E k supp ρ k µ(ρ k ) = µ(e k ). Necessrily µ(e k ) is non zero for ech k (see remrk fter definition 2.1) nd the bsolute continuity hypothesis on µ implies tht the E k s re not negligible. It follows tht E 1 < < E n nd det[µ(ρ 1 ),, µ(ρ n )] = det[µ(e 1 ),, µ(e n )] > 0. We shll denote by Γ k the subset Γ k = { (x 1,, x k ) R k : x 1 x k b }. Definition 2.4. The mesure µ is sid to be loclly oriented if for ech n tuple x of Γ n there exists neighbourhood V = V 1 V n of x such tht for ech k tuple of mesurble sets A 1 < < A k stisfying A 1 A k V 1 V k, the determinnt µ 1 (A 1 ) µ 1 (A k )..... µ k (A 1 ) µ k (A k ) is positive. As curiosity, we prove the following Proposition 2.5. A loclly oriented mesure on [, b] is oriented on [, b]. Proof. Let µ be loclly oriented mesure. The compct set Γ n cn be covered by finite fmily of open sets (V i ) i Υ where V i = I1 i In i nd (Ik i ) i Υ re subintervls of 1 k n [, b] in such wy tht for ech k tuple of mesurble sets A 1 < < A k stisfying A 1 A k V i for some i Υ, the determinnt formed with the first k components of the vectors µ(a 1 ),, µ(a k ) is positive. Let (J l ) l Σ be the finite fmily of the toms of the lgebr generted by the sets (Ik i, i Υ, 1 k n) (thus the J l s re exctly the sets of the form i,k:x I I i k i k for some x [, b]). Let us remrk tht for ech (l 1,, l k ) in Σ k, the product J l1 J lk is contined in some product I i 0 1 Ii 0 k. In fct J l1 J lk I1 i Ik i 7 i Υ

8 so tht there exits i 0 such tht J l1 J lk I i 0 1 Ii 0 k is not empty. It follows tht J l1 I i 0 1,, J l k I i 0 k nd by the very construction of the sets J l s we obtin J l1 I i 0 1,, J l k I i 0 k. We denote by ˆµ k the mesure ˆµ k = σ Σ k ɛ(σ)µ σ(1) µ σ(k). Let (A 1,, A k ) be k tuple of mesurble sets such tht A 1 < < A k. The product A 1 A k is the disjoint union of the sets (A 1 A k ) (J l1 J lk ) when (l 1,, l k ) vries in Σ k. Let now (l 1,, l k ) belong to Σ k. Either (A 1 A k ) (J l1 J lk ) is empty (nd thus hs zero ˆµ k mesure) or it is not empty nd necessrily, J l1 < < J lk. Proceeding s in the proof of theorem 2.2, we show tht ˆµ k is positive mesure on the set (J l1 J lk ) whence ˆµ k ((A 1 A k ) (J l1 J lk )) is non negtive. Since the set A 1 A k is not negligible, t lest one of these sets is not negligible. Let (A 1 A k ) (J l1 J lk ) be such set. It s subset of one of the V i s nd moreover (A 1 J l1 ) < < (A k J lk ) whence ˆµ k ((A 1 J l1 ) (A k J lk )) is positive. Thus ˆµ k (A 1 A k ) is positive. 3. Oriented mesures with densities Orienttion condition. We sy tht n rel functions f 1,, f n verify condition on n intervl [, b] if for ech k in {1,, n}, the determinnt f 1 (x 1 ) f 1 (x k ) f 2 (x 1 ) f 2 (x k )..... f k (x 1 ) f k (x k ) is positive whenever the x i s in [, b] re such tht x 1 < x 2 < < x k b. Remrk. In our previous pper [2], we didn t impose the sign of the bove determinnt to be positive. When deling with continuous functions, connectedness rgument shows immeditely tht the sign is constnt on the set Γ k. In our present frmework (t the mesure level), we find it convenient to work with this slightly more restrictive condition. Exmples. For n = 1, condition sttes tht the function f 1 is positive; for n = 2, the functions f 1, f 2 stisfy if nd only if f 1 is positive nd f 2 /f 1 is strictly incresing. The functions f i (t) = t i 1 (i 1) stisfy condition on R (the corresponding determinnts re Vndermonde determinnts). Proposition 3.1. Let f 1,, f n be n functions in L 1 ([, b]) stisfying the orienttion condition on [, b]. Let µ i be the mesure on [, b] whose density with respect to the Lebesgue mesure is f i. Then the mesure µ = (µ 1,, µ n ) is oriented. Proof. Let A 1 < < A k be k mesurble sets of [, b]. Since the determinnt is 8

9 multiliner continuous form, we cn write f 1 f 1 A 1 A k..... = f k f k A 1 A k A 1 A k By condition, the integrnd is positive on A 1 A k. f 1 (s 1 ) f 1 (s k ) f 2 (s 1 ) f 2 (s k ).... ds 1 ds k.. f k (s 1 ) f k (s k ) If f 1,, f k re of clss C k 1 on [, b] we will denote their Wronskin by W (f 1,, f k ). The following opertionl criterion for the fulfilment of the orienttion condition hs been used in [2]. Proposition 3.2. Let f 1,, f n C n 1 ([, b]) be such tht t [, b] W (f 1 )(t) > 0,, W (f 1,, f n )(t) > 0. Then f 1,, f n stisfy the orienttion condition on [, b]. 4. Nottions nd preliminry lemms Let us introduce some nottions. If u 1,, u n re vectors of R n, their determinnt is sometimes denoted by det [u 1,, u n ]. Let A be n n mtrix with rel coefficients; by det A or A we denote its determinnt. For ech i, j {1,, n}, by A ij we men the (n 1) (n 1) mtrix obtined by removing the i th row nd the j th column from A. Surprisingly, the following simple lgebric trick will ply n essentil role in the existence prt of the proof of theorem 1. Lemm 4.1. be such tht If det A nn 0 then Let A = ( ij ) 1 i,j n be n n n mtrix with rel coefficients. Let x 1,, x n 1,1 x ,n 1 x n 1 + 1,n x n = 0 2,1 x ,n 1 x n 1 + 2,n x n = n 1,1 x n 1,n 1 x n 1 + n 1,n x n = 0 n1 x nn x n = Proof. Crmer rule pplied to the bove system yields A A nn x n. i {1,, n 1} x i = ( 1)n+i A ni A nn 9 x n

10 so tht n1 x nn x n = n i=1 ( 1)n+i ni A ni x n = A nn A A nn x n since n i=1 ( 1)n+i ni A ni is the development of the determinnt A long the first row. The next lemms involve strongly the notion of oriented mesure. Lemm 4.2. Let F nd G be two distinct subsets of [, b] which re the union of l nd m disjoint closed intervls l m F = I i, G = nd let µ = (µ 1,, µ n ) be n oriented mesure. Assume µ(f ) = µ(g). Then n < l + m; moreover if F G then n < l + m 1. i=1 j=1 Proof. Let us first remrk tht the symmetric difference ( (I 1 I l ) (J 1 J m ) = i,j J j ) ( (I i J j ) \ i,j ) (I i J j ) is the union of t most l + m non trivil intervls nd tht whenever t lest two intervls hve common boundry point then this number is smller thn l + m 1. Since the intervls I 1,, I l re disjoint, s well s J 1,, J m, we hve (I 1 I l ) (J 1 J m )\(I 1 J 1 ) = (I 1 J 1 )\(I 1 J 1 ) (I 2 I l ) (J 2 J m ). Now, the set (I 2 I l ) (J 2 J m ) is union of t most l + m 2 disjoint intervls. Either I 1 J 1 = or I 1 J 1 nd (I 1 J 1 ) is n intervl. In both cses (I 1 J 1 )\(I 1 J 1 ) is the union of t most two intervls (t most one if I 1 nd J 1 hve boundry point in common). A strightforwrd induction gives the result. Since the sets F nd G re distinct, F G is not empty. Let A 1 < < A p be the connected components of F G. For k in {1,, p} we hve A k = (A k F ) (A k G), (A k F ) (A k G) A k (F G) (F G) (F G) = ; the set A k being connected, either A k F \ G or A k G \ F. Put { +1 if Ak F \ G λ k = 1 if A k G \ F 10

11 so tht the equlity µ(f ) = µ(g) cn be rewritten s λ 1 µ 1 (A 1 )+ +λ p µ 1 (A p )= λ 1 µ n (A 1 )+ +λ p µ n (A p )= 0 Suppose n p; the first p equtions imply tht the determinnt µ 1 (A 1 ) µ 1 (A p )..... µ p (A 1 ) µ p (A p ) vnishes, which contrdicts the fct tht µ is oriented. The following nottions will be used throughout the pper. Nottions 4.3. We shll denote by Γ k the set Γ k = { (γ 1,, γ k ) R k : γ 1 γ k b }. To ech k tuple γ = (γ 1,, γ k ) belonging to Γ k we ssocite the two sets E γ = 0 i k i odd where by convention γ 0 =, γ k+1 = b. [γ i, γ i+1 ], E + γ = 0 i k [γ i, γ i+1 ] Lemm 4.4 (Uniqueness). Let µ be n dimensionl oriented mesure on [, b]. Assume the n tuples γ = (γ 1,, γ n ) nd δ = (δ 1,, δ n ) of Γ n stisfy µ(eγ ) = µ(e δ ) (respectively µ(e γ + ) = µ(e + δ )). Then E γ = E δ (resp. E γ + = E + δ ). Proof. Assume Eγ, E δ re distinct nd µ(e γ ) = µ(e δ ). Now, two possible cses my occur ccording to the prity of n. If n = 2r the sets Eγ nd E δ re the union of t most r intervls. Lemm 4.2 implies n < r + r which is bsurd. If n = 2r + 1 the sets Eγ nd E δ re the union of t most r + 1 intervls. However b is common boundry point. Lemm 4.2 implies n < (r + 1) + (r + 1) 1 which is bsurd. The dul cse µ(e γ + ) = µ(e + δ ) cn be treted similrly. The following essentil lemm will be used repetedly. 11

12 Lemm 4.5. Let µ = (µ 1,, µ n ) be n oriented mesure on the intervl [, b] nd I 0 < I 1 < < I n be n + 1 subintervls of [, b]. Then, given positive ɛ, there exist n + 1 positive rel numbers λ 0,, λ n such tht l {0,, n} 0 < λ l < ɛ nd n ( 1) l λ l µ(i l ) = 0. l=0 Proof. Consider the n n liner system λ 0 µ(i 0 ) λ 1 µ(i 1 ) + + ( 1) n 1 λ n 1 µ(i n 1 ) = ( 1) n 1 λ n µ(i n ). where λ n is prmeter. The determinnt of the system is ω n = ( 1) n(n 1) 2 det [µ(i 0 ),, µ(i n 1 )]. The mesure µ being oriented, ω n is not zero. Moreover, for ech i in {0,, n 1}, µ 1 (I 0 ) ( 1) i 2 µ 1 (I i 2 ) ( 1) n 1 µ 1 (I n ) ( 1) i µ 1 (I i ) ( 1) n 1 µ 1 (I n 1 ) µ 2 (I 0 ) ( 1) i 2 µ 2 (I i 2 ) ( 1) n 1 µ 2 (I n ) ( 1) i µ 2 (I i ) ( 1) n 1 µ 2 (I n 1 ) ω i = µ n (I 0 ) ( 1) i 2 µ n (I i 2 ) ( 1) n 1 µ n (I n ) ( 1) i µ n (I i ) ( 1) n 1 µ n (I n 1 ) i.e. ω i = ( 1) n(n 1) 2 det [µ(i 0 ),, µ(i i 2 ), µ(i i ),, µ(i n )]. By Crmer formul, λ i equls λ n ω i /ω n. The mesure µ being oriented ω i nd ω n hve the sme sign so tht λ i is positive whenever λ n is positive. Choosing λ n such tht 0 < λ n < min( ω n ω n ɛ,, ɛ, ɛ) ω 0 ω n 1 we obtin n (n + 1) tuple which solves the problem. 5. Min result The sttement of the min result involves the nottions

13 Theorem 5.1. Let µ be n oriented mesure on [, b] nd let ρ be mesurble function defined on [, b] with vlues in [0, 1]. There exist two n tuples α = (α 1,, α n ) nd β = (β 1,, β n ) in Γ n such tht µ(e α ) = ρ dµ = µ(e + β ). If in ddition 0 < ρ < 1 then α nd β in Γ n stisfying ( ) re unique nd verify ( ) < α 1 < < α n < b, < β 1 < < β n < b. Remrk. The mesure µ being non tomic we don t cre bout boundry points of intervls nd we might write µ(α, β) for the mesure of the intervl µ([α, β]). Proof. We consider first the cse 0 < ρ < 1 nd we prove the result by induction on n. n=1. The mesure µ being oriented on [, b], the mps α µ([α, b]) nd β µ([, β]) re continuous nd strictly monotonic on [, b]. It follows tht there exist unique rel numbers α 1 nd β 1 such tht µ([α 1, b]) = ρ dµ = µ([, β 1 ]). Assume the result is true t rnk n 1. We del only with the n tuple β: existence of the n tuple α corresponding to ρ t rnk n follows from the fct tht it coincides with the n tuple β corresponding to 1 ρ. Define for ech k in {1,, n} µ k (ρ) = ρ dµ k nd for ech n tuple β in Γ n θ k (β) = µ k (E + β ). The inductive ssumption yields the existence of two (n 1) tuples ᾱ = (ᾱ 1,, ᾱ n 1 ) nd β = ( β 1,, β n 1 ) such tht < ᾱ 1 < < ᾱ n 1 < b, < β 1 < < β n 1 < b nd for ech k in {1,, n 1} θ k (, ᾱ 1,, ᾱ n 1 ) = θ k ( β 1,, β n 1, b) = 0 i n 1 i odd 0 i n 1 13 µ k (ᾱ i, ᾱ i+1 ) = µ k (ρ), µ k ( β i, β i+1 ) = µ k (ρ). ( )

14 Put S = { β = (β 1,, β n ) Γ n : β 1 β 1, k {1,, n 1} θ k (β) = µ k (ρ) }. Since ( β 1,, β n 1, b) nd (, ᾱ 1,, ᾱ n 1 ) belong to S, the set S is not empty. We show now tht either θ n ( β 1,, β n 1, b) < µ n (ρ) < θ n (, ᾱ 1,, ᾱ n 1 ) or θ n (, ᾱ 1,, ᾱ n 1 ) < µ n (ρ) < θ n ( β 1,, β n 1, b). The equlities ( ) yield for ech k in {1,, n 1} 0 i n 1 Put for k, j in {1,, n} βi+1 β i (1 ρ) dµ k 0 i n 1 i odd βi+1 β i ρ dµ k = 0. x β j = ( 1)j+1, β kj = βj β j 1 ρ β j dµ k, A β = ( ) β kj 1 k,j n where ρ β j = { ρ if j is even, 1 ρ if j is odd. With these nottions the bove equlities become k {1,, n 1} n β kj xβ j = 0. j=1 Since the mesure µ is oriented then the determinnt A β nn does not vnish by theorem 2.2. We re thus in the position to pply lemm 4.1: θ n ( β 1,, β n 1, b) µ n (ρ) = Similrly if we define for k, j in {1,, n} n j=1 β nj xβ j = Aβ A β nn ( 1)n+1. x α j = ( 1) j, α kj = ᾱj ρ α j dµ k, ᾱ j 1 14 A α = ( α ) kj 1 k,j n

15 where we hve ρ α j = { ρ if j is odd, 1 ρ if j is even, θ n (, ᾱ 1,, ᾱ n 1 ) µ n (ρ) = Aα A α nn ( 1)n. The mesure µ being oriented, the determinnts A α nd A β hve the sme sign, s do A α nn nd A β nn. It follows tht θ n ( β 1,, β n 1, b) µ n (ρ) nd θ n (, ᾱ 1,, ᾱ n 1 ) µ n (ρ) hve opposite signs. At this stge, we prove tht the set S is the grph of continuous function, this will imply tht S is connected. Let β 1 belong to [, β 1 ]. We re looking for (n 1) tuple (β 2,, β n ) stisfying for ech k in {1,, n 1} µ k (, β 1 ) + 2 i n µ k (β i, β i+1 ) = µ k (ρ) = µ k (, β 1 ) + 2 i n 1 µ k ( β i, β i+1 ) or equivlently k {1,, n 1} 2 i n µ k (β i, β i+1 ) = µ k (β 1, β 1 ) + 2 i n 1 µ k ( β i, β i+1 ) Suppose first β 1 = β 1. The bove equtions become k {1,, n 1} 2 i n µ k (β i, β i+1 ) = We put β = (β 2,, β n 1, β n ) nd ˆβ = ( β 2,, β n 1, b). If n is odd then if n is even then 2 i n 1 µ k ( β i, β i+1 ). E β = [β 2, β 3 ] [β n 1, β n ], E ˆβ = [ β 2, β 3 ] [ β n 1, b]; E β = [β 2, β 3 ] [β n, b], E ˆβ = [ β 2, β 3 ] [ β n 2, β n 1 ]. In both cses the preceding formule cn be rewritten s k {1,, n 1} µ k (E β ) = µ k(e ˆβ ); 15

16 lemm 4.4 implies tht E β = E ˆβ. Since in ddition β 2 < < β n 1 < b then necessrily β 2 = β 2,, β n 1 = β n 1, β n = b. Suppose now β < β 1. Since β 1 < β 1 < < β n 1 < b then lemm 4.5 yields the existence of n rel numbers λ 1,, λ n in ]0, 1/2[ such tht for ech k in {1,, n 1} λ 1 µ k (β 1, β 1 ) + 1 i n 1 ( 1) i+1 λ i+1 µ k ( β i, β i+1 ) = 0. The function ρ = (1 λ 1 )χ [β1, β 1 ] + 1 i n 1 i odd λ i+1 χ [ βi, β i+1 ] + 2 i n 1 (1 λ i+1 )χ [ βi, β i+1 ] stisfies 0 < ρ < 1 on [β 1, b] nd for ech k in {1,, n 1} β 1 ρ dµ k = µ k (β 1, β 1 ) + 2 i n 1 µ k ( β i, β i+1 ). We re thus led to find (n 1) tuple (β 2,, β n ) such tht (β 1 )β 2 β n ( b) nd for ech k in {1,, n 1} 2 i n µ k (β i, β i+1 ) = or equivlently, if we put β = (β 2,, β n ), β 1 ρ dµ k, k {1,, n 1} µ k (E β ) = β 1 ρ dµ k. Existence nd uniqueness of β follow from the inductive ssumption t rnk n 1. In ddition, since 0 < ρ < 1 on [β 1, b], we hve β 1 < β 2 < < β n < b. We cn thus define mp ψ : [, β 1 ] R n 1 such tht for ll n tuple (β 1,, β n ) in Γ n (β 1,, β n ) S (β 2,, β n ) = ψ(β 1 ). Thus S is the grph of ψ. By the continuity of the mesure µ, the mps θ k, 1 k n 1, re continuous so tht the set S is closed; moreover, the function ψ tkes its vlues in the compct set [, b] n 1. It follows tht ψ is continuous. Henceforth S is connected. As consequence, 16

17 the mp θ n, being continuous on S, reches ll the vlues between θ n ( β 1,, β n 1, b) nd θ n (, ᾱ 1,, ᾱ n 1 ). In prticulr, there exists n tuple β in S such tht θ n (β) = µ n (ρ). This n tuple β solves the problem. Since θ n (, ᾱ 1,, ᾱ n 1 ) µ n (ρ) nd θ n ( β 1,, β n 1, b) µ n (ρ) then < β 1 < β 1 so tht < β 1 < β 2 < < β n < b. Uniqueness of β follows from lemm 4.4. Consider now the cse 0 ρ 1. Let (ρ m ) m N be sequence of mesurble functions such tht 0 < ρ m < 1 nd ρ m converges to ρ in L 1 µ([, b]). For ech function ρ m there exists unique n-tuple β m such tht µ(e + β m ) = ρ m dµ. By compctness, we my ssume tht β m converges to some n tuple β of Γ n. Pssing to the limit, we obtin µ(e + β ) = µ(ρ). 6. The rnge of n oriented mesure Let µ be n oriented mesure on [, b]. We denote by R the rnge of µ i.e. R = { µ(a) : A mesurble subset of [, b] }. Lemm 6.1. Let ρ be mesurble function on [, b], 0 ρ 1. Suppose there exist non trivil intervl I of [, b] nd positive rel number ɛ such tht ɛ ρ 1 ɛ on I. Then the set { } b ρ dµ : ρ = νχ I + ρ, ν L 1 µ(i), ν < ɛ is neighbourhood of ρ dµ in Rn. Proof. Let I 1 < < I n be n non trivil subintervls of I. The mesure µ being oriented, the vectors µ(i 1 ),, µ(i n ) form bsis of R n. The mp Λ : (λ 1,, λ n ) R n λ i µ(i i ) R n is liner isomorphism nd is thus open. Let 1 i n V ɛ = { (λ 1,, λ n ) : mx 1 i n λ i < ɛ }. Since Λ(V ɛ ) is neighbourhood of the origin nd is contined in the set { } ν dµ : ν L 1 µ(i), ν < ɛ, I 17

18 then the conclusion follows. Remrk. The hypothesis ɛ ρ 1 ɛ implies tht µ( ρ) belongs to the interior of R. Remrk. The conclusion of lemm 6.1 does not hold for n rbitrry vector mesure: consider for instnce the n dimensionl Lebesgue mesure. Let θ : Γ n R be the function defined by θ(γ) = µ(e γ ). The interior of Γ n is the set Γ n = { (γ 1,, γ n ) R n : < γ 1 < < γ n < b }. Corollry 6.2. The set θ(γ n) is contined in R. Lemm 6.3. The set θ(γ n) coincides with the set { } b F = ρ dµ : 0 < ρ < 1. Proof. Existence prt of theorem 5.1 implies tht F is contined in θ(γ n). Conversely, let γ = (γ 1,, γ n ) belong to Γ n; pplying lemm 4.5 to µ, γ nd ɛ < 1/2, we obtin (n + 1) tuple (λ 0,, λ n ) such tht i {0,, n} 0 < λ i < ɛ nd n ( 1) i λ i µ(γ i, γ i+1 ) = 0. i=0 Put ρ = 0 i n λ i χ [γi,γ i+1 ] + By construction we hve 0 < ρ < 1 nd 0 i n i odd (1 λ i )χ [γi,γ i+1 ]. so tht θ(γ) belongs to F. We hve the following ρ dµ = µ(e γ ) = θ(γ) Theorem 6.4. The rnge of θ coincides with R; the mp θ induces n homeomorphism from Γ n onto R nd mps Γ n onto R. Proof. The surjectivity of θ follows directly from theorem 5.1. Injectivity of the restriction of θ to Γ n is consequence of the uniqueness prt of theorem 5.1 together with lemm 6.3. We clim tht θ(γ n) is open. Let γ belong to Γ n. Lemm 4.5 llows s usul to find 18

19 piecewise constnt function ρ such tht 0 < ρ < 1 nd µ( ρ) = θ(γ). Clerly there exist positive ɛ nd subintervl I of [, b] on which ɛ ρ 1 ɛ. Put Lemm 6.1 implies tht the set V I,ɛ ρ = { νχ I + ρ : ν L 1 µ(i), ν < ɛ }. µ(v I,ɛ ρ ) = { ρ dµ : ρ V I,ɛ ρ is neighbourhood of µ( ρ) in R n. Since ech element ρ of V ρ I,ɛ stisfies 0 < ρ < 1 then ) is entirely contined in F. Moreover F coincides with θ(γ n) nd thus θ(γ n) is µ(v I,ɛ ρ neighbourhood of θ(γ). Now ech open convex set in R n is the interior of its closure; by lemm 6.3, the set θ(γ n) is convex nd its closure is R, whence θ(γ n) = R. Finlly we show tht the mp θ is proper (i.e. tht the inverse imge of compct subset is compct). Let K be compct subset of F nd (γ m ) m N be sequence in θ 1 (K) such tht θ(γ m ) converges to µ(ρ) for some ρ, 0 < ρ < 1. Since the sequence (γ m ) m N is contined in Γ n, by compctness, we my ssume tht γ m converges to γ in Γ n. By the continuity of θ, we hve θ(γ) = µ(e γ ) = ρ dµ. Uniqueness prt of theorem 5.1 implies tht γ belongs to Γ n. The mp θ is proper nd thus closed. It follows tht its inverse θ 1 is continuous. The equlity θ( Γ n ) = R is consequence of the inclusion θ(γ n) R nd the fct tht θ is one to one. We refer to [7] for the definitions of clssicl notions ssocited with convex sets. We hve the following Theorem 6.5. The rnge R of n oriented mesure is strictly convex. Proof. Let µ(e), µ(f ) be two distinct points of R. By theorem 5.1 we my ssume tht the sets E nd F re finite unions of closed intervls. Let λ ]0, 1[ nd put ρ = λχ E +(1 λ)χ F. Assume for instnce E \ F. Then there exists non trivil intervl I such tht x I ρ(x) = λχ E (x) + (1 λ)χ F (x) = λ. } Put ɛ = min(λ, 1 λ). Lemm 6.1 pplied to ρ, I, ɛ shows tht µ( ρ) belongs to R. 19

20 Corollry 6.6. Let E be mesurble subset of [, b]. Then µ(e) belongs to the boundry of R if nd only if there exists set F which is finite union of intervls such tht χ F hs less thn n 1 discontinuity points nd E F is µ negligible (such set hs lso zero Lebesgue mesure). Proof. We first remrk tht the fmily of the sets which re finite union of intervls nd whose chrcteristic function hs less thn n 1 discontinuity points coincides with the fmily { Eγ : γ Γ n }. Theorem 6.4 shows tht µ(f ) belongs to R whenever F = Eγ for some γ Γ n. Conversely let E be such tht µ(e) belongs to R. Theorem 6.4 yields the existence of n tuple γ belonging to Γ n such tht µ(eγ ) = µ(e); consequence of theorem 6.5 is tht µ(e) is n extreme point of R. Olech Theorem [5, Th. 1] implies tht E Eγ is µ negligible. Our pproch discloses the recursive structure of the boundry of the rnge of n oriented mesure. For k belonging to {0,, n} let R k = { µ(e γ ) : γ Γ k }, R + k = { µ(e+ γ ) : γ Γ k }. Notice tht Γ 0 =, R 0 = {0}, R+ 0 = {µ(, b)}. Proposition 6.7. The function γ Γ k µ(eγ ) R k (resp. γ Γ k µ(e γ + ) R + k ) is homeomorphism from Γ k onto its rnge which coincides with R k (resp. R + k ). Proof. Injectivity follows directly from corollry 6.6. The rest of the proof uses the techniques of the proof of theorem 6.4. Remrk. For ech k in {1,, n 1}, the set R k \ R k 1 is prtitioned into two connected components R k, R + k. However, for k = n, R n = R + n = R. These results yield the following Proposition 6.8. The boundry of the rnge R of n oriented n dimensionl mesure dmits the decomposition R = R n 1 R 1 {0} {µ(, b)} R + 1 R + n 1. Let T be the symmetry with respect to µ(, b)/2 (so tht for ech mesurble subset A of [, b], T (µ(a)) = µ([, b] \ A)). Then for ech k belonging to {0,, n} we hve T (R k ) = R + k, T (R k) = R k. 20

21 References 1. R. Cerf C. Mricond, On bng bng constrined solutions of control system, SIAM J. Control nd Optimiztion 33 no. 2 (1995), R. Cerf C. Mricond, Oriented mesures with continuous densities nd the bng bng principle, Jour. Funct. Anl. 126 no. 2 (1994), H. Hlkin, On generliztion of theorem of Lypounov, Jour. Mth. Anl. Appl. 10 (1965), P. Hlmos, The rnge of vector mesure, Bull. Amer. Mth. Soc. 54 (1948), C. Olech, The Lypunov theorem: its extensions nd pplictions, in Lecture Notes in Mthemtics: Methods of nonconvex nlysis, Vol (1989), W. Rudin, Rel nd complex nlysis, McGrw Hill (1966). 7. F.A. Vlentine, Convex sets, McGrw Hill (1964). 21