Portmanteau theorem for unbounded measures

Portmanteau theorem for unbounded measures M´ aty´ as Barczy and Gyula Pap University of Debrecen, Hungary Abstract We prove an analogue of the portmanteau theorem on weak convergence of probability measures allowing measures which are unbounded on an underlying metric space but finite on the complement of any Borel neighbourhood of a fixed element. 1 Introduction Weak convergence of probability measures on a metric space has a very important role in probability theory. The well known portmanteau theorem due to A. D. Alexandroff (see for example Theorem 11.1.1 in Dudley [1]) provides useful conditions equivalent to weak convergence of probability measures; any of them could serve as the definition of weak convergence. Proposition 1.2.13 in the book of Meerschaert and Scheffler [3] gives an analogue of the portmanteau theorem for bounded measures on Rd. Moreover, Proposition 1.2.19 in [3] gives an analogue for special unbounded measures on Rd, more precisely, for extended real valued measures which are finite on the complement of any Borel neighbourhood of 0∈Rd. By giving counterexamples we show that the equivalences of (c) and (d) in Propositions 1.2.13 and 1.2.19 in [3] are not valid (see our Remarks 2.2 and 2.3). We reformulate Proposition 1.2.19 in [3] in a more detailed form adding new equivalent assertions to it (see Theorem 2.1). Moreover, we note that Theorem 2.1 generalizes the equivalence of (a) and (b) in Theorem 11.3.3 of [1] in two aspects. On the one hand, the equivalence is extended allowing not necessarily finite measures which are finite on the complement of any Borel neighbourhood of a fixed element of an underlying metric space. On the other hand, we do not assume the separability of the underlying metric space to prove the equivalence. But we mention that this latter possibility is hiddenly contained in Problem 3, p. 312 in [1]. For completeness we give a detailed proof of Theorem 2.1. Our proof goes along the lines of the proof of the original portmanteau theorem and differs from the proof of Proposition 1.2.19 in [3]. Key words and phrases: Weak convergence of bounded measures; portmanteau theorem; L´evy measure The first author has been supported by the Hungarian Scientific Research Fund under Grant No. OTKA– F046061/2004. The authors have been supported by the Hungarian Scientific Research Fund under Grant No. OTKA–T048544/2005. 1 To shed some light on the sense of a portmanteau theorem for unbounded measures, let us consider the question of weak convergence of infinitely divisible probability measures µn, n∈Ntowards an infinitely divisible probability measure µ0in case of the real line R. Theorem VII.2.9 in Jacod and Shiryayev [2] gives equivalent conditions for weak convergence µn w →µ0. Among these conditions we have ZR fdηn→ZR fdη0for all f∈ C2(R), (1.1) where ηn,n∈Z+are nonnegative, extended real valued measures on Rwith ηn({0}) = 0 and RR(x2∧1) dηn(x)<∞(i.e., L´evy measures on R) corresponding to µn, and C2(R) is the set of all real valued bounded continuous functions fon Rvanishing on some Borel neighbourhood of 0 and having a limit at infinity. Theorem 2.1 is about equivalent reformulations of (1.1) when it holds for all real valued bounded continuous functions on R vanishing on some Borel neighbourhood of 0. 2 An analogue of the portmanteau theorem Let Nand Z+be the set of positive and nonnegative integers, respectively. Let (X, d) be a metric space and x0be a fixed element of X. Let B(X) denote the σ-algebra of Borel subsets of X. A Borel neighbourhood Uof x0is an element of B(X) for which there exists an open subset e Uof Xsuch that x0∈e U⊂U. Let Nx0denote the set of all Borel neighbourhoods of x0, and the set of bounded measures on Xis denoted by Mb(X). The expression ”a measure µon X” means a measure µon the σ-algebra B(X). Let C(X), Cx0(X) and BLx0(X) denote the spaces of all real valued bounded continuous functions on X, the set of all elements of C(X) vanishing on some Borel neighbourhood of x0, and the set of all real valued bounded Lipschitz functions vanishing on some Borel neighbourhood of x0, respectively. For a measure ηon Xand for a Borel subset B∈ B(X), let η|Bdenote the restriction of ηonto B, i.e., η|B(A) := η(B∩A) for all A∈ B(X). Let µn,n∈Z+be bounded measures on X. We write µn w →µif µn(A)→µ(A) for all A∈ B(X) with µ(∂A) = 0. This is called weak convergence of bounded measures on X. Now we formulate a portmanteau theorem for unbounded measures. Theorem 2.1 Let (X, d)be a metric space and x0be a fixed element of X. Let ηn, n∈Z+, be measures on Xsuch that ηn(X\U)<∞for all U∈ Nx0and for all n∈Z+. Then the following assertions are equivalent: (i) RX\Ufdηn→RX\Ufdη0for all f∈ C(X),U∈ Nx0with η0(∂U) = 0, (ii) ηn|X\U w →η0|X\Ufor all U∈ Nx0with η0(∂U) = 0, 2 (iii) ηn(X\U)→η0(X\U)for all U∈ Nx0with η0(∂U) = 0, (iv) RXfdηn→RXfdη0for all f∈ Cx0(X), (v) RXfdηn→RXfdη0for all f∈BLx0(X), (vi) the following inequalities hold: (a) lim sup n→∞ ηn(X\U)⩽η0(X\U)for all open neighbourhoods Uof x0, (b) lim inf n→∞ ηn(X\V)⩾η0(X\V)for all closed neighbourhoods Vof x0. Proof. (i)⇒(ii): Let Ube an element of Nx0with η0(∂U) = 0. Note ηn|X\U∈ Mb(X), n∈Z+. By the equivalence of (a) and (b) in Proposition 1.2.13 in [3], to prove ηn|X\U w →η0|X\Uit is enough to check RXfdηn|X\U→RXfdη0|X\Ufor all f∈ C(X). For this it suffices to show that for all real valued bounded measurable functions hon X, for all A∈ B(X) and for all n∈Z+we have ZX hdηn|A=ZA hdηn.(2.1) By Beppo-Levi’s theorem, a standard measure-theoretic argument implies (2.1). (ii)⇒(iii): Let Ube an element of Nx0with η0(∂U) = 0. By (ii), we have ηn|X\U w → η0|X\U. Since η0|X\U(∂X) = η0|X\U(∅) = 0, we get ηn(X\U) = ηn|X\U(X)→η0|X\U(X) = η0(X\U), as desired. (iii)⇒(ii): Let Ube an element of Nx0with η0(∂U) = 0 and let B∈ B(X) be such that η0|X\U(∂B) = 0. We have to show ηn|X\U(B)→η0|X\U(B). Since B∩(X\U) = X\[X\(B∩(X\U))] and ηn|X\U(B) = ηn(B∩(X\U)), n∈Z+, by (iii), it is enough to check η0¡∂¡X\(B∩(X\U))¢¢= 0. First we show ∂¡B∩(X\U)¢⊂¡∂B ∩(X\U)¢∪∂U for all subsets B,Uof X. (2.2) Let xbe an element of ∂¡B∩(X\U)¢and (yn)n⩾1, (zn)n⩾1be two sequences such that limn→∞ yn= limn→∞ zn=xand yn∈B∩(X\U), zn∈X\(B∩(X\U)), n∈N. Then for all n∈Nwe have one or two of the following possibilities: •yn∈B, yn∈X\Uand zn∈X\B, •yn∈B, yn∈X\Uand zn∈U. Then we get x∈¡∂B ∩((X\U)∪∂U)¢∪¡∂U ∩(B∪∂B)¢∪¡∂B ∩∂U¢. Since ∂B ∩ ((X\U)∪∂U)⊂(∂B ∩(X\U)) ∪∂U, we have x∈¡∂B ∩(X\U)¢∪∂U, as desired. Using (2.2) we get η0¡∂¡X\(B∩(X\U))¢¢⩽η0¡∂B∩(X\U)¢+η0(∂U) = 0. Indeed, by the assumptions η0¡∂B∩(X\U)¢= 0 and η0(∂U) = 0. Hence η0¡∂¡X\(B∩(X\U))¢¢= 0. (ii)⇒(i): Using again the equivalence of (a) and (b) in Proposition 1.2.13 in [3] and (2.1) we obtain (i). (iii)⇒(iv): Let fbe an element of Cx0(X). Then there exists A∈ Nx0such that f(x) = 0 for all x∈Aand η0(∂A) = 0. Indeed, the function t7→ η0¡{x∈X: 3 d(x, x0)⩾t}¢from (0,+∞) into Ris monotone decreasing, hence the set ©t∈(0,+∞) : η0({x∈X:d(x, x0) = t})>0ªof its discontinuities is at most countable. Consequently, for all e U∈ Nx0there exists some t > 0 such that U:= {x∈X:d(x, x0)< t}∈Nx0, U⊂e Uand η0(∂U) = 0. (At this step we use that an element e Uof Nx0contains an open subset of Xcontaining x0.) This implies the existence of A. We show that the set D:= ©t∈R:η0¡{x∈X:f(x) = t}¢>0ªis at most countable. The function F:R→[0, η0(X\A)], defined by F(t) := η0¡{x∈X\A:f(x)< t}¢, t ∈R, is monotone increasing and left continuous. (Note that η0(X\A)<∞,by the assumption on η0.) Hence it has at most countably many discontinuity points, and t0∈Ris a discontinuity point of Fif and only if F(t0+ 0) > F(t0),i.e., η0¡{x∈X\A:f(x) = t0}¢>0. If t06= 0,then {x∈X:f(x) = t0}={x∈X\A:f(x) = t0}, thus t06= 0 is a discontinuity point of Fif and only if η0({x∈X:f(x) = t0})>0. Hence if t∈Dthen t= 0 or t is a discontinuity point of F, consequently Dis at most countable. Since fis bounded and Dis at most countable, there exists a real number M > 0 such that −M, M /∈D and |f(x)|< M for x∈X. Let ε > 0. Choose real numbers ti, i = 0, . . . , k such that −M=t0< t1<···< tk=M,ti/∈D, i = 0, . . . , k and max0⩽i⩽k−1(ti+1 −ti)< ε. The countability of Dimplies the existence of ti,i= 0, . . . , k. Let Bi:= f−1¡[ti, ti+1)¢∩(X\A) = nx∈X\A:ti⩽f(x)< ti+1o for all i= 0, . . . , k −1. Then Bi,i= 0, . . . , k −1, are pairwise disjoint Borel sets and X\A=Sk−1 i=0 Bi. Since fis continuous, the boundary ∂(f−1(H)) of the set f−1(H) is a subset of the set f−1(∂H) for all subsets Hof R. Using (2.2) this implies ∂(X\Bi) = ∂Bi⊂f−1({ti})∪f−1({ti+1})∪∂A for all i= 0, . . . , k −1. Since ti/∈D, i= 0, . . . , k, η0(∂A) = 0 and η0(∂(X\Bi)) ⩽η0¡{x∈X:f(x)=ti}¢+η0¡{x∈X:f(x)=ti+1}¢+η0(∂A), we get η0(∂(X\Bi)) = 0, i= 0, . . . , k −1. Since A⊂X\Bi, we have X\Bi∈ Nx0 for all i= 0, . . . , k −1. Hence condition (iii) implies that ηn(Bi)→η0(Bi) as n→ ∞, i= 0, . . . , k −1. By the triangle inequality ¯¯¯ZX fdηn−ZX fdη0¯¯¯=¯¯¯ZX\A fdηn−ZX\A fdη0¯¯¯ ⩽2 max 0⩽i⩽k−1(ti+1 −ti) + ¯¯¯ k−1 X i=0 ti¡ηn(Bi)−η0(Bi)¢¯¯¯. Hence lim supn→∞ ¯¯RXfdηn−RXfdη0¯¯⩽2 max0⩽i⩽k−1(ti+1 −ti)<2ε. Since ε > 0 is arbitrary, (iv) holds. (iv)⇒(v): It is trivial, since BLx0(X)⊂ Cx0(X). (v)⇒(vi): First let Ube an open neighbourhood of x0. Let ε > 0. We show the existence of a closed neighbourhood Uεof x0such that Uε⊂Uand η0(U\Uε)< ε, 4 and of a function f∈BLx0(X) such that f(x) = 0 for x∈Uε,f(x) = 1 for x∈X\U and 0 ⩽f(x)⩽1 for x∈X. For all B∈ B(X) and for all λ > 0 we use notation Bλ:= ©x∈X:d(x, B)< λª, where d(x, B) := inf{d(x, z) : z∈B}. Since Uis open, we get U=S∞ n=1 Fn, where Fn:= X\(X\U)1/n,n∈N. Then Fn⊂Fn+1,n∈N,Fnis a closed subset of X for all n∈Nand T∞ n=1(X\Fn) = X\U. We also have η0(X\FN)<∞for some sufficiently large N∈Nand X\Fn⊃X\Fn+1 for all n∈N, and hence the continuity of the measure η0implies that limn→∞ η0(X\Fn) = η0(X\U). Since η0(X\U)<∞, there exists some n0∈Nsuch that η0(X\Fn0)−η0(X\U)< ε. Set Uε:= Fn0. Since η0(X\Fn0)−η0(X\U) = η0¡(X\Fn0)\(X\U)¢=η0(U\Fn0), the set Uεis a closed neighborhood of x0,Uε⊂Uand η0(U\Uε)< ε. We show that the function f:X→R,defined by f(x) := min(1, n0d(x, Uε)), x∈X, is an element of BLx0(X), f(x) = 0 for x∈Uε,f(x) = 1 for x∈X\Uand 0 ⩽f(x)⩽1 for x∈X. Note that if x∈Uεthen d(x, Uε) = 0, hence f(x) = 0. And if x∈X\Uthen d(x, Uε)⩾d(X\U, Uε)⩾1/n0, hence f(x) = 1. The fact that 0 ⩽f(x)⩽1, x ∈Xis obvious. To prove that fis Lipschitz, we check that |f(x)−f(y)|⩽n0d(x, y) for all x, y ∈X. If x, y ∈Xwith d(x, y)⩾1/n0then |f(x)−f(y)|⩽1⩽n0d(x, y). If x, y ∈Xwith d(x, y)<1/n0then we have to consider the following four cases apart from changing the role of xand y:x∈X\U,y∈U\Uε;x∈Uε,y∈U\Uε;x, y ∈U\Uεand the case x, y ∈Uεor x, y ∈X\U. Let us consider the case when x, y ∈U\Uεand f(x) = 1, f(y) = n0d(y, Uε). Then d(x, Uε)⩾1/n0, d(y, Uε)⩽1/n0and we get |f(x)−f(y)|= 1 −n0d(y, Uε)⩽n0d(x, y). Indeed, 1/n0⩽d(x, Uε)⩽d(x, y) + d(y, Uε). The case x, y ∈U\Uεand f(y)=1 f(x) = n0d(x, Uε) can be handled similarly. If x, y ∈U\Uεand f(x) = n0d(x, Uε), f(y) = n0d(y, Uε) then |f(x)−f(y)|=n0|d(x, Uε)−d(y, Uε)|⩽n0d(x, y). Indeed, since Uεis closed, we have |d(x, Uε)−d(y, Uε)|⩽d(x, y). If x, y ∈U\Uεand f(x) = f(y) = 1 then |f(x)−f(y)|= 0 ⩽n0d(x, y). The other cases can be handled similarly. Hence f∈BLx0(X) and we get ZX fdη0=ZX\Uε fdη0⩽η0(X\Uε) = η0(X\U) + η0(U\Uε)< η0(X\U)+ε, and RXfdηn⩾RX\Ufdηn=ηn(X\U). Hence by condition (v) we have lim sup n→∞ ηn(X\U)⩽lim sup n→∞ ZX fdηn=ZX fdη0< η0(X\U) + ε. Since ε > 0 is arbitrary, we get (a). 5 Now let Vbe a closed neighbourhood of x0. Let ε > 0. As in case of an open neighbourhood of x0, one can show that there exist an open neighbourhood Vεof x0 such that V⊂Vεand η0(Vε\V)< ε and a function f∈BLx0(X) such that f(x)=0 for x∈V,f(x) = 1 for x∈X\Vεand 0 ⩽f(x)⩽1 for x∈X. Then we get ZX fdη0=ZX\V fdη0=η0(X\Vε) + ZVε\V fdη0 ⩾η0(X\V)−η0(Vε\V)> η0(X\V)−ε, and RXfdηn=RX\Vfdηn⩽ηn(X\V). Hence by condition (v) we have lim inf n→∞ ηn(X\V)⩾lim inf n→∞ ZX fdηn=ZX fdη0> η0(X\V)−ε. Since ε > 0 is arbitrary, we obtain (b). (vi)⇒(iii): The proof can be carried out similarly to the proof of the corresponding part of Theorem 11.1.1 in Dudley [1]. 2 Remark 2.1 Assertion (v) in Theorem 2.1 can be replaced by ZX fdηn→ZX fdη0for all f∈ Cu x0(X), where Cu x0(X) denotes the set of all uniformly continuous functions in Cx0(X). Remark 2.2 By giving a counterexample we show that (a) and (b) in condition (vi) of Theorem 2.1 are not equivalent. For all n∈Nlet ηnbe the Dirac measure δ2on Rconcentrated on 2 and let η0be the Dirac measure δ0on Rconcentrated on 0. Then η0(R\V) = 0 for all closed neighbourhoods Vof 0, hence (b) in condition (vi) of Theorem 2.1 holds. But (a) in condition (vi) of Theorem 2.1 is not satisfied. Indeed, U:= (−1,1) is an open neighbourhood of 0, η0(R\U) = 0, but ηn(R\U) = ηn¡(−∞,−1] ∪[1,+∞)¢= 1, n ∈N, hence lim supn→∞ ηn(R\U) = 1. This counterexample also implies that the equivalence of (c) and (d) in Proposition 1.2.19 in [3] is not valid. Remark 2.3 By giving a counterexample we show that the equivalence of (c) and (d) in Proposition 1.2.13 in [3] is not valid. For all n∈Nlet µnbe the measure 2δ1/n on Rand µbe the Dirac measure δ0on R. We have µ(A)⩽lim infn→∞ µn(A) for all open subsets Aof Rbut there exists some closed subset Fof Rsuch that lim supn→∞ µn(F)> µ(F). If Ais an open subset of Rsuch that 0 ∈Athen µ(A) = 1 and µn(A) = 2 for all sufficiently large n, which implies µ(A)⩽lim infn→∞ µn(A). If Ais an open subset of Rsuch that 0 /∈Athen µ(A) = 0, hence µ(A)⩽lim infn→∞ µn(A) is valid. Let F be the closed interval [−1,1]. Then µ(F) = 1 and µn(F) = 2, n∈N, which yields lim supn→∞ µn(F) = 2. Hence lim supn→∞ µn(F)> µ(F). 6 References [1] R. M. Dudley:Real Analysis and Probability. The Wadsworth & Brooks Cole Mathematics Series, Pacific Grove, California, 1989. [2] J. Jacod and A. N. Shiryayev:Limit Theorems for Stochastic Processes. Springer– Verlag, Berlin, Heidelberg, New York, London, Paris, Tokyo, 1987. [3] M. M. Meerschaert and H.–P. Scheffler:Limit Distributions for Sums of Independent Random Vectors. Heavy Tails in Theory and Practice. John Wiley & Sons, Inc., New York, 2001. M´aty´as Barczy Faculty of Informatics University of Debrecen Pf.12 H–4010 Debrecen Hungary [email protected]u Gyula Pap Faculty of Informatics University of Debrecen Pf.12 H–4010 Debrecen Hungary [email protected]u 7