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Portmanteau theorem for unbounded measures

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Portmanteau theorem for unbounded measures

Author: Barczy, Mátyás; Pap, Gyula
Year: 2006
Source: https://dea.lib.unideb.hu/bitstreams/a52ae2a0-5dee-4e30-b7f7-88db5cae26f9/download
Po man eau heo em o unbounded measu es
M´
a y´
as Ba czy and Gyula Pap
Uni e si y o Deb ecen, Hunga y
Abs ac
We p o e an analogue o he po man eau heo em on weak con e gence o p oba-
bili y measu es allowing measu es which a e unbounded on an unde lying me ic space
bu ini e on he complemen o any Bo el neighbou hood o a ixed elemen .
1 In oduc ion
Weak con e gence o p obabili y measu es on a me ic space has a e y impo an ole in
p obabili y heo y. The well known po man eau heo em due o A. D. Alexand o (see
o example Theo em 11.1.1 in Dudley [1]) p o ides use ul condi ions equi alen o weak
con e gence o p obabili y measu es; any o hem could se e as he de ini ion o weak
con e gence. P oposi ion 1.2.13 in he book o Mee schae and Sche le [3] gi es an analogue
o he po man eau heo em o bounded measu es on Rd. Mo eo e , P oposi ion 1.2.19 in
[3] gi es an analogue o special unbounded measu es on Rd, mo e p ecisely, o ex ended
eal alued measu es which a e ini e on he complemen o any Bo el neighbou hood o
0∈Rd.
By gi ing coun e examples we show ha he equi alences o (c) and (d) in P oposi-
ions 1.2.13 and 1.2.19 in [3] a e no alid (see ou Rema ks 2.2 and 2.3). We e o mula e
P oposi ion 1.2.19 in [3] in a mo e de ailed o m adding new equi alen asse ions o i (see
Theo em 2.1). Mo eo e , we no e ha Theo em 2.1 gene alizes he equi alence o (a) and
(b) in Theo em 11.3.3 o [1] in wo aspec s. On he one hand, he equi alence is ex ended
allowing no necessa ily ini e measu es which a e ini e on he complemen o any Bo el
neighbou hood o a ixed elemen o an unde lying me ic space. On he o he hand, we do
no assume he sepa abili y o he unde lying me ic space o p o e he equi alence. Bu we
men ion ha his la e possibili y is hiddenly con ained in P oblem 3, p. 312 in [1]. Fo
comple eness we gi e a de ailed p oo o Theo em 2.1. Ou p oo goes along he lines o he
p oo o he o iginal po man eau heo em and di e s om he p oo o P oposi ion 1.2.19
in [3].
Key wo ds and ph ases: Weak con e gence o bounded measu es; po man eau heo em; L´e y measu e
The i s au ho has been suppo ed by he Hunga ian Scien i ic Resea ch Fund unde G an No. OTKA–
F046061/2004. The au ho s ha e been suppo ed by he Hunga ian Scien i ic Resea ch Fund unde G an
No. OTKA–T048544/2005.
1
To shed some ligh on he sense o a po man eau heo em o unbounded measu es, le
us conside he ques ion o weak con e gence o in ini ely di isible p obabili y measu es µn,
n∈N owa ds an in ini ely di isible p obabili y measu e µ0in case o he eal line R.
Theo em VII.2.9 in Jacod and Shi yaye [2] gi es equi alen condi ions o weak con e gence
µn
w
→µ0. Among hese condi ions we ha e
ZR
dηn→ZR
dη0 o all ∈ C2(R), (1.1)
whe e ηn,n∈Z+a e nonnega i e, ex ended eal alued measu es on Rwi h ηn({0}) = 0
and RR(x2∧1) dηn(x)<∞(i.e., L´e y measu es on R) co esponding o µn, and C2(R)
is he se o all eal alued bounded con inuous unc ions on R anishing on some
Bo el neighbou hood o 0 and ha ing a limi a in ini y. Theo em 2.1 is abou equi alen
e o mula ions o (1.1) when i holds o all eal alued bounded con inuous unc ions on R
anishing on some Bo el neighbou hood o 0.
2 An analogue o he po man eau heo em
Le Nand Z+be he se o posi i e and nonnega i e in ege s, espec i ely. Le (X, d)
be a me ic space and x0be a ixed elemen o X. Le B(X) deno e he σ-algeb a o
Bo el subse s o X. A Bo el neighbou hood Uo x0is an elemen o B(X) o which
he e exis s an open subse e
Uo Xsuch ha x0∈e
U⊂U. Le Nx0deno e he se
o all Bo el neighbou hoods o x0, and he se o bounded measu es on Xis deno ed by
Mb(X). The exp ession ”a measu e µon X” means a measu e µon he σ-algeb a
B(X).
Le C(X), Cx0(X) and BLx0(X) deno e he spaces o all eal alued bounded con inuous
unc ions on X, he se o all elemen s o C(X) anishing on some Bo el neighbou hood
o x0, and he se o all eal alued bounded Lipschi z unc ions anishing on some Bo el
neighbou hood o x0, espec i ely.
Fo a measu e ηon Xand o a Bo el subse B∈ B(X), le η|Bdeno e he
es ic ion o ηon o B, i.e., η|B(A) := η(B∩A) o all A∈ B(X).
Le µn,n∈Z+be bounded measu es on X. We w i e µn
w
→µi µn(A)→µ(A)
o all A∈ B(X) wi h µ(∂A) = 0. This is called weak con e gence o bounded measu es
on X.
Now we o mula e a po man eau heo em o unbounded measu es.
Theo em 2.1 Le (X, d)be a me ic space and x0be a ixed elemen o X. Le ηn,
n∈Z+, be measu es on Xsuch ha ηn(X U)<∞ o all U∈ Nx0and o all
n∈Z+. Then he ollowing asse ions a e equi alen :
(i) RX U dηn→RX U dη0 o all ∈ C(X),U∈ Nx0wi h η0(∂U) = 0,
(ii) ηn|X U
w
→η0|X U o all U∈ Nx0wi h η0(∂U) = 0,
2
(iii) ηn(X U)→η0(X U) o all U∈ Nx0wi h η0(∂U) = 0,
(i ) RX dηn→RX dη0 o all ∈ Cx0(X),
( ) RX dηn→RX dη0 o all ∈BLx0(X),
( i) he ollowing inequali ies hold:
(a) lim sup
n→∞
ηn(X U)⩽η0(X U) o all open neighbou hoods Uo x0,
(b) lim in
n→∞ ηn(X V)⩾η0(X V) o all closed neighbou hoods Vo x0.
P oo . (i)⇒(ii): Le Ube an elemen o Nx0wi h η0(∂U) = 0. No e ηn|X U∈ Mb(X),
n∈Z+. By he equi alence o (a) and (b) in P oposi ion 1.2.13 in [3], o p o e
ηn|X U
w
→η0|X Ui is enough o check RX dηn|X U→RX dη0|X U o all ∈ C(X).
Fo his i su ices o show ha o all eal alued bounded measu able unc ions hon X,
o all A∈ B(X) and o all n∈Z+we ha e
ZX
hdηn|A=ZA
hdηn.(2.1)
By Beppo-Le i’s heo em, a s anda d measu e- heo e ic a gumen implies (2.1).
(ii)⇒(iii): Le Ube an elemen o Nx0wi h η0(∂U) = 0. By (ii), we ha e ηn|X U
w
→
η0|X U. Since η0|X U(∂X) = η0|X U(∅) = 0, we ge ηn(X U) = ηn|X U(X)→η0|X U(X) =
η0(X U), as desi ed.
(iii)⇒(ii): Le Ube an elemen o Nx0wi h η0(∂U) = 0 and le B∈ B(X) be such
ha η0|X U(∂B) = 0. We ha e o 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), i is enough o check η0¡∂¡X (B∩(X U))¢¢= 0. Fi s we show
∂¡B∩(X U)¢⊂¡∂B ∩(X U)¢∪∂U o all subse s B,Uo X. (2.2)
Le xbe an elemen o ∂¡B∩(X U)¢and (yn)n⩾1, (zn)n⩾1be wo sequences such
ha limn→∞ yn= limn→∞ zn=xand yn∈B∩(X U), zn∈X (B∩(X U)), n∈N.
Then o all n∈Nwe ha e one o wo o he ollowing possibili ies:
•yn∈B, yn∈X Uand zn∈X B,
•yn∈B, yn∈X Uand zn∈U.
Then we ge x∈¡∂B ∩((X U)∪∂U)¢∪¡∂U ∩(B∪∂B)¢∪¡∂B ∩∂U¢. Since ∂B ∩
((X U)∪∂U)⊂(∂B ∩(X U)) ∪∂U, we ha e x∈¡∂B ∩(X U)¢∪∂U, as desi ed.
Using (2.2) we ge η0¡∂¡X (B∩(X U))¢¢⩽η0¡∂B∩(X U)¢+η0(∂U) = 0. Indeed, by
he assump ions η0¡∂B∩(X U)¢= 0 and η0(∂U) = 0. Hence η0¡∂¡X (B∩(X U))¢¢= 0.
(ii)⇒(i): Using again he equi alence o (a) and (b) in P oposi ion 1.2.13 in [3] and
(2.1) we ob ain (i).
(iii)⇒(i ): Le be an elemen o Cx0(X). Then he e exis s A∈ Nx0such ha
(x) = 0 o all x∈Aand η0(∂A) = 0. Indeed, he unc ion 7→ η0¡{x∈X:
3
d(x, x0)⩾ }¢ om (0,+∞) in o Ris mono one dec easing, hence he se © ∈(0,+∞) :
η0({x∈X:d(x, x0) = })>0ªo i s discon inui ies is a mos coun able. Consequen ly,
o all e
U∈ Nx0 he e exis s some > 0 such ha U:= {x∈X:d(x, x0)< }∈Nx0,
U⊂e
Uand η0(∂U) = 0. (A his s ep we use ha an elemen e
Uo Nx0con ains
an open subse o Xcon aining x0.) This implies he exis ence o A. We show ha
he se D:= © ∈R:η0¡{x∈X: (x) = }¢>0ªis a mos coun able. The unc ion
F:R→[0, η0(X A)], de ined by
F( ) := η0¡{x∈X A: (x)< }¢, ∈R,
is mono one inc easing and le con inuous. (No e ha η0(X A)<∞,by he assump ion on
η0.) Hence i has a mos coun ably many discon inui y poin s, and 0∈Ris a discon inui y
poin o Fi and only i F( 0+ 0) > F( 0),i.e., η0¡{x∈X A: (x) = 0}¢>0. I
06= 0, hen {x∈X: (x) = 0}={x∈X A: (x) = 0}, hus 06= 0 is a discon inui y
poin o Fi and only i η0({x∈X: (x) = 0})>0. Hence i ∈D hen = 0 o
is a discon inui y poin o F, consequen ly Dis a mos coun able. Since is bounded
and Dis a mos coun able, he e exis s a eal numbe M > 0 such ha −M, M /∈D
and | (x)|< M o x∈X. Le ε > 0. Choose eal numbe s i, i = 0, . . . , k such ha
−M= 0< 1<···< k=M, i/∈D, i = 0, . . . , k and max0⩽i⩽k−1( i+1 − i)< ε. The
coun abili y o Dimplies he exis ence o i,i= 0, . . . , k. Le
Bi:= −1¡[ i, i+1)¢∩(X A) = nx∈X A: i⩽ (x)< i+1o
o all i= 0, . . . , k −1. Then Bi,i= 0, . . . , k −1, a e pai wise disjoin Bo el se s
and X A=Sk−1
i=0 Bi. Since is con inuous, he bounda y ∂( −1(H)) o he se
−1(H) is a subse o he se −1(∂H) o all subse s Ho R. Using (2.2) his implies
∂(X Bi) = ∂Bi⊂ −1({ i})∪ −1({ i+1})∪∂A o all i= 0, . . . , k −1. Since i/∈D,
i= 0, . . . , k, η0(∂A) = 0 and
η0(∂(X Bi)) ⩽η0¡{x∈X: (x)= i}¢+η0¡{x∈X: (x)= i+1}¢+η0(∂A),
we ge η0(∂(X Bi)) = 0, i= 0, . . . , k −1. Since A⊂X Bi, we ha e X Bi∈ Nx0
o all i= 0, . . . , k −1. Hence condi ion (iii) implies ha ηn(Bi)→η0(Bi) as n→ ∞,
i= 0, . . . , k −1. By he iangle inequali y
¯¯¯ZX
dηn−ZX
dη0¯¯¯=¯¯¯ZX A
dηn−ZX A
dη0¯¯¯
⩽2 max
0⩽i⩽k−1( i+1 − i) + ¯¯¯
k−1
X
i=0
i¡ηn(Bi)−η0(Bi)¢¯¯¯.
Hence lim supn→∞ ¯¯RX dηn−RX dη0¯¯⩽2 max0⩽i⩽k−1( i+1 − i)<2ε. Since ε > 0 is
a bi a y, (i ) holds.
(i )⇒( ): I is i ial, since BLx0(X)⊂ Cx0(X).
( )⇒( i): Fi s le Ube an open neighbou hood o x0. Le ε > 0. We show he
exis ence o a closed neighbou hood Uεo x0such ha Uε⊂Uand η0(U Uε)< ε,
4
and o a unc ion ∈BLx0(X) such ha (x) = 0 o x∈Uε, (x) = 1 o x∈X U
and 0 ⩽ (x)⩽1 o x∈X.
Fo all B∈ B(X) and o all λ > 0 we use no a ion Bλ:= ©x∈X:d(x, B)< λª,
whe e d(x, B) := in {d(x, z) : z∈B}. Since Uis open, we ge U=S∞
n=1 Fn, whe e
Fn:= X (X U)1/n,n∈N. Then Fn⊂Fn+1,n∈N,Fnis a closed subse o X
o all n∈Nand T∞
n=1(X Fn) = X U. We also ha e η0(X FN)<∞ o some
su icien ly la ge N∈Nand X Fn⊃X Fn+1 o all n∈N, and hence he con inui y
o he measu e η0implies ha limn→∞ η0(X Fn) = η0(X U). Since η0(X U)<∞,
he e exis s some n0∈Nsuch ha η0(X Fn0)−η0(X U)< ε. Se Uε:= Fn0. Since
η0(X Fn0)−η0(X U) = η0¡(X Fn0) (X U)¢=η0(U Fn0), he se Uεis a closed
neighbo hood o x0,Uε⊂Uand η0(U Uε)< ε.
We show ha he unc ion :X→R,de ined by (x) := min(1, n0d(x, Uε)), x∈X,
is an elemen o BLx0(X), (x) = 0 o x∈Uε, (x) = 1 o x∈X Uand 0 ⩽ (x)⩽1
o x∈X.
No e ha i x∈Uε hen d(x, Uε) = 0, hence (x) = 0. And i x∈X U hen
d(x, Uε)⩾d(X U, Uε)⩾1/n0, hence (x) = 1. The ac ha 0 ⩽ (x)⩽1, x ∈Xis
ob ious. To p o e ha is Lipschi z, we check ha
| (x)− (y)|⩽n0d(x, y) o all x, y ∈X.
I x, y ∈Xwi h d(x, y)⩾1/n0 hen | (x)− (y)|⩽1⩽n0d(x, y). I x, y ∈Xwi h
d(x, y)<1/n0 hen we ha e o conside he ollowing ou cases apa om changing he
ole o xand y:x∈X U,y∈U Uε;x∈Uε,y∈U Uε;x, y ∈U Uεand he case
x, y ∈Uεo x, y ∈X U.
Le us conside he case when x, y ∈U Uεand (x) = 1, (y) = n0d(y, Uε). Then
d(x, Uε)⩾1/n0, d(y, Uε)⩽1/n0and we ge | (x)− (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 (y)=1
(x) = n0d(x, Uε) can be handled simila ly. I x, y ∈U Uεand (x) = n0d(x, Uε),
(y) = n0d(y, Uε) hen
| (x)− (y)|=n0|d(x, Uε)−d(y, Uε)|⩽n0d(x, y).
Indeed, since Uεis closed, we ha e |d(x, Uε)−d(y, Uε)|⩽d(x, y). I x, y ∈U Uεand
(x) = (y) = 1 hen | (x)− (y)|= 0 ⩽n0d(x, y).
The o he cases can be handled simila ly. Hence ∈BLx0(X) and we ge
ZX
dη0=ZX Uε
dη0⩽η0(X Uε) = η0(X U) + η0(U Uε)< η0(X U)+ε,
and RX dηn⩾RX U dηn=ηn(X U). Hence by condi ion ( ) we ha e
lim sup
n→∞
ηn(X U)⩽lim sup
n→∞ ZX
dηn=ZX
dη0< η0(X U) + ε.
Since ε > 0 is a bi a y, we ge (a).
5

Now le Vbe a closed neighbou hood o x0. Le ε > 0. As in case o an open
neighbou hood o x0, one can show ha he e exis an open neighbou hood Vεo x0
such ha V⊂Vεand η0(Vε V)< ε and a unc ion ∈BLx0(X) such ha (x)=0
o x∈V, (x) = 1 o x∈X Vεand 0 ⩽ (x)⩽1 o x∈X. Then we ge
ZX
dη0=ZX V
dη0=η0(X Vε) + ZVε V
dη0
⩾η0(X V)−η0(Vε V)> η0(X V)−ε,
and RX dηn=RX V dηn⩽ηn(X V). Hence by condi ion ( ) we ha e
lim in
n→∞ ηn(X V)⩾lim in
n→∞ ZX
dηn=ZX
dη0> η0(X V)−ε.
Since ε > 0 is a bi a y, we ob ain (b).
( i)⇒(iii): The p oo can be ca ied ou simila ly o he p oo o he co esponding
pa o Theo em 11.1.1 in Dudley [1]. 2
Rema k 2.1 Asse ion ( ) in Theo em 2.1 can be eplaced by
ZX
dηn→ZX
dη0 o all ∈ Cu
x0(X),
whe e Cu
x0(X) deno es he se o all uni o mly con inuous unc ions in Cx0(X).
Rema k 2.2 By gi ing a coun e example we show ha (a) and (b) in condi ion ( i)
o Theo em 2.1 a e no equi alen . Fo all n∈Nle ηnbe he Di ac measu e δ2on
Rconcen a ed on 2 and le η0be he Di ac measu e δ0on Rconcen a ed on 0.
Then η0(R V) = 0 o all closed neighbou hoods Vo 0, hence (b) in condi ion ( i)
o Theo em 2.1 holds. Bu (a) in condi ion ( i) o Theo em 2.1 is no sa is ied. Indeed,
U:= (−1,1) is an open neighbou hood o 0, η0(R U) = 0, bu
ηn(R U) = ηn¡(−∞,−1] ∪[1,+∞)¢= 1, n ∈N,
hence lim supn→∞ ηn(R U) = 1. This coun e example also implies ha he equi alence o
(c) and (d) in P oposi ion 1.2.19 in [3] is no alid.
Rema k 2.3 By gi ing a coun e example we show ha he equi alence o (c) and (d) in
P oposi ion 1.2.13 in [3] is no alid. Fo all n∈Nle µnbe he measu e 2δ1/n on Rand
µbe he Di ac measu e δ0on R. We ha e µ(A)⩽lim in n→∞ µn(A) o all open subse s
Ao Rbu he e exis s some closed subse Fo Rsuch ha lim supn→∞ µn(F)> µ(F).
I Ais an open subse o Rsuch ha 0 ∈A hen µ(A) = 1 and µn(A) = 2 o all
su icien ly la ge n, which implies µ(A)⩽lim in n→∞ µn(A). I Ais an open subse o
Rsuch ha 0 /∈A hen µ(A) = 0, hence µ(A)⩽lim in n→∞ µn(A) is alid. Le F
be he closed in e al [−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
Re e ences
[1] R. M. Dudley:Real Analysis and P obabili y. The Wadswo h & B ooks Cole Ma h-
ema ics Se ies, Paci ic G o e, Cali o nia, 1989.
[2] J. Jacod and A. N. Shi yaye :Limi Theo ems o S ochas ic P ocesses. Sp inge –
Ve lag, Be lin, Heidelbe g, New Yo k, London, Pa is, Tokyo, 1987.
[3] M. M. Mee schae and H.–P. Sche le :Limi Dis ibu ions o Sums o In-
dependen Random Vec o s. Hea y Tails in Theo y and P ac ice. John Wiley & Sons,
Inc., New Yo k, 2001.
M´a y´as Ba czy
Facul y o In o ma ics
Uni e si y o Deb ecen
P .12
H–4010 Deb ecen
Hunga y
[email p o ec ed]u
Gyula Pap
Facul y o In o ma ics
Uni e si y o Deb ecen
P .12
H–4010 Deb ecen
Hunga y
[email p o ec ed]u
7