DIANA M ˘ARGINEAN PETROVAI
LetE be a semisaturated set (with respect to a proper sub-Markovian resolvent U of kernels on a Lusin measurable space) andτ is a Ray topology onE (with respect toU). We show thatξ∈ExcU(the set of all excessive measures associated withU) is a potential if and only if for any closed subsetsF1, F2 ofE such that F1∩F2=∅we have∗RF1ξf∗RF2ξ= 0.Also forξ∈PotU, whereξ=µ◦U,for any Ray compact subsetF of E we haveξF =f{ξG∩E |G∈τ, G⊃F}if and only ifµξ.
AMS 2000 Subject Classification: 31D05, 60J45.
Key words: excessive measure, excessive function, Ray topology.
INTRODUCTION
We consider a proper sub-Markovian resolvent U = (Uα)α>0 of kernels on a Lusin measurable space (E,B) such that the set EU of allB-measurable, U-excessive functions on E which areU-a.e. finite is min-stable, contains the positive constant functions, and generates the σ-algebra B.
We suppose that the setE is semisaturated (see [3]) with respect to U, i.e., any U-excessive measure dominated by a potential also is a potential.
(E is called saturated provided that every m ∈ExcU with L(m,1)< ∞ is a potential, where L is the energy functional associated with U).We denote by ExcU the set of all U-excessive measures on E (see [11], [12]). The specific order on ExcU is defined as follows: for ξ, ξ0 ∈ExcU we haveξ ξ0 if and only if there exists η ∈ExcU such thatξ+η=ξ0.
In this paper we deepen the study of the question when a measure ξ ∈ ExcU is a potential. This study has been initiated in [11], [12]. For every σ-finite measure µ on (E,B) such that there exists ξ ∈ExcU with µ≤ξ, we putR(µ) =V{ξ ∈ExcU |µ≤ξ}.
IfA∈ Bu and ξ∈ExcU, we put∗RAξ=R(1A·ξ) andξA=∗ RAξfξ.
Recall that the topology onEgenerated by a Ray cone on (E,B) is called a Ray topology.
In Section 2 we show (Theorem 2.1) that if ξ is a potential and (Fn)n is a decreasing sequence of finely closed subsets of E such that T
nFn = ∅, then ξFn ↓0.
MATH. REPORTS10(60),4 (2008), 337–345
Using this fact we show (Theorem 2.2) that ifξ ∈ExcU and τ is a Ray topology onE the following assertions are equivalent:
(1)ξ is potential;
(2) ifF1, F2are closed subsets ofEwith respect toτ such thatF1∩F2=∅, then ∗RF1ξf∗RF2ξ= 0;
(3) ifF1, F2are closed subsets ofEwith respect toτ such thatF1∩F2=∅, then ξF1 fξF2 = 0.
In Section 3, we show (Theorem 3.1) that a measureξ ∈ExcU is regular if and only if there existsµ∈ M+(E) such thatξ=µ◦U,whereµis aσ-finite measure on E which does not charge any B-measurable semipolar set.
1. PRELIMINARIES
Throughout the paperU = (Uα)α>0 is a proper sub-Markovian resolvent on (E,B) as in the introduction. If A ⊂E and s∈ EU, then the r´eduite of s on A is the functionRAson E defined by RAs:= inf{t∈ EU |t≥sonA}. If A∈ B ands∈ EU, thenRAsis universally measurable (see [5]) and we denote by BAs its U-excessive regularization. (The fine toplogy is the topology on E generated byEU.) The function BAs is called the baleyage of s on A. For every A∈ Bu we set
A∗ =n
x∈A|lim
n infnUn(1A)(x) = 1o ,
where 1A is the characteristic function of A. Clearly, A∗ ∈ Bu and A∗ ∈ B provided that A∈ B.
Theorem 1. For every A∈ Bu the following assertions hold:
(1)if A is finely open then A=A∗;
(2) the setA\A∗ isU-negligible and for each s∈ EU the functionRA∗s is U-excessive and there exists a sequence(fn)n in bpBu such that
(1) U fn↑RA∗s and fn= 0 on E\A∗;
(3) if A ∈ B and s ∈ EU, then RA∗s ∈ EU and there exists a sequence (fn)n in bpB such that (1)holds.
For theproof see Theorem 1.3.8 in [6].
Theorem2. For anyξ∈ExcU, s∈ EU andA∈ Buwe haveL(∗RAξ, s) = L(ξ, RA∗s) (L is the energy functional associated withU).
For theproof see Theorem 1.4.12 in [6].
It is known that ifA∈ B thenBAs=RAson E\A(see [5]).
For ξ ∈ ExcU, a subset A of E is called ξ-polar if there exists a U- excessive function son E such thats= +∞ on A and s <∞ ξ-a.e. Ifµis a
σ-finite measure onE such thatµ◦U ∈ExcU, then we sayµ-polar instead of µ◦U polar. A set A is called polar if it is ξ-polar for everyξ∈ExcU.
A subsetM ofEis called nearlyB-measurable of for every finite measure µon (E,B) there exists aB-measurable setM0 ⊂M such that the setM\M0
is µ-polar andµ-negligible.
We denote byBnthe family of all nearlyB-measurable subsets ofE (see [6]). Obviously, Bn is a σ-algebra on E and B ⊂ Bn ⊂ Bu (the set of all universally B-measurable subsets ofE).
Recall that a set A ∈ B is said to be thin at a point x ∈ E if there exists s ∈ EU such that BAs(x) < s(x). A subset M of E is called thin at x if there exists A ∈ B such that M ⊂ A, which is thin at x. A set M is called totally thin if it is thin at any point of E.For any subset M of E the set b(M) :={x ∈E |M is not thin at x} is usually called the base of M. It is a finely closed set and b(M) = b(Mf) ⊂ Mf, where Mf denotes the fine closure of M. If M is nearly B-measurable and p := U f0 is bounded with f0
B-measurable and, 0< f0 ≤1, then
M is thin atx⇔BMp(x)< p(x); b(M) = [BMp=p].
A subset M of E is called basic (respectively subbasic) if b(M) = M (respectively M ⊂b(M)).If M is subbasic, then Mf is basic.
A subset of E is called semipolar if it is a countable union of totally thin sets.
Remark. SinceRA∗s∈ EU for anyA∈ Bu(see [5]), we haveRA∗s=BA∗s for all s∈ EU. Hence BA∗U f0 = U f0 on A∗, where 0< f0 ≤1,and f0 is B- measurable such thatU f0 is bounded. Therefore,A∗ is a subbasic set.
For ξ ∈ ExcU, a set A ∈ B is called ξ-semipolar if it is a union of the form A =A0∪A1, where A0, A1 ∈ B, A0 isξ-polar and A1 is semipolar. An arbitrary subset of E is calledξ-semipolar if it is a subset of aξ-semipolar set from B.
Let (E1,B1) be a second measurable space withE ⊂E1 and B1|E =B.
A proper sub-Markovian resolvent U1 = (Uα1)α>0 on (E1,B1) is called an extension ofU = (Uα)α>0 to (E1,B1) if
(1) (Uα1f)|E =Uα(f|E) for all α >0 and f ∈pB1; (2)Uα1(1A) = 0 for all α >0 andA∈ B1, A⊂E1\E;
(3) the set EU1 is min-stable, contains the positive constant functions, and generates B1.
It is known that there exists a sub-Markovian resolvent U1(Uα1)α>0 on a Lusin measurable space (E1,B1), extension ofU such thatE1 is saturated with respect to U1 (see [6]).
The resolvent U being proper, there exists a bounded resolvent V = (Vα)α>0 on (E,B) such that the V-excessive functions coincide with the U- excesive ones.
A Ray cone on (E,B) is a convex coneR of bounded B-measurable U- excessive functions such that there exists a bounded sub-Markovian resolvent V = (Vα)l>0 on (E,B) withEU =EV and such that
(1) the cone Rcontains the positive constant functions;
(2) the cone Ris min-stable;
(3)V0(p(R − R)⊂ R,where p(R − R) ={u∈ R − R, u≥0};
(4)Vα(R)⊂ R,for all α >0;
(5) the cone Ris separable with respect to the uniform norm;
(6) theσ-algebra on E generated by Rcoincides withB.
We say that the Ray coneRis associated withV.The topology onEgenerated by a Ray cone is called a Ray topology.
For every Ray coneRon (E,B), there exists a compact metrizable space Y such thatEis a dense subset ofY,for everyf ∈ Rthere exists a continuous function ˜f on Y with ˜f|E = f and the set ˜R = {f˜| f ∈ R} separates the points of Y. We say that Y is the Ray compactification ofE with respect to R.For the proof see Theorem 1.5.2 in [6].
2. SOME PROPERTIES OF PotU
As usual, PotU ={ξ ∈ExcU |ξ=µ◦U, µ∈ M+(E)},whereM+(E) is the set of all positive measures on E.
Note that PotU is increasingly dense in ExcU with respect to the natural order and is a solid subset of ExcU with respect to the specific order (see [7]).
Remark. Ifξ =µ◦U ∈PotU and A∈ Bu then
∗RA(µ◦U)(f) =L(∗RA(µ◦U), U f) =L(µ◦U, RA∗U f)
=µ(RA∗U f) = ((µ◦RA∗)◦U)(f)
for all f ∈ pB. Hence ∗RA(µ◦U) = (µ◦RA∗)◦U and ∗RA(µ◦U) ≤µ◦U.
Therefore,∗RA(µ◦U)∈PotU andµ◦RA∗ is carried byAifAis a finely closed subset of E.
Theorem3. Let E be a semisaturated set(with respect to U), ξ ∈PotU
and(Fn)na decreasing sequence of finely closed subsets ofEsuch thatT
nFn=
∅.Then ξFn ↓0.
Proof. Letξ =µ◦U.It follows from the previous remark that ∗RFn(µ◦ U) = (µ◦RFn∗)◦U =µn◦U,where µnis carried by Fn.
Hence ξFn = νn◦U, where νn is carried by Fn and so, fξFn = ν ◦U, where νV
nνn, and so ν = 0.Therefore, ξFn ↓0.
Proposition 1. If ξ∈ExcU then∗R[f >0]ξ(f) =ξ(f) for allf ∈pB.
Proof. For all f ∈ pB we have ∗R[f >0]ξ(f) = L(∗R[f >0]ξ, U f) = L(ξ, R[f >0]∗U f). Because A∗ is a subbasic set and U(1A\A∗) = 0, we have R[f >0]∗U f R[f >0]∗U(f ·1[f >0]∗) = U(f ·1[f >0]∗)U f for all f ∈ pB. It fol- lows that L(ξ, R[f >0]∗U f) = L(ξ, U f) = ξ(f) for all f ∈ pB. Therefore,
∗R[f >0]ξ(f) =ξ(f) for allf ∈pB.
Theorem 4. Let ξ ∈ ExcU and let τ be a Ray toplogy on E. Assume that E is semisaturated (with respect to U). Then the following assertions are equivalent:
(1)ξ is a potential;
(2)IfF1, F2are closed subsets ofEwith respect toτ such thatF1∩F2 =∅, then ∗RF1ξf∗RF2ξ= 0;
(3)IfF1, F2are closed subsets ofEwith respect toτ such thatF1∩F2 =∅, then ξF1 fξF2 = 0.
Proof. (1) ⇒(2). Assume thatξ ∈PotU,i.e., there exists µ∈ Mσf+(E) (the set of positive σ-finite measures onE) withξ=µ◦U.Let F1, F2 be two closed subsets of E with respect to τ such that F1∩F2 =∅. We show that
∗RF1ξf∗RF2ξ = 0.
From the previous remark we have ∗RFiξ =µ◦RFi∗◦U, i = 1,2, hence
∗RF1ξf∗RF2ξ = (µ◦RF1∗∧µ◦RF2∗)◦U, and sinceµ◦RF1∗(respectivelyµ◦RF2∗) is a measure carried byF1(respective byF2), we deduce thatµ◦RF1∗∧µ◦RF2∗ = 0, therefore, ∗RF1ξf∗RF2ξ= 0.
(2)⇒(3).Trivial.
(3) ⇒ (1). Assume first that L(ξ,1) < ∞. Let R be a Ray cone such that τ is associated with R.Denote by (Y, τ) the Ray compactification of E with respect to R, and let E1 ⊂ Y, E ⊂ E1 such that there exists a proper sub-Markovian resolvent U1(Uα1)α>0 on (E1,B1), where B1 =B(Y)|E1, which is an extension of U = (Uα)α>0 and such that E1 is saturated with respect to U1. It is known thatE is a finely dense subset of E1 (with respect to the fine topology generated by U1), and every s ∈ EU has a unique extension
˜
s ∈ EU1. Also, ExcU = ExcU1 and for any ξ ∈ ExcU and s ∈ EU we have L(ξ, s) =LU1(ξ,˜s).
SinceL(ξ,1)<∞, we haveξ=µ◦U1,whereµis a measure on (E1,B1).
It is sufficient to show that µ(E1\E) = 0.
Assume that there exists a compact setK⊂E1\E such thatµ(K)>0.
We substitute the excessive measure ξ by the measure µ|K ◦U1 and have µ|K◦U1ξ, hence µ|K◦U1 satisfies the property of (3).
Let now s∈ EU such that s∈ Rand is a generator of EU.Remark that if D is an open set of Y and K ⊂ D, then the fine closure of D∩E in E1
includes K.Choose a decreasing sequence of closed sets (Fn)n of Y such that Fn+1 ⊂
◦
Fn,T
n
Fn = K and inf
n R
◦
Fn∩Es≥ s on K. Remark that the set T = S
n
(Fn\F◦n+1) is a closed subset ofE.Indeed, let (xn)nbe a sequence inT which converges to a pointx∈E.Then there existsn0 ∈Nsuch that for anyn > n0 we havexn6∈F◦n0+1.On the other hand, there exists a sequence (pn)n∈Nwith xn∈Fpn−F◦pn+1for alln∈N.From the above we deduce thatpn≤n0.Hence xn∈
n0
S
k=1
Fkfor alln≥n0.Because
n0
S
k=1
(Fk−F◦k+1) =
n0
S
k=1
Fk\F◦n0+1, we have xn∈
n0
S
k=1
(Fk−
◦
Fk+1),for alln > n0and so,x∈
n0
S
k=1
(Fk−
◦
Fk+1).It follows that x∈T.Applying the above considerations to the sequences (F2n)nand (F3n)n, we deduce that the sets T1 =S
n
(F2n−F◦2n+1) andT2 =S
n
(F3n−F◦3n+1) are closed subsets of E such that T1 ∩T2 = ∅ and RT1s ≥ RKs, RT2s ≥ RKs.
It follows that µ|K(RT1s) = µ|K(RT2s) = µ|K(s), i.e., ∗RT1(µ|K ◦ U1) =
∗RT2(µ|K ◦U1) = µ|K ◦U1. From (3) we deduce that µ|K ◦U1 = 0, i.e., µ|K = 0.
In general, let f0 ∈ pB, with 0 < f0 ≤ 1 such that ξ(f0) < 0 and U f0 is bounded. Denote q := U f0 and let W = (Wα)α>0 be a sub-Markovian resolvent with the initial kernel W f = U(fq0f) for all f ∈ pB. A positive measure η on (E,B) is U-excessive if and only if η0 = f0 ·η is W-excessive.
Also, s∈ E(W) if and only ifq·s∈ EU.We haveLU(η, s) =LW(f0·η,sq) for every η∈ExcU,hence η(f0) =LU(η, U f0)LW(f0·η,1).
Since ξ(f0) < ∞, f0 ·ξ has the property LW(f0·ξ,1) < ∞. If τR is a Ray topology generated by a Ray cone R with respect to U, which contains U f0, thenRq ={r/q|r∈ R}is a Ray topology onE associated withW and τR=τRq.
Let now F ⊂ E be a closed set in the Ray topology τR. We have f0U ∗RFξ =W∗ RF(f0·ξ) and alsoξ =µ◦U if and only if f0·ξ = (q·µ)◦W.
Indeed, we have
f0·(µ◦U)(f) =µ(U(f0·f))(q·µ)
U(f0·f) q
=q·µ(W f), for all f ∈pB.Also, for η1, η2 ∈ExcU we have
η1 ≤η2 ⇔f0·η1≤f0·η2, η1η2⇔f0·η1 f0·η2.
It follows from the above that (ξhas the property thatU ∗RF1ξfU ∗RF2ξ= 0 for every τR-closed subsets F1, F2 of E such that F1∩F2 =∅) if and only
if (ξ0 = f0·ξ has the property that W∗RF1ξ0 fW∗ RF2ξ0 = 0 for every τR- closed subsets F1, F2 of E such that F1∩F2 =∅). From the above and from LW(ξ0,1)<∞, we deduce that ξ∈PotU iff ξ0 ∈PotW.
3. THE REGULAR EXCESSIVE MEASURES
As usual, for anyξ ∈ExcU set Excξ:={η ∈ExcU |ηξ}.
We recall that an elementξ ∈ExcU is calledregular if for any increasing sequence (ξn)n∈N⊂ExcU such thatW
n∈Nξn=ξ we have V
n∈NR(ξ−ξn) = 0 (see [3]). Note (see [5], [6]) that a U-excessive measure ξ = µ◦U is regular iff µ is a σ-finite measure on E which does not charge any B-measurable semipolar set.
For theproof see Theorem 3.4.5 in [6].
Theorem 5. Let E be a semisaturated set with respect to U. Then a measure ξ ∈ExcU is regular if and only if there exists µ∈ M+(E) such that ξ = µ◦U, where µ is a σ-finite measure on E which does not charge any B-measurable semipolar set.
Proof. Let (µn)n a sequence in M+(E) such that µn◦U ↑ξ. Becauseξ is regular, R(ξ−µn◦U) ↓0.Putting ξn=R(ξ−µn◦U) we getξnξ and the excessive measure ηn:=ξ−ξn is such thatηn ≤µn◦U, i.e., there exists νn∈ M+(E) withηn=νn◦U andηnξ.Fromξn↓0 we haveηn↑ξ.Denote ξ0 = gnηn. Because ηn+ξn = ξ, we have gnk=1ηk+fnk=1ξk = ξ. But from ξn↓ 0 we have gnηn=ξ. Therefore, ξ0 =ξ. Puttinggk≤nηk =ηn0, it follows that there exists νn0 ∈ M+(E) such thatη0n=νn0 ◦U.Because gnηn =gnηn0, putting µ = W
nνn0, we have ξ =µ◦U. Therefore, ξ is regular if and only if µ◦U is regular, i.e., iffµdoes not charge anyB-measurable semipolar set.
Proposition 2. Let ξ ∈ PotU such that ξ =µ◦U. Then the following assertions are equivalent:
(1)for any Ray compact subset F of E we have ξF =f{ξG∩E |G Ray open,G⊃F};
(2)µξ.
Proof. Let F be a Ray compact subset ofE. We then have f{ξG∩E |G Ray open G⊃F}=µ|F ◦U.Indeed,
∗RG∩Eξ(f) =L(∗RG∩Eξ, U f) =L(ξ, RG∩EU f) = (µ◦RG∩E)(U f) for all f ∈pB and so,∗RG∩Eξ= (µ◦RG∩E)◦U.It follows that
ξG∩E =ξf∗RG∩Eξ= (µ∧µ◦RG∩E)◦U.
Becauseµµ|
G∩Ef+µ0,whereµ0 =µ|
E\G∩Ef, we haveµ◦RG∩E =µ◦RG∩Ef = µ|G∩Ef +µ0 ◦RG∩E and so, µ∧(µ◦RG∩E) = µ|G∩Ef +µ0 ∧(µ0 ◦RG∩E).
Since µ0◦RG∩E is carried by G∩Ef, we have µ0∧µ0 ◦RG∩E = 0 and so, µ∧(µ◦RG∩E) =µ|
G∩Ef,i.e. ξG∩E =µ|
G∩Ef ◦U.
Therefore,
f{ξG∩E |GRay open,G⊃F}=
^
GRay openG⊃F
µ|G∩Ef
◦U =µ|F ◦U.
Hence condition (1) is equivalent toξF =µ|F ◦U.ButξF =ξf∗RFξ= (µ∧(µ◦RF∗))◦U, hence condition (1) is equivalent to µ|F =µ∧(µ◦RF∗).
(1)⇒(2).IfF is a Ray compact set such thatξ(F) = 0, then∗RFξ = 0 and so, µ◦RF∗= 0.It follows that µ|F = 0 i.e. µ(F) = 0.Therefore,µξ.
(2)⇒(1).Assume thatµξ and letF be a Ray compact subset of E.
Since F\F∗ isU-negligible, it isξ-negligible, henceµ(F\F∗) = 0.Therefore, µ|F =µ|F∗.Obviously
µ|F∗◦RF∗=µ|F∗ =µ|F, µ=µ|F +µE\F, µ◦RF∗ =µ|F +µ|E\F ◦RF∗ and so, µ∧µ◦RF∗ = µ|F, since µ|E\F ◦RF∗ is carried by F and µE\F is carried by E\F.
Corollary. Letξ∈PotU, ξ=µ◦U.Assume that for every Ray compact subset F of E we have ξF = f{ξG∩E | G Ray open, G ⊃ F}. Then ξ is a regular element of ExcU.
Proof. From Proposition 2 we deduce that µ ξ, hence µ(A) = 0 for all ξ-semipolar subset Aof E. It follows that µ◦U is regular in Excξ and so, in ExcU.
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Received 14 April 2008 “Petru Maior” University
Str. N. Iorga nr. 1 540088 Tg. Mure¸s, Romania
dianapetrovai@yahoo.com
