ON REGULAR AND SIGMA-SMOOTH TWO VALUED MEASURES AND LATTICE GENERATED TOPOLOGIES

. Let X be an abstract set and L a lattice of subsets of X. I(L) denotes the non-trivial zero one valued finitely additive measures on A(L), the algebra generated by L, and IR(L) those elements of I(L) that are L-regular. It is known that I(L)=IR(L) if and only if L is an algebra. We first give several new proofs of this fact and a number of characterizations of this in topologicial terms. Next we consider, I(o*,L) the elements of I(L) that are o-smooth on L, and IR(o,L) those elements of I(o*,L) that are L-regular. We then obtain necessary and sufficent conditions for I(o*,L)=IR(o,L), and in particuliar ,we obtain conditions in terms of topologicial demands on associated Wallman spaces of the lattice.


INTRODUCTION
In this paper we wish to determine when certain classes of measures are equal, and to obtain necessary and sufficent conditions for such equality to hold,emphazing topologicial characterizations.
To be specific let X be an abstract set,L a lattice of subsets of X.Let A(L) denote the algebra generated by the lattice L,and I(L) the collection of non-trivial zero-one valued fintely additive measures on A(L).IR(L) will denote measures in I(L) that are L-regular on A(L),i.e. if I.telR(L) and Be.A(L) then there exists a LeL st BL and I.t(B)=(L).I(o*,L) will denote those elements of I(L) that are sigma- smooth on L,i.e. if LneL n=l,2..... and Ln,l.o then for I.teI(t*,L), limt.t(Ln)=0.IR(t,L) will denote those measures in I(r*,L) that are L-regular.
The first area of concern is when I(L)=IR(L).It is well known that this is true iff L is an algebra .We give several proofs of this,highlighting topologicial considerations ,to be more precise I(L)=IR(L) is equivalent to the following: a) The lattice V(L) (see below for definitions) in the space I(L) is regular.b) The topology of closed sets xV(L) in I(L) is T 1. c) The lattice of sets V(L) in I(L) is disjunctive.
The second main area of concern is determining conditions for I(o*,L)=IR(o,L),and conversely what this implies for the lattice.We show (see below for definitions) that I(o*,L)=IR(o,L) is equivalent to: The lattice V(o,L) in the space I(o*,L) is regular.We also show that if L is disjunctive and W(o,L) is prime complete or L is normal and countably compact and if I(o*,L),V(o,L)is T then I(o*,L)=IR(o,L).Also suppose xLEA(L)L then if :L is Lcb or more generally if E (and thus A(L)) is Lcb and a) Either S(L)o(L) (Where S(L) are the lattice Souslin derived sets.In particuliar if p(L)=o(L) .)and L is delta or b) If L is complement generated (and not necessarily delta) then IR(o,L)=I(o*,L).

BACKROUND AND NOTATION
We begin by reviewing some notation and terminology which is fairly standard (see Alexandroff ],Frolik [4],andSzeto [7] ).We supply backround material for the readers convenience.
Let X be abstract set,and L a lattice of subsets of X st X,OeL.A delta latdce is one that is closed under countable intersections,and the delta lattice generated by L is denoted/5(L).In addition L is complement generated iff for every element LeL.there exists a sequence of subsets AieL i=1,2.., and L=nA i' i=1,2.... (where 'denotes complement ).L is countably paracompact if for every sequence LneL and LnJ.O there exists Bn'eL' st Bn',l.and Bn'Ln for every n.A tau lattice is one that is closed under arbitrary intersections,and the tau lattice generated by L is denoted by xL.A(L) will denote the algebra generated by the lattice L.
Let I(L) denote the set of non-trivial two valued {0,1 finitely additive measures on the algebra generated by L,and let I(o*,L) denote those elements of I(L) that are sigma-smooth on L,i.e. if {L n }eL, Ln,l,o and I.teI(o*,L),liml.t(Ln)=0n-**.I(o,L) denotes those elements of I(L) that are sigma-smooth on A(L),i.e. if {An}cA(L), An,l,i,and I.teI(o,L) liml.t(An)=0 as n-->**.This is equivalent to countably additivity on A(L).IR(L) will stand for the measures on A(L) that are L-regular on A(L),i.e.I.telR(L) t(A)=supl.t(L)LeL A.L and AeA(L).This is equivalent to being L-regular on L'.IR(o,L) denotes the set of I.telR(L) that are o-smooth on L.The obvious relations hold,I(L)I(o*,L)I(o,L)IR(o,L) and I(L)IR(L).The support of a measure S(t),II(L) is defined as S()=n{LeL la(L)=l }.
Let L1 and L2 be two lattices of sets of X st L2L1 then L2 is L1 countably bounded (cb) if for L2neL2 and L2n,l,(, then there exists L1 neL 1,L n,l,o and L nL2n. A lattice is said to be disjunctive_if for any xeX and LeL such that x L there exists L"eL such that xeL~and LoL~=O.L is said to be regular_if for xeX x L I.,eL then there exists L1 ,L2eL xeL 1' L2'L and LI'L2'---O.L is said to be normal if for L1,L2eL and LlL2---O,there exist L3,L4eL st L3';L1 L4'gL2 and L3'L4'=O.L is said to be coutably compact if for any {Ln}eL and OLn=O n=1,2...** ,then there exists a finite subindexing st nLni=O ni--1 to N.A lattice is said to be T1 if for x,yeX there exist L 1,L2eL st xeL 1' Y L l'and yeL2' x L2'.p(L) denotes the smallest collection of sets that is closed under countable unions and intersections and contains L.o(L) will stand for the smallest o-algebra containing L.S(L) will stand for the collection of Souslin sets generated by L.
Note: For g,gleI(L),wc write g-<gl (L) if g(L)_<gl(L) for all LL.Wc now note some measure equivalences of topologicial properties: 1) L is disjunctive iff for all xeX gxelR(o,L) where gx is the point mass measure,i.e,gx(A)=l if xeA ,gx(A)=0 if xe A ,AeA(L).2) L is regular iff g_< (L) where g,g 1el(L) implies S(g)=S(g 1)-3) L is normal iff geI(L) and g 1,g2elR(L) implies that if g<_g (L) and g-<g2 (L) then g 1=g2.4)L is countably compact iff gel (L) implies that geI(o*,L).Thc proofs arc not difficult and thus wc will only prove the third result for the sake of completeness.Further facts about regular and normal lattices appear in Eid [3] and Grassi [5].
Note: A fact wc will use in the second part of the proof and in the proceeding parts of the paper is that there exists a one-one correspondence between prime L-filters and elements of I(L),and a one-one correspondence between L-ultrafiltcrs and elements of IR(L).This correspondence is set up by letting geI(L) and H={ LeL g(L)=l }.Then H is a prime L-filter and conversely if H is a prime L-filter there exists a measure geI(L) associated with H st if LeH, g(L)=I.A similiar correspondence holds for H and gelR(L) in which case H is an L-ultrafiltcr.Now wc return to the proof of the theorem,conversely,let gel(L) and g<g (L),g 1<g2 (L) for g 1,g2eIR(L) imply gl=g2,and assume L is not normal.Then there exists L 1,L 1-eL st L 11~=O and H={ LeL L'L1 or L'LI'" has the finite intersection property and thus there exists an associated measure geI(L') associated with the filter base H st g(L')=l LeH.Now let L2eL and suppose that g(L2')=0, then L2' does not contain L1 thus LlcL2;J.Sincc the collection {LlcL2 gCL2)=I L2eL} has the tip thus there exists a gleIR(L) st gl(L1)=l and g-<g1 (L).By similiar reasoning there exists a g2eIR(L) st g-<g2 (L) g2(L 1")=1.By hypothesis g l=g2.Hencc g I(L 1)=g (L 1~)=1.Therefore g (Llr'xL1-)= 1.But L lC'L1"'--O thus g I(L lc'L l"-)=0,a contradiction.Thus L must bc normal.
Wc now prove a result that will be useful in the sequel.
Next wc consider various sets of measures defined on the algebra generated by a lattice L.For example consider I(L),I(o*,L),IR(L),or IR(o,L).Denote such sets by I.Also consider the collection of sets H,(L) where fl,(L)={K(L) LzL} and K(L)={geI g(L)=l}.Thcn the following hold: a) fl,(AB)=K(A)H,(B) for A,BeL.b)K(A)cK(B)=H,(AnB)A,BeL.c)fl,(A)'=fl,(A') for AeL.d)If AB then K(A)_z:(B) A,BeL.e)If L is disjunctive (if necessary) and (A);;(B) then AB A,BeL.f) The collection K(L) is a lattice and K(A(L))=A(K(L)).
We will assume in discussing x,,(L) for convenience,that l is disjunctive,although it will be clear that this assumption is not always needed.IfixeI(L) then define a measure on A(H(L))IX^eI(H(L))by IX^(t(A))=It(A) for AeA(L).Conversely if Ix^eI((/)) define a measure on A(L)gel(L)by I(A)eA(H,(L)).Then the following hold: Theorem 2.3."If L is disjunctive (if necessary) then there is a 1-1 correspondence between the sets I(/) and I((L)) given by Ix<-->it^.Further gel(L) is o-smooth or regular iff It^eI(H,(L)) is o-smooth or H,(/) regular.
These sets are topologized by taking H,(L) ,.(L)e.J(L) as a basis for the closed sets, and will be referred to as generalized Wallman spaces 3 THE SPACES IR(L) AND I(L) In this section we investigate a variety of conditions which are equivalent to IR(/)=I(L) both abstractly and from a topologicial point of view with respect to the space I(L),xV(l).This will useful for ottt subsequent analysis of I(o*,L),as well as being interesting in its own right.
Theorem 3.1: Let L be a lattice of subsets of X,then the following are equivalent: a) e) The lattice of sets V(L) in I(L) is disjunctive f) L is an algebra.
Next we wish to show IR(L)=I(L) iff the lattice V(L) in the space I(L) is regular.
Next wc show IR(L)=I(L) iff the topology of closed sets xV(L) in I(L) is T 1.
Finally,now we claim I(L)=IR(L) iff L is an algebra,i.e.L=L'.
Let L be an algebra and IxeI(L) then since L=L' IX is trivially regular and IR(L)--I(L).Conversely let I(L)=IR(L) and assume that LL',i.e. that L is not an algebra.Thus there exists a L-EL st L~' L and look at H={ L L_L~or I..L~'}.Then H has the tip and thus there exists a I.tel(L) st tt(L)=l,LeH.For IX(LI')=I LIEL implies that L1 does not contain L~or L~'.Thus there exists I.tlEIR(L st IXI(L~)=I <IXl (L) and also a I.t2elR(L') st IX2(L-")=I and Ix<l.t2 (L').But since I(L)=IR(L) and this implies from above that IR(L')=IR(L),IX (L-)=IX (L")=I or IX (L~cL~')= =IX (t)=0,a contradiction.L=L' and L is an algebra.
Because of the importance of the last result we present an alternative approach which is of importance because of its relevance to lattice separation properties.
Theorem 3.2: Suppose L1,L2 are lattices of subsets of X st L2L 1.If L2 is disjunctive and L is normal and if v:IR(L2)---IR(L 1) where V is the restriction map,i.e.V(v)=IX the restriction of v to A(L1),then L1 semi-separates L2.
Proof:Set L I=L and L2=A(L) .Since I(L)=IR(L), L is normal.Then the hypotheses of the theorem hold,thus L ss A(L).Let L'cL---LeL,then since L'eA(L) and L ss A(L),this implies that L'EL L=L',i.e.L is an algebra.