Volume 2008, Article ID 896480,13pages doi:10.1155/2008/896480
Research Article
On Constructing Finite, Finitely Subadditive Outer
Measures, and Submodularity
Charles Traina
Department of Mathematics & Computer Science, St. John’s University, 8000 Utopia Parkway Queens, NY 11439, USA
Correspondence should be addressed to Charles Traina,[email protected]
Received 1 August 2008; Accepted 3 December 2008
Recommended by Andrei Volodin
Given a nonempty abstract setX, and a covering classC, and a finite, finitely subadditive outer measureν, we construct an outer measureνand investigate conditions forνto be submodular. We then consider several other set functions associated withνand obtain conditions for equality of these functions on the lattice generated byC. Lastly, we describe a construction of a finite, finitely subadditive outer measure given an arbitrary family of subsets,B, ofXand a nonnegative, finite set functionτdefined onB.
Copyrightq2008 Charles Traina. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
LetXbe a nonempty set, and letCbe a covering class of subsets ofXsuch that∅, X∈ C, with
Cclosed under finite intersections, and letν be a finite, finitely subadditive outer measure defined for all subsets ofX.In1,2, we considered a finite, finitely subadditive outer measure
νdefined by means ofν.The outer measureνgeneralized results obtained by3for the case of a 0-1 valued, finitely additive measureμdefined on the algebra of subsets generated by a lattice of subsetsLofX.In investigating properties ofν, the hypothesis thatνbe submodular is often present. We determine conditions forνto be submodular.
We also consider for a finite, finitely subadditive outer measureνand covering class
Cunder the hypothesis thatC⊂ Sν, the finite, finitely additive measureμν/AL∈ML,
whereLLCis the lattice generated byC. There are several set functions associated with
Lastly, we consider the general problem of constructing a finite, finitely subadditive outer measure from an arbitrary family, B, of subsets of X,with ∅, X ∈ B and B closed under finite unions, and a nonnegative, finite set functionτ defined on B, withτ∅ 0. Assuming thatτis finitely subadditive onB, the functionλE inf{τB/E⊂B, B∈ B}for
E⊂Xis a finite, finitely subadditive outer measure defined for all subsets ofX.We obtain a characterization of those setsE⊂ Xthat areλ-measurable. This provides us with a condition for the sets ofBto beλ-measurable without requiring thatλbe submodular.
2. Background and notation
We introduce the necessary outer measure, and covering class definitions, and note some known properties that we shall need.
The definitions and notations are standard and are consistent with those found in, for example,1,4–6. We collect the ones we need and some of their properties for the reader’s convenience.
Throughout this section and the rest of the paper,X will denote a nonempty abstract set. ForE⊂ X,Ewill denote the complement ofE.The power set ofXwill be denoted by
PX.We will be concerned with covering classes of subsets ofX.By a covering class, we will always mean a nonempty class Cof subsets ofX such that∅, X∈ Cand such that for any
E⊂ X,there is a finite family of setsC1, C2, . . . , Cn ∈ CwithE⊂
n
i1Ci.We always assume
thatCis closed under finite intersections. We define L LC to be the lattice generated byC.LCis the set of all finite unions of sets fromC. The algebra generated byLisAL. It is known thatAL AC, the algebra generated byC. The set of complements of the members ofCis denoted byC.
Definition 2.1. A covering classC is said to coallocate itself if whenever C, C1, C2 ∈ C,and C⊂C
1∪C2, there are setsA, B∈ CwithCA∪BandA⊂C1, B⊂C2.
Definition 2.2. A covering classCis said to be normal if for anyA, B∈ CwithA∩B∅, there existC, D∈ CwithA⊂C, B⊂D, andC∩D∅.
Definition 2.3. LetC1 and C2 be covering classes with C1 ⊂ C2.C1 is said to coseparateC2 if
wheneverB1, B2 ∈ C2, B1∩B2 ∅,there exist A1, A2 ∈ C1 with B1 ⊂ A1, B2 ⊂ A2,and A
1∩A2∅.
It is known that if a covering class coallocates itself, then it is a normal covering class. Letνbe a finite, finitely subadditive outer measure defined for all subsets ofXand let
Cbe a covering class of the subsets ofX.In general, there are several set functions that we can associate with this outer measureν.ForE⊂X, we defineρE νX−νE.ρis finite valued and in general is not an inner measure. Conditions whenρis an inner measure have been thoroughly investigated in4,5. We also define a set functionνby
νE infνC/E⊂C, C∈ C forE⊂X, 2.1
whereνis always a finite, finitely subadditive outer measure, and so it has associated with it aρ, which may be expressed as
It is always true that ρ ≤ ρ ≤ ν ≤ ν onPX.When a set functionν is an outer measure, the family ofν-measurable sets is denoted bySν,that is,
SνM⊂X/νA νA∩M νA∩M, ∀A⊂X. 2.3
It is well known thatSν is an algebra and the restriction ofν toSν, that is,ν/Sν is
a finite, finitely additive measure. We also defineSν {E ⊂ X/νE ρE}.Sνis not in general an algebra of sets, and we always haveSν⊂ Sν.
An outer measureν satisfies Condition A: ifν ρonC.νsatisfies Condition B: if
νC sup{νA/A⊂C, A∈ C}, forC∈ C.
Definition 2.4. LetE be a collection of subsets ofX and letf be a real-valued set function defined onE.
If for allA, B∈ EwithA∪B, A∩B∈ E, we have
fA∪B fA∩B≤fA fB, 2.4
then f is said to be submodular on E. If the reverse inequality holds, then f is said to be supermodular onE.fis said to be modular onEif equality holds. If
fA∪B≥fA fB wheneverA, B∈ EwithA∩B∅, 2.5
thenfis said to be superadditive onE.
IfEPX,we usually sayfis submodular.
Let L be a lattice of subsets of X such that ∅, X ∈ L. AL denotes the algebra generated byL, and MLdenotes the set of all nontrivial, nonnegative, finitely additive measures onAL.MRLdenotes the set of all thoseμ∈MLwhich areL-regular, that is,
thoseμ∈MLwith the property that for anyA∈ ALμA sup{μL/L⊂A, L∈ L}.
ML,MRL, and several other subsets ofMLhave been extensively studied in
the literature; we cite just a few recent papers4–8. Forμ∈ML, the set functionμdefined forE⊂Xby
μE infμL/E⊂L, L∈ L 2.6
is a finite, finitely subadditive outer measure onPX.We also define the set functionμiby
μiE μX−μE supμL/L⊂E, L∈ L. 2.7
3. Submodularity of the outer measureν
In3, a 0-1 measureμwas used to give necessary and sufficient conditions for a lattice to be normal, by constructing a 0-1, finitely subadditive outer measureμ.In1,2, the measure
measures. In this section, we focus on the latter set functions, and determine conditions forν to be submodular.
We begin by recalling the basic construction and properties ofν.For emphasis, letXbe a nonempty set and letCbe a covering class of subsets ofX,with∅, X∈ CandCclosed under finite intersections. Letνbe a finite, finitely subadditive outer measure defined for all subsets ofX.Assume thatCcoallocates itself and the associated set functionρis finitely subadditive onPX.Sinceρis finitely subadditive onPX,the set functionρis also finitely subadditive onPX see, e.g.,2. ForE⊂Xdefine
νE infρC/E⊂C, C∈ C, 3.1
whereνis a finite, finitely subadditive outer measure and
1νρonC,
2ρνonC,
3ρ≤ρ≤ν≤νonPX,
4νsatisfies condition A, that is,νρonCwhich is equivalent toρνonC.
The functions ν, ρ, ρ are the usual ones associated with the outer measure ν as defined for a general finite, finitely subadditive outer measure inSection 2.
We now determine conditions for ν to be submodular. It will be shown that by requiring that ν or its associated set functions satisfy certain conditions on either Cor C, andCis a lattice, the submodularity ofνwill follow. The importance ofνbeing submodular can be seen from the following known facts2.
It is known that ifνis submodular then
1C ⊂ SνSν,
2νisCregular onL,whereLLC,
3ν/AL∈MRL,
4ν≤νonC.
Theorem 3.1. IfC is a lattice, and ifνis supermodular onC, thenρis submodular onCand soν is submodular onC.
Proof. LetC1, C2∈ C. Sinceνis supermodular on the latticeC,
νC 1∪C2
νC 1∩C2
≥νC 1
νC 2
. 3.2
Therefore,
νX−νC 1∪C2
νX−νC 1∩C2
≤νX−νC 1
νX−νC 2
, 3.3
and so
ρC 1∩C2
ρC 1∪C2
≤ρC 1
ρC 2
SinceνρonC, the above inequality shows
νC 1∩C2
νC 1∪C2
≤νC 1
νC 2
. 3.5
SinceC1, C2 ∈ Care arbitrary, this shows thatρis submodular onCandνis submodular
onC.
Theorem 3.2. SupposeCis a lattice. Ifνis submodular onC, thenνis submodular.
Proof. LetA, B⊂Xand let >0 be given and arbitrary. By definition ofνthere existC1, C2∈ C
such thatA⊂C1, B⊂C2 and
νA≤ρC 1
< νA
2, νB≤ρ
C 2
< νB
2. 3.6
NowA∪B⊂C1∪C2, A∩B⊂C1∩C2,
νA∪B νA∩B≤νC 1∪C2
νC 1∩C2
. 3.7
Sinceνis submodular onC,
νC 1∪C2
νC 1∩C2
≤νC 1
νC 2
< ρC 1
ρC 2
. 3.8
By3.7,3.8,3.6, we have
νA∪B νA∩B< νA νB . 3.9
Since >0 is arbitrary,3.9shows
νA∪B νA∩B≤νA νB, 3.10
and soνis submodular.
Combining Theorems3.1and3.2, we have the following theorem.
Theorem 3.3. IfC is a lattice, and ifνis supermodular onCthenνis submodular.
We recall that given an outer measureν, a measurable cover for a setE ⊂ X is a set
M∈ Sνsuch thatνE νM.
Definition 3.4. Let ν be a finite, finitely subadditive outer measure, and let Sν be the ν
-measurable sets. We define forE⊂X
ν0E infνM/E⊂M∈ S
ν, 3.11
ν0is a finite, finitely subadditive, submodular outer measure, andν0M νMfor M∈ Sν.
It is known that when a finite, finitely subadditive outer measureνis approximately regular thenνis submodular5.
Theorem 3.5. LetC be a covering class and letνbe a finite, finitely subadditive outer measure. If everyC∈ Chas a measurable cover with respect toν,thenνis approximately regular and therefore submodular.
Proof. We first show that the hypothesis that eachC∈ Chas a measurable cover with respect toνimplies thatνν0onC.
LetC ∈ C. We always have for anyM ∈ SνwithC ⊂M, νC ≤νM.Therefore, by the definition ofν0C,
νC≤ν0C. 3.12
Now by hypothesis, there exists an M0 ∈ Sν such that C ⊂ M0 and νC νM0.By
monotonicity ofν0,
ν0C≤νM 0
νC. 3.13
By3.12and3.13,ν0C νC.SinceC∈ Cis arbitrary,νν0onC, so thatνis approximately regular onC.
We next show that when ν is approximately regular on C implies that ν is approximately regular.
LetE ⊂ X.LetM ∈ Sνwith E ⊂ M.By monotonicity ofν, νE ≤ νM.It follows from the definition that
νE≤ν0E. 3.14
Suppose now thatC∈ CandE⊂C.By monotonicity ofν0,
ν0E≤ν0C νC. 3.15
We always haveνρonC, so3.15shows that
ν0E≤ρC. 3.16
SinceCis any set fromCwithE⊂C,3.16shows that
ν0E≤νE. 3.17
An important question to consider is whether or notC ⊂ Sνwithout requiring thatν is submodular. If this is true, then sinceSν is an algebra,ALC AL ⊂ Sν and thus
ν/AL ∈ML. Our next theorem shows that this is indeed the case whenCis a lattice and ρis superadditive onC.
Theorem 3.6. IfCis a lattice, and ifρis superadditive onC, thenC ⊂ Sν.
Proof. It suffices to show by a theorem in2that eachC ∈ Csplits all sets ofCadditively with respect toν.
LetC∈ CandA∈ C.We always haveνA≤νA∩C νA∩C.For the reverse inequality, we proceed as follows. Let >0 be given and arbitrary. There existsB, D∈ Csuch that
B⊂A∩C, ρA∩C−
2 < ρB,
D⊂A∩B, ρA∩B−
2 < ρD.
3.18
NowB∩D ⊂B∩A∩B ∅,andB∪D ⊂ A∩C∪A∩B⊂ A.SinceCis a lattice,B∪D∈ C. We haveνA ρA, so by the definition ofρA, we have
νA> ρA≥ρB∪D. 3.19
By the superadditivity ofρonC,
ρB∪D≥ρB ρD. 3.20
Using3.19,3.20, and3.18, we have
νA> ρA∩CρA∩B−. 3.21
Again, sinceCis a lattice andA, B, C∈ CandνρonC, we have by3.21
νA> νA∩CνA∩B−. 3.22
SinceA∩B⊃A∩A∪C A∩C, by monotonicity ofν,
νA∩B> νA∩C. 3.23
Therefore, by3.22and3.23, we getνA> νA∩C νA∩C−.Since >0 is arbitrary, we getνA≥νA∩C νA∩C.Therefore,νA νA∩C νA∩C, andCsplitsAadditively with respect toν.SinceA ∈ Cis arbitrary, it follows thatCsplits all sets ofCadditively with respect toνand soC∈ Sν.The arbitrariness ofC∈ Cshows that C ⊂ Sν.
4. Equality ofνand its associated set functions onL
In this section, we again consider a covering class C of subsets of a nonempty set X, C being closed under finite intersections, and such that ∅, X ∈ C, and ν a finite, finitely subadditive outer measure defined for all subsets ofX. We assume thatC ⊂ Sν,the set of
ν-measurable sets. SinceSνis an algebra of sets,LLC⊂ Sν, henceAL⊂ Sν.Therefore,
μ ν/AL ∈ ML. We consider the following set functions obtained from ν,which are
defined inSection 2:μ, μi, ρ, ν, ρ.We obtain an inequality between all these set functions
that holds for all subsets ofX,and determine conditions for equality of these functions onL. Equality onLbecomes useful, since for instance we will then haveμ∈MRL.
Theorem 4.1. ρ≤μi≤ρ≤ν≤μ≤νonPX.
Proof. LetE⊂Xbe fixed but arbitrary. We always have
ρE≤νE, 4.1
so it is the other inequalities that we must establish.
LetL ∈ LwithE ⊂ L.The definition ofμEshows thatμE ≤ μL νL.By monotonicity ofν, we haveνE ≤ νL μLwheneverE ⊂ L, L ∈ L. Thus, νEis a lower bound for the set{μL/E⊂L, L∈ L}. Therefore,
νE≤μE. 4.2
Now supposeC∈ CandE⊂C.SinceC∈ L,μE≤μC νC.Therefore,μEis a lower bound for the set{νC/E⊂C, C∈ C}. The definition ofνEgives
μE≤νE. 4.3
Next, consider anyC∈ CwithC⊂E.By the definition ofμiE, we haveμC≤μiE.
SinceμC νC,we haveνC ≤ μiEfor allC ⊂ Ewith C∈ C. SinceρC ≤νC, we haveρC ≤μiEfor allC∈ C, withC ⊂E.Therefore,μiEis an upper bound for the set {ρC/C⊂E, C∈ C}. The definition ofρEthus gives
ρE≤μ
iE. 4.4
To show thatμiE≤ρE, we argue by contradiction. Thus suppose thatρE< μiE.
Then there exists anL0 ∈ Lsuch thatL0 ⊂ EandρE < μL0 ≤μiE.By monotonicity of
ρ, ρL0≤ρE,so we haveρL0< μL0.NowρL0 νX−νL0 μX−μL0, hence
μX< μL0 μL
0.This contradicts the finite additivity ofμ. Therefore, it must be
μiE≤ρE. 4.5
Combining inequalities4.1through4.5, we have
ρE≤μ
iE≤ρE≤νE≤μE≤νE. 4.6
We now determine conditions when the inequality ofTheorem 4.1will be an equality onL. Before doing so, we indicate what happens when this is true. We always have thatμ is submodular. Assuming ρ μi ρ ν μ νonL, supposeL ∈ L. ThenμiL
μX−μL μX−μL μLsinceμ∈ML. NowμiL sup{μK/K⊂L, K∈ L}
so we haveμL sup{μK/K ⊂ L, K ∈ L}. By5, this shows sinceL ∈ Lis arbitrary thatμ∈MRL. It follows from this thatL ⊂ Sμ,and soμ0μ,so thatμis approximately
regular, hence submodular.
Theorem 4.2. IfC is a lattice,νsatisfies condition A, andνis submodular onC, thenρμiρ
νμνonL.
Proof. We show thatν is submodular. Sinceν satisfies condition A,C ⊂ Sν.Let E, F ⊂ X and let >0 be arbitrary. By the definition ofν, there existC1, C2 ∈ Csuch thatE⊂ C1and νE≤νC
1< νE /2, andF⊂C2and
νF≤νC 2
< νF
2. 4.7
NowE∪F⊂C1 ∪C2,E∩F⊂C1∩C2, whereC1∩C2andC1∪C2belong to the lattice
C. Therefore,
νE∪F νE∩F≤νC 1∪C2
νC 1∩C2
. 4.8
Sinceνis submodular onC,νC1∪C2 νC1 ∩C2≤νC1 νC2.Combining this last inequality with4.8and4.7, we have
νE∪F νE∩F< νE νF . 4.9
Since > 0 is arbitrary, this shows that νE∪F νE∩F ≤ νE νF, where
E, F ⊂Xare arbitrary. Therefore,νis submodular. Hence,Sν Sν whereSν is an algebra.
By our initial observation,C ⊂ SνSν,and soL ⊂ Sν. Thus for allL∈ L,νL ρL, and
so byTheorem 4.1, we haveρμiρνμνonL.
Theorem 4.3. Ifν ν0onC,CcoseparatesLC, andνsatisfies condition B, thenρ μ
i ρ
νμνonL.
Proof. The hypothesis thatνν0onCshown by6thatνis submodular and soSν Sν .
We show that the hypotheses thatCcoseparatesLCand νsatisfies condition B giveρ
μiρνμνonC.
SinceC ⊂ Sν⊂ Sν,νC ρCfor allC∈ C. Since for anyC∈ C,ρC ρC μiC,
we have
ρμ
iρν onC. 4.10
μC0 ≤ μL
0 < νC0. SinceC0∩L0 ∅ with C0, L0 ∈ L and C coseparates L, there
existA, B∈ CwithC0⊂A, L0⊂BandA∩B∅. Therefore,C0⊂A⊂B⊂L0and
μC0≤μA μA≤μB≤μL 0
< νC0. 4.11
We also haveμA νA.Thus, we have anA ∈ C ⊂ L, withC0 ⊂AandνA < νC0.This contradicts the definition ofνC0.Therefore, we must haveμνonC.
We next showνμonC. We reason by contradiction. Suppose there is aC∈ Cwith
νC< μC. Therefore,νX−νC> νX−μC.But we also haveνX νC νCso νC νX−νCandνX μX μX.Therefore,νC> μX−μC μ
iC.Since
νsatisfies condition B,νC sup{νA/A ⊂C, A∈ C}> μiC. Thus, there is anA0 ∈ C
withA0 ⊂CandμiC< νA0≤νC,andA0∩C∅.SinceCcoseparatesL, there exist B, D∈ Csuch thatA0⊂B, C⊂DandB∩D∅.Therefore,A0⊂B⊂D⊂C.This gives
μiC< νA≤νB≤νD≤νC. 4.12
ButνD μD, so we have aD ∈ C ⊂ L withD ⊂ C and μiC < μD.This contradicts the definition of μiC. Therefore, we must haveν μ on C. Thus we have
ρ μ
i ρ ν μ ν onC. Sinceρ ν onC,C ⊂ Sν.By the submodularity ofν, Sν
Sν.Therefore,C ⊂ Sν and soL ⊂ Sν.Therefore, for allL ∈ L,ρL νLand so ρμ
iρνμνonL.
Thus far, in order to have the equality
ρμ
iρνμν onL, ∗
the property of submodularity ofνorνonCorCplayed a major role. A natural question to ask is whether this equality can be achieved without submodularity. To obtain some insight into how one may proceed, we make the following observation. For the outer measureν, it is always true thatSν ⊂ Sν,whereSν is an algebra of sets. IfC ⊂ Sν, thenLLC⊂ Sν,
and consequentlyC ⊂ Sν, whenceνL ρLfor allL ∈ L. It follows from this that the equality∗holds onL. It seems reasonable then to seek conditions forC ⊂ Sν.
We have the following theorem.
Theorem 4.4. Suppose eachC∈ Csplits all sets ofCadditively with respect toν. ThenC ⊂ Sνand ρμ
iρνμνonL.
The result follows from2.
The results of this section prove useful in studying the restriction of a finite, finitely subadditive outer measure to the algebra generated by the covering class, when the sets of the covering class are measurable, which will be investigated in a subsequent paper . Although Theorem 4.4may be difficult to implement in practice, it leads naturally to the content of the next section.
5. An outer measure construction
nonnegative set function τ on B. By the standard method, we construct a finite, finitely subadditive outer measureλfromτandB. We seek a characterization of those sets that split all sets ofBadditively with respect toλ.
For emphasis, letXbe a nonempty set and letBbe an arbitrary family of subsets ofX with∅, X∈ BandBis closed under finite unions. Letτbe a finite, nonnegative set function defined onBwithτ∅ 0,andτsubadditive onB. For E⊂Xlet
λE infτB/E⊂B, B∈ B. 5.1
It follows by a standard argument thatλis a finite, finitely subadditive outer measure defined for all subsets ofX.LetSλbe the algebra ofλ-measurable subsets ofX.We determine conditions when a set or a family of sets splits all sets ofBadditively with respect toλ,this set or family will split all subsets ofX additively with respect toλ.The importance of this was hinted at in the last section.
We observe that ifB∈ B, thenλB≤τB.Also, ifτ is monotone onB, thenλB
τBfor allB∈ B.
The following theorem is known, and its proof can be found in2, for instance.
Theorem 5.1. IfE⊂Xsplits all sets ofBadditively with respect toλ, thenE∈ Sλ.
Our motivation for what follows comes from the characterization of the measurable sets for aσ-finite measure which can be found in8,9. We begin with the observation that there are always at least two sets inB that split all sets ofB additively with respect to λ, namely,∅andXthey may be the only ones. We seek a characterization of such sets.
Theorem 5.2. LetB0 ∈ Bsplit all sets ofBadditively with respect toλ.Let N ∈ Bbe such that τN 0.ThenEB0−Nsplits all sets ofBadditively with respect toλ, and soE∈ Sλ.
Proof. LetB∈ Bbe arbitrary. We can writeB B∩E∪B∩E.Now,B∩EB∩B0∩N⊂ B∩B0andB∩EB∩B0∪N⊂B∩B0∪N.By the monotonicity and finite subadditivity
ofλ,
λB∩E λB∩E≤λB∩B 0
λB∩B 0
λN. 5.2
SinceN∈ B,λN≤τN 0,soλN 0 and5.2becomes
λB∩E λB∩E≤λB∩B 0
λB∩B 0
. 5.3
The hypothesis onB0givesλB∩B0λB∩B0 λBso by5.3,λB∩EλB∩E≤ λB.The reverse inequality is always true by finite subadditivity ofλ.Thus byTheorem 5.1,
E∈ Sλ. 5.4
Thus we see that a sufficient condition for a set E to be λ-measurable is:E must be expressible as the difference of a set fromBthat splits all sets ofBadditively with respect to
An example may help to illustrate the theorem.
Example 5.3. LetX {a, b, c}and takeB {∅, X,{a},{b},{a, b}}.Defineτ onBbyτ∅
τ{a} 0, τ{b} τ{a, b} 1, τX 2.The outer measureλdetermined byBandτ isλ∅ λ{a} 0,λ{b} λ{a, b} 1,λ{c} λ{a, c} λ{b, c} λX 2.Now
Sλ{∅, X,{a},{b, c}}.We see that∅X−∅,X X−∅,{a}{a} −∅,{b, c}X− {a}.
We return to characterizing those setsEthat split all sets ofBadditively with respect toλ.In the style of Munroe10, we have the following theorem.
Theorem 5.4. LetE⊂Xbe any set. There is a sequence of sets{Bn}fromBsuch thatE⊂Bnall n andλE λ∞n1Bn.
Proof. LetE⊂Xbe arbitrary. By the definition ofλE,for eachn∈Nthere is aBn∈ Bsuch
thatE⊂Bnand
λE≤τBn< λE n1. 5.5
LetB∞n1Bn.ThenE⊂B⊂Bnall n. By monotonicity ofλ,
λE≤λB≤λBn alln. 5.6
NowλBn≤τBnall n, so by5.6and5.5we getλE≤λB< λE 1/nfor any
n∈N.Thus for any given >0,we can find ann∈Nlarge enough so that 0<1/n < .For any suchn, λE≤λB< λE .Since >0 is arbitrary, it follows thatλE λB.
Observation 1. Suppose E ⊂ X splits all sets A ⊂ X additively with respect to λ. By Theorem 5.4, there exist Bn ∈ B, n ∈ N with E ⊂ ∞n1Bn and λE λ∞n1Bn. Let
B ∞
n1Bn.We can writeB E∪B∩E.SinceE ∈ Sλ, λB λE λB∩E, and
since all quantities are finite,λB∩E λB−λE 0.SinceEB−B−E,letNB−E so thatλN 0.
This observation establishes the following theorem.
Theorem 5.5. IfE∈ Sλ, then there is a sequence of sets Bn∈ B, and a setN⊂Xsuch thatE⊂Bn
all n, andE ∞n1Bn−N, whereλN 0.
We see that our result is analogous to what happens in the case of constructing an outer measure from a measure in the general case, see for instance Munroe10.
In summary, to obtain a setEthat splits all sets ofBadditively with respect toλ: let
B0 ∈ Bbe a set that splits all sets ofBadditively with respect toλfor instanceXand let N⊂XwithτN 0.ThenEB0−Nwill split all sets ofBadditively with respect toλ,so
thatE∈ Sλ.
IfE∈ Sλ, then there is a sequence of sets{Bn}fromBsuch thatE⊂Bnall n,and a set
N⊂Xsuch thatλN 0 andE ∞n1Bn−N.
Acknowledgments
The author wishes to thank the referee for several useful suggestions leading to the improved clarity of the paper, and to the editor for his courtesy. The questions considered for this paper were presented to the author by the late Dr. George Bachman, the author’s teacher and friend.
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