# On Constructing Finite, Finitely Subadditive Outer Measures, and Submodularity

Authors: Charles Traina
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

Given a nonempty abstract setX, and a covering class 𝒞, and a finite, finitely subadditive outer measure ν, we construct an outer measure ν¯ and investigate conditions for ν¯ to be submodular. We then consider several
other set functions associated with ν and obtain conditions for equality of these
functions on the lattice generated by 𝒞. Lastly, we describe a construction of a finite, finitely subadditive outer measure given an arbitrary family of subsets, ℬ, of X and a nonnegative, finite set function τ defined on ℬ.


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## Body

## 1. Introduction

LetX be a nonempty set, and let 𝒞 be a covering class of subsets of X such that ∅,X∈𝒞,
with 𝒞 closed under finite intersections, and let ν be a finite, finitely subadditive outer
measure defined for all subsets of X. In [1, 2], we considered a finite, finitely
subadditive outer measure ν¯ defined by means of ν.  The outer measure ν¯ generalized results obtained by [3] for the
case of a  0-1 valued, finitely
additive measure μ defined on the algebra of subsets generated by
a lattice of subsets ℒ of X. In investigating properties of ν¯, the hypothesis that ν¯ be submodular is often present. We determine
conditions for ν¯ to be submodular.We also
consider for a finite, finitely subadditive outer measureν and covering class 𝒞 under the hypothesis that C⊂𝒮ν, the finite, finitely additive measure μ=ν/𝒜(ℒ)∈M(ℒ),
where ℒ=ℒ(𝒞) is the lattice generated by 𝒞.  There are several set functions associated
with μ and ν, namely, ν′,μ′,μi,ρ,ρ′ and it is shown that the inequality ρ′≤μi≤ρ≤ν≤μ′≤ν′ always holds for all subsets of X. We determine when an equality of  all these set functions holds on ℒ.  This is useful since for instance it allows
one to conclude that μ is an ℒ-regular measure on 𝒜(ℒ).
Once again, the condition of submodularity plays an essential role.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 function τ defined on ℬ,
with τ(∅)=0. Assuming that τ is finitely subadditive on ℬ,
the function λ(E)=inf{τ(B)/E⊂B,B∈ℬ} for E⊂X is a finite, finitely subadditive outer
measure defined for all subsets of X. We obtain a characterization of those sets E⊂X that are λ-measurable. This provides us with a condition for the sets
of ℬ to be λ-measurable  without requiring that λ be 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. For E⊂X, E′  will denote the complement of E. The power set of X will be denoted by 𝒫(X). We will be concerned with covering classes of subsets of X. By a covering class, we will always
mean a nonempty class 𝒞 of subsets of X such that ∅,X∈𝒞 and such that for any E⊂X, there is a finite family of sets C1,C2,…,Cn∈𝒞 with E⊂⋃i=1nCi. We always assume that C is closed under finite intersections. We
define ℒ=ℒ(𝒞) to be the lattice generated by 𝒞.  ℒ(𝒞) is the set of all finite unions of sets from 𝒞.  The algebra generated by ℒ is 𝒜(ℒ).  It is known that 𝒜(ℒ)=𝒜(𝒞), the algebra generated by 𝒞.  The set of complements of the members of 𝒞 is denoted by 𝒞′.Definition 2.1.
A covering class𝒞  is said to  coallocate itself if
whenever C,C1,C2∈𝒞, and C⊂C1′∪C2′, there are sets A,B∈𝒞  with C=A∪B and A⊂C1′,B⊂C2′.Definition 2.2.
A covering class𝒞 is said to be normal if for any A,B∈𝒞 with A∩B=∅,
there exist C,D∈𝒞 with A⊂C′,B⊂D′,
and C′∩D′=∅.Definition 2.3.
Let𝒞1 and 𝒞2 be covering classes with 𝒞1⊂𝒞2. 𝒞1 is said to coseparate𝒞2 if whenever B1,B2∈𝒞2,B1∩B2=∅, there exist A1,A2∈𝒞1 with B1⊂A1′,B2⊂A2′, and A1′∩A2′=∅.It is
known that if a covering class coallocates itself, then it is a normal
covering class.Letν be a finite, finitely subadditive outer
measure defined for all subsets of X and let 𝒞 be a covering class of the subsets of X. In general, there are several set functions
that we can associate with this outer measure ν. For E⊂X,
we define ρ(E)=ν(X)−ν(E′). ρ  is finite valued and in general is not an
inner measure. Conditions when ρ is an inner measure have been thoroughly
investigated in [4, 5]. We also define a set function ν′ by (2.1)ν′(E)=inf{ν(C′)/E⊂C′,C∈𝒞}forE⊂X,
where
ν′  is always a finite, finitely subadditive outer
measure, and so it has associated with
it a ρ′,
which may
be expressed as (2.2)ρ′(E)=sup{ρ(C)/C⊂E,C∈𝒞},E⊂X.It is always true thatρ′≤ρ≤ν≤ν′ on 𝒫(X). When a set function ν is an outer measure, the family of ν-measurable sets is denoted by 𝒮ν, that is, (2.3)𝒮ν={M⊂X/ν(A)=ν(A∩M)+ν(A∩M′),∀A⊂X}.It is well known that𝒮ν is an algebra and the restriction of ν to 𝒮ν,
that is, ν/𝒮ν is a finite, finitely additive measure.  We also define 𝒮ν={E⊂X/ν(E)=ρ(E)}.𝒮ν is not in general an algebra of sets, and we
always have 𝒮ν⊂𝒮ν.An outer measureν satisfies Condition A: if ν=ρ′ on 𝒞′. ν satisfies Condition B: if ν(C′)=sup{ν(A)/A⊂C′,A∈𝒞}, for C∈𝒞.Definition 2.4.
Letℰ be a collection of subsets of X and let f be a real-valued set function defined on ℰ.
If for allA,B∈ℰ with A∪B,A∩B∈ℰ,
we have (2.4)f(A∪B)+f(A∩B)≤f(A)+f(B), then f is said to be submodular onℰ.  If the reverse inequality holds, then f is said to be supermodular onℰ. f is said to be modular onℰ if equality holds. If (2.5)f(A∪B)≥f(A)+f(B)wheneverA,B∈ℰwithA∩B=∅,     then f is said to be superadditive onℰ.
Ifℰ=𝒫(X), we usually say f is submodular.Letℒ be a lattice of subsets of X such that ∅,X∈ℒ.  𝒜(ℒ) denotes the algebra generated by ℒ,
and M(ℒ) denotes the set of all nontrivial, nonnegative, finitely additive measures on 𝒜(ℒ). MR(ℒ) denotes the set of all those μ∈M(ℒ) which are ℒ-regular,
that is, those μ∈M(ℒ) with the property that for any A∈𝒜(ℒ)μ(A)=sup{μ(L)/L⊂A,L∈ℒ}.M(ℒ), MR(ℒ),
and several other subsets of M(ℒ) have been extensively studied in the
literature; we cite just a few recent papers [4–8]. For μ∈M(ℒ),
the set function μ′ defined for E⊂X by (2.6)μ′(E)=inf{μ(L′)/E⊂L′,L∈ℒ} is a finite, finitely subadditive outer measure on 𝒫(X). We also define the set function μi by (2.7)μi(E)=μ′(X)−μ′(E′)=sup{μ(L)/L⊂E,L∈ℒ}.


## 3. Submodularity of the Outer Measureν¯

In
[3], a 0-1 measure μ was used to give necessary and sufficient
conditions for a lattice to be normal, by constructing a 0-1, finitely
subadditive outer measure μ¯. In [1, 2], the measure μ was replaced by a finite, finitely subadditive
outer measure ν, and the outer measure μ¯ was generalized by a finite, finitely
subadditive outer measure ν¯. However, we now need to impose a separation
property on the covering class, and require submodularity of the outer
measures. In this section, we focus on the latter set functions, and determine
conditions for ν¯ to be submodular.We
begin by recalling the basic construction and properties ofν¯. For emphasis, let X be a nonempty set and let 𝒞 be a covering class of subsets of X, with ∅,X∈𝒞 and 𝒞 closed under finite intersections. Let ν be a finite, finitely subadditive outer
measure defined for all subsets of X. Assume that 𝒞 coallocates itself and the associated set
function ρ is finitely subadditive on 𝒫(X). Since ρ is finitely subadditive on 𝒫(X), the set function ρ′ is also finitely subadditive on 𝒫(X) (see, e.g., [2]).  For E⊂X define (3.1)ν¯(E)=inf{ρ′(C′)/E⊂C′,C∈𝒞},
where
ν¯ is a finite, finitely subadditive outer
measure and(1)
ν¯=ρ′  on 𝒞′,(2)
ρ¯=ν′  on 𝒞,(3)
ρ¯′≤ρ¯≤ν¯≤ν¯′   on 𝒫(X),(4)
ν¯ satisfies  condition
A, that is, ν¯=ρ¯′  on 𝒞′ which is
equivalent to  ρ¯=ν¯′  on 𝒞.(The functionsν¯′,ρ¯,ρ¯′ are the usual ones associated with the outer
measure ν¯ as defined for a general finite, finitely
subadditive outer measure in Section 2).We now
determine conditions forν¯ to be submodular. It will be shown that by
requiring that ν or its associated set functions satisfy
certain conditions on either 𝒞 or 𝒞′,  and 𝒞 is a lattice, the submodularity of ν¯ will follow. The importance of ν¯ being submodular can be seen from the
following known facts [2].It is known
that ifν¯ is submodular then(1)
𝒞⊂𝒮ν¯=𝒮ν¯,(2)
ν¯  is 𝒞 regular on ℒ′,  where ℒ=ℒ(𝒞),(3)
ν¯/𝒜(ℒ)∈MR(ℒ),(4)
ν≤ν¯ on 𝒞.Theorem 3.1.
If𝒞   is a lattice, and if  ν′   is supermodular on  𝒞,
then  ρ′    is submodular on  𝒞′  and so  ν¯  is submodular on  𝒞′.Proof.
LetC1,C2∈𝒞.
Since ν′  is supermodular on the lattice 𝒞, (3.2)ν′(C1∪C2)+ν′(C1∩C2)≥ν′(C1)+ν′(C2). Therefore, (3.3)ν′(X)−ν′(C1∪C2)+ν′(X)−ν′(C1∩C2)≤ν′(X)−ν′(C1)+ν′(X)−ν′(C2), and so (3.4)ρ′(C1′∩C2′)+ρ′(C1′∪C2′)≤ρ′(C1′)+ρ′(C2′).  Since ν¯=ρ′ on 𝒞′,
the above inequality shows (3.5)ν¯(C1′∩C2′)+ν¯(C1′∪C2′)≤ν¯(C1′)+ν¯(C2′). Since C1,C2∈𝒞 are arbitrary, this shows that ρ′ is submodular on 𝒞′ and ν¯ is submodular on 𝒞′.Theorem 3.2.
Suppose𝒞  is a lattice. If ν¯   is submodular on 𝒞′,
then  ν¯    is submodular.Proof.
LetA,B⊂X and let ϵ>0 be given and arbitrary. By definition of ν¯ there exist C1,C2∈𝒞 such that A⊂C1′,B⊂C2′ and (3.6)ν¯(A)≤ρ′(C1′)<ν¯(A)+ϵ2,ν¯(B)≤ρ′(C2′)<ν¯(B)+ϵ2. Now A∪B⊂C1′∪C2′,A∩B⊂C1′∩C2′, (3.7)ν¯(A∪B)+ν¯(A∩B)≤ν¯(C1′∪C2′)+ν¯(C1′∩C2′). Since ν¯ is submodular on 𝒞′, (3.8)ν¯(C1′∪C2′)+ν¯(C1′∩C2′)≤ν¯(C1′)+ν¯(C2′)<ρ′(C1′)+ρ′(C2′). By (3.7), (3.8), (3.6), we have (3.9)ν¯(A∪B)+ν¯(A∩B)<ν¯(A)+ν¯(B)+ϵ. Since ϵ>0 is arbitrary, (3.9) shows (3.10)ν¯(A∪B)+ν¯(A∩B)≤ν¯(A)+ν¯(B), and so ν¯ is submodular.Combining
Theorems3.1 and 3.2, we have the following
theorem.Theorem 3.3.
If𝒞   is a lattice, and if  ν′   is supermodular on  𝒞  then  ν¯   is submodular.We recall that given an outer measureν, a measurable cover for a set E⊂X is a set M∈𝒮ν such that ν(E)=ν(M).Definition 3.4.
Letν be a finite, finitely subadditive outer measure,
and let 𝒮ν be the
ν-measurable sets. We define for E⊂X(3.11)ν0(E)=inf{ν(M)/E⊂M∈𝒮ν}, and say that ν is approximately regular if ν=ν0.
ν0 is a finite, finitely subadditive, submodular
outer measure, and ν0(M)=ν(M) for M∈𝒮ν.It is known
that when a finite, finitely subadditive outer measureν is approximately regular then ν is submodular [5].Theorem 3.5.
Let𝒞   be a covering class and let  ν  be a finite, finitely subadditive outer
measure. If every  C′∈𝒞′  has a measurable cover with respect to  ν¯,  then  ν¯
is approximately regular and therefore
submodular.Proof.
We first
show that the hypothesis that eachC′∈𝒞′ has a measurable cover with respect to ν¯ implies that ν¯=ν¯0 on 𝒞′.
LetC′∈𝒞′.  We always have for any M∈𝒮ν¯ with C′⊂M,ν¯(C′)≤ν¯(M). Therefore, by the definition of ν¯0(C′),(3.12)ν¯(C′)≤ν¯0(C′). Now by hypothesis, there exists an M0∈𝒮ν¯ such that C′⊂M0 and ν¯(C′)=ν¯(M0). By monotonicity of ν¯0,  (3.13)ν¯0(C′)≤ν¯(M0)=ν¯(C′).
By (3.12) and (3.13),	ν¯0(C′)=ν¯(C′). Since C′∈𝒞′ is arbitrary, ν¯=ν¯0 on 𝒞′,
so that ν¯ is approximately regular on 𝒞′.
We next show that
whenν¯
is approximately
regular on 𝒞′  implies that ν¯ is approximately regular.
LetE⊂X. Let M∈𝒮ν¯ with E⊂M. By monotonicity of ν¯,ν¯(E)≤ν¯(M). It follows from the definition that (3.14)ν¯(E)≤ν¯0(E). Suppose now that C′∈𝒞′ and E⊂C′. By monotonicity of ν¯0,(3.15)ν¯0(E)≤ν¯0(C′)=ν¯(C′). We always have ν¯=ρ′ on 𝒞′,
so (3.15) shows that (3.16)ν¯0(E)≤ρ′(C′). Since 𝒞′ is any set from 𝒞′ with E⊂C′,
(3.16) shows that (3.17)ν¯0(E)≤ν¯(E). By (3.14) and (3.17), we have ν¯(E)=ν¯0(E) for any E⊂X. Therefore, ν¯=ν¯0,
so that ν¯ is approximately regular. It follows from a
result in [5] that ν¯ is submodular.An important
question to consider is whether or not𝒞⊂𝒮ν¯ without requiring that ν¯ is submodular. If this is true, then since 𝒮ν¯ is an algebra, 𝒜(ℒ(𝒞))=𝒜(ℒ)⊂𝒮ν¯ and thus ν¯/𝒜(ℒ)∈M(ℒ).
Our next theorem shows that this is indeed the case when 𝒞 is a lattice and ρ is superadditive on 𝒞.Theorem 3.6.
If𝒞   is a lattice, and if  ρ  is superadditive on 𝒞, then  𝒞⊂𝒮ν¯.Proof.
It suffices
to show by a theorem in [2] that each C∈𝒞 splits all sets of 𝒞′ additively with respect to ν¯.
LetC∈𝒞 and A′∈𝒞′.  We always have ν¯(A′)≤ν¯(A′∩C)+ν¯(A′∩C′).  For the reverse inequality, we proceed as
follows. Let ϵ>0 be given and arbitrary.  There exists B,D∈𝒞 such that
(3.18)B⊂A′∩C′,ρ′(A′∩C′)−ϵ2<ρ(B),D⊂A′∩B′,ρ′(A′∩B′)−ϵ2<ρ(D).
NowB∩D⊂B∩(A′∩B′)=∅, and B∪D⊂(A′∩C′)∪(A′∩B′)⊂A′.  Since 𝒞 is a lattice, B∪D∈𝒞.
We have ν¯(A′)=ρ′(A′),
so by the definition of ρ′(A′),
we have (3.19)ν¯(A′)>ρ′(A′)≥ρ(B∪D). By the superadditivity of ρ on 𝒞, (3.20)ρ(B∪D)≥ρ(B)+ρ(D). Using (3.19), (3.20), and  (3.18), we have (3.21)ν¯(A′)>ρ′(A′∩C′)+ρ′(A′∩B′)−ϵ. Again, since 𝒞 is a lattice and A,B,C∈𝒞 and ν¯=ρ′ on 𝒞′,
we have by (3.21)
(3.22)ν¯(A′)>ν¯(A′∩C′)+ν¯(A′∩B′)−ϵ. Since A′∩B′⊃A′∩(A∪C)=A′∩C,
by monotonicity of ν¯, (3.23)ν¯(A′∩B′)>ν¯(A′∩C).
Therefore, by (3.22) and (3.23), we get ν¯(A′)>ν¯(A′∩C′)+ν¯(A′∩C)−ϵ. Since ϵ>0 is arbitrary, we get ν¯(A′)≥ν¯(A′∩C′)+ν¯(A′∩C). Therefore, ν¯(A′)=ν¯(A′∩C)+ν¯(A′∩C′),
and C splits A′ additively with respect to ν¯. Since A′∈𝒞′ is arbitrary, it follows that C splits all sets of 𝒞′ additively with respect to ν¯ and so C∈𝒮ν¯. The arbitrariness of C∈𝒞 shows that 𝒞⊂𝒮ν¯.
Consequently,𝒜(ℒ)⊂𝒮ν¯ and so ν¯/𝒜(ℒ)∈M(ℒ).


## 4. Equality ofν and Its Associated Set Functions on ℒ

In
this section, we again consider a covering class𝒞 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 of  X. We assume that 𝒞⊂𝒮ν, the set of
ν-measurable sets. Since 𝒮ν is an algebra of sets, ℒ=ℒ(𝒞)⊂𝒮ν, hence 𝒜(ℒ)⊂𝒮ν.  Therefore, μ=ν/𝒜(ℒ)∈M(ℒ).
We consider the following set functions obtained from ν, which are defined in Section 2: μ′,μi,ρ,ν′,ρ′. We obtain an inequality between all these set
functions that holds for all subsets of X, and determine conditions for equality of these
functions on ℒ.
Equality on ℒ becomes useful, since for instance we will
then have μ∈MR(ℒ).Theorem 4.1.
ρ′≤μi≤ρ≤ν≤μ′≤ν′    on  𝒫(X).Proof.
LetE⊂X be fixed but arbitrary.  We always have (4.1)ρ(E)≤ν(E),so it is the other inequalities that we must establish.
LetL∈ℒ with E⊂L′. The definition of μ′(E) shows that μ′(E)≤μ′(L′)=ν(L′). By monotonicity of ν,
we have ν(E)≤ν(L′)=μ(L′) whenever E⊂L′,L∈ℒ.
Thus, ν(E) is a lower bound for the set {μ(L′)/E⊂L′,L∈ℒ}. Therefore, (4.2)ν(E)≤μ′(E).
Now supposeC∈𝒞 and E⊂C′. Since C∈ℒ, μ′(E)≤μ(C′)=ν(C′). Therefore, μ′(E) is a lower bound for the set {ν(C′)/E⊂C′,C∈𝒞}.  The definition of ν′(E) gives (4.3)μ′(E)≤ν′(E).
Next, consider anyC∈𝒞 with C⊂E. By the definition of μi(E),
we have μ(C)≤μi(E). Since μ(C)=ν(C), we have ν(C)≤μi(E) for all C⊂E with C∈𝒞.
Since ρ(C)≤ν(C),
we have ρ(C)≤μi(E) for all C∈𝒞,
with C⊂E. Therefore, μi(E) is an upper bound for the set {ρ(C)/C⊂E,C∈𝒞}.  The definition of ρ′(E) thus gives (4.4)ρ′(E)≤μi(E).
To show thatμi(E)≤ρ(E),
we argue by contradiction.  Thus suppose
that ρ(E)<μi(E). Then there exists an L0∈ℒ such that L0⊂E and ρ(E)<μ(L0)≤μi(E). By monotonicity of ρ,ρ(L0)≤ρ(E), so we have ρ(L0)<μ(L0). Now ρ(L0)=ν(X)−ν(L0′)=μ(X)−μ(L0′),
hence μ(X)<μ(L0)+μ(L0′). This contradicts the finite additivity of μ.
Therefore, it must be (4.5)μi(E)≤ρ(E).
Combining inequalities (4.1) through (4.5), we have (4.6)ρ′(E)≤μi(E)≤ρ(E)≤ν(E)≤μ′(E)≤ν′(E).
SinceE⊂X was arbitrary, we have shown that ρ′≤μi≤ρ≤ν≤μ′≤ν′ on 𝒫(X).We now
determine conditions when the inequality of Theorem4.1 will be an equality on ℒ.
Before doing so, we indicate what happens when this is true.  We always have that μ′ is submodular.
Assuming ρ′=μi=ρ=ν=μ′=ν′ on ℒ, suppose L∈ℒ.
Then μi(L′)=μ′(X)−μ′(L)=μ(X)−μ(L)=μ(L′) since μ∈M(ℒ).  Now μi(L′)=sup{μ(K)/K⊂L′,K∈ℒ} so we have μ(L′)=sup{μ(K)/K⊂L′,K∈ℒ}.  By [5], this shows since L∈ℒ is arbitrary that μ∈MR(ℒ).  It follows from this that ℒ⊂𝒮μ′, and so (μ′)0=μ,′ so that μ′ is approximately regular, hence submodular.Theorem 4.2.
If𝒞   is a lattice,  ν   satisfies condition  A, and  ν   is submodular on  𝒞′, then  ρ′=μi=ρ=ν=μ′=ν′  on  ℒ.Proof.
We show thatν′  is submodular. Since ν satisfies condition A,
𝒞⊂𝒮ν′. Let E,F⊂X and let ϵ>0 be arbitrary. By the definition of ν′,
there exist C1,C2∈𝒞 such that E⊂C1′ and ν′(E)≤ν(C1′)<ν′(E)+ϵ/2, and F⊂C2′ and (4.7)ν′(F)≤ν(C2′)<ν′(F)+ϵ2.
NowE∪F⊂C1′∪C2′, E∩F⊂C1′∩C2′,
where C1∩C2 and C1∪C2 belong to the lattice 𝒞.  Therefore, (4.8)ν′(E∪F)+ν′(E∩F)≤ν(C1′∪C2′)+ν(C1′∩C2′).
Sinceν is submodular on 𝒞′, ν(C1′∪C2′)+ν(C1′∩C2′)≤ν(C1′)+ν(C2′).  Combining this last inequality with (4.8) and
(4.7), we have (4.9)ν′(E∪F)+ν′(E∩F)<ν′(E)+ν′(F)+ϵ.
Sinceϵ>0 is arbitrary, this shows that ν′(E∪F)+ν′(E∩F)≤ν′(E)+ν′(F),
where E,F⊂X are arbitrary.
Therefore, ν′ is submodular. Hence, 𝒮ν′=𝒮ν′ where 𝒮ν′ is an algebra.
By our initial observation, 𝒞⊂𝒮ν′=𝒮ν′,  and so ℒ⊂𝒮ν′.  Thus for all L∈ℒ, ν′(L)=ρ′(L),
and so by Theorem 4.1, we have ρ′=μi=ρ=ν=μ′=ν′ on ℒ.Theorem 4.3.
Ifν=ν0   on  𝒞′, 𝒞  coseparates  ℒ(𝒞), and  ν   satisfies condition B, then  ρ′=μi=ρ=ν=μ′=ν′   on   ℒ.Proof.
The
hypothesis thatν=ν0 on 𝒞′ shown by [6] that ν′ is submodular and so 𝒮ν′=𝒮ν′. We show that the hypotheses that 𝒞 coseparates ℒ(𝒞) and ν satisfies condition B
give
ρ′=μi=ρ=ν=μ′=ν′ on 𝒞.
Since𝒞⊂𝒮ν⊂𝒮ν,  ν(C)=ρ(C) for all C∈𝒞.  Since for any C∈𝒞, ρ′(C)=ρ(C)=μi(C),
we have (4.10)ρ′=μi=ρ=νon𝒞.
We next showμ′=ν′ on 𝒞.
To do so, we argue by contradiction. Suppose there is a C0∈𝒞 such that μ′(C0)<ν′(C0).
Then there is an L0∈ℒ such that C0⊂L0′ and μ′(C0)≤μ(L0′)<ν′(C0). Since C0∩L0=∅ with C0,L0∈ℒ and 𝒞 coseparates ℒ, there exist A,B∈𝒞 with C0⊂A′,L0⊂B′ and A′∩B′=∅.  Therefore, C0⊂A′⊂B⊂L0′ and (4.11)μ′(C0)≤μ′(A′)=μ(A′)≤μ(B)≤μ(L0′)<ν′(C0).
We also haveμ(A′)=ν(A′). Thus, we have an A∈𝒞⊂ℒ,
with C0⊂A′ and ν(A′)<ν′(C0). This contradicts the definition of ν′(C0). Therefore, we must have μ′=ν′ on 𝒞.
We next showν=μ′ on 𝒞.  We reason by contradiction.  Suppose there is a C∈𝒞 with ν(C)<μ′(C).  Therefore, ν(X)−ν(C)>ν(X)−μ′(C). But we also have ν(X)=ν(C)+ν(C′) so ν(C′)=ν(X)−ν(C) and ν(X)=μ(X)=μ′(X). Therefore, ν(C′)>μ′(X)−μ′(C)=μi(C′).  Since ν satisfies
condition B,
ν(C′)=sup{ν(A)/A⊂C′,A∈𝒞}>μi(C′).  Thus, there is an A0∈𝒞 with A0⊂C′ and μi(C′)<ν(A0)≤ν(C′), and A0∩C=∅. Since 𝒞 coseparates ℒ,
there exist B,D∈𝒞 such that A0⊂B′,C⊂D′ and B′∩D′=∅. Therefore, A0⊂B′⊂D⊂C′. This gives (4.12)μi(C′)<ν(A)≤ν(B′)≤ν(D)≤ν(C′).
Butν(D)=μ(D),
so we have a D∈𝒞⊂ℒ with D⊂C′ and μi(C′)<μ(D). This contradicts the definition of μi(C′). Therefore, we must have ν=μ′ on 𝒞.  Thus we have ρ′=μi=ρ=ν=μ′=ν′ on 𝒞.  Since ρ′=ν′ on 𝒞, 𝒞⊂𝒮ν′. By the submodularity of ν′, 𝒮ν′=𝒮ν′. Therefore, 𝒞⊂𝒮ν′ and so ℒ⊂𝒮ν′. Therefore, for all L∈ℒ, ρ′(L)=ν′(L) and so ρ′=μi=ρ=ν=μ′=ν′ on ℒ.Thus far, in
order to have the equality(*)ρ′=μi=ρ=ν=μ′=ν′onℒ, the property of submodularity  of ν or ν′ on 𝒞 or 𝒞′ played 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 measure ν′,
it is always true that 𝒮ν′⊂𝒮ν′, where 𝒮ν′ is an algebra of sets.  If 𝒞⊂𝒮ν′,
then ℒ=ℒ(𝒞)⊂𝒮ν′, and consequently 𝒞⊂𝒮ν′,
whence ν′(L)=ρ′(L) for all L∈ℒ.  It follows from this that the equality (*)
holds on ℒ.  It seems reasonable then to seek conditions
for 𝒞⊂𝒮ν′.We have the
following theorem.Theorem 4.4.
Suppose
eachC∈𝒞   splits all sets of 𝒞′  additively with respect to ν′. Then 𝒞⊂𝒮ν′  and ρ′=μi=ρ=ν=μ′=ν′   on ℒ.The result follows from [2].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 Theorem4.4 may be difficult to
implement in practice, it leads naturally to the content of the next section.


## 5. An Outer Measure Construction

In this
section, we consider a nonempty setX and the construction of a finite, finitely
subadditive outer measure given an arbitrary family of subsets ℬ of X, and a finite, nonnegative set function τ on ℬ.  By the standard method, we construct a
finite, finitely subadditive outer measure λ from τ and ℬ.  We seek a characterization of those sets that
split all sets of ℬ additively with respect to λ.For
emphasis, letX be a nonempty set and let ℬ be an arbitrary family of subsets of X with ∅,X∈ℬ and ℬ is closed under finite unions.  Let τ be a finite, nonnegative set function defined
on ℬ with τ(∅)=0, and τ subadditive on ℬ.  For E⊂X let (5.1)λ(E)=inf{τ(B)/E⊂B,B∈ℬ}.It follows by a standard argument thatλ is a finite, finitely subadditive outer
measure defined for all subsets of X. Let 𝒮λ be the algebra of λ-measurable subsets of X. We determine conditions when a set or a family
of sets splits all sets of ℬ additively with respect to λ, this set or family will split all subsets of X additively with respect to λ. The importance of this was hinted at in the
last section.We observe that ifB∈ℬ,
then λ(B)≤τ(B). Also, if τ is monotone on ℬ,
then λ(B)=τ(B) for all B∈ℬ.The
following theorem is known, and its proof can be found in [2], for instance.Theorem 5.1.
IfE⊂X  splits all sets of  ℬ   additively with respect to  λ, then E∈𝒮λ.Our
motivation for what follows comes from the characterization of the measurable
sets for aσ-finite measure which can be found in [8, 9].
We begin with the observation that there are always at least two sets in ℬ that split all sets of ℬ additively with respect to λ, namely, ∅ and X (they may be the only ones). We seek a
characterization of such sets.Theorem 5.2.
LetB0∈ℬ   split all sets
of  ℬ  additively with respect to  λ.    Let  N∈ℬ  be such that  τ(N)=0.   Then  E=B0−N  splits all sets of  ℬ  additively with respect to  λ,   and so  E∈𝒮λ.Proof.
LetB∈ℬ be arbitrary.
We can write B=(B∩E)∪(B∩E′). Now, B∩E=B∩(B0∩N′)⊂B∩B0  and B∩E′=B∩(B0′∪N)⊂(B∩B0′)∪N. By the monotonicity and finite subadditivity
of λ,(5.2)λ(B∩E)+λ(B∩E′)≤λ(B∩B0)+λ(B∩B0′)+λ(N).
SinceN∈ℬ, λ(N)≤τ(N)=0, so λ(N)=0 and (5.2) becomes (5.3)λ(B∩E)+λ(B∩E′)≤λ(B∩B0)+λ(B∩B0′).
The hypothesis onB0 gives λ(B∩B0)+λ(B∩B0′)=λ(B) so by (5.3), λ(B∩E)+λ(B∩E′)≤λ(B). The reverse inequality is always true by
finite subadditivity of λ. Thus by Theorem 5.1, (5.4)E∈𝒮λ.Thus we see that a sufficient condition for a setE to be λ-measurable is: E must be expressible as the difference of a set
from ℬ that splits all sets of ℬ additively with respect to λ and a subset of X for which τ is 0.An example
may help to illustrate the theorem.Example 5.3.
LetX={a,b,c} and take ℬ={∅,X,{a},{b},{a,b}}.  Define τ on ℬ by τ(∅)=τ({a})=0,τ({b})=τ({a,b})=1,τ(X)=2. The outer measure λ determined by ℬ and τ is λ(∅)=λ({a})=0, λ({b})=λ({a,b})=1, λ({c})=λ({a,c})=λ({b,c})=λ(X)=2. Now 𝒮λ={∅,X,{a},{b,c}}. We see that ∅=X−∅,  X=X−∅, {a}={a}−∅,{b,c}=X−{a}.We return to
characterizing those setsE that split all sets of ℬ additively with respect to λ. In the style of Munroe [10], we have the
following theorem.Theorem 5.4.
LetE⊂X  be any set. There is a sequence of sets {Bn}  from ℬ  such that E⊂Bn  all n and λ(E)=λ(⋂n=1∞Bn).Proof.
LetE⊂X be arbitrary. By the definition of λ(E), for each n∈ℕ there is a Bn∈ℬ such that E⊂Bn and (5.5)λ(E)≤τ(Bn)<λ(E)+1n.
LetB=⋂n=1∞Bn. Then E⊂B⊂Bn all n. By monotonicity of λ, (5.6)λ(E)≤λ(B)≤λ(Bn)alln.
Nowλ(Bn)≤τ(Bn) all n, so by (5.6) and (5.5)  we get λ(E)≤λ(B)<λ(E)+1/n for any n∈ℕ. Thus for any given ϵ>0, we can find an n∈N large enough so
that 0<1/n<ϵ. For any such n,λ(E)≤λ(B)<λ(E)+ϵ. Since ϵ>0 is arbitrary, it follows that λ(E)=λ(B).Observation.
SupposeE⊂X splits all sets A⊂X additively with respect to λ. By Theorem 5.4, there exist Bn∈ℬ,n∈ℕ with E⊂⋂n=1∞Bn and λ(E)=λ(⋂n=1∞Bn). Let B=⋂n=1∞Bn. We can write B=E∪(B∩E′). Since E∈𝒮λ,λ(B)=λ(E)+λ(B∩E′),
and since all quantities are finite, λ(B∩E′)=λ(B)−λ(E)=0.  Since E=B−(B−E′), let N=B−E so that λ(N)=0.
This observation establishes the following theorem.Theorem 5.5.
IfE∈𝒮λ, then there is a sequence of sets Bn∈ℬ, and a set N⊂X  such that E⊂Bn  all n, and E=(⋂n=1∞Bn)−N, where λ(N)=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 Munroe [10].In
summary, to obtain a setE that splits all sets of ℬ additively with respect to λ: let B0∈ℬ be a set that splits all sets of ℬ additively with respect to λ (for instance X) and let N⊂X with τ(N)=0. Then E=B0−N will split all sets of ℬ additively with respect to λ, so that E∈𝒮λ.IfE∈𝒮λ,
then there is a sequence of sets {Bn} from ℬ such that E⊂Bn all n , and a set N⊂X such that λ(N)=0 and E=(⋂n=1∞Bn)−N.We note
that ifℬ is a
δ-lattice that is, a lattice closed under countable
intersections, then the set B=⋃n=1∞Bn∈ℬ.



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*Source: 896480-2009-04-22.xml*