International Journal of Mathematics and Mathematical Sciences

VolumeΒ 2008Β (2008), Article IDΒ 896480, 13 pages

http://dx.doi.org/10.1155/2008/896480

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

Department of Mathematics & Computer Science, St. John's University, 8000 Utopia Parkway Queens, New York, NY 11439, USA

Received 1 August 2008; Accepted 3 December 2008

Academic Editor: AndreiΒ Volodin

Copyright Β© 2008 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.

#### Abstract

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

#### 1. Introduction

Let be a nonempty set, and let be a covering class of subsets of such that , with closed under finite intersections, and let be a finite, finitely subadditive outer measure defined for all subsets of In [1, 2], we considered a finite, finitely subadditive outer measure defined by means of The outer measure generalized results obtained by [3] for the case of a 0-1 valued, finitely additive measure defined on the algebra of subsets generated by a lattice of subsets of In investigating properties of , the hypothesis that be submodular is often present. We determine conditions for to be submodular.

We also
consider for a finite, finitely subadditive outer measure and covering class under the hypothesis that , the finite, finitely additive measure ,
where is the lattice generated by . There are several set functions associated
with and namely, and it is shown that the inequality always holds for all subsets of We determine when an equality of all these set functions holds on . This is useful since for instance it allows
one to conclude that is an -*regular* measure on .
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, ,
of subsets of with and closed under finite unions, and a nonnegative,
finite set function defined on ,
with Assuming that is finitely subadditive on ,
the function for is a finite, finitely subadditive outer
measure defined for all subsets of We obtain a characterization of those sets that are -*measurable.* This provides us with a condition for the sets
of to be -*measurable* without requiring that 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, 4β6]. 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, will denote a nonempty abstract set. For , will denote the complement of The power set of will be denoted by We will be concerned with covering classes of subsets of By a *covering class*, we will always
mean a nonempty class of subsets of such that and such that for any there is a finite family of sets with We always assume that is closed under finite intersections. We
define to be the lattice generated by . is the set of all finite unions of sets from . The algebra generated by is . It is known that , the algebra generated by . The set of complements of the members of is denoted by

*Definition 2.1. * A covering class is said to *coallocate itself* if
whenever and , there are sets with and

*Definition 2.2. * A covering class is said to be *normal* if for any with ,
there exist with ,
and

*Definition 2.3. * Let and be covering classes with . is said to *coseparate* if whenever there exist with and

It is known that if a covering class coallocates itself, then it is a normal covering class.

Let be a finite, finitely subadditive outer measure defined for all subsets of and let be a covering class of the subsets of In general, there are several set functions that we can associate with this outer measure For , we define . is finite valued and in general is not an inner measure. Conditions when is an inner measure have been thoroughly investigated in [4, 5]. We also define a set function by where is always a finite, finitely subadditive outer measure, and so it has associated with it a , which may be expressed as

It is always true that on When a set function is an outer measure, the family of -*measurable sets* is denoted by that is,

It is well known that is an algebra and the restriction of to , that is, is a finite, finitely additive measure. We also define is not in general an algebra of sets, and we always have

An outer measure satisfies Condition A: if on . satisfies Condition B: if , for .

*Definition 2.4. * Let be a collection of subsets of and let be a real-valued set function defined on .

If for all with ,
we have then is said to be *submodular on*. If the reverse inequality holds, then is said to be *supermodular on*. is said to be *modular on* if equality holds. If then is said to be *superadditive on*.

If we usually say is *submodular.*

Let be a lattice of subsets of such that . denotes the algebra generated by ,
and denotes the set of all nontrivial, nonnegative, finitely additive measures on . denotes the set of all those which are -*regular*,
that is, those with the property that for any .

, , and several other subsets of have been extensively studied in the literature; we cite just a few recent papers [4β8]. For , the set function defined for by is a finite, finitely subadditive outer measure on We also define the set function by

#### 3. Submodularity of the Outer Measure

In [3], a 0-1 measure was used to give necessary and sufficient conditions for a lattice to be normal, by constructing a 0-1, finitely subadditive outer measure In [1, 2], the measure was replaced by a finite, finitely subadditive outer measure and the outer measure was generalized by a finite, finitely subadditive outer measure 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 to be submodular.

We begin by recalling the basic construction and properties of For emphasis, let be a nonempty set and let be a covering class of subsets of with and closed under finite intersections. Let be a finite, finitely subadditive outer measure defined for all subsets of Assume that coallocates itself and the associated set function is finitely subadditive on Since is finitely subadditive on the set function is also finitely subadditive on (see, e.g., [2]). For define where is a finite, finitely subadditive outer measure and

(1) on (2) on ,(3) on (4) satisfies condition A, that is, on which is equivalent to on

(The functions are the usual ones associated with the outer measure as defined for a general finite, finitely subadditive outer measure in Section 2).

We now determine conditions for to be submodular. It will be shown that by requiring that or its associated set functions satisfy certain conditions on either or , and is a lattice, the submodularity of will follow. The importance of being submodular can be seen from the following known facts [2].

It is known that if is submodular then

(1)(2) is regular on where ,(3),(4) on

Theorem 3.1. * If is a lattice, and if is supermodular on ,
then is submodular on and so is submodular on .*

*Proof. *Let .
Since is supermodular on the lattice , Therefore, and so Since on ,
the above inequality shows Since are arbitrary, this shows that is submodular on and is submodular on .

Theorem 3.2. * Suppose is a lattice. If is submodular on ,
then is submodular.*

*Proof. *Let and let be given and arbitrary. By definition of there exist such that and Now , Since is submodular on , By (3.7), (3.8), (3.6), we have Since is arbitrary, (3.9) shows and so is submodular.

Combining Theorems 3.1 and 3.2, we have the following theorem.

Theorem 3.3. * If is a lattice, and if is supermodular on then is submodular.*

We recall that given an outer measure , a *measurable cover* for a set is a set such that

*Definition 3.4. * Let be a finite, finitely subadditive outer measure,
and let be the
-measurable sets. We define for and say that is *approximately regular* if .

is a finite, finitely subadditive, submodular
outer measure, and for

It is known that when a finite, finitely subadditive outer measure is approximately regular then is submodular [5].

Theorem 3.5. * Let be a covering class and let be a finite, finitely subadditive outer
measure. If every has a measurable cover with respect to then
is approximately regular and therefore
submodular.*

*Proof. *We first
show that the hypothesis that each has a measurable cover with respect to implies that on

Let . We always have for any with Therefore, by the definition of Now by hypothesis, there exists an such that and By monotonicity of ,

By (3.12) and (3.13), Since is arbitrary, on ,
so that is approximately regular on .

We next show that
when
is approximately
regular on implies that is approximately regular.

Let Let with By monotonicity of It follows from the definition that Suppose now that and By monotonicity of We always have on ,
so (3.15) shows that Since is any set from with ,
(3.16) shows that By (3.14) and (3.17), we have for any Therefore, ,
so that is approximately regular. It follows from a
result in [5] that is submodular.

An important question to consider is whether or not without requiring that is submodular. If this is true, then since is an algebra, and thus . Our next theorem shows that this is indeed the case when is a lattice and is superadditive on .

Theorem 3.6. * If is a lattice, and if is superadditive on , then *

*Proof. *It suffices
to show by a theorem in [2] that each splits all sets of additively with respect to

Let and We always have For the reverse inequality, we proceed as
follows. Let be given and arbitrary. There exists such that

Now and Since is a lattice, .
We have ,
so by the definition of ,
we have By the superadditivity of on , Using (3.19), (3.20), and (3.18), we have Again, since is a lattice and and on ,
we have by (3.21)
Since ,
by monotonicity of ,

Therefore, by (3.22) and (3.23), we get Since is arbitrary, we get Therefore, ,
and splits additively with respect to Since is arbitrary, it follows that splits all sets of additively with respect to and so The arbitrariness of shows that

Consequently, and so

#### 4. Equality of and Its Associated Set Functions on

In this section, we again consider a covering class of subsets of a nonempty set , being closed under finite intersections, and such that , and a finite, finitely subadditive outer measure defined for all subsets of We assume that the set of -measurable sets. Since is an algebra of sets, , hence Therefore, . We consider the following set functions obtained from which are defined in Section 2: We obtain an inequality between all these set functions that holds for all subsets of and determine conditions for equality of these functions on . Equality on becomes useful, since for instance we will then have .

Theorem 4.1. * on *

*Proof. *Let be fixed but arbitrary. We always have so it is the other inequalities that we must establish.

Let with The definition of shows that By monotonicity of ,
we have whenever .
Thus, is a lower bound for the set . Therefore,

Now suppose and Since , Therefore, is a lower bound for the set . The definition of gives

Next, consider any with By the definition of ,
we have Since we have for all with .
Since ,
we have for all ,
with Therefore, is an upper bound for the set . The definition of thus gives

To show that ,
we argue by contradiction. Thus suppose
that Then there exists an such that and By monotonicity of so we have Now ,
hence This contradicts the finite additivity of .
Therefore, it must be

Combining inequalities (4.1) through (4.5), we have

Since was arbitrary, we have shown that on

We now determine conditions when the inequality of Theorem 4.1 will be an equality on . Before doing so, we indicate what happens when this is true. We always have that is submodular. Assuming on , suppose . Then since . Now so we have . By [5], this shows since is arbitrary that . It follows from this that and so so that is approximately regular, hence submodular.

Theorem 4.2. * If is a lattice, satisfies condition A, and is submodular on , then on .*

*Proof. *We show that is submodular. Since satisfies condition A,
Let and let be arbitrary. By the definition of ,
there exist such that and , and and

Now , ,
where and belong to the lattice . Therefore,

Since is submodular on , Combining this last inequality with (4.8) and
(4.7), we have

Since is arbitrary, this shows that ,
where are arbitrary.
Therefore, is submodular. Hence, where is an algebra.
By our initial observation, and so . Thus for all , ,
and so by Theorem 4.1, we have on .

Theorem 4.3. * If on , coseparates , and satisfies condition B, then on .*

*Proof. *The
hypothesis that on shown by [6] that is submodular and so We show that the hypotheses that coseparates and satisfies condition B
give
on .

Since ,β for all . Since for any ,β,
we have

We next show on .
To do so, we argue by contradiction. Suppose there is a such that .
Then there is an such that and Since with and coseparates , there exist with and . Therefore, and

We also have Thus, we have an ,
with and This contradicts the definition of Therefore, we must have on .

We next show on . We reason by contradiction. Suppose there is a with . Therefore, But we also have so and Therefore, Since satisfies
condition B,
. Thus, there is an with and and Since coseparates ,
there exist such that and Therefore, This gives

But ,
so we have a with and This contradicts the definition of Therefore, we must have on . Thus we have on . Since on , By the submodularity of , Therefore, and so Therefore, for all , and so on .

Thus far, in order to have the equality the property of submodularity of or on or 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 , it is always true that where is an algebra of sets. If , then and consequently , whence for all . It follows from this that the equality (*) holds on . It seems reasonable then to seek conditions for

We have the following theorem.

Theorem 4.4. * Suppose
each splits all sets of additively with respect to . Then and on .*

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 Theorem 4.4 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 set and the construction of a finite, finitely subadditive outer measure given an arbitrary family of subsets of and a finite, nonnegative set function on . By the standard method, we construct a finite, finitely subadditive outer measure from and . We seek a characterization of those sets that split all sets of additively with respect to

For emphasis, let be a nonempty set and let be an arbitrary family of subsets of with and is closed under finite unions. Let be a finite, nonnegative set function defined on with and subadditive on . For let

It follows by a standard argument that is a finite, finitely subadditive outer
measure defined for all subsets of Let be the algebra of -*measurable* subsets of We determine conditions when a set or a family
of sets splits all sets of additively with respect to this set or family will split all subsets of additively with respect to The importance of this was hinted at in the
last section.

We observe that if , then Also, if is monotone on , then for all .

The following theorem is known, and its proof can be found in [2], for instance.

Theorem 5.1. * If splits all sets of additively with respect to , then *

Our motivation for what follows comes from the characterization of the measurable sets for a -finite measure which can be found in [8, 9]. We begin with the observation that there are always at least two sets in that split all sets of additively with respect to namely, and (they may be the only ones). We seek a characterization of such sets.

Theorem 5.2. * Let split all sets
of additively with respect to Let be such that Then splits all sets of additively with respect to and so *

*Proof. *Let be arbitrary.
We can write Now, and By the monotonicity and finite subadditivity
of

Since , so and (5.2) becomes

The hypothesis on gives so by (5.3), The reverse inequality is always true by
finite subadditivity of Thus by Theorem 5.1,

Thus we see that a sufficient condition for a set to be -*measurable* is: must be expressible as the difference of a set
from that splits all sets of additively with respect to and a subset of for which is 0.

An example may help to illustrate the theorem.

*Example 5.3. *Let and take Define on by The outer measure determined by and is , , Now We see that , ,

We return to characterizing those sets that split all sets of additively with respect to In the style of Munroe [10], we have the following theorem.

Theorem 5.4. * Let be any set. There is a sequence of sets from such that all n and *

*Proof. *Let be arbitrary. By the definition of for each there is a such that and

Let Then all *n*. By monotonicity of ,

Now all *n*, so by (5.6) and (5.5) we get for any Thus for any given we can find an large enough so
that For any such Since is arbitrary, it follows that

*Observation. *Suppose splits all sets additively with respect to By Theorem 5.4, there exist with and Let We can write Since ,
and since all quantities are finite, Since let so that

This observation establishes the following theorem.

Theorem 5.5. * If , then there is a sequence of sets , and a set such that all n, and , where *

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 set that splits all sets of additively with respect to : let be a set that splits all sets of additively with respect to (for instance ) and let with Then will split all sets of additively with respect to so that

If , then there is a sequence of sets from such that all n and a set such that and

We note that if is a -lattice that is, a lattice closed under countable intersections, then the set .

#### 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 authorβs teacher and friend.

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