Deficient topological measures on locally compact spaces
Svetlana V. Butler

TL;DR
This paper introduces and studies deficient topological measures on locally compact spaces, exploring their properties, variations, and conditions under which they become measures or topological measures, thereby extending the theory of measures.
Contribution
It defines deficient topological measures, investigates their properties, and provides criteria for when they are measures or topological measures, advancing the understanding of non-linear functionals.
Findings
Deficient topological measures generalize measures and topological measures.
Positive, negative, and total variations of signed set functions are deficient topological measures.
Conditions are established for deficient topological measures to be measures or topological measures.
Abstract
Topological measures and quasi-linear functionals generalize measures and linear functionals. We define and study deficient topological measures on locally compact spaces. A deficient topological measure on a locally compact space is a set function on open and closed subsets which is finitely additive on compact sets, inner regular on open sets, and outer regular on closed sets. Deficient topological measures generalize measures and topological measures. First we investigate positive, negative, and total variation of a signed set function that is only assumed to be finitely additive on compact sets. These positive, negative, and total variations turn out to be deficient topological measures. Then we examine finite additivity, superadditivity, smoothness, and other properties of deficient topological measures. We obtain methods for generating new deficient topological measures. We…
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Deficient topological measures on locally compact spaces
S. V. Butler, University of California Santa Barbara
Abstract.
Topological measures and quasi-linear functionals generalize measures and linear functionals. We define and study deficient topological measures on locally compact spaces. A deficient topological measure on a locally compact space is a set function on open and closed subsets which is finitely additive on compact sets, inner regular on open sets, and outer regular on closed sets. Deficient topological measures generalize measures and topological measures. First we investigate positive, negative, and total variation of a signed set function that is only assumed to be finitely additive on compact sets. These positive, negative, and total variations turn out to be deficient topological measures. Then we examine finite additivity, superadditivity, smoothness, and other properties of deficient topological measures. We obtain methods for generating new deficient topological measures. We provide necessary and sufficient conditions for a deficient topological measure to be a topological measure and to be a measure. The results presented are necessary for further study of topological measures, deficient topological measures, and corresponding non-linear functionals on locally compact spaces.
Key words and phrases:
deficient topological measure, topological measure, positive variation, superadditivity
2010 Mathematics Subject Classification:
Primary 28C15; Secondary 28C99
1. Introduction
This paper is one in a series by the author devoted to the study of quasi-linear functionals and topological measures on locally compact spaces.
Topological measures (initially called quasi-measures) were introduced by J. F. Aarnes in [1]. These generalizations of measures and corresponding generalizations of linear functionals are connected to the problem of linearity of the expectational functional on the algebra of observables in quantum mechanics. Despite the fact that topological measures lack many features of measures, such as algebraic structure of the domain, subadditivity, etc., many results typical for measures still hold for topological measures. There are many papers now devoted to topological measures and quasi-linear functionals. In the last 12 years numerous applications of quasi-linear functionals and topological measures to symplectic topology have been found. In fact, the first paper [6] on the connections between these two fields has been cited over 100 times, and was followed by many articles and a chapter in a monograph [11]. Deficient topological measures were first defined and used by A. Rustad and O. Johansen in [10]. They were later independently rediscovered by M. Svistula in [12] (where they were called ”functions of class ”) and further developed in [13]. In all previous works, deficient topological measures were defined as real-valued functions on a compact space. In this paper we study deficient topological measures on locally compact spaces as functions into extended real numbers.
On locally compact spaces deficient topological measures are set functions defined on open and closed subsets of a space, that are finitely additive on compact sets, inner compact regular on open sets, and outer regular for closed sets. Thus, they generalize topological measures and measures. It is interesting that many properties of topological measures and even measures still hold for deficient topological measures. Many results for topological measures and deficient topological measures on compact spaces remain valid for deficient topological measures on locally compact spaces. This is remarkable, since in the compact setting we work with open sets and closed sets, and use the fact that they are complements of each other; while in the locally compact setting the main sets used are compact sets and open sets. Finally, results transfer nicely from real-valued functions in a compact setting to functions into extended reals in a locally compact setting. In this paper we generalize for locally compact spaces many existing results and obtain some new ones. The results of this paper are necessary for further study of topological measures and corresponding non-linear functionals on locally compact spaces.
The paper is organized as follows. In Section 2 we define the positive, negative, and total variation of a signed set function that is finitely additive on compact sets. We then study various properties of positive, negative, and total variation. In Section 3 we define a deficient topological measure on a locally compact space. The positive variation is the unique smallest deficient topological measure that on compact sets is larger than its defining signed set function. We study finite additivity, superadditivity, smoothness, and other properties of deficient topological measures. The main focus of Section 4 is finding conditions under which a deficient topological measure is a topological measure or a measure. In Section 5 we discuss obtaining new deficient topological measures by restricting and extending, and also by using continuous functions. Section 6 is devoted to examples.
In this paper is a locally compact, connected space. By a component of a set we always mean a connected component. We denote by the closure of a set . A set is called bounded if is compact. A set is called solid if is connected, and has only unbounded connected components. We denote by a union of disjoint sets. When we consider set functions into extended real numbers, we consider functions that are not identically or .
Several collections of sets will be used often. They include: , the collection of open subsets of ; the collection of closed subsets of ; the collection of compact subsets of ; . By we denote the collection of finite unions of disjoint compact connected sets, by the collection of compact connected sets, and by the collection of compact solid sets. is the power set of .
Definition 1**.**
Let be a topological space and be a nonnegative set function on , a family of subsets of that contains . We say that
- •
is inner regular (or inner compact regular) at if
- •
is outer regular at if
- •
is inner regular if is inner regular at for every
- •
is outer regular if is outer regular at for every
- •
is regular if is both inner and outer regular
- •
is compact-finite if for any .
- •
is smooth on compact sets if implies .
- •
is smooth on open sets if implies .
- •
is simple if it only assumes values [math] and .
We consider set functions that are not identically or .
Definition 2**.**
A Radon measure on is a Borel measure that is compact-finite, outer regular on all Borel sets, and inner regular on all open sets, i.e. for every Borel set
[TABLE]
and for every open set
[TABLE]
Remark 3**.**
Here is a useful observation which follows, for example, from Corollary 3.1.5 in [5].
- (i)
If where , and and at least one of are compact, then there exists such that for all .
- (ii)
If where then there exists such that for all .
The following lemma can be found, for example, in [8] (see Chapter X, par. 50, Theorem A).
Lemma 4**.**
Let be locally compact. If , where is compact and are open, then there exist compact sets and such that .
We would like to note the following fact. (See, for example, [4], Chapter XI, 6.2)
Lemma 5**.**
Let in a locally compact space . Then there exists a bounded open sets such that
[TABLE]
We will also need the following two results (see, for example, section 2 in [3]).
Lemma 6**.**
Let in a locally compact, locally connected space . If either or is connected there exist a bounded open connected set and a compact connected set such that
[TABLE]
One may take .
Lemma 7**.**
Let be a locally compact and locally connected space. Suppose . Then there exists such that .
2. Positive, negative and total variation for a signed set function
Definition 8**.**
Given signed set function which assumes at most one of we define two set functions on , the positive variation and the total variation , as follows:
for an open subset let
[TABLE]
[TABLE]
and for a closed subset let
[TABLE]
[TABLE]
We define the negative variation associated with a signed set function as a set function .
Remark 9**.**
Let be two signed set functions as in Definition 8. We see that , , and for any open set . Thus, and .
For any open set we have . It follows that . Similarly, for any real number .
If then .
Lemma 10**.**
Let be locally compact. Let which assumes at most one of be a signed set function which is finitely additive on compact sets. Then
- (s1)
* Also, and .* 2. (s2)
For an open set , if then , and if then . 3. (s3)
* and are monotone, i.e. if then and .* 4. (s4)
* and are finitely additive on open sets.* 5. (s5)
* and for any compact . In particular, if on , then . Also, .* 6. (s6)
If are disjoint, is closed and all are compact, then
[TABLE]
and
[TABLE] 7. (s7)
* and are inner compact regular, i.e. and for any * 8. (s8)
If are disjoint, is closed and all are compact, then
[TABLE]
and
[TABLE]
In particular, and are finitely additive on compact sets. 9. (s9)
* and are superadditive, i.e. if where , and at most one of the closed sets is not compact, then*
[TABLE]
In particular, if , where is compact and is open, then
[TABLE] 10. (s10)
** 11. (s11)
.
Proof.
- (s1)
Since is not identically (or ), from finite additivity of on compact sets it follows that . Then , and the remaining statements easily follow. Note that is non-negative, since . 2. (s2)
If for an open set , then for any compact , and then by finite additivity of on compact sets we have:
[TABLE]
Therefore, if then , and if then . 3. (s3)
Let . Monotonicity of and is clear in the case , and then it is also easy for the cases , and . 4. (s4)
Let be disjoint. For any with we have by additivity of on compact sets and Definition 8:
[TABLE]
Then by Definition 8
[TABLE]
For the converse inequality, note that given compact set we have , where . Then giving
[TABLE]
So we have finite additivity of on open sets. If and , where then
[TABLE]
which by Definition 8 shows that
[TABLE]
On the other hand, given any with we may write , where . Then
[TABLE]
which shows that
[TABLE]
This gives the finite additivity of on open sets. 5. (s5)
Let be compact. For any open set containing we have and . Taking infimum over all such we obtain and . Since , we see that if on , then . It is also easy to see that : for instance, if then . 6. (s6)
Consider first . It is enough to show that where is closed and is compact. Take any open set containing . Since is completely regular, find disjoint open sets such that and . By monotonicity and finite additivity of on open sets
[TABLE]
Passing to the infimum over we get A similar argument gives the statement for . 7. (s7)
Let be open. By monotonicity of and , for any compact set we have and . If , then there are compact sets such that . Then by part (s5) , and . Assume now that . By Definition 8, given , choose compact such that . Then by part (s5)
[TABLE]
which shows that
[TABLE]
Now we shall show the inner regularity for . If , then for each there is a finite disjoint family of compact sets such that . Let . Then and by parts (s5) and (s6)
[TABLE]
for each , so . Assume now that . For let be such that . Let , so compact . Since , we have , showing that
[TABLE] 8. (s8)
It is enough to show that where is closed and is compact. By monotonicity of we may assume that . For , since is completely regular, we may find disjoint open sets such that Then using parts (s3) and (s4)
[TABLE]
Together with part (s6) this gives . A similar argument proves the statement for . 9. (s9)
Since , it is enough to assume that is finite. By inner compact regularity of (respectively, of ) we may take all sets to be disjoint closed with at most one of them not compact. The assertion follows from part (s6) and the monotonicity of (respectively, ). 10. (s10)
By part (s5) for any compact set , so using parts (s8) and (s9) it is easy to see that for any open set , and then for any closed set . Thus, . 11. (s11)
By part (s8) is finitely additive on compact sets, and by part (s1) so we may define . If is open, from part (s7) it is easy to see that . Then .
∎
Remark 11**.**
If is a signed measure such that its total variation (as defined for measures) is inner regular, then it is not hard to show that is a total variation of restricted to .
3. Deficient topological measures
Definition 12**.**
A deficient topological measure on a locally compact space is a set function on which is finitely additive on compact sets, inner compact regular, and outer regular, i.e. :
- (DTM1)
if then ; 2. (DTM2)
for ; 3. (DTM3)
for .
We denote by the collection of all deficient topological measures on .
For a closed set , iff for every open set containing .
Remark 13**.**
Let be a deficient topological measure on . If is locally compact and locally connected then by Lemma 7 for each open set
[TABLE]
If is locally compact, connected, and locally connected, then from Lemma 6
[TABLE]
and considering for a compact connected set its solid hull (see section 3 in [3] for detail), we also obtain
[TABLE]
Remark 14**.**
The argument as in part (s3) of Lemma 10 shows that a deficient topological measure is monotone. Since is not identically , from monotonicity and finite additivity on compact sets it follows that . If and are deficient topological measures that agree on , then by regularity . If and are deficient topological measures such that on (or on ) then .
Remark 15**.**
It is easy to see that if and are deficient topological measures, then so is , and so is for any . Thus, deficient topological measures constitute a positive cone.
In Definition 12 we do not require that for every compact set . By analogy with semi-finite measures (see, for example, [2], 1.12.132) we define semi-finite deficient topological measures:
Definition 16**.**
A deficient topological measure is semi-finite if for every set with there exists such that and .
The deficient topological measure in part (iii) of Example 59 below is semi-finite. Examples of compact-finite deficient topological measures are provided in Example 53 below.
Lemma 17**.**
If is a deficient topological measure, then . A deficient topological measure is finitely additive on open sets, monotone, and superadditive, in the sense of part (s9) of Lemma 10, i.e. if where , and at most one of the closed sets is not compact, then .
Proof.
From Remark 14 , and so we may apply Lemma 10. From Definition 8 we see that and then from part (s2) of Lemma 10 it follows that . Again by Lemma 10, any deficient topological measure is finitely additive on open sets, monotone, and superadditive. ∎
Lemma 18**.**
Let be a set function which is inner compact regular on open sets and outer regular on closed sets, i.e. for each open set , and for each closed set . (In particular, this holds for a deficient topological measure.) Then
- (i)
* is monotone.* 2. (ii)
* is -smooth on open sets; in particular, is countably additive on .* 3. (iii)
When is locally compact is finitely additive on compact sets iff it is finitely additive on open sets.
Proof.
- (i)
Use argument as in part (s3) of Lemma 10. 2. (ii)
Suppose . Let compact By Remark 3, there is such that for all . Then for all , and we see from the inner regularity (whether or ) that . 3. (iii)
Assume is finitely additive on open sets. We need to show that
[TABLE]
for any compact sets . By monotonicity, (5) holds if at least one of or . So let and . For pick disjoint open sets such that . Then
[TABLE]
which shows that . To show the opposite inequality, note that it trivially holds when , so we assume that . For let be an open set such that and . Pick disjoint open sets such that . Now
[TABLE]
which gives . Thus, , and is finitely additive on compact sets.
One can prove that finite additivity on compact sets implies finite additivity on open sets in a similar manner, using the simple fact that if compact then , where .
∎
Proposition 19**.**
Let be locally compact. Suppose which assumes at most one of is a signed set function which is finitely additive on compact sets. Then
- (I)
The set functions defined as in Definition 8 are deficient topological measures, and . 2. (II)
* is the unique smallest deficient topological measure such that for every and is the unique largest deficient topological measure such that for every .* 3. (III)
If then and are real-valued deficient topological measures, .
Proof.
- (I)
Outer regularity follows from Definition 8, the rest follows from Lemma 10. 2. (II)
Suppose that is a deficient topological measure such that for all . By inner compact regularity of , we have for any open set , i.e on . Both and are deficient topological measures, so by Remark 14 . Thus, is the unique smallest deficient topological measure such that on . Since , the last assertion follows. 3. (III)
Clearly, . By part (s10) of Lemma 10 .
∎
Lemma 20**.**
Let be locally compact and let be a deficient topological measure on . Given an open set with and there exists such that for any compact or open we have .
Proof.
For an open by regularity choose compact with . Then by monotonicity and superadditivity of , for any compact or open we have . ∎
Proposition 21**.**
A nonnegative set function has a unique extension to a deficient topological measure on if and only if is finitely additive on compact sets and satisfies the following condition:
For any compact with and any there exists an open set containing such that
[TABLE]
In this case .
Proof.
Necessity and condition (6) follow from the definition and monotonicity of a deficient topological measure. To prove sufficiency, observe that by Proposition 19, the set function is the unique smallest deficient topological measure such that on . If is compact with , then also . For with and , let be the set given by condition (6). Then
[TABLE]
which implies that on . So is the desired extension . ∎
Remark 22**.**
Proposition 21 is the generalization to a locally compact setting of Corollary 2 in [13].
Lemma 23**.**
A deficient topological measure on a locally compact space is smooth on compact sets and smooth on open sets. In particular, a deficient topological measure is additive on open sets.
Proof.
Let be a deficient topological measure. To show the smoothness on compact sets, let . It is enough to consider the case . For let be an open set such that and By Remark 3, there exists such that for all . Then for all , and the smoothness on compact sets follows. smoothness on open sets follows from Lemma 18. ∎
4. Deficient topological measures, topological measures, and measures
In this section we discuss the relationship between deficient topological measures, topological measures, and measures.
Definition 24**.**
A topological measure on is a set function satisfying the following conditions:
- (TM1)
if then 2. (TM2)
for ; 3. (TM3)
for .
We denote by the collection of all topological measures on .
When is compact , and we simplify Definition 12 and Definition 24 accordingly. On a compact space we consider real-valued deficient topological measures.
Remark 25**.**
The immediate difference between the definitions of a topological measure and a deficient topological measure is that a deficient topological measure is additive on compact sets, while a topological measure is additive on .
By we denote the collection of all Borel measures on that are inner regular on open sets and outer regular (restricted to ).
Remark 26**.**
Let be locally compact. In general,
[TABLE]
The proper inclusion follows from examples in the last section in [3]. The proper inclusion follows from examples in the last section of this paper. We would like to know when a deficient topological measure is a measure from or a topological measure.
Theorem 27**.**
- (I)
Let be compact, and a deficient topological measure. The following are equivalent:
- (a)
* is a topological measure* 2. (b)
** 3. (c)
** 2. (II)
Let be locally compact, and a deficient topological measure. The following are equivalent:
- (a)
* is a topological measure* 2. (b)
** 3. (c)
**
Proof.
- (I)
(a) (b) follows from the definition of a topological measure on a compact space. Superadditivity of a deficient topological measure implies equivalence of (b) and (c). Since a deficient topological measure is finitely additive on closed and open sets (see Lemma 18), to show that (b) implies (a) it is enough to check that when are not both closed or both open. We may also assume that and are not both closed or both open. Then the sets and are both closed or both open, and disjoint. By additivity on open (or closed) sets
[TABLE]
so by (b)
[TABLE]
i.e.
[TABLE] 2. (II)
(a) (b) follows from the definition of a topological measure on a locally compact space. Superadditivity of a deficient topological measure gives equivalence of (b) and (c). Finite additivity of a deficient topological measure on open sets (see Lemma 18) and Proposition 37 in [3] show that (b) implies (a).
∎
Now we are interested in finding out when a deficient topological measure is a measure.
Definition 28**.**
For a deficient topological measure on a locally compact space define a set function as follows:
[TABLE]
for any subset of .
Note that for any open set .
Lemma 29**.**
Let be a deficient topological measure on a locally compact space which is finitely subadditive on open sets, i.e. if are open subsets of , then . Let be as in Definition 28. Then
- (i)
* is an outer measure on .* 2. (ii)
A subset of is -measurable iff for each open set .
[TABLE] 3. (iii)
Each closed subset of is -measurable.
Proof.
- (i)
Clearly, is monotone and finitely subadditive. To show that is an outer measure, we only need to check that it is countably subadditive. From finite subadditivity of and Lemma 23 it is easy to see that is countably subadditive on open sets. We shall show countable additivity of . By monotonicity we may assume for each . For arbitrary and sets choose open sets such that and for . Then
[TABLE]
so is countably subadditive. 2. (ii)
Assume (7) holds. The proof is basically as in [8, p.234, Theorem D] and proceeds as follows. Let be an arbitrary set, and take any open containing . Since
[TABLE]
taking infimum over all open containing we see that
[TABLE]
Since is also subadditive, is -measurable. 3. (iii)
The argument is essentially as in [8, p.235, Theorem E] and proceeds as follows. Let be closed. We need to show that for any open set , and it is enough to consider the case . Let be open. Let be a compact subset of an open set , and let be a compact subset of an open set . Then and . Taking supremum over all compacts we see that
[TABLE]
Taking supremum over all compacts we have:
[TABLE]
which finishes the proof of the statement.
∎
Theorem 30**.**
Let be a deficient topological measure on a locally compact space which is finitely subadditive on open sets, i.e. if are open subsets of , then . Then admits a unique extension to an inner regular on open sets, outer regular Borel measure on the Borel -algebra of subsets of . is a Radon measure iff is compact-finite. If is finite then is a regular Borel measure.
Proof.
By Lemma 29, is an outer measure. The collection of all -measurable sets is a -algebra, and the restriciton of to is a measure. By Lemma 29, the -algebra contains the Borel -algebra . Let be the restriction of to . Thus, is a Borel measure. For each Borel set
[TABLE]
showing that is outer regular, and this gives the uniqueness of extension of to an outer regular Borel measure on .
For a closed set we see that . For an open set we have . Thus, on . Also, is inner regular on open sets.
We see that is compact-finite iff is. Thus, is a Radon measure iff is compact-finite. If is finite, it is easy to show that outer regularity of is equivalent to inner regularity of , hence, is a regular Borel measure. ∎
Remark 31**.**
This proof is very similar to the proof in [8] that a content on a locally compact space extends to a compact regular Borel measure. The reason is that a deficient topological measure that is subadditive and finite on compact sets is a content in the classical definition (see [8], 53). The proof in the case where is compact, is a topological measure, and the extension is to the Borel algebra was first given by Wheeler in [15]; the proof in the case where is compact and is a deficient topological measure was first given by Svistula in [12]. If is a compact-finite deficient topological measure on a locally compact space, the existence of a unique extension of to an inner regular on open sets, outer regular Borel measure on the Borel -algebra of subsets of follows also from Theorem 7.11.1 in [2], vol.2. Note that for a locally compact space we no longer require the deficient topological measure to be real-valued on compacts. See Example 59, part (iii).
We are now ready to present necessary and sufficient conditions for a deficient topological measure to be a measure.
Theorem 32**.**
Let be a deficient topological measure on a locally compact space . The following are equivalent:
- (a)
If are compact subsets of , then .
- (b)
If are open subsets of , then .
- (c)
* admits a unique extension to an inner regular on open sets, outer regular Borel measure on the Borel -algebra of subsets of . is a Radon measure iff is compact-finite. If is finite then is a regular Borel measure.*
Proof.
. Let be open in , and let . By Lemma 4 write , where . Since
[TABLE]
taking supremum over all compacts we have .
. This is Lemma 30.
. . Note that is neither open nor closed in general, so need not be defined. ∎
5. New deficient topological measures
In this section we gives some methods for obtaining new deficient topological measures from known ones.
Proposition 33**.**
Let be a compact space and be a locally compact space. If is a deficient topological measure on and is continuous then the set function on is a deficient topological measure on . If is a topological measure on , then is a topological measure on .
Proof.
First, we shall show that is inner regular. Let be open in . If , for each find a compact in set such that . Then for the set , compact in , we have: , and , so . If , for an arbitrary choose a compact such that . For compact set we have: and , showing again that .
Now we shall show the outer regularity. Let be closed in . By monotonicity of it is enough to assume that . For choose such that and . The set is open in and contains . Also, , so , showing that .
It is easy to see the finite additivity of on open sets. By Lemma 18, is then finitely additive on compact sets. Thus, is a deficient topological measure on .
If is a topological measure on , using Lemma 27 it is easy to see that is in fact a topological measure on . ∎
Remark 34**.**
Proposition 33 holds for a locally compact space if we require in addition to be a closed mapping.
Remark 35**.**
Proposition 33 was first proved for both spaces compact and finite in [10].
If is a deficient topological measure on , and , one can not obtain a deficient topological measure on by simply restricting to , i.e., considering . One simple reason for this is that the intersection of two arbitrary sets from does not in general belong to . Nevertheless, we may obtain a deficient topological measures on closed and open sets as restrictions.
Let be a locally compact space, and be a closed subset. Consider as a space with the subspace topology. Note that open subsets of are described as follows:
[TABLE]
closed subsets of have the form where , a compact in is also a compact in , and so compact subsets of are
[TABLE]
Proposition 36**.**
Let be locally compact, and be a deficient topological measure on . Let be a closed subset equipped with the subspace topology. Then there exists a deficient topological measure on with the subspace topology such that for any
[TABLE]
Proof.
Define on to be the restriction of to . Then is finitely additive on . Let be such that . Since is a deficient topological measure on , it satisfies condition (6), so for there exists containing such that for any we have . Let . Then and for any such that we have and so
[TABLE]
Thus, condition (6) holds for , and by Proposition 21 extends to a required deficient topological measure. ∎
Proposition 37**.**
Let be a deficient topological measure on , and be a closed subset of . Define on open and closed subsets of by letting . Then is a deficient topological measure on .
Proof.
The proof is easy and is left as an exercise. ∎
Definition 38**.**
We will call the deficient topological measure in Proposition 36 the restriction of to , and the deficient topological measure in Proposition 37 the extension of to .
We have an analog of Proposition 36 for an open set.
Proposition 39**.**
Let be locally compact, and be a deficient topological measure on . Let be an open subset equipped with the subspace topology, Then there exists a deficient topological measure on with the subspace topology such that for any
[TABLE]
Proof.
Let . We consider the space with the subspace topology. Then . Define a set function on by for and for . Then is finitely additive on and outer regular. We need to show the inner regularity of . Let . Suppose first that . For pick such that . By Lemma 5 let be a bounded open set such that . Note that and
[TABLE]
Thus, .
Now suppose . For pick such that . Then . For any such that we have , and so . Then . ∎
Definition 40**.**
Let be locally compact, be a deficient topological measure on . Let . Define a set function on by letting
[TABLE]
and
[TABLE]
Remark 41**.**
If then , because
[TABLE]
Theorem 42**.**
Let be locally compact. Let be a deficient topological measure on . Then defined in Definition 40 is a deficient topological measure on .
Proof.
By its definition, is outer regular. By Remark 41, if then . We shall show that is inner compact regular. Let . Assume first that . For choose such that . Then by Remark 41 and so . Now assume . Given find such that and . Then , and the inner compact regularity follows. Finite additivity of on open sets follows from the same property of . By Lemma 18 is finitely additive on compact sets. Hence, is a deficient topological measure. ∎
Lemma 43**.**
Let be locally compact, be a deficient topological measure on , and . Then for any
[TABLE]
where is a deficient topological measure on from Theorem 42. If then (8) holds for any .
Proof.
Let , where . Then , so . The equality is easy to see if .
Now suppose that . Given , let be such that and . Since , whether or , by complete regularity of we may choose disjoint open sets such that and . Let . Then and . Now
[TABLE]
and the statement follows. ∎
Theorem 44**.**
Let be locally compact, and let . There exists a deficient topological measure on such that
[TABLE]
[TABLE]
If then for every .
Proof.
The existence of required can be proved, for example, by applying Proposition 37 to a deficient topological measure on given by Proposition 36. We shall show now that if is compact, then for every closed set . We need to check that . For any open set such that we have
[TABLE]
so
[TABLE]
It is enough to show the opposite inequality for the case . By Lemma 43 . Given choose such that and . Note that for any with we have . Then using Definition 40
[TABLE]
This gives
[TABLE]
and the proof is complete. ∎
Remark 45**.**
When is compact, topological measures (deficient topological measures) that are restrictions of topological measures (deficient topological measures) to sets appeared in several papers, including [7], [9], [13], [14]. Proposition 36, Theorem 42, and Lemma 43 are generalizations to a locally compact case of Proposition 3 in [13], Proposition 5.1 in [14], and the stated without proof part (4) of Proposition 5.2 in [14].
6. Examples of deficient topological measures
Proposition 19 allows us to build examples of deficient topological measures from simple set functions. Results in section 5 allow us to obtain new deficient topological measures from existing ones. We will now present some concrete examples.
Example 46**.**
Let be locally compact, and let be a connected compact subset of . Define a set function on by setting if and otherwise, for any . If , where then by connectedness either or , and this implies finite additivity of on . It is also easy to check the inner and outer regularity of : for example, if , then there exists a point , and taking we see that , so outer regularity is satisfied for . By Definition 12, is a deficient topological measure. Note that if is a singleton, then is a point mass. If has more than one element, then is a deficient topological measure, but not a topological measure: consider , where , and . Then , and by Definition 24 is not a topological measure.
Example 47**.**
Let be a finite family of disjoint bounded nonempty connected subsets of . Let be the family of closed subsets of that contain at least one set from , and be some set function on . For a closed set we set
[TABLE]
In particular, . The components of two disjoint compact sets give the components of their union. It follows that is finitely additive on .
- (i)
Suppose . By Proposition 19 the function generates deficient topological measure . We claim that if each set in has more than one point and , then is a deficient topological measure, but not a topological meausure. Indeed, taking a point from each set in we obtain a compact set . It is easy to check that , but . By Theorem 27 is not a topological measure.
Note that if we repeat the same construction, starting with as a finite family of disjoint unbounded connected subsets of , then we obtain on , and then . 2. (ii)
Suppose attains at most one of . By Proposition 19 the function generates deficient topological measure . As in part (i), if each set in has more than one point and , then is a deficient topological measure, but not a topological meausure.
We can say more about deficient topological measure in Example 47, part (i) under additional assumptions on . We first need two lemmas.
Lemma 48**.**
Suppose is a finite family of subsets of and for each . Then there is an open set such that for each .
Proof.
The proof as in [13] (Lemma 1): construct a closed set by taking one point in each . Take . ∎
Lemma 49**.**
Let be locally compact. Assume that is a component of a compact set and . Then there are disjoint compact sets such that and .
Proof.
We consider as a subspace of . is compact, and we know (see, for example, [5], paragraph 6.1.23) that its component is a quasi-component, i.e. the intersection of all clopen subsets of containing . Using, for example, paragraph 3.1.5 in [5], it is easy to obtain a clopen subset of such that . Take . ∎
Proposition 50**.**
Let be locally compact. Suppose is a nonnegative set function on , which is monotone on , finitely additive on , and outer regular on . Let be a set function on defined as in (11) in Example 47 for some finite family of disjoint bounded nonempty connected subsets of and defined as in Definition 8. Then for all . Thus, if is compact, then for all .
Proof.
The result follows from Proposition 21 applied to , if we check condition (6). Let be compact, . If contains no sets from , then by Lemma 48 there exists an open set such that and does not contain elements of . Then for any compact , and satisfies condition (6). Now assume that contains sets from . Let . Recall that where are components of that contain sets from . Since the sets s are disjoint and is monotone on open sets and outer regular, we may find disjoint open sets such that and . By Lemma 49 for each find disjoint compact sets such that , and . Let . Then is compact, for each , and . Since are disjoint, choose disjoint open sets such that and for . Note that implies for each , so , being a subset of , does not contain any element from . By Lemma 48 we may assume that also does not contain elements of . Let . We shall show that is the set needed in condition (6). So let be compact, be its components that contain sets from . By connectedness each for some . Set for , and consider nonempty . By outer regularity and finite additivity of on we have: . Now
[TABLE]
Thus, condition (6) is satisfied for , and this finishes the proof. ∎
Remark 51**.**
Lemma 49 and Proposition 50 are generalizations of Lemma 2 and part of an argument on p. 733 in [13].
Example 52**.**
Let be a deficient topological measure. Let and let . We apply Proposition 50. For a compact disjoint from we see that , whether or not . Thus, in general .
Example 53**.**
Let be a non-trivial deficient topological measure on a locally compact space . We may have or . Suppose there is a point in for which . (For example, we may take to be the Lebesque measure or a point mass at on . We may also take to be any topological measure from the last section in [3].) Let the family consist of one set, . Defined as in Example 47 the set function generates a deficient topological measure , and by Proposition 50 on . Then . For any compact we have , and then . Since , by Theorem 27, is not a topological measure. Note that if, for instance, is the Lebesque measure and , then the range of deficient topological measure is . If is a compact-finite topological measure, then is also compact-finite.
Remark 54**.**
Example 53 is easy to generalize to the case when consists of one set, for which .
Example 55**.**
Suppose is a deficient topological measure on a locally compact space . Let the family , , and . For example, we may take to be a Lebesque measure or a point mass at on , or any topological measure from the last section in [3]. The set function as in Example 47 generates a deficient topological measure , and by Proposition 50 on . As in Example 53 we have , while . By Theorem 27, is not a topological measure. Again, the range of could be .
Remark 56**.**
In Example 53 and Example 55 is not a topological measure even though no set in contains more than 1 point. Compare to Example 47.
If is also locally connected we may strengthen Proposition 50 and specify how acts on compact and open sets.
Proposition 57**.**
Let be locally compact and locally connected. Suppose is a nonnegative set function on , which is monotone on , finitely additive on , and outer regular on . Suppose is a set function on defined as in (11) in Example 47 for some finite family of disjoint bounded nonempty connected subsets of , and is defined as in Definition 8. Then for
[TABLE]
where are the components of containing at least one set from ; if there are no such components, then .
Proof.
For compact, the statement follows from Proposition 50. Now let be open. If does not contain any sets from then for any compact by Proposition 50 we have , and hence, by inner regularity . Now assume that are the components of that contain at least one set from . Let , and be components of that contain at least one set from . By connectedness, each for some . Set for , and consider nonempty . By outer regularity of we have: . Then which implies that
[TABLE]
We shall show the equality. It is enough to consider the case for each , for if for some then for each there is a compact such that , and then , so equality holds. Let . Since each is open connected, and is inner regular and monotone, using Lemma 6 we may find a compact connected set such that it contains elements of that are contained in and . Then for compact we have: and
[TABLE]
It follows that . ∎
Remark 58**.**
Examples 46, part (i) of Example 47, and Example 53 are generalizations of Example 1, Example 2, and Example (c) on p. 733 in [13].
Example 59**.**
Let be a countable subset of . Let be a finitely additive set function on with that does not assume both and . Define a set function on as follows: if then
[TABLE]
where is compact, and are components of that intersect . If then . Again, is finitely additive on . By Proposition 19, is a deficient topological measure on .
- (i)
Let for a finite , and otherwise. Then counts how many points from are contained in . 2. (ii)
If is a finite family of disjoint connected subsets of (as in Example 47), is a set where we take one point from each set in , and for , then counts how many sets from are contained in . 3. (iii)
Let be or . Set , where is the set and is the set of points with all natural coordinates, and let be as in part (i). Then for any compact containing , and for any compact not intersecting . It is easy to see that is not compact-finite, but is semifinite.
Remark 60**.**
Part (ii) in Example 59 implies Example (a2) on p. 732 in [13].
Acknowledgments: This work was conducted at the Department of Mathematics at the University of California Santa Barbara. The author would like to thank the department for its hospitality and supportive environment.
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