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Solid-set functions and topological measures on locally compact spaces

A topological measure on a locally compact space is a set function on open and closed subsets which is finitely additive on the collection of open and compact sets, inner regular on open sets, and outer regular on closed sets. Almost all works devoted to topological measures, corresponding non-linear functionals, and their applications deal with compact spaces. The present paper is one in a series that investigates topological measures and corresponding non-linear functionals on locally compact spaces. Here we examine solid and semi-solid sets on a locally compact space. We then give a method of constructing topological measures from solid-set functions on a locally compact, connected, locally connected space. The paper gives examples of finite and infinite topological measures on locally compact, non-compact spaces and presents an easy way to generate topological measures on spaces whose one-point compactification has genus 0 from existing examples of topological measures on compact spaces.


INTRODUCTION
The study of topological measures (initially called quasi-measures) began with papers by J. F. Aarnes [1], [2], and [3]. There are now many papers devoted to topological measures and corresponding non-linear functionals; their application to symplectic topology has been studied in numerous papers (beginning with [11]) and a monograph ( [12]).
To date, however, almost all these works deal with topological measures on compact spaces. In [4] J. F. Aarnes gives a definition of a topological measure on a locally compact space, presents a procedure for obtaining topological measures from solid set functions on a locally compact, connected, locally connected space, and constructs some examples. While [4] contains many interesting ideas, it is not entirely satisfactory. It contains incomplete proofs and sometimes asks the reader to adapt lengthy proofs from other papers to its subject matter. In addition, the approach in [4] makes heavy use of sets that are connected and coconnected (i.e. have connected complements). We do not think this is the right approach for the non-compact setting. For example, using these sets one may end up constructing trivial topological measures (see Example 6.2 in [4]). Finally, the paper has never been published in a refereed mainstream journal.
The construction technique employed by Aarnes for a compact space X in [2] was later nicely simplified by D. J. Grubb, who used semi-solid sets in a compact space. Grubb presented his elegant construction in a series of lectures in 1998, but, unfortunately, never published it. Influenced by ideas of Aarnes and Grubb, we have developed an approach for constructing topological measures on locally compact spaces. Instead of sets that are connected and have connected complements we use sets that are connected and whose complement has finitely many bounded and unbounded components. Our approach allows us to extend a solid set function (see Definition 39) to a topological measure on X when X is a locally compact, connected, and locally connected space; the restriction of a topological measure to solid sets with compact closure is a solid set function that uniquely determines the topological measure. We obtain an easy way to construct topological measures on non-compact locally compact spaces whose one-point compactification has genus 0. (See [2], and section 9 for more information about genus.) Thus, we are able to produce a variety of topological measures on R n , half-spaces, punctured balls, etc.. The paper is organized as follows. In section 2 we give necessary topological preliminaries. In section 3 we study the structure of solid and semi-solid sets. In section 4 we give a definition and basic properties of topological measures on locally compact spaces, and in section 5 we do the same for solid-set functions.
In section 6 on a locally compact, connected, and locally connected space we extend a solid-set function from bounded solid sets to compact connected and bounded semi-solid sets. In section 7 the extension is done to the finite unions of disjoint compact connected sets, and in section 8 the extension produces a topological measure that is uniquely defined by a solid set function (see Theorem 48 and Theorem 51). In section 9 we give examples and present an easy way (Theorem 64) to generate topological measures on locally compact, connected, and locally connected spaces whose one-point compactification has genus 0 from existing examples of topological measures on compact spaces. In this paper by a component of a set we always mean a connected component. We denote by E the closure of a set E and by ∂E the boundary of E. We denote by a union of disjoint sets.

Definition 1. A set A ⊆ X is called bounded if A is compact. A set A is solid if
A is connected, and X \A has only unbounded components. A set A is semi-solid if A is connected, and X \ A has only finitely many components.
Several collections of sets will be used often. They include: O(X), the collection of open subsets of X; C (X), the collection of closed subsets of X; and K (X), the collection of compact subsets of X. P(X) is the power set of X. We consider set functions that are not identically ∞.

PRELIMINARIES
This section contains necessary topological preliminaries. Some results in this section are known, but sometimes we give proofs for the reader's convenience.
Remark 3. An easy application of compactness (see, for example, Corollary 3.1.5 in [8]) shows that (i) If K α ց K, K ⊆ U, where U ∈ O(X), K, K α ∈ C (X), and K and at least one of K α are compact, then there exists α 0 such that K α ⊆ U for all α ≥ α 0 .

Remark 4.
(a) SupposeX is connected, U ∈ O c (X) and F ∈ C c (X) are disjoint sets. If U ∩ F = ∅ then U ⊔ F is connected.
(b) If X is locally compact and locally connected, for each x ∈ X and each We would like to note the following fact. (See, for example, [7], Chapter XI, In the spirit of this result we can say more, given connectedness. Lemma 7. Let X be a locally compact, locally connected space, K ⊆ U, K ∈ K (X), U ∈ O(X). If either K or U is connected there exist a set V ∈ O * c (X) and a set C ∈ K c (X) such that One may take C = V .
is a bounded open connected set for which is open connected and Again, let C = V .
Proof. Let U = i∈I ′ U i be the decomposition into connected components.
Since X is locally connected, each U i is open, and by compactness of K there exists a finite set

(ii) If A is closed then B is open and
(iv) B is connected and co-connected.
Proof. The proof of the first two parts is not difficult. For the third part, observe that by Remark 4 A B is connected for each component B of X \ A. To prove the last part, let X \ A = s∈S B s be a decomposition into connected components. For each t ∈ S connected component B t is also co-connected, because is a connected set by the previous part.
Then at most a finite number of connected components of X\ K are not contained in U.
Proof. Let X\K = s∈S W s be the decomposition of X\K into connected com- Assume that there are infinitely many components of X \K that are not contained in U. Then we may choose components W i , i = 1, 2, . . ., such that W i ∩ U = ∅ and W i ∩ (X \ U) = ∅ for each i. Connectivity of W i implies that W i ∩ ∂U = ∅ for each i. Let x i ∈ W i ∩ ∂U. By compactness of ∂U, let x 0 ∈ ∂U be the limit point of (x i ). Then x 0 ∈ X \ U ⊆ X \ K = s∈S W s , i.e. x 0 ∈ W t for some t ∈ S. But then all but finitely many x i must also be in W t , which is impossible, Corollary 11. Let X be a connected, locally connected space. Let K ∈ K (X) and let W be the union of bounded components of X \ K. Then W ∈ O * (X).
Proof. By Lemma 7 pick V ∈ O * c (X) such that K ⊆ V . From Lemma 10 it follows that W is bounded. By Lemma 9 W is open. Proof. Suppose first that A ⊆ K, K ∈ K (X). The first assertion is Remark 12.
Now let E be an unbounded component of X \ A which contains no unbounded components of X \ K. Then E is contained in the union of K and all bounded components of X \ K. By Corollary 11 this union is a bounded set, and so is E, which leads to a contradiction. Therefore, each unbounded component of X \ A must contain an unbounded component of X \ K.
Lemma 14. Let X be locally compact, locally connected. Let A ∈ A * (X). Then the number of unbounded components of X \ A is finite.
Proof. Suppose first that A ∈ K (X). By Lemma 7 let U ⊆ O * c (X) be such that A ⊆ U. Then the assertion follows from Lemma 10. Now suppose that A ∈ O * (X). Then A ∈ K (X), so the number of unbounded components of X \ A is finite. From Lemma 13 it follows that the number of unbounded components of X \ A is also finite, since it does not exceed the number of unbounded components of X \ A.
Lemma 15. Let X be locally compact, connected, locally connected. Suppose D ⊆ U where D ∈ K (X), U ∈ O * (X). Let C be the intersection of the union of bounded components of X \D with the union of bounded components of X \U.
Then C is compact and U ⊔ C is open.
where V is the union of bounded components of X \ D, and W is the union of unbounded components of X \ D. Also write where B is the union of bounded components of X \ U, and F is the union of unbounded components of X \ U. By Lemma 14 F is a closed set. Let Clearly, C and U are disjoint. To see that U ⊔ C is open, note first that U ⊔ B = X \ F is an open set. Hence, is also an open set. Now we shall show that C is closed, i.e. that X \ C is open.
Note that F ⊆ W by Remark 12. The set W is open by Lemma 9. Now is an open set. By Corollary 11 the set C is bounded.

SOLID AND SEMI-SOLID SETS
where n ∈ N and E i 's are unbounded connected components.
Lemma 17. Let X be locally compact, locally connected. If A ∈ A * (X) then Proof. Let be the decomposition of X \ A into components, where B i , i ∈ I are bounded components, and D j , j ∈ J are unbounded ones. Pick a bounded component B k . Then Note that the set on the right hand side is connected by Lemma 9 and unbounded.
Hence, B k is solid.
A set A ∈ A * c (X) may not be solid. But we may make it solid by filling in the "holes" by adding to A all bounded components of X \ A. More precisely, we have the following result.
Lemma 18. Let X be locally compact, locally connected.
be the unbounded components of X \ A and {B s } s∈S be the bounded components of X \ A. Then the setÃ Proof. The setÃ is connected by Lemma 9. It is clear that X \Ã has only unbounded components.
Definition 19. Let X be locally compact, locally connected. For A ∈ A * c (X) let {A i } n i=1 be the unbounded components of X \ A and {B s } s∈S be the bounded components of X \ A. We say thatÃ Lemma 20. Let X be locally compact, connected, locally connected.
(a2)Ã is a bounded solid set, A ⊆Ã, and A is solid iff A =Ã.
(a3)Ã =Ã. Lemma 21. Let X be locally compact, connected, locally connected. If K ⊆ U, K ∈ K (X), U ∈ O * s (X) then there exists C ∈ K s (X) such that Proof. One may take C to be the solid hull of the set V from Lemma 7. Then C ⊆ U by Lemma 20.
Lemma 22. Let X be locally compact, connected, locally connected.
Proof. By Lemma 7 we may choose U ∈ O * c (X) such that be the decomposition into connected components. Each V j is an unbounded open connected set. Since X \ U ⊆ X \ K, for each j = 1, . . . , n let E j be the union of all bounded components of X \ U contained in V j , and let F j be the union of (finitely many by Lemma 14) unbounded components of X \ U contained in V j .
By Lemma 9 each F j is closed. By Lemma 13 each F j is non-empty. Then by Lemma 9 non-empty set F j ∩ U ⊆ V j and F j ∩ U ∈ K (X). Now, E j ⊆Ũ , so E j is bounded. Note that X = K ⊔ n j=1 V j , and a limit point x of E j can not be in V i for i = j; and it can not be in K, since in this case a neighborhood U of x Then B j is connected because from (2) one sees that D j intersects every component comprising F j . Thus, each B j is an unbounded closed connected set, By (1) we see that and we may take W =Õ.
Remark 23. The closure of a solid set need not be solid. For example, in the Note that T ′ is a finite index set. Now let us index by S ′ the family of all bounded components of X \ U each of which contains a component V t for some t ∈ T ′ . Note that S ′ is a finite index set and ss (X) (as the first equality indicates that X \ V has finitely many components), and that U ⊆ V . Now consider Then D is compact. By Lemma 9 D is connected. We have so X \ D has finitely many components, and V ⊆ D. Thus, D ∈ K ss (X). Also, Therefore, D ⊆ W . Then we have: ss (X) and D ∈ K ss (X).
Let V be an open subset of X endowed with the subspace topology. Let D ⊆ V . By D V we denote the closure of D in V with the subspace topology. As before, D stands for the closure of D in X.
Remark 26. Let V ∈ O * (X) be endowed with the subspace topology. From The next two results give relations between being a solid set in a subspace of X and being a solid set in X.
is a disjoint union of unbounded connected sets, the proof is complete. Now we shall take a closer look at the structure of an open solid or semi-solid set that contains a closed solid or closed connected set.
Lemma 29. Let X be locally compact, connected, locally connected. Let C ⊆ V, C ∈ K s (X). Proof.
be the decomposition into connected components, so S is a finite index set and each F s is unbounded. If V \ C is connected then taking W = V \ C we see that has finitely many components, i.e. W ∈ O * ss (X). Now assume that V \ C is not connected. By Lemma 27 and Remark 26 C ∈ C s (V ) and is bounded in V . The set V \ C is also disconnected in V , so using Remark 16 let we only need to check that the components of X \ V i are unbounded. For simplicity, we shall show it for V 1 . For 2 ≤ j ≤ n by Lemma 27 and Remark 26 V j intersects X \ V , hence, intersects some F s . Let S 1 = {s ∈ S : F s ∩ V j = ∅ for some 2 ≤ j ≤ n}. By Remark 4 and Lemma 9 the set and let k j=1 F j be the components of X \ V . By Lemma 27 and Remark 26 C ∈ C s (V ) and is bounded in V . Let ss (X) we only need to check that X \ V i has finitely many components. For simplicity, we shall show it for V 1 . We have: Since X \ V 1 is a finite disjoint union of connected sets, the number of Lemma 30. Let X be locally compact, connected, locally connected. Suppose If U \C is connected then Proof. Note first thatC ∈ K s (X) andC ⊆ U by Lemma 20. Assume that U \C is disconnected. By Lemma 29 we may write . After reindexing, one may write The proof for the case when U \C is connected follows similarly from Lemma 29.
Lemma 31. Let X be locally compact, connected, locally connected. Suppose Proof. The proof is by induction on m. Let m = 1. Using Lemma 29 we have Since sets V i and U t are connected, T must be finite. Now let V = m j=1 C j ⊔ t∈T U t and assume that the result holds for any bounded open semi-solid set which contains less than m compact solid sets. Using Lemma 29 we see that where V i ∈ O * ss (X). All involved sets are connected, so each set V i is the disjoint union of sets from the collection {C 2 , . . . , C m , U t , t ∈ T }. By the induction hypothesis each V i contains finitely many sets, and it follows that T is finite.
. If the cardinality |T | > 1 then there must be a compact solid set among A t , and the result follows from Lemma 31. Assume now that A ∈ K s (X) and write Then from Lemma 29 we may write and by Lemma 31 T is finite.
Remark 33. Lemma 14, Lemma 20, Lemma 22, and Lemma 27 are close to Lemmas 3.5, 3.6, 3.8, 3.9, and 4.2 in [4]. Lemma 15 is related to a part in the proof of Lemma 5.9 in [4]. The case " V \ C is disconnected" in the first part of Lemma 29 is Lemma 4.3 in [4], and Lemma 32 is an expanded (to compact sets as well) version of Lemma 4.4 in [4]. In all instances our proofs are modified, expanded, or different, compared to the proofs in [4].

DEFINITION AND BASIC PROPERTIES OF TOPOLOGICAL MEASURES ON LOCALLY COMPACT SPACES
Definition 34. A topological measure on X is a set function µ : satisfying the following conditions: The following result gives some immediate properties of topological measures on locally compact spaces.
Lemma 36. The following is true for a topological measure: In particular, µ is additive on O(X).
(t7) If X is connected then If X is locally compact, connected, locally connected then also Proof.
(t3) Easy to see since µ is not identically ∞.
Proof. Our proof is an expanded version of the proof of Proposition 2.2 in [13] where the result first appeared for compact-finite topological measures. Suppose that µ is a set function satisfying (TM2), (TM3) as well as conditions (c1) and (c2). We need to show that µ satisfies (TM1). X is completely regular, so it is evident from (c1) and (TM3) that µ is finitely additive on O(X) and on K (X).
Hence, we only need to check (TM1) in the situation when A ∈ K (X), B ∈ O(X), and A⊔B is either compact or open.
If A⊔B is open then using condition (c2) we get: Now suppose A ⊔ B ∈ K (X). Note that (TM3) implies monotonicity of µ on K (X). Let C ∈ K (X), C ⊆ B. Then finite additivity and monotonicity of µ on K (X) gives: Applying (TM2) and then condition (c2) we see that This finishes the proof.

SOLID SET FUNCTIONS
Our goal now is to extend a set function defined on a smaller collection of subsets of X than O(X) ∪ C (X) to a topological measure on X. One such convenient collection is the collection of solid bounded open and solid compact sets, and the corresponding set function is a solid set function.
Lemma 40. Let X be locally compact, connected, locally connected. Suppose λ is a solid set function on X. Then Since s∈S λ(A s ) = sup{ s∈S ′ λ(A s ) : S ′ ⊆ S, S ′ is finite }, it is enough to assume that S is finite. By regularity in Definition 39 we may take all sets A s to be disjoint compact solid. If also A ∈ K s (X), the assertion is just part (s1) of Definition 39. If A ∈ O * s (X) then there exists C ∈ K s (X) such that s∈S A s ⊆ C ⊆ A by Lemma 21. Now the assertion follows from parts (s1) and (s2) of Definition 39.

EXTENSION TO
We start with a solid set function λ : A * s (X) → [0, ∞) on a locally compact, connected, locally connected space X. Our goal is to extend λ to a topological measure on X. We shall do this in steps, each time extending the current set function to a new set function defined on a larger collection of sets.
Definition 41. Let X be locally compact, connected, locally connected. For A ∈ A * ss (X) ∪ K c (X) define where {B i : i ∈ I} is the family of bounded components of X \ A.
By Lemma 17 each B i ∈ A * s (X). If A ∈ A * ss (X)∪K c (X) then i∈I B i ⊆Ã and by Lemma 40 Lemma 42. The set function λ 1 : A * ss (X) ∪ K c (X) → [0, ∞) defined in Definition 41 satisfies the following properties: s (X) with at most finitely many of B s ∈ K s (X). Then Proof.
. We may assume that A ∈ A * s (X), since the inequality where {D t : t ∈ T } is the disjoint family of bounded components of X \ A, and by Lemma 17 each D t ∈ A * s (X). The proof is by induction on n. For n = 0 the statement is Lemma 40. Suppose now n ≥ 1 and assume the result is true for any disjoint collection (contained in a bounded solid set) of bounded semi-solid or compact connected sets among which there are less than n non-solid sets.
Assume now that we have n disjoint sets A 1 , . . . , A n from the collection A * ss (X) ∪ K c (X). Consider a partial order on {A 1 , A 2 , . . . , A n } where A i ≤ A j iffÃ i ⊆Ã j . (See Lemma 20.) Let A 1 , . . . , A p where p ≤ n be maximal elements in {A 1 , A 2 , . . . A n } with respect to this partial order.
For a maximal element A k , k ∈ {1, . . . , p} define the following index sets: Let {E α } α∈H be the disjoint family of bounded components of X \A k .
Then we may say that The set I k and each set I k,α has cardinality < n. The set E α is a solid set according to Lemma 17, and i∈I k,α By induction hypothesis i∈I k,α

It follows that
Then using part (i) and Definition 41 we have: Notice thatÃ 1 , . . . ,Ã p , being the maximal elements, are all disjoint by part (a5) of Lemma 20. This also implies that the sets I k , k = 1, . . . , p are disjoint (otherwise, if i ∈ I k and also i ∈ I m , 1 ≤ k, m ≤ p theñ A k ∩Ã m = ∅). Similarly, the sets S k , k = 1, . . . , p are also disjoint.

Consider the index set
Note that {1, . . . , n} = {1, . . . , p} ⊔ p k=1 I k . Indeed, if i ∈ {1, . . . , n} \ {1, . . . , p} we must have A i ⊆Ã i ⊆Ã k for some maximal element A k (where k ∈ {1, . . . , p}), and since A i and A k are disjoint, A i must be contained in a bounded component of A k , i.e. i ∈ I k . Now we have: The first inequality is by formula (6), and for the last inequality we ap- (iii) The proof is almost identical to the proof of the previous part, and we keep the same notations. Again, we may assume that A ∈ A * s (X), since the inequalities (3) and (4) become equalities. The proof is by induction on n, and the case n = 0 is given by Lemma 32 and part (s4) of Definition 39. The inequalities in the induction step become equalities once one observes that (5) above becomes i∈I k,

EXTENSION TO K 0 (X)
Our goal now is to extend the set function λ 1 to a set function λ 2 defined on K 0 (X). Recall that K ∈ K 0 (X) if K = n i=1 K i where n ∈ N and K i ∈ K c (X) for i = 1, . . . , n.
(ii) λ 2 is finitely additive on K 0 (X) Proof. The first part easily follows from the definition of λ 2 and Lemma 42.
The second part is obvious. To prove the third one, let C ⊆ K, where C, K ∈ where the sets C i (i = 1, . . . , n) and K j (j = 1, . . . , m) are compact connected. By connectivity, each C i is contained in one of the sets K j . Consider index sets I j = {i : C i ⊆ K j } for each j = 1, . . . , m. By Lemma 42 we have i∈I j λ 1 (C i ) ≤ λ 1 (K j ). Then We are now ready to extend the set function λ 2 to a set function µ defined on O(X) ∪ C (X). (p2) µ(A) < ∞ for each A ∈ A * (X), so µ is compact-finite.
In particular, µ is finitely additive on compact sets.
(p6) Let U 1 , U 2 ∈ O(X) be disjoint. For any C i , C 2 ∈ K 0 (X) with C i ⊆ U i , i = 1, 2 we have by Lemma 44 and Definition 45 Then by Definition 45 we obtain For the converse inequality, note that given C ⊆ U 1 ⊔U 2 , C ∈ K 0 (X) we have C i = C ∩ U i ∈ K 0 (X), i = 1, 2 (since each connected component of C must be contained either in U 1 or in U 2 ) and C = C 1 ⊔ C 2 . Then (p7) Let C 1 , C 2 be a compact and a closed set that are disjoint. Given U ∈ O(X), C 1 ⊔ C 2 ⊆ U we may find disjoint open sets U 1 , U 2 such that Then by parts (p6) and (p1) so using Definition 45 we have For the converse inequality, observe that for any Then by parts (p6) and (p1) which gives by Definition 45 for all i ∈ I. By parts (p6) and (p1) for any finite Then i∈I µ(V i ) ≤ µ(V ). To prove the opposite inequality, first assume By Lemma 44 and part (p3) we have: Now suppose µ(V ) = ∞. For n ∈ N find a compact K ⊆ V such that µ(K) > n. Choose a finite index set I n ⊆ I such that K ⊆ ⊔ i∈In V i .
Lemma 47. For the set function µ in Definition 45 Proof. We shall prove the statement in steps. Recall that µ = λ 1 on K c (X) and µ = λ 2 on K 0 (X) by part (p11) of Lemma 46.
The proof is by induction on n. Suppose n = 1, i.e. C ∈ K s (X). By Lemma 29 where each U i ∈ O * ss (X). By Lemma 42 Then where the last equality follows from additivity of µ on O(X) in Lemma 46.
Suppose that result holds for all C ⊆ U, U ∈ O * ss (X) where C is the disjoint union of less than n sets C j ∈ K s (X). Now let C = C 1 ⊔ C 2 ⊔ . . . ⊔ C n , where each C j ∈ K s (X). By Lemma 29 where each U i ∈ O * ss (X). By connectivity each C j , j = 2, . . . , n is contained in one of the sets U i . For i = 1, . . . , m let K i be the disjoint union of those C j , j ∈ {2, . . . , n} that are contained in U i . Notice that each K i is the union of no more than n − 1 disjoint sets, and m i=1 K i = n j=2 C j . By induction hypothesis, By finite additivity of µ on compact sets in Lemma 46 Also we have By the first part of the induction proof µ(U) = µ(U \ C 1 ) + µ(C 1 ).
Suppose that the result holds for all C ⊆ U, U ∈ O * s (X), C ∈ K 0 (X) where C is the disjoint union of less than n sets C i ∈ K c (X). Now assume that C = C 1 ⊔ . . . ⊔ C n , C i ∈ K c (X). As in the proof of Lemma 42, consider partial order on {C 1 , . . . , C n } where C i ≤ C j iffC i ⊆C j . Some parts of the argument here are as in the proof of Lemma 42. Let C 1 , . . . , C p , p ≤ n be maximal elements in {C 1 , . . . , C n } with respect to this partial order. ThenC 1 , . . . ,C p are disjoint.
This implies that the family is a disjoint family of sets. Each W s ∈ O * s (X) by Lemma 17 and s∈S W s ∈ O * (X), because s∈S W s ⊆ U. Let I = {1, . . . , n} \ {1, . . . , p}. For each i ∈ I C i is non-maximal element, and there exists k ∈ {1, . . . , p} such that C i ⊆ C k . In other words, each non-maximal set C i , i ∈ I is contained in a bounded component of X \ C k for some maximal element C k (for some k ∈ {1, . . . , p}), that is C i ⊆ W s for some s ∈ S. Let S 1 be a finite subset of S such that for s ∈ S 1 the set W s contains some C i , i ∈ I. Let S ′ = S \ S 1 . For each s ∈ S 1 let C s be the disjoint union of those sets C i , i ∈ I that are contained in W s . Since |I| ≤ n − 1, each C s is the union of no more than n − 1 disjoint sets, and by induction hypothesis for each s ∈ S 1 µ(W s ) = µ(W s \ C s ) + µ(C s ).
Note also that Then using Definition 19 and (13) we see that: is an open bounded set (since W ⊆ s∈S W s ⊆ U). Using Definition 19 and Definition 41, (12), (13), finite additivity of µ on K (X) and additivity of µ on O(X) in Lemma 46 we have:

It follows that
Since U ⊔ C is open, by Lemma 22 we may find W ′ ∈ O * s (X) such that Then According to part (p4) of Lemma 46, pick K 1 ∈ K 0 (X) such that Let U = i∈I U i be the decomposition of U into connected components, and let I ′ be a finite subset of I such that K ⊆ i∈I ′ U i . For each i ∈ I ′ the set By Step 3 we know that Then using (19), finite additivity of µ on K (X) and additivity of µ on O(X) in Lemma 46, and (20) we have: First assume that µ(U) < ∞. Given ǫ > 0 by Definition 45 find C ∈ K (X) such that K ⊆ C and µ(U) − µ(C) < ǫ. Using Lemma 6 find V ∈ O * (X) such Then µ(U)−ǫ < µ(C) ≤ µ(U \K)+µ(K). Therefore, µ(U) ≤ µ(U \K)+µ(K).
The opposite inequality is part (p10) of Lemma 46. Therefore, if µ(U) < ∞ then Now assume µ(U) = ∞. For n ∈ N choose C ∈ K (X) such that K ⊆ C and µ(C) > n. By Lemma 6 find V ∈ O * (X) such that Using again (21) we have: To show that the extension from a solid set function to a topological measure is unique suppose ν is a topological measure on X such that µ = ν = λ on A * s (X). If A ∈ K c (X) then by Definition 19 A =Ã\( s∈S B s ), whereÃ, B s ∈ A * s (X), so from Definition 41 it follows that µ = ν on K c (X), and, hence, on K 0 (X). From part (t6) of Lemma 36 it then follows that µ = ν on O(X), so µ = ν.
Remark 49. We will summarize the extension procedure for obtaining a topological measure µ from a solid set function λ on a locally compact, connected, locally connected space. First, for a compact connected set C we have: whereC is the solid hull of C and B i (open solid sets) are bounded components of X \ C. For C ∈ K 0 (X), i.e. for a compact set which is the union of finitely many disjoint compact connected sets C 1 , . . . , C n , we have: For an open set U we have: and for a closed set F let Theorem 51. The restriction λ of a compact-finite topological measure µ to A * s (X) is a solid set function, and µ is uniquely determined by λ.
Proof. Let λ be the restriction of µ to A * s (X). Monotonicity of a topological measure (see Lemma 36) and (TM1) of Definition 34 show that λ satisfies conditions (s1) and (s4) of Definition 39. For U ∈ O * s (X) by (TM2) let K ∈ K (X) be such that µ(U) − µ(K) < ǫ and by Lemma 21 we may assume that K ∈ K s (X).
Remark 52. Lemma 40, Lemma 42, and Lemma 46 give us some additional properties of topological measures. For example, by part (p7) of Lemma 46, if a closed set F and a compact K are disjoint, then µ(F ⊔ K) = µ(F ) + µ(K).

EXAMPLES
When X is compact, a set is called solid if it and its complement are both connected. For a compact space X we define a certain topological characteristic, genus. See [2] for more information about genus g of the space. We are particularly interested in spaces with genus 0. One way to describe the "g = 0" condition is the following: if the union of two open solid sets in X is the whole space, their intersection must be connected. (See [9].) Intuitively, X does not have holes or loops. In the case where X is locally path connected, g = 0 if the fundamental group π 1 (X) is finite (in particular, if X is simply connected). Knudsen [10] was able to show that if H 1 (X) = 0 then g(X) = 0, and in the case of CW-complexes the converse also holds.
The following two remarks for a compact space follow from results in [2]: Remark 55. When X is compact, a solid-set function on X extends in a unique way to a topological measure on X. For precise definitions and extension procedure see [2].
The reason that the majority of existing examples of topological measures on compact spaces are given for the spaces with genus 0 is the following. To obtain a topological measure it is enough to define a solid-set function. When a space has genus 0, in the definition of a solid-set function the hardest condition to verify, the irreducible partition condition, becomes easy to verify. When X is locally compact, the hardest condition in Definiton 39 to verify is the condition (s4) that deals with solid partitions. But, as we shall see in this section, it turns out that this condition holds trivially for spaces whose one-point compactification has genus 0.
In this section we denote byX the one-point compactification of X.
Lemma 57. Let X be locally compact andX be its one-point compactification.
It is easy to see that each E i is connected inX. Thus,X \ A is connected, and so A is solid inX.
Lemma 58. Let X be a locally compact space whose one-point compactification X has genus 0. If A ∈ A * s (X) then any solid partition of A is the set A itself.
Proof. Suppose first that V ∈ O * s (X) and its solid partition is given by where each C i ∈ K s (X) and each U j ∈ O * s (X). From Lemma 57 it follows that X \ V and each C i are closed solid sets inX. SinceX has genus 0, by Remark U j must be connected inX. It follows that m = 1 and we may write Then {U 1 ,X \ V, C 1 , . . . , C n } is a solid partition ofX, and it has only one open set. By Remark 54 this solid partition also has only one closed set in it, and it must beX \ V . So each C i = ∅, and the solid partition of V is V = U 1 , i.e. the set itself.
Now suppose that C ∈ K s (X) and its solid partition is given by

SOLID-SET FUNCTIONS AND TOPOLOGICAL MEASURES ON LOCALLY COMPACT SPACES 39
where each C i ∈ K s (X) and each U j ∈ O * s (X). Then {X\C, U 1 , . . . , U m , C 1 , . . . , C n } is a solid partition ofX. Again by Remark 53 must be connected inX. It follows that U j = ∅ for j = 1, . . . , m. Then by connectivity of C we see that the solid partition of C must be the set itself.
Remark 59. From Lemma 58 it follows that for any locally compact space whose one-point compactification has genus 0 the last condition of Definition 39 holds trivially. This is true, for example, for X = R n , half-plane in R n with n ≥ 2, or for a punctured ball in R n with the relative topology.
Example 60. Lemma 58 may not be true for spaces whose one-point compactification has genus greater than 0. For example, let X be an infinite strip R ×[0, 1] without the ball of radius 1/4 centered at (−1/2, 1/2), soX has genus greater than 0. It is easy to give an example of a solid partition of a bounded solid set Example 61. Let X be a locally compact space whose one-point compactification has genus 0. Let λ be a real-valued topological measure on X (or, more generally, a real-valued deficient topological measure on X; for definition and properties of deficient topological measures on locally compact spaces see [6]).
Let P be a set of two distinct points. For each A ∈ A * s (X) let ν(A) = 0 if ♯A = 0, ν(A) = λ(A) if ♯A = 1, and ν(A) = 2λ(X) if ♯A = 2, where ♯A is the number of points in A ∩ P . We claim that ν is a solid set function. By Remark 59 we only need to check the first three conditions of Definition 39. The first one is easy to see. Using Lemma 21 and Lemma 22 it is easy to verify conditions (s2) and (s3) of Definition 39. The solid set function ν extends to a unique finite topological measure on X. Suppose, for example, that λ is the Lebesgue measure on X = R 2 , the set P consists of two points p 1 = (0, 0) and p 2 = (2, 0). Let K i be the closed ball of radius 1 centered at p i for i = 1, 2. Then K 1 , K 2 and C = K 1 ∪ K 2 are compact solid sets, ν(K 1 ) = ν(K 2 ) = π, ν(C) = 4π. Since ν is not subadditive, ν is a topological measure that is not a measure.
The next two examples are adapted from Example 2.2 in [4] and are related to Example 56.
Example 62. Let X be the unit disk on the plane with removed origin. X is a locally compact Hausdorff space with respect to the relative topology. Any subset of X whose closure in R 2 contains the origin is unbounded in X. For A ∈ A * s (X) (since A is also solid subset of the unit disk by Lemma 57) we define µ ′ (A) = µ(A) where µ is the solid set function on the unit disk from Example 56. From Remark 59, Lemma 21, Lemma 22 and the fact that µ is a solid set function onX we see that µ ′ is a solid-set function on X. By Theorem 48 µ ′ extends uniquely to a topological measure on X, which we also call µ ′ .
Then U 1 , U 2 are open (unbounded) in X and F is a closed (unbounded) set in X consisting of two disjoint segments. Note that X = F ∪ U 1 ∪ U 2 . Using Remark 49 we calculate µ ′ (F ) = µ ′ (U 1 ) = µ ′ (U 2 ) = 0. The boundary of the disk, C, is a compact connected set, X \ C is unbounded in X, so C ∈ K s (X). Since µ ′ (C) = 1, we have µ ′ (X) = 1. Thus, µ ′ is not subadditive, so it is not a measure.
This example also shows that on a locally compact space finite additivity of topological measures holds on K (X) ∪ O(X) by Definition 34, but fails on C (X) ∪ O(X). This is in contrast to topological measures on compact spaces, where finite additivity holds on C (X) ∪ O(X).
Example 63. Let X = R 2 , l be a straight line and p a point of X not on the line l. For A ∈ A * s (X) define µ(A) = 1 if A ∩ l = ∅ and p ∈ A; otherwise, let µ(A) = 0. Using Lemma 21 and Lemma 22 it is easy to verify the first three conditions of Definition 39. From Remark 59 it follows that µ is a solid set function on X. By Theorem 48 µ extends uniquely to a topological measure on X, which we also call µ. Note that µ is simple. We claim that µ is not a measure.
Let F be the closed half-plane determined by l which does not contain p. Then using Remark 49 we calculate µ(F ) = µ(X \ F ) = 0, and µ(X) = 1. Failure of subadditivity shows that µ is not a measure.
The sets F and X \ F are both unbounded. Now take a bounded open disk V around p that does not intersect l. Then where V ∈ O * (X), µ(V ) = µ(X \ V ) = 0, while µ(X) = 1.
This example also shows that on a locally compact space finite additivity of topological measures holds on K (X) ∪ O(X) by Definition 34, but fails on C (X) ∪ O(X). It fails even in the situation X = V ⊔ F , where V is a bounded open set, and F is a closed set.
The last two examples suggest that having a topological measure onX helps us to get a topological measure on X. In fact, we have the following result.
Theorem 64. Let X be a locally compact, connected, locally connected space whose one-point compactificationX has genus 0. Suppose ν is a solid set function onX. For A ∈ A * s (X) define µ(A) = ν(A). Then µ is a solid set function on X and, thus, extends uniquely to a topological measure on X.
Proof. Let A ∈ A * s (X). By Lemma 57, A is a solid set inX. Using Lemma 21, Lemma 22, the fact that ν is a solid set function onX, and that a bounded solid set does not contain ∞ it is easy to verify the first three conditions of Definition 39. By Remark 59 µ is a solid set function on X. Example 65. Let X be a locally compact space whose one-point compactification has genus 0. Let n be a natural number. Let P be the set of distinct 2n + 1 points. For each A ∈ A * s (X) let ν(A) = i/n if ♯A = 2i or 2i + 1, where ♯A is the number of points in A ∩ P . When X is compact, a set function defined in this way is a solid-set function (see Example 2.1 in [3], Examples 4.14 and 4.15 in [5]). By Theorem 64 ν is a solid-set function on X; it extends to a unique topological measure on X that assumes values 0, 1/n, . . . , 1.
We conclude with an example of another infinite topological measure.
From Remark 59 it follows that µ is a solid set function on X. By Theorem 48 µ extends uniquely to a topological measure on X, which we also call µ. Note that µ(X) = ∞. µ is not subadditive, for we may cover a compact ball with Lebesque measure greater than 1 by finitely many balls of Lebesque measure less than 1.
Hence, µ is not a measure.