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147 (1995)

Construction of non-subadditive measures and discretization of Borel measures

by

Johan F. A a r n e s (Trondheim)

Abstract. The main result of the paper provides a method for construction of regular non-subadditive measures in compact Hausdorff spaces. This result is followed by several examples. In the last section it is shown that “discretization” of ordinary measures is possible in the following sense. Given a positive regular Borel measure λ, one may construct a sequence of non-subadditive measures µ

n

, each of which only takes a finite set of values, and such that µ

n

converges to λ in the w

-topology.

1. Introduction. In this paper we continue the study of non-subadditive measures undertaken in [1], [2] and [5], called there “quasi-measures”. They are set-functions defined on the open and on the closed subsets of a locally compact Hausdorff space X, and represent a genuine generalization of reg- ular Borel measures in such spaces. This paper is devoted to showing how they arise and may be constructed when X is compact, and to giving some applications.

Non-subadditive measures (NSA-measures), as the name indicates, are generally not subadditive. Indeed, if they are, then they turn out to be ordinary regular Borel measures. This lack of subadditivity is what makes NSA-measures different, and in some respects more interesting than ordinary measures. Instead of weighing effects or events on an additive scale, the NSA-measures register a cumulative effect of events. To produce a certain result, several other results must occur simultaneously. This is of course a very superficial description, and only future development and applications can substantiate what we indicate here.

Even if NSA-measures by definition are generalizations of ordinary mea- sures, their existence is not an obvious matter, and turns out to be closely linked to properties of the underlying topological space. The existence of NSA-measures was first established in the author’s paper [1]. In [2] we gave

1991 Mathematics Subject Classification: 28C15.

[213]

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a description of the basic properties of extremal NSA-measures (those taking only the values 0 and 1). In [5] Knudsen gave a procedure for the construc- tion of such extremal NSA-measures in certain spaces. The present paper is devoted to the construction of general NSA-measures. Our main result (Theorem 5.1) shows how all NSA-measures arise and may be constructed (in certain spaces). We believe that even when applied to ordinary measures this is a new result of some interest. The main result is followed by several examples of construction. In the last section of the paper we show that “dis- cretization” of ordinary measures is possible in the following sense: Given a positive, regular Borel measure λ (for instance Lebesgue measure on the unit sphere in R

3

), we construct a sequence of NSA-measures µ

n

, each of which takes only a finite set of values, and such that µ

n

converges to λ in the w

-topology.

1.1. Notation and basic concepts. Throughout X denotes a compact Hausdorff space and A = C(X) is the algebra of real-valued continuous functions on X. For a ∈ A we let A(a) denote the smallest uniformly closed subalgebra of A containing a and 1. A function % : A → R satisfying

%(1) = 1, %(a) ≥ 0 if a ≥ 0 and such that % is linear on A(a) for each a ∈ A is called a non-linear state (previously called a quasi-state).

Let C denote the collection of closed subsets of X, let O denote the collection of open subsets of X and put A = C ∪ O. A real-valued, non- negative function µ on A is called a NSA-measure in X if the following conditions are satisfied:

(Q0) µ(K) + µ(X \ K) = µ(X) for K ∈ C, (Q1) K

1

⊂ K

2

⇒ µ(K

1

) ≤ µ(K

2

) for K

1

, K

2

∈ C,

(Q2) K

1

∩ K

2

= ∅ ⇒ µ(K

1

∪ K

2

) = µ(K

1

) + µ(K

2

) for K

1

, K

2

∈ C, (Q3) µ(U ) = sup{µ(K) : K ⊂ U, K ∈ C} for U ∈ O.

µ is normalized if µ(X) = 1. For simplicity we shall assume that all NSA-measures in this paper are normalized.

R e m a r k. A NSA-measure µ which is also subadditive is called a regular content. In Halmos ([4], §54, Theorem A) it is shown that a regular content has a unique extension to a regular Borel measure in X.

In [1] we established that there is a 1-1 correspondence between non- linear states and normalized NSA-measures. The set of all non-linear states is a convex set, denoted by Q, which is compact in the topology of pointwise convergence on A.

A subset D of X is co-connected if X \ D is connected. D is solid if it is connected and co-connected.

In what follows a subscript s indicates “solid”, and a subscript c indicates

“connected”, so that for instance A

s

is the collection of all solid sets that

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are either open or closed and C

c

is the family of closed connected sets. C

0

is the family of closed sets with only a finite number of connected components.

O

0

= {U ∈ O : X \ U ∈ C

0

}, and A

0

= C

0

∪ O

0

.

2. Fundamentals. In this section we introduce the main object of study in this paper: the solid set-function. Some preliminaries are needed. From now on we assume that X is a compact Hausdorff space which is connected and locally connected. By convention the empty set ∅ is connected, so ∅ and X both belong to C

s

.

Definition 2.1. A partition of X is a collection of mutually disjoint, non-void sets {A

i

}

i∈I

⊆ A

s

, where at most finitely many of the A

i

are closed, and such that X = S

i∈I

A

i

. The number of closed sets in a partition P is called the order of P.

Any connected and locally connected space X with more than one point has a partition. For if x ∈ X, then {x}

c

is open and has a non-void connected component V ∈ O

s

(cf. Lemma 3.2 in the next section). Hence C = V

c

∈ C

s

and {C, V } is a partition of X of order 1. Accordingly, partitions of order 1 are called trivial.

Let {A

i

}

i∈I

be a non-trivial partition, and let I

0

= {i ∈ I : A

i

is closed}.

Definition 2.2. {A

i

}

i∈I

is irreducible if the following two conditions hold:

(i) S

i∈I0

A

i

is not co-connected.

(ii) For any proper subset I

0

of I

0

, S

i∈I0

A

i

is co-connected.

Necessarily, any irreducible partition has order ≥ 2, and any partition of order 2 is irreducible. For a given space X, let n denote the maximal order of any irreducible partition. If n is finite, let g = n − 1. If X only permits trivial partitions, put g = 0.

Definition 2.3. A function µ : A

s

→ [0, 1] satisfying the conditions (A), (B) and (C) below is called a solid set-function.

(A) For any finite collection of disjoint sets {C

1

, . . . , C

n

} ⊆ C

s

such that C

j

⊆ C ∈ C

s

for j = 1, . . . , n we have

X

n j=1

µ(C

j

) ≤ µ(C).

(B) For all U ∈ O

s

we have

µ(U ) = sup{µ(C) : C ⊆ U, C ∈ C

s

}.

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(C) For any trivial or irreducible partition {A

i

}

i∈I

of X we have X

i∈I

µ(A

i

) = 1.

N o t e. If g = 0 then X has no irreducible partitions, so (C) reduces to the condition that µ(A) + µ(A

c

) = 1 for each set A ∈ A

s

.

R e m a r k. It is important to realize that the restriction of a NSA- measure µ to A

s

is always a solid set-function. Indeed, since µ is additive and monotone on C, (A) is clearly true. If K is closed and contained in a solid open set U , there is a solid closed set C such that K ⊆ C ⊆ U (see Section 3), and (B) follows. Finally, (C) is a consequence of Proposition 2.1 of [1] and Corollary 2.1 of [2].

We next point out that any NSA-measure (or any regular Borel measure) is uniquely determined by its restriction to C

s

. For suppose µ is a NSA- measure in X. If the values of µ on C

s

are known, then they are also known on the complements of these sets, i.e. on O

s

. But then, by virtue of Corollary 2.1 quoted above, and Lemma 3.2 of this paper, it follows that µ is determined on the family of closed connected sets, and hence also on C

0

. Now let K be a closed set contained in an open set U , and let U = S

i∈I

U

i

be the decomposition of U into its connected components. By compactness of K there is a finite index set I

0

⊆ I such that K ⊆ S

i∈I0

U

i

. Let K

i

= K T U

i

(i ∈ I

0

). By Lemma 3.1 of this paper there are connected closed sets C

i

such that K

i

⊆ C

i

⊆ U

i

for each i ∈ I

0

. But then C = S

C

i

belongs to C

0

and K ⊆ C ⊆ U . By (Q3) in the definition of NSA-measures it therefore follows that µ is determined on the open sets by the values it takes on the class C

0

. Taking complements again we see that the uniqueness claim follows.

Example. Let X = S

2

and let p

1

, . . . , p

5

be five distinct points in X.

For C ∈ C

s

define µ(C) to be 0 if C contains at most one of these points, to be 1/2 if C contains two or three of the points, and to be 1 if C contains four or five points. It is easily seen that (A) and (B) hold, and (C) is true because S

2

only permits trivial partitions.

The main purpose of this paper is to show that a solid set-function has a unique extension to a NSA-measure in X. In the example just given this extension turns out to be an extreme point in the convex set of all normalized NSA-measures in X (cf. [2], Example 2.1).

To begin with, we record some of the basic properties of solid set- functions.

Proposition 2.1. If µ is a solid set-function in X then 1. µ(∅) = 0, µ(X) = 1.

2. A

1

⊆ A

2

⇒ µ(A

1

) ≤ µ(A

2

) for all A

1

, A

2

∈ A

s

.

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3. If U

1

, . . . , U

n

belong to O

s

, where n is an arbitrary integer ≥ 1, U

i

∩ U

j

= ∅, i 6= j, and U

i

⊆ U ∈ O

s

for all i = 1, . . . , n, then

X

n i=1

µ(U

i

) ≤ µ(U ).

The proof is left to the reader.

R e m a r k. Let µ be a solid set-function in X, and suppose {C

i

}

i∈I

⊆ C

s

is an arbitrary collection of mutually disjoint sets. From (A) it immediately follows (let C = X) that the subfamily {C

i

: µ(C

i

) > 0} is at most countably infinite. When it comes to summation we may therefore replace the index set I by Z

+

or N. This is implicit when we write P

i∈I

µ(C

i

). If C

i

⊆ C ∈ C

s

for all i ∈ I we must clearly have P

i∈I

µ(C

i

) ≤ µ(C). Analogously, in view of Proposition 2.1.3 the corresponding statements hold for open sets, and will be utilised without further comments.

For later use we include the following:

Lemma 2.1. Let F = {C

j

}

j∈J

⊆ C

s

(n ≥ 1) be a finite family of mutually disjoint sets such that S

j∈J

C

j

is not co-connected. Then F has a subfamily F

0

such that each connected component U

i

of U = ( S

{C

j

∈ F

0

})

c

belongs to O

s

(i ∈ I = some index set) and F

0

∪ {U

i

}

i∈I

is an irreducible partition of X.

P r o o f. F contains a subfamily F

0

= {C

j

}

j∈J0

such that U = ( S

j∈J0

C

j

)

c

is not connected, but for any proper subset J

00

of J

0

, the set ( S

j∈J00

C

j

)

c

is connected. Let U = S

i∈I

U

i

be the decomposition of U into its connected components. Then card I ≥ 2 and each U

i

is in O

c

. We claim that each U

i

is also co-connected. So pick an arbitrary U

0

∈ {U

i

}

i∈I

, and let I

0

= I \ {0}.

We must show that U

0c

= ( S

j∈J0

C

j

) ∪ ( S

i∈I0

U

i

) is connected. Suppose not.

Then there are non-void, disjoint, closed sets K

1

and K

2

in X such that K

1

∪ K

2

= U

0c

. By connectedness, K

1

(and K

2

) must contain all or nothing of each of the sets C

j

(j ∈ J

0

) and U

i

(i ∈ I

0

). Let

J

k

= {j ∈ J

0

: C

j

⊆ K

k

}, I

k

= {i ∈ I

0

: U

i

⊆ K

k

}, k = 1, 2.

We must have J

1

6= ∅ and J

2

6= ∅ and at least one of the sets I

k

is non- empty since card I ≥ 2. Suppose I

1

6= ∅. Since K

1

is closed, its complement K

1c

= K

2

∪ U

0

is open. But then also

V = (K

2

∪ U

0

) ∪  [

i∈I1

U

i



is open and not connected. But V

c

= S

j∈J1

C

j

, which is a contradiction,

since J

1

is a proper subset of J

0

. Hence U

0c

is connected and the proof is

complete.

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Proposition 2.2. For a space X we have g = 0 if and only if S

{C ∈ F}

is co-connected for any finite family F ⊆ C

s

of disjoint sets.

P r o o f. This follows from Lemma 2.1.

R e m a r k. In [5] Knudsen has shown that if X is also locally pathwise connected and H

1

(X, Q) = (0) then g = 0. Hence S

n

(the unit n-sphere) and B

n

(the unit n-ball) have g = 0, the torus and an annulus in the plane have g = 1, etc.

3. Topological preliminaries. The results in this section are more or less known. Some proofs are included for the convenience of the reader.

Good references are [3] and [6].

Throughout we assume that X is a compact Hausdorff space which is connected and locally connected.

Lemma 3.1. Let K ∈ C, U ∈ O and K ⊆ U . If either K or U is connected, then there is a set V ∈ O

c

such that K ⊆ V ⊆ V ⊆ U .

Lemma 3.2. Let K ∈ C

c

. Then each connected component of V = X \ K belongs to O

s

.

Lemma 3.3. Let K ∈ C

s

, U ∈ O and K ⊆ U . Then there is a set V ∈ O

s

such that K ⊆ V ⊆ V ⊆ U .

P r o o f o f L e m m a 3.3. By Lemma 3.1 we may assume that U is connected. Let F = X \ U , so F ⊆ X \ K, which is open and connected.

Hence again by Lemma 3.1 there is an open set V ∈ O

c

such that F ⊆ V ⊆ V and V ∩ K = ∅. Let W = X \ V so K ⊆ W ⊆ U . Since V ∈ C

c

it now follows that each connected component of W belongs to O

s

(Lemma 3.2), and K must be contained in exactly one of them. The proof is complete.

3.1. The solid hull of a closed connected set. Let K ∈ C

c

, V ∈ O

s

and suppose K ⊆ V . Let X \ K = W = S

i∈I

W

i

be the decomposition of W into disjoint connected components. Then C = X \ V is connected and is contained in one of them, say W

0

, called the exterior component of W with respect to V . Each W

i

with i 6= 0 is then called an interior component. Since each W

i

is in O

s

the set e K = X \ W

0

belongs to C

s

. We have

K = e [

{W

i

: i 6= 0} ∪ K

and K ⊆ e K ⊆ V. e K is called the solid hull of K with respect to V .

Lemma 3.4. Suppose C

1

, C

2

are disjoint, closed, connected sets, both

contained in V ∈ O

s

. Then either

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(i) e C

1

∩ e C

2

= ∅, or

(ii) e C

1

⊆ e C

2

or e C

2

⊆ e C

1

(proper inclusion).

P r o o f. Let W

j

= X \ C

j

(j = 1, 2), and let W

j

= S

i∈Ij

W

ij

as above, with W

0j

as the exterior component of W

j

with respect to V . Now C

1

∩ C

2

= ∅ so

C

1

⊆ W

2

and C

2

⊆ W

1

.

Since C

1

and C

2

are connected, there are unique components W

i2

and W

k1

such that C

1

⊆ W

i2

and C

2

⊆ W

k1

.

(i) Suppose C

1

⊆ W

02

and C

2

⊆ W

01

. Then C

2

⊆ e C

2

= X \ W

02

⊆ W

1

by the first inclusion. Since e C

2

is connected the second inclusion implies that C e

2

⊆ W

01

= X \ e C

1

, so e C

1

∩ e C

2

= ∅.

(ii) Suppose C

2

⊆ W

k1

for some k 6= 0. Then W

k1

⊆ V so X \ V ⊆ X \ W

k1

⊆ W

2

. Since X \ W

k1

is connected and contains X \ V we must have X \ W

k1

⊆ W

02

, so that e C

2

= X \ W

02

⊆ W

k1

⊆ e C

1

. Similarly, if C

1

⊆ W

i2

for some i 6= 0, then e C

1

⊆ e C

2

. The proof is complete.

R e m a r k. We may note that in the above situation one of the sets is always contained in the exterior component associated with the other set.

4. Extension to C

0

. Let µ be a solid set-function. Our goal in this section is to extend µ to a function on C

0

to [0, 1] such that

(Q1)

0

K

1

⊆ K

2

⇒ µ(K

1

) ≤ µ(K

1

) for K

1

, K

2

∈ C

0

.

(Q2)

0

K

1

∩ K

2

= ∅ ⇒ µ(K

1

∪ K

2

) = µ(K

1

) + µK

2

) for K

1

, K

2

∈ C

0

. (Q3)

0

For all C ∈ C

0

and all ε > 0 there is C

0

∈ C

0

such that C ∩ C

0

= ∅ and µ(C) + µ(C

0

) > 1 − ε.

If K ∈ C

c

we know by Lemma 3.2 that X \ K = V = S

i∈I

V

i

, where the sets V

i

are open, connected, co-connected and mutually disjoint. From Proposition 2.1.3 it therefore follows that the set {V

i

: µ(V

i

) > 0} is at most countable, and that P

i∈I

µ(V

i

) ≤ 1. We may then define

(4.1) µ(K) = 1 − X

i∈I

µ(V

i

).

Next, if K = K

1

∪ . . . ∪ K

n

∈ C

0

, where the K

j

are connected and mutually disjoint, we put µ(K) = P

n

j=1

µ(K

j

).

Proposition 4.1. The function µ as defined above maps C

0

into [0, 1]

and satisfies conditions (Q1)

0

, (Q2)

0

and (Q3)

0

.

We start with (Q1)

0

: If K

1

, K

2

∈ C

0

, then K

1

⊆ K

2

⇒ µ(K

1

) ≤ µ(K

2

).

Suppose first that K

1

is connected, while K

2

= C

1

∪ . . . ∪ C

n

with the

C

j

in C

c

and mutually disjoint. Then we must have K

1

⊆ C

j

for some j, say

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K

1

⊆ C

1

. Let S

i∈I

U

i

= X \K

1

and S

i∈J

V

j

= X \C

1

be the decompositions of these sets into their solid components. Each V

j

must be contained in some U

i

. Let

J

i

= {j ∈ J : V

j

⊆ U

i

}.

Then S

i∈I

J

i

= J, and by Proposition 2.1.3 we get X

j∈J

µ(V

j

) = X

i∈I

X

j∈Ji

µ(V

j

) ≤ X

i∈I

µ(U

i

).

Hence µ(K

1

) = 1 − P

i∈I

µ(U

i

) ≤ 1 − P

j∈J

µ(V

j

) ≤ µ(C

1

).

Next, assume that K

1

= C

10

∪ . . . ∪ C

m0

is the decomposition of K

1

into its connected components. Since K

1

⊆ K

2

each C

i0

is contained in C

j

for some j ∈ {1, . . . , n}.

Suppose C

i01

, . . . , C

i0k

⊆ C

j

. We need to show that P

k

l=1

µ(C

i0l

) ≤ µ(C

j

).

To simplify the notation, assume that C

0

, C

1

, . . . , C

n

are connected, closed, disjoint and contained in C ∈ C

c

. Let X \C

0

= V = S

i∈I

V

i

. Since all the C

j

are connected, each C

j

for j ≥ 1 is contained in some V

i

, and only finitely many V

i

will contain some C

j

. Let

J

i

= {j : C

j

⊆ V

i

}, i ∈ I.

Let U = X \ C = S

i∈K

U

k

. Then U ⊆ V and we put K

i

= {k ∈ K : U

k

⊆ V

i

}.

The U

k

are disjoint and obviously also disjoint from all the C

j

. We want to show that

(4.2)

X

n j=0

µ(C

j

) ≤ µ(C),

which is equivalent to 1 − X

i∈I

µ(V

i

) + X

n j=1

µ(C

j

) ≤ 1 − X

k∈K

µ(U

k

), or

(4.3)

X

n j=1

µ(C

j

) + X

k∈K

µ(U

k

) ≤ X

i∈I

µ(V

i

).

Now (4.3) will follow if we can show that

(4.4) X

j∈Ji

µ(C

j

) + X

k∈Ki

µ(U

k

) ≤ µ(V

i

)

for all i ∈ I. Therefore, what we really need is the following result, formally

stated as a lemma.

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Lemma A. Let C

1

, . . . , C

n

∈ C

c

and U

1

, . . . , U

k

, . . . ∈ O

s

be mutually disjoint subsets of V ∈ O

s

. Then

(4.5)

X

n j=1

µ(C

j

) + X

k=1

µ(U

k

) ≤ µ(V ).

P r o o f. We employ induction on n.

S t e p 1: n = 1. Let X \ C

1

= W = S

i∈I

W

i

. Now X \ V is connected and contained in W , and is therefore contained in one of W ’s components, which we denote by W

0

, i.e. X \ V ⊆ W

0

and W

0

is the exterior component of W with respect to V (cf. Section 3). Let I

0

= I \ {0} and put e C

1

= C

1

∪ ( S

i∈I0

W

i

) = X \ W

0

, i.e. e C

1

is the solid hull of C

1

with respect to V . We want to show that

µ(C

1

) + X

k=1

µ(U

k

) ≤ µ(V ), or

1 − X

i∈I0

µ(W

i

) − µ(W

0

) + X

k=1

µ(U

k

) ≤ µ(V ), i.e.

(4.6) µ( e C

1

) + X

k=1

µ(U

k

) ≤ µ(V ) + X

i∈I0

µ(W

i

).

We have U

k

⊆ W for all k = 1, 2, . . . (since all U

k

are disjoint from C

1

) and we let

K

i

= {k ∈ Z

+

: U

k

⊆ W

i

}, i ∈ I.

Then we have, using Proposition 2.1.3 again, X

k=1

µ(U

k

) = X

k∈K0

µ(U

k

) + X

i∈I0

h X

k∈Ki

µ(U

k

) i

X

k∈K0

µ(U

k

) + X

i∈I0

µ(W

i

).

Hence (4.6) will follow if we can show that

(4.7) µ( e C

1

) + X

k∈K0

µ(U

k

) ≤ µ(V ),

where we observe that if k ∈ K

0

then U

k

is disjoint from e C

1

. In turn, (4.7)

will follow from the next lemma (which in fact is just a weakened version of

Lemma A).

(10)

Lemma B. Let C

1

, . . . , C

m

∈ C

s

and U

1

, . . . , U

k

, . . . ∈ O

s

be mutually disjoint subsets of V ∈ O

s

. Then

(4.8)

X

m j=1

µ(C

j

) + X

k=1

µ(U

k

) ≤ µ(V ).

P r o o f. Let ε > 0 be arbitrary. Select N ∈ Z

+

such that P

k>N

µ(U

k

) <

ε/2. For k = 1, . . . , N take C

k0

⊆ U

k

such that C

k0

∈ C

s

and µ(C

k0

) >

µ(U

k

) − ε/2

k+1

by (B). Then X

k=1

µ(U

k

) = X

N k=1

µ(U

k

) + X

k>N

µ(U

k

)

<

X

N k=1

µ(C

k0

) + ε X

N k=1

1 2

k+1

+ ε

2 <

X

N k=1

µ(C

k0

) + ε.

Let C = X \ V , so C ∈ C

s

and C ∩ C

j

= C ∩ C

k0

= ∅ for all j and k. Then by (A) we get

X

m j=1

µ(C

j

) + X

N k=1

µ(C

k0

) + µ(C) ≤ 1.

Hence

X

m j=1

µ(C

j

) + X

k=1

µ(U

k

) ≤ µ(V ) + ε.

Since ε > 0 was arbitrary, the assertion follows.

The inequality (4.7) now follows (taking m = 1) since e C

1

∈ C

s

. We have now established Lemma A for n = 1.

S t e p 2. Suppose inductively that Lemma A has been verified for k = 1, . . . , n − 1. For j = 1, . . . , n let X \ C

j

= W

j

= S

{W

ij

: i ∈ I

j

}, and let W

0j

be the exterior component of W

j

with respect to V . Let e C

j

= X \ W

0j

denote the solid hull of C

j

with respect to V .

C a s e 1: e C

j

∩ e C

l

= ∅ for all j, l with j 6= l. The assertion (4.5) is equivalent to

(4.9)

X

n j=1

µ( e C

j

) + X

k=1

µ(U

k

) ≤ µ(V ) + X

n j=1

X

i∈Ij0

µ(W

ij

),

where I

j0

= I

j

\ {0}. Now observe that X \ S

n

j=1

C e

j

= T

n

j=1

W

0j

, so that if

(11)

we put

K

0

= n

k ∈ Z

+

: U

k

\

n j=1

W

0j

o

,

K

j

= {k ∈ Z

+

: U

k

⊆ e C

j

}, j = 1, . . . , n,

then we exhaust all possibilities for the sets U

k

in mutually exclusive cases.

By Lemma B we now get (4.10)

X

n j=1

µ( e C

j

) + X

k∈K0

µ(U

k

) ≤ µ(V ).

If U

k

⊆ e C

j

then U

k

⊆ W

ij

for some i ∈ I

j0

so that (using Proposition 2.1.3 again)

(4.11) X

k∈Kj

µ(U

k

) ≤ X

i∈I0j

(W

ij

), j = 1, . . . , n.

Combining (4.10) and (4.11) we obtain (4.9), which establishes the assertion (4.5) in Case 1.

C a s e 2: e C

j

∩ e C

j

6= ∅ for some pair j 6= l. By Lemma 3.4 either e C

j

⊆ e C

l

or e C

l

⊆ e C

j

(proper inclusion). We may therefore re-index the sets C

j

(j = 1, . . . , n) as follows:

1. e C

10

, e C

20

, . . . , e C

m0

(m < n) are mutually disjoint.

2. e C

jl

⊆ e C

j0

for l = 0, . . . , m

j

and 1 ≤ j ≤ m, where the inclusions are proper if l ≥ 1.

3. m + P

m

j=1

m

j

= n.

The verification of this is a simple induction argument based on Lemma 3.4, and is left to the reader. We return to the proof of (4.9), which after the re-indexing takes the form

(4.12)

X

m j=1

mj

X

l=0

µ( e C

jl

) + X

k=1

µ(U

k

) ≤ µ(V ) + X

m j=1

mj

X

l=0

X

i∈Ijl0

µ(W

ijl

),

where X \ C

jl

= S

{W

ijl

: i ∈ I

jl

}.

Let K

0

, K

1

, . . . , K

n

be defined as in Case 1. Then again by Lemma B we immediately get

(4.13)

X

m j=1

µ( e C

j0

) + X

k∈K0

µ(U

k

) ≤ µ(V ),

(12)

so to establish (4.12) it remains to show that (4.14)

X

m j=1

mj

X

l=1

µ( e C

jl

) + X

m j=1

X

k∈Kj

µ(U

k

) ≤ X

m j=1

X

mj l=0

X

i∈Ijl0

µ(W

ijl

),

which will follow if we can show that for each j = 1, . . . , m we have (4.15)

mj

X

l=1

µ( e C

jl

) + X

k∈Kj

µ(U

k

) ≤ X

mj

l=0

X

i∈Ijl0

µ(W

ijl

).

Now µ( e C

jl

) = µ(C

jl

) + P

i∈Ijl0

µ(W

ijl

), so (4.15) is equivalent to (4.16)

X

mj l=1

µ(C

jl

) + X

k∈Kj

µ(U

k

) ≤ X

i∈I0j0

µ(W

ij0

).

Let

J

ji

= {l ≤ m

j

: C

jl

⊆ W

ij0

}, i ∈ I

j00

, K

ji

= {k ∈ K

j

: U

k

⊆ W

ij0

}, i ∈ I

j00

.

Since C

jl

⊆ W

ij0

for some i ∈ I

j00

if 1 ≤ l ≤ m

j

, and U

k

⊆ W

ij0

for some i ∈ I

j00

if k ∈ K

j

, we exhaust all possibilities for l and k in this manner.

Hence (4.16) will follow if we can show that for each i ∈ I

j00

,

(4.17) X

l∈Jji

µ(C

jl

) + X

k∈Kji

µ(U

k

) ≤ µ(W

ij0

).

The number of elements in J

ji

is ≤ m

j

≤ m < n so (4.17) now follows by the induction hypothesis. This concludes the proof of Lemma A.

This also concludes the proof of condition (Q1)

0

in Proposition 4.1. As for (Q2)

0

, if K

1

, K

2

∈ C

0

and K

1

∩ K

2

= ∅ let K

1

= C

1

∪ . . . ∪ C

n

and K

2

= C

10

∪. . .∪C

m0

be their decompositions into connected components. Then K

1

∪K

2

= C

1

∪. . .∪C

n

∪C

10

∪. . .∪C

m0

and the right-hand side represents the decomposition of K

1

∪ K

2

into connected components. From the definition of µ on C

0

it now immediately follows that µ(K

1

∪ K

2

) = µ(K

1

) + µ(K

2

), and (Q2)

0

follows.

It remains to verify the regularity property (Q3)

0

.

Let K ∈ C

0

be arbitrary. Then K = C

0

∪ C

1

∪ . . . ∪ C

n

, where C

j

∈ C

c

and C

j

∩ C

k

= ∅ if j 6= k. Let X \ C

0

= V = S

i∈I

V

i

, where the V

i

are mutually disjoint and belong to O

s

. Each C

j

(j = 1, . . . , n) is contained in some V

i

, and we let

J

i

= {j ∈ {1, . . . , n} : C

j

⊆ V

i

}, i ∈ I.

From Lemma A it follows that P

j∈Ji

µ(C

j

) ≤ µ(V

i

) for all i ∈ I. Let ε > 0 be arbitrary, and choose N such that P

i>N

µ(V

i

) < ε/2 (since

(13)

{i ∈ I : µ(V

i

) > 0} is countable, we may assume that I = Z

+

). Now suppose for the moment that there are sets C

i0

∈ C

0

for i = 1, . . . , N such that

(4.18)

 C

i0

⊆ V

i

and C

i0

∩ C

j

= ∅ if j ∈ J

i

, µ(C

i0

) > µ(V

i

) − P

j∈Ji

µ(C

j

) − ε/2

i+1

. Then let C

0

= S

N

i=1

C

i0

, so C

0

∈ C

0

and C

0

∩ K = ∅. Moreover (by (Q2)

0

), µ(C

0

) =

X

N i=1

µ(C

i0

) >

X

N i=1

µ(V

i

) − X

N i=1

X

j∈Ji

µ(C

j

) − X

N i=1

ε 2

i+1

>

X

i=1

µ(V

i

) − ε 2

X

n j=1

µ(C

j

) − ε 2

= 1 − µ(C

0

) − X

n j=1

µ(C

j

) − ε = 1 − µ(K) − ε,

which establishes (Q3)

0

under the assumption that (4.18) holds. So what we need is the following.

Lemma C. Let C

1

, . . . , C

n

∈ C

c

be mutually disjoint and contained in V ∈ O

s

. For each ε > 0 there is a C

0

∈ C

0

such that C

0

∩ C

j

= ∅ for all j, C

0

⊆ V and

X

n j=1

µ(C

j

) + µ(C

0

) > µ(V ) − ε.

P r o o f. We first consider the special case where all the sets C

1

, . . . , C

n

belong to C

s

, and V = X. The proof in this case goes by induction on n. For n = 1 the assertion is covered by (B) in the definition of a solid set-function.

Now assume n ≥ 2 and that the assertion is true for all k < n. Let ε > 0 be arbitrary. Combining (B) and Lemma 3.3, we may find a family of mutually disjoint sets V

1

, . . . , V

n

in O

s

such that C

j

⊆ V

j

and

µ(V

j

) < µ(C

j

) + ε/n, j = 1, . . . , n.

Next, choose W

j

∈ O

s

such that

C

j

⊆ W

j

⊆ W

j

⊆ V

j

, j = 1, . . . , n

(Lemma 3.3). Let F

j

be the solid hull of W

j

with respect to V

j

. Then C

j

⊆ W

j

⊆ F

j

⊆ V

j

, F

j

∈ C

s

,

for j = 1, . . . , n.

At this point we distinguish between two subcases:

(i) U = X \ S

n

j=1

F

j

is connected. Then C

0

= U is connected and con- tained in X \ S

n

j=1

W

j

, so C

0

∩ C

j

= ∅ for j = 1, . . . , n. On the other hand,

(14)

X \ C

0

S

n

j=1

V

j

. If X \ C

0

= S

i∈I

O

i

is the decomposition of X \ C

0

into its disjoint connected components, then each O

i

belongs to O

s

and each O

i

is contained in some V

j

. But then, by Proposition 2.3.3, it follows that P

i∈I

µ(O

i

) ≤ P

n

j=1

µ(V

j

). Consequently, we get µ(C

0

) = 1 − X

i∈I

µ(O

i

) ≥ 1 − X

n j=1

µ(V

j

) > 1 − X

n j=1

(C

j

) − ε, which yields the assertion in this case.

(ii) X \ S

n

j=1

F

j

is not connected. Let m be the minimal number of sets in any subcollection F

0

⊆ F = {F

1

, . . . , F

n

} such that S

{F

j

: F

j

∈ F

0

} is not co-connected. Let F

1

be such a subfamily of F with m elements, 2 ≤ m ≤ n.

Let U = ( S

{F

j

: F

j

∈ F

1

})

c

. By Lemma 2.1 we have U = S

i∈I

U

i

, U

i

disjoint, U

i

∈ O

s

. Again we distinguish between two cases:

a) m = n. Let N be such that P

i>N

µ(U

i

) < ε/2. Then for i ≤ N choose K

i

⊆ U

i

such that K

i

∈ C

s

and

µ(K

i

) > µ(U

i

) − ε 2N .

The family {F

j

, U

i

: j = 1, . . . , n, i ∈ I} is an irreducible partition of X so that by (C),

X

n j=1

µ(F

j

) + X

i∈I

µ(U

i

) = 1, and hence

X

n j=1

µ(C

j

) + X

N i=1

µ(K

i

) >

X

n j=1

µ(F

j

) − ε + X

N i=1

µ(U

i

) − ε

>

X

n j=1

F

j

+ X

i∈I

µ(U

i

) −

2 = 1 − 2 , which proves the assertion when we take C

0

= S

N

i=1

K

i

.

b) m < n. The family {F

j

, U

i

: j = 1, . . . , m, i ∈ I} is an irreducible partition of X. If F

j

6∈ F

1

then F

j

⊆ U

i

for some i ∈ I. Since m ≥ 2 we have n − m < n − 1, so that for any i ∈ I the collection

F

i

= {F

k

∈ F : F

k

⊆ U

i

} ∪ {X \ U

i

}

will have at most n − 1 elements. By the induction hypothesis it therefore follows that there is a K

i

∈ C

0

such that F

k

∩K

i

= ∅ for all F

k

∈ F

i

, K

i

⊆ U

i

and X

Fk∈Fi

µ(F

k

) + µ(K

i

) + µ(X \ U

i

) > 1 − ε

N ,

(15)

so X

Fk∈Fi

µ(F

k

) + µ(K

i

) > µ(U

i

) − ε

N , i = 1, . . . , N.

Then we get X

n j=1

µ(C

j

) + X

N i=1

µ(K

i

) >

X

n j=1

µ(F

j

) − ε + X

N i=1

µ(K

i

)

X

m j=1

µ(F

j

) + X

N i=1

X

Fk∈Fi

µ(F

k

) + X

N i=1

µ(K

i

) − ε

>

X

m j=1

µ(F

j

) + X

N i=1

µ(U

i

) − 2ε

>

X

m j=1

µ(F

j

) + X

i∈I

µ(U

i

) −

2 = 1 − 2 , which proves the assertion in case b).

Finally, if C

1

, . . . , C

n

⊆ V ∈ O

s

, let C

0

= X \ V . Applying the argument above to the family C

0

, C

1

, . . . , C

n

now yields Lemma C when all the sets C

1

, . . . , C

n

are solid.

Next, we turn to the general case, assuming that all the sets C

j

(j = 1, . . . , n) belong to C

c

. Again we use induction on n.

S t e p 1: n = 1. Let C ⊆ V ∈ O

s

and C ∈ C

c

. Let X \ C = S

i∈I

W

i

with W

i

∈ O

s

. Let W

0

be the exterior component with respect to V , and put e C = X \ W

0

as usual. We have e C ⊆ V and by the first part of the proof, there is a K

0

∈ C

0

such that e C ∩ K

0

= ∅, K

0

⊆ V and µ( e C) + µ(K

0

) > µ(V ) − ε/2. Choose N such that P

i>N

µ(W

i

) < ε/4 (restricting to a countable subfamily again). Next, choose K

i

⊆ W

i

such that K

i

∈ C

s

and µ(K

i

) > µ(W

i

) − ε/(4 · 2

i

) for i = 1, . . . , N . All the K

i

(i = 0, 1, . . . , N ) are now disjoint from C, mutually disjoint (K

0

⊆ W

0

∩ V ) and contained in V . Take K = S

N

i=0

K

i

, so K ∈ C

0

, K ∩ C = ∅ and K ⊆ V . Moreover, µ(K) =

X

N i=0

µ(K

i

) > µ(V ) − µ( e C) − ε 2 +

X

N i=1

µ(W

i

) − ε 4

> µ(V ) − (1 − µ(W

0

)) + X

i∈I

µ(W

i

) − ε

= µ(V ) − µ(C) − ε, which establishes the assertion for n = 1.

S t e p 2: Assume inductively that the assertion is true for k=1, . . . , n−1.

(16)

C a s e 1: e C

1

, . . . , e C

n

are mutually disjoint. By the first part of the proof there is K

0

∈ C

0

such that K

0

⊆ V, K

0

∩ e C

j

= ∅ (j = 1, . . . , n) and µ(K

0

) >

µ(V ) − P

n

j=1

µ( e C

j

) − ε/2.

Let X \ C

j

= W

j

= S

i∈Ij

W

ij

, where W

0j

is the exterior component of W

j

with respect to V (i.e. e C

j

= X \ W

0j

). Let N

j

∈ Z

+

be such that P

i>Nj

µ(W

ij

) < ε/(4n). For i = 1, . . . , N

j

pick K

ij

⊆ W

ij

such that K

ij

∈ C

s

and

µ(K

ij

) > µ(W

ij

) − ε 4n · 1

2

i

, j = 1, . . . , n.

Since the e C

j

are mutually disjoint, all the sets K

ij

are mutually disjoint and also disjoint from K

0

T

n

j=1

W

0j

. Let K = K

0

S

{K

ij

: 1 ≤ i ≤ N

j

, j = 1, . . . , n} so K ∈ C

0

, K ∩ C

j

= ∅ (j = 1, . . . , n) and K ⊆ V . We now obtain

µ(K) = µ(K

0

) + X

n j=1

Nj

X

i=1

µ(K

ij

)

> µ(V ) − X

n j=1

µ( e C

j

) − ε 2 +

X

n j=1

Nj

X

i=1



µ(W

ij

) − ε 4n · 1

2

i



> µ(V ) − X

n j=1

h

µ( e C

j

) − X

i∈Ij0

µ(W

ij

) i

− ε

= µ(V ) − X

n j=1

µ(C

j

) − ε, which proves the assertion in Case 1.

C a s e 2: e C

j

∩ e C

k

6= ∅ for some pair j 6= k. We proceed as in the proof of Lemma A, and re-index the sets C

1

, . . . , C

n

so that

1. e C

10

, e C

20

, . . . , e C

m0

(m < n) are mutually disjoint.

2. e C

jl

⊆ e C

j0

for l = 0, . . . , m

j

and 1 ≤ j ≤ m, where the inclusions are proper if l ≥ 1.

3. m + P

m

j=1

m

j

= n.

As before, let X \ C

jl

= S

{W

ijl

: i ∈ I

jl

} = W

jl

, and let W

0jl

denote the exterior component of W

jl

with respect to V . Again, by the first part of the proof, there is K

0

∈ C

0

such that K

0

⊆ V, K

0

∩ e C

j0

= ∅ (j = 1, . . . , m), and (4.19) µ(K

0

) > µ(V ) −

X

m j=1

µ( e C

j0

) − n − m n ε.

For each j ∈ {1, . . . , m} choose N

j

∈ Z

+

such that P

i>Nj

µ(W

ij0

) < ε/(2n).

(17)

When l ≥ 1 each set C

jl

is contained in some interior component W

ij0

of W

j0

, i.e. i 6= 0. We let

J

ji

= {l : 1 ≤ l ≤ m

j

, C

jl

⊆ W

ij0

}, 1 ≤ i ≤ N

j

. If J

ji

= ∅ there is by (B) a set K

ij

⊆ W

ij0

such that K

ij

∈ C

s

and (4.20) µ(K

ij

) > µ(W

ij0

) − ε

2N

j

n .

If J

ji

6= ∅ there is, by the induction hypothesis, since m

j

< n, a set K

ij

W

ij0

such that K

ij

∈ C

0

and

(4.21) µ(K

ij

) > µ(W

ij0

) − X

l∈Jji

µ(C

jl

) − ε 2N

j

n . Let K

j

= S

Nj

i=1

K

ij

∈ C

0

for j = 1, . . . , m. Then, adding the inequalities (4.20) and (4.21), we get

µ(K

j

) =

Nj

X

i=1

µ(K

ij

)

>

Nj

X

i=1

µ(W

ij0

) −

Nj

X

i=1

X

l∈Jji

µ(C

jl

) − ε 2n

>

X

i=1

µ(W

ij0

) −

mj

X

l=1

µ(C

jl

) − ε n

= µ( e C

j0

) − µ(C

j0

) −

mj

X

l=1

µ(C

jl

) − ε n , so

(4.22) µ(K

j

) > µ( e C

j0

) −

mj

X

l=0

µ(C

jl

) − ε

n , j = 1, . . . , n.

Now let K = S

m

j=0

K

j

∈ C

0

. By addition of (4.19) and the equations (4.22) we get

µ(K) = X

m j=0

µ(K

j

)

> µ(V ) − X

m j=1

µ(C

j0

) − (n − m) ε n +

X

m j=1

µ( e C

j0

) − X

m j=1

mj

X

l=0

µ(C

jl

) − n

= µ(V ) − X

m j=1

mj

X

l=0

µ(C

jl

) − ε,

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