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
nconverges 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]
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
nconverges 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
sis the collection of all solid sets that
are either open or closed and C
cis the family of closed connected sets. C
0is 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
iare 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}
cis open and has a non-void connected component V ∈ O
s(cf. Lemma 3.2 in the next section). Hence C = V
c∈ C
sand {C, V } is a partition of X of order 1. Accordingly, partitions of order 1 are called trivial.
Let {A
i}
i∈Ibe a non-trivial partition, and let I
0= {i ∈ I : A
iis closed}.
Definition 2.2. {A
i}
i∈Iis irreducible if the following two conditions hold:
(i) S
i∈I0
A
iis not co-connected.
(ii) For any proper subset I
0of I
0, S
i∈I0
A
iis 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
ssuch that C
j⊆ C ∈ C
sfor j = 1, . . . , n we have
X
n j=1µ(C
j) ≤ µ(C).
(B) For all U ∈ O
swe have
µ(U ) = sup{µ(C) : C ⊆ U, C ∈ C
s}.
(C) For any trivial or irreducible partition {A
i}
i∈Iof 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
sis 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
sare 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
ibe 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
isuch that K
i⊆ C
i⊆ U
ifor each i ∈ I
0. But then C = S
C
ibelongs to C
0and 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
2and let p
1, . . . , p
5be five distinct points in X.
For C ∈ C
sdefine 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
2only 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.
3. If U
1, . . . , U
nbelong to O
s, where n is an arbitrary integer ≥ 1, U
i∩ U
j= ∅, i 6= j, and U
i⊆ U ∈ O
sfor all i = 1, . . . , n, then
X
n i=1µ(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
sis 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
sfor 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
jis not co-connected. Then F has a subfamily F
0such that each connected component U
iof U = ( S
{C
j∈ F
0})
cbelongs to O
s(i ∈ I = some index set) and F
0∪ {U
i}
i∈Iis an irreducible partition of X.
P r o o f. F contains a subfamily F
0= {C
j}
j∈J0such that U = ( S
j∈J0
C
j)
cis not connected, but for any proper subset J
00of J
0, the set ( S
j∈J00
C
j)
cis connected. Let U = S
i∈I
U
ibe the decomposition of U into its connected components. Then card I ≥ 2 and each U
iis in O
c. We claim that each U
iis 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
1and K
2in 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
16= ∅ and J
26= ∅ and at least one of the sets I
kis non- empty since card I ≥ 2. Suppose I
16= ∅. Since K
1is closed, its complement K
1c= K
2∪ U
0is open. But then also
V = (K
2∪ U
0) ∪ [
i∈I1
U
iis open and not connected. But V
c= S
j∈J1
C
j, which is a contradiction,
since J
1is a proper subset of J
0. Hence U
0cis connected and the proof is
complete.
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
sof 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
csuch 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
ssuch 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
csuch that F ⊆ V ⊆ V and V ∩ K = ∅. Let W = X \ V so K ⊆ W ⊆ U . Since V ∈ C
cit 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
sand suppose K ⊆ V . Let X \ K = W = S
i∈I
W
ibe 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
iwith i 6= 0 is then called an interior component. Since each W
iis in O
sthe set e K = X \ W
0belongs 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
2are disjoint, closed, connected sets, both
contained in V ∈ O
s. Then either
(i) e C
1∩ e C
2= ∅, or
(ii) e C
1⊆ e C
2or 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
ijas above, with W
0jas the exterior component of W
jwith respect to V . Now C
1∩ C
2= ∅ so
C
1⊆ W
2and C
2⊆ W
1.
Since C
1and C
2are connected, there are unique components W
i2and W
k1such that C
1⊆ W
i2and C
2⊆ W
k1.
(i) Suppose C
1⊆ W
02and C
2⊆ W
01. Then C
2⊆ e C
2= X \ W
02⊆ W
1by the first inclusion. Since e C
2is 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
k1for some k 6= 0. Then W
k1⊆ V so X \ V ⊆ X \ W
k1⊆ W
2. Since X \ W
k1is 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
i2for 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
0to [0, 1] such that
(Q1)
0K
1⊆ K
2⇒ µ(K
1) ≤ µ(K
1) for K
1, K
2∈ C
0.
(Q2)
0K
1∩ K
2= ∅ ⇒ µ(K
1∪ K
2) = µ(K
1) + µK
2) for K
1, K
2∈ C
0. (Q3)
0For all C ∈ C
0and all ε > 0 there is C
0∈ C
0such that C ∩ C
0= ∅ and µ(C) + µ(C
0) > 1 − ε.
If K ∈ C
cwe know by Lemma 3.2 that X \ K = V = S
i∈I
V
i, where the sets V
iare 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
jare connected and mutually disjoint, we put µ(K) = P
nj=1
µ(K
j).
Proposition 4.1. The function µ as defined above maps C
0into [0, 1]
and satisfies conditions (Q1)
0, (Q2)
0and (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
1is connected, while K
2= C
1∪ . . . ∪ C
nwith the
C
jin C
cand mutually disjoint. Then we must have K
1⊆ C
jfor some j, say
K
1⊆ C
1. Let S
i∈I
U
i= X \K
1and S
i∈J
V
j= X \C
1be the decompositions of these sets into their solid components. Each V
jmust 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
m0is the decomposition of K
1into its connected components. Since K
1⊆ K
2each C
i0is contained in C
jfor some j ∈ {1, . . . , n}.
Suppose C
i01, . . . , C
i0k⊆ C
j. We need to show that P
kl=1
µ(C
i0l) ≤ µ(C
j).
To simplify the notation, assume that C
0, C
1, . . . , C
nare connected, closed, disjoint and contained in C ∈ C
c. Let X \C
0= V = S
i∈I
V
i. Since all the C
jare connected, each C
jfor j ≥ 1 is contained in some V
i, and only finitely many V
iwill 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
kare disjoint and obviously also disjoint from all the C
j. We want to show that
(4.2)
X
n j=0µ(C
j) ≤ µ(C),
which is equivalent to 1 − X
i∈I
µ(V
i) + X
n j=1µ(C
j) ≤ 1 − X
k∈K
µ(U
k), or
(4.3)
X
n j=1µ(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.
Lemma A. Let C
1, . . . , C
n∈ C
cand U
1, . . . , U
k, . . . ∈ O
sbe mutually disjoint subsets of V ∈ O
s. Then
(4.5)
X
n j=1µ(C
j) + X
∞ k=1µ(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
0and W
0is 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
1is the solid hull of C
1with respect to V . We want to show that
µ(C
1) + X
∞ k=1µ(U
k) ≤ µ(V ), or
1 − X
i∈I0
µ(W
i) − µ(W
0) + X
∞ k=1µ(U
k) ≤ µ(V ), i.e.
(4.6) µ( e C
1) + X
∞ k=1µ(U
k) ≤ µ(V ) + X
i∈I0
µ(W
i).
We have U
k⊆ W for all k = 1, 2, . . . (since all U
kare 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
0then U
kis 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).
Lemma B. Let C
1, . . . , C
m∈ C
sand U
1, . . . , U
k, . . . ∈ O
sbe mutually disjoint subsets of V ∈ O
s. Then
(4.8)
X
m j=1µ(C
j) + X
∞ k=1µ(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
ksuch that C
k0∈ C
sand µ(C
k0) >
µ(U
k) − ε/2
k+1by (B). Then X
∞k=1
µ(U
k) = X
N k=1µ(U
k) + X
k>N
µ(U
k)
<
X
N k=1µ(C
k0) + ε X
N k=11 2
k+1+ ε
2 <
X
N k=1µ(C
k0) + ε.
Let C = X \ V , so C ∈ C
sand C ∩ C
j= C ∩ C
k0= ∅ for all j and k. Then by (A) we get
X
m j=1µ(C
j) + X
N k=1µ(C
k0) + µ(C) ≤ 1.
Hence
X
m j=1µ(C
j) + X
∞ k=1µ(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
0jbe the exterior component of W
jwith respect to V . Let e C
j= X \ W
0jdenote the solid hull of C
jwith 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=1µ( e C
j) + X
∞ k=1µ(U
k) ≤ µ(V ) + X
n j=1X
i∈Ij0
µ(W
ij),
where I
j0= I
j\ {0}. Now observe that X \ S
nj=1
C e
j= T
nj=1
W
0j, so that if
we put
K
0= n
k ∈ Z
+: U
k⊆
\
n j=1W
0jo
,
K
j= {k ∈ Z
+: U
k⊆ e C
j}, j = 1, . . . , n,
then we exhaust all possibilities for the sets U
kin mutually exclusive cases.
By Lemma B we now get (4.10)
X
n j=1µ( e C
j) + X
k∈K0
µ(U
k) ≤ µ(V ).
If U
k⊆ e C
jthen U
k⊆ W
ijfor some i ∈ I
j0so 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
j6= ∅ for some pair j 6= l. By Lemma 3.4 either e C
j⊆ e C
lor 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
j0for l = 0, . . . , m
jand 1 ≤ j ≤ m, where the inclusions are proper if l ≥ 1.
3. m + P
mj=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=1mj
X
l=0
µ( e C
jl) + X
∞ k=1µ(U
k) ≤ µ(V ) + X
m j=1mj
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
nbe defined as in Case 1. Then again by Lemma B we immediately get
(4.13)
X
m j=1µ( e C
j0) + X
k∈K0
µ(U
k) ≤ µ(V ),
so to establish (4.12) it remains to show that (4.14)
X
m j=1mj
X
l=1
µ( e C
jl) + X
m j=1X
k∈Kj
µ(U
k) ≤ X
m j=1X
mj l=0X
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
mjl=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=1µ(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
ij0for some i ∈ I
j00if 1 ≤ l ≤ m
j, and U
k⊆ W
ij0for some i ∈ I
j00if 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
jiis ≤ 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)
0in Proposition 4.1. As for (Q2)
0, if K
1, K
2∈ C
0and K
1∩ K
2= ∅ let K
1= C
1∪ . . . ∪ C
nand K
2= C
10∪. . .∪C
m0be their decompositions into connected components. Then K
1∪K
2= C
1∪. . .∪C
n∪C
10∪. . .∪C
m0and the right-hand side represents the decomposition of K
1∪ K
2into connected components. From the definition of µ on C
0it now immediately follows that µ(K
1∪ K
2) = µ(K
1) + µ(K
2), and (Q2)
0follows.
It remains to verify the regularity property (Q3)
0.
Let K ∈ C
0be arbitrary. Then K = C
0∪ C
1∪ . . . ∪ C
n, where C
j∈ C
cand C
j∩ C
k= ∅ if j 6= k. Let X \ C
0= V = S
i∈I
V
i, where the V
iare 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
{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
0for i = 1, . . . , N such that
(4.18)
C
i0⊆ V
iand 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
Ni=1
C
i0, so C
0∈ C
0and C
0∩ K = ∅. Moreover (by (Q2)
0), µ(C
0) =
X
N i=1µ(C
i0) >
X
N i=1µ(V
i) − X
N i=1X
j∈Ji
µ(C
j) − X
N i=1ε 2
i+1>
X
∞ i=1µ(V
i) − ε 2 −
X
n j=1µ(C
j) − ε 2
= 1 − µ(C
0) − X
n j=1µ(C
j) − ε = 1 − µ(K) − ε,
which establishes (Q3)
0under the assumption that (4.18) holds. So what we need is the following.
Lemma C. Let C
1, . . . , C
n∈ C
cbe mutually disjoint and contained in V ∈ O
s. For each ε > 0 there is a C
0∈ C
0such that C
0∩ C
j= ∅ for all j, C
0⊆ V and
X
n j=1µ(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
nbelong 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
nin O
ssuch that C
j⊆ V
jand
µ(V
j) < µ(C
j) + ε/n, j = 1, . . . , n.
Next, choose W
j∈ O
ssuch that
C
j⊆ W
j⊆ W
j⊆ V
j, j = 1, . . . , n
(Lemma 3.3). Let F
jbe the solid hull of W
jwith 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
nj=1
F
jis connected. Then C
0= U is connected and con- tained in X \ S
nj=1
W
j, so C
0∩ C
j= ∅ for j = 1, . . . , n. On the other hand,
X \ C
0⊆ S
nj=1
V
j. If X \ C
0= S
i∈I
O
iis the decomposition of X \ C
0into its disjoint connected components, then each O
ibelongs to O
sand each O
iis contained in some V
j. But then, by Proposition 2.3.3, it follows that P
i∈I
µ(O
i) ≤ P
nj=1
µ(V
j). Consequently, we get µ(C
0) = 1 − X
i∈I
µ(O
i) ≥ 1 − X
n j=1µ(V
j) > 1 − X
n j=1(C
j) − ε, which yields the assertion in this case.
(ii) X \ S
nj=1
F
jis 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
1be 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
idisjoint, 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
isuch that K
i∈ C
sand
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=1µ(F
j) + X
i∈I
µ(U
i) = 1, and hence
X
n j=1µ(C
j) + X
N i=1µ(K
i) >
X
n j=1µ(F
j) − ε + X
N i=1µ(U
i) − ε
>
X
n j=1F
j+ X
i∈I
µ(U
i) − 5ε
2 = 1 − 5ε 2 , which proves the assertion when we take C
0= S
Ni=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
j6∈ F
1then F
j⊆ U
ifor 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
0such that F
k∩K
i= ∅ for all F
k∈ F
i, K
i⊆ U
iand X
Fk∈Fi
µ(F
k) + µ(K
i) + µ(X \ U
i) > 1 − ε
N ,
so X
Fk∈Fi
µ(F
k) + µ(K
i) > µ(U
i) − ε
N , i = 1, . . . , N.
Then we get X
n j=1µ(C
j) + X
N i=1µ(K
i) >
X
n j=1µ(F
j) − ε + X
N i=1µ(K
i)
≥ X
m j=1µ(F
j) + X
N i=1X
Fk∈Fi
µ(F
k) + X
N i=1µ(K
i) − ε
>
X
m j=1µ(F
j) + X
N i=1µ(U
i) − 2ε
>
X
m j=1µ(F
j) + X
i∈I
µ(U
i) − 5ε
2 = 1 − 5ε 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
nnow yields Lemma C when all the sets C
1, . . . , C
nare 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
sand C ∈ C
c. Let X \ C = S
i∈I
W
iwith W
i∈ O
s. Let W
0be the exterior component with respect to V , and put e C = X \ W
0as usual. We have e C ⊆ V and by the first part of the proof, there is a K
0∈ C
0such 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
isuch that K
i∈ C
sand 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
Ni=0
K
i, so K ∈ C
0, K ∩ C = ∅ and K ⊆ V . Moreover, µ(K) =
X
N i=0µ(K
i) > µ(V ) − µ( e C) − ε 2 +
X
N i=1µ(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.
C a s e 1: e C
1, . . . , e C
nare mutually disjoint. By the first part of the proof there is K
0∈ C
0such that K
0⊆ V, K
0∩ e C
j= ∅ (j = 1, . . . , n) and µ(K
0) >
µ(V ) − P
nj=1
µ( e C
j) − ε/2.
Let X \ C
j= W
j= S
i∈Ij
W
ij, where W
0jis the exterior component of W
jwith 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
jpick K
ij⊆ W
ijsuch that K
ij∈ C
sand
µ(K
ij) > µ(W
ij) − ε 4n · 1
2
i, j = 1, . . . , n.
Since the e C
jare mutually disjoint, all the sets K
ijare mutually disjoint and also disjoint from K
0⊆ T
nj=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=1Nj
X
i=1
µ(K
ij)
> µ(V ) − X
n j=1µ( e C
j) − ε 2 +
X
n j=1Nj
X
i=1
µ(W
ij) − ε 4n · 1
2
i> µ(V ) − X
n j=1h
µ( e C
j) − X
i∈Ij0
µ(W
ij) i
− ε
= µ(V ) − X
n j=1µ(C
j) − ε, which proves the assertion in Case 1.
C a s e 2: e C
j∩ e C
k6= ∅ for some pair j 6= k. We proceed as in the proof of Lemma A, and re-index the sets C
1, . . . , C
nso that
1. e C
10, e C
20, . . . , e C
m0(m < n) are mutually disjoint.
2. e C
jl⊆ e C
j0for l = 0, . . . , m
jand 1 ≤ j ≤ m, where the inclusions are proper if l ≥ 1.
3. m + P
mj=1
m
j= n.
As before, let X \ C
jl= S
{W
ijl: i ∈ I
jl} = W
jl, and let W
0jldenote the exterior component of W
jlwith respect to V . Again, by the first part of the proof, there is K
0∈ C
0such that K
0⊆ V, K
0∩ e C
j0= ∅ (j = 1, . . . , m), and (4.19) µ(K
0) > µ(V ) −
X
m j=1µ( 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).
When l ≥ 1 each set C
jlis contained in some interior component W
ij0of 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
ij0such that K
ij∈ C
sand (4.20) µ(K
ij) > µ(W
ij0) − ε
2N
jn .
If J
ji6= ∅ there is, by the induction hypothesis, since m
j< n, a set K
ij⊆ W
ij0such that K
ij∈ C
0and
(4.21) µ(K
ij) > µ(W
ij0) − X
l∈Jji
µ(C
jl) − ε 2N
jn . Let K
j= S
Nji=1
K
ij∈ C
0for 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=1µ(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
mj=0
K
j∈ C
0. By addition of (4.19) and the equations (4.22) we get
µ(K) = X
m j=0µ(K
j)
> µ(V ) − X
m j=1µ(C
j0) − (n − m) ε n +
X
m j=1µ( e C
j0) − X
m j=1mj
X
l=0
µ(C
jl) − mε n
= µ(V ) − X
m j=1mj
X
l=0