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We study capacity-like properties of the AM–modulus, the M1–modulus, and the 1–plan content. In many respects, both the M1– and the AM–modulus behave quite differently from the Mp–modulus, p > 1. Contrary to p > 1 neither the M1– nor the AM–modulus define a Choquet capacity in M+(X). We show that the AM– modulus is continuous under (all) increasing families of measures in M+(X) only in trivial cases and, moreover, we provide a counterexample consisting of families of curves. In spite of this fact, the M1– and the AM–modulus have many capacity like properties which are studied in Section 4 together with the interplay of the M1– and the AM–modulus. We employ these properties to show the M1–modulus is not a Choquet capacity. On the other hand, the 1-plan content is continuous with respect to increasing families Choquet capacity, as shown in Section 5. The fact that continuity with respect to increasing families is a subtle property can be demonstrated on the Hausdorff content on a metric space Y , for which it holds. Already Federer’s proof under the assumption that the underlying space is boundedly compact is difficult, see [11, Thm. 2.10.22]. The general case is very deep and due to Davies [8, Thm. 8]. Another property required in the definition of Choquet capacity is continuity with respect to decreasing families of compact sets. In Proposition 6.6 we show that the AM–modulus, the M1–modulus, and the 1–plan content satisfy this property + on (M (X), τA). For the 1–plan content this completes the verification that it is a Choquet capacity. In fact, the continuity with respect to decreasing families of compact sets of these functions is obtained for all 1 ≤ p < ∞. For τb the range p> 1 follows already from [4], for other topologies it is new. In Section 6 we start comparison of AM and the 1-plan–content. It is shown that the M1– and the AM–modulus, as well as the 1–plan content, coincide on τA- compact families. This is a partial counterpart of the equality result in [4]. However, the passing from compact families to general families is not as straightforward as for p> 1. The reason consists in the above mentioned failure of continuity on increasing families. Nevertheless, in Section 7 a connection between the AM–modulus and a special content in locally compact Polish spaces X is obtained. (Note that local compactness is satisfied in most applications.) Namely, in Theorem 7.1 we show that, for an arbitrary family E⊂M+(X),
AM(E) = inf{Ct(G): G⊃E is Fσ in τ0}, where Ct is the 1–plan content. As the expression on the right hand side can be regarded as a natural outer-capacity-like function on families of measures created from plans, we achieve the goal that the modulus approach and plan approach meet on arbitrary families of measures also for p = 1.
2. Choquet capacity theory In this section we recall the basic notions of Choquet’s capacity theory [7] in- cluding the famous Choquet capacitability theorem. In next sections, we will study capacity-like set functions on sets of measures, + ∗ i.e. our choice of Y will M (X) embedded into the positive cone of Cb(X) , where X is a metric measure space. In this setting, Ambrosio, Di Marino and Savar´e[4] proved that the p-modulus is a Choquet capacity, see Theorem 4.10. We will demonstrate that the situation for p = 1 is not as simple and show that some of properties listed below fail although one could expect them to hold. The classical motivation for Choquet capacity theory comes from the potential theory where the fundamental example is that of Newtonian capacity. Let Y be a Hausdorff topological space. PLANS ON MEASURES AND AM-MODULUS 3
Definition 2.1. Let cap: 2Y → [0, ∞] be a set function. Let us list some properties that such functions may (or may not) have. (i) For each A, B ⊂ Y , A ⊂ B =⇒ cap(A) ≤ cap(B) (monotonicity), (ii) for each A1, A2, ···⊂ Y , ∞ A1 ⊂ A2 ⊂ · · · =⇒ cap Aj = lim cap(Aj ), j→∞ j=1 [ (iii) for each compact K1,K2, ···⊂ Y , ∞ K1 ⊃ K2 ⊃ · · · =⇒ cap Kj = lim cap(Kj ), j→∞ j=1 \ (iv) For each E ⊂ Y , cap(E) = inf(cap(G): G ⊃ E, G open) (outer regularity), (v) For each K ⊂ Y compact, cap(K) = inf(cap(G): G ⊃ K, G open), (vi) For each E ⊂ Y , cap(E) = sup(cap(K): K ⊂ E,K compact) (inner regularity). We say that cap is a Choquet capacity if it satisfies (i)–(iii) and Y is a Polish space. Remark 2.2. If the set function cap satisfies (i) and (v), it also satisfies (iii). Conversely, (i) and (iii) imply (v) if Y is metrizable and locally compact. Remark 2.3. Set functions that appear in applications are seldom both outer and inner regular. Classical capacities satisfy (i)–(v), but not (vi). For them, the outer regular capacity cap in consideration, has an “associated” inner regular “inner capacity” cap∗, which satisfies (i) and (vi) and coincide with cap at least on Borel sets. Thus, cap∗ satisfies the formula from (ii) if we restrict our attention to sets on which cap∗ = cap. Definition 2.4. Let cap: 2Y → [0, ∞] be a set function and E ⊂ Y . We say that E is cap-capacitable if cap(E) = sup(cap(K): K ⊂ E,K compact). Let us emphasize that we use this terminology even if cap is not a (Choquet) capacity. Theorem 2.5 (Choquet capacitability theorem [7]). Let cap be a Choquet capacity on a Polish space Y . Then every Souslin set E ⊂ X is cap-capacitable. Proof. See also e.g. [2, Appendix] or [16, Theorem 30.13]. Proposition 2.6 (cf. Example 30.B.1 in [16]). Let µ be a locally finite Borel measure on a Polish space Y and let µ∗ be the induced outer measure, this means µ∗(E) = inf{µ(A): A Borel , A ⊃ E}. Then µ∗ is a Choquet capacity. In particular, µ∗(A) = sup{µ(K): K ⊂ A compact} holds for any Souslin set A, so that µ is a Radon measure. 4VENDULA HONZLOVA´ EXNEROVA,´ ONDREJˇ F.K. KALENDA, JAN MALY,´ AND OLLI MARTIO
3. Topologies on measures In this section, X will be a fixed Polish space. First we introduce some terminology in a more general setting. If µ is a Borel measure on a Hausdorff completely regular space Y , we say that µ is a Radon measure if µ is finite on compact sets and inner regular with respect to compact sets. (Note that there is a discrepancy in literature what should be the meaning of “Radon measure”.) By M(Y ) we denote the vector space of all finite signed Borel measures on Y and M+(Y ) stands for the cone formed by positive measures from M(Y ). In the full generality, we distinguish M(Y ) (all finite signed Borel measures) and M(Y,t) (finite signed Radon measures, the letter t comes from tight which is another term used for the inner regularity). Since M(Y,t) is canonically embedded ∗ to the dual space Cb(Y ) (where Cb(Y ) is the space of all bounded continuous functions on Y endowed with the supremum norm), it may be considered with the ∗ inherited weak topology. This topology will be denoted by τb in the sequel. In our setting of Polish spaces, it follows from Proposition 2.6 that any locally finite Borel measure is Radon, in particular, M+(X)= M+(X,t). Conversely, any Radon measure on a metrizable space X is locally finite. (As- sume that µ is Radon and not locally finite. This means that there is some x ∈ X such that µ(U)= ∞ for each open set U containing x. Fix a metric d on X. Using the assumption and inner regularity we find a sequence (Kn)n of compact subsets 1 of X such that Kn ⊂ B(x, n )\({x}∪ k
(i) The topology τb is the weakest one in which µ 7→ µ(T ) is continuous and µ 7→ µ(G) is lower semi-continuous for each open set G ⊂ T . (ii) If S ⊂ T , then M+(S,t) is a topological subspace of M+(T,t). Proof. Assertion (i) follows from the equivalence (v)⇔(viii) in [24, Theorem 8.1]. Assertion (ii) is an easy consequence of (i) as remarked in [14, Lemma 1].
Now, we return to the generality of Polish spaces. + The space (M (X), τb) is Polish (see Proposition 3.6 below). However, in case X is locally compact there is another natural topology. In this case the space M(X) (equipped with the total variation norm) is, due to the Riesz theorem, isometric ∗ to the dual space C0(X) (where C0(X) is the space of continuous functions on X vanishing at infinity equipped with the supremum norm which coincides with the closure of continuous functions with compact support in Cb(X). The respective ∗ weak topology will be denoted by τ0. By Banach-Alaoglu theorem M(X) (and + hence also M (X)) is σ-compact in τ0. If X is even compact, then clearly τb = τ0. However, if X is not compact, + these two topologies are different. In particular, (M (X), τ0) is not metrizable + and (M (X), τb) is not σ-compact (see Proposition 3.6 below). In order to cover these two interesting cases and possibly some other cases as well we adopt an abstract approach. Let us call a closed subspace A(X) of Cb(X) acceptable if it satisfies the following two properties.
(A1) A(X) is a sublattice of Cb(X). PLANS ON MEASURES AND AM-MODULUS 5
(A2) For any closed subset F ⊂ X and x ∈ X \ F there exists g ∈ A(X) such that 0 ≤ g ≤ 1, g(x) = 1 and g|F = 0. In the following lemma we collect several equivalent formulations of the property (A2). The property (A4) will the the key one.
Lemma 3.2. Let A(X) be a closed linear sublattice of Cb(X). Then the following assertions are equivalent. (A2) For any closed subset F ⊂ X and x ∈ X \ F there exists g ∈ A(X) such that g(x)=1 and g|F =0. + (A3) For any f ∈ Cb(X) we have f(x) = sup{g(x): g ∈ A(X), 0 ≤ g ≤ f}, x ∈ X. + + (A4) For any f ∈ Cb(X) there exists a sequence (gj )j of functions from A(X) such that gj ր f. (A5) The family {x ∈ X : g(x) > 0}, g ∈ A(X) form a base of the topology of X. Proof. The equivalence (A2)⇔(A5) is obvious. + (A2)⇒(A3): Fix f ∈ Cb(X) . It is enough to prove that, given x ∈ X and c> 0 such that f(x) >c, there is g ∈ A(X)+ with 0 ≤ g ≤ f and g(x) ≥ c. To this end set F = {y ∈ X : f(y) ≤ c}. Then F is closed and x∈ / F . Hence there is h ∈ A(X) with 0 ≤ h ≤ 1, h|F = 0 and h(x) = 1. The choice g = ch does the job. + + (A3)⇒(A4) Fix f ∈ Cb(X) and set M = {g ∈ A(X) ; g ≤ f}. It follows that f = sup M. For g ∈ M and n ∈ N let 1 U(g,n)= {x ∈ X : g(x) >f(x) − }. n For each n ∈ N the family {U(g,n): g ∈ M} is an open cover of X. It follows that there is a countable subcover, i.e., a countable set Mn ⊂ M such that {U(g,n): g ∈ Mn} covers X. It follows that f = sup Mn. n N [ N Now enumerate n Mn = {hm : m ∈ } and set gn = max{h1,...,hn} for n ∈ . (A4)⇒(A2) Let F ⊂ X be a closed subset and x ∈ X \ F . Then there is S f ∈ Cb(X) such that 0 ≤ f ≤ 1, f(x) = 1, f|F = 0. Let (gn) be a sequence in + A(X) with gn ր f. Then gn|F = 0 for each n and gn(x) ր 1. Hence, there is some n with gn(x) > 0. A suitable multiple of gn does the job.
Lemma 3.3. Let A(X) be an acceptable subset of Cb(X). Then the following assertions are true. (i) There exists h ∈ A(X) such that h> 0 on X. + + (ii) If u: X → R is lsc, then there is a sequence (gk)k of functions from A(X) such that gk ր u. + Proof. (i) By (A4) there is a sequence (hj ) in A(X) such that hj ր 1. Set −j h = j 2 hj . (ii) Since A(X) is a lattice, it is enough to find a countable subset M ⊂ A(X)+ P with sup M = u. But this is easy using (A4) and the fact that there is a sequence + (fk) in Cb(X) with fk ր u (see e.g. [17, Proposition 3.4]). Let us collect some examples of acceptable subspaces: 6VENDULA HONZLOVA´ EXNEROVA,´ ONDREJˇ F.K. KALENDA, JAN MALY,´ AND OLLI MARTIO
• Cb(X) itself is acceptable, i.e., we may have A(X)= Cb(X). • If X is locally compact, then A(X)= C0(X) is acceptable. • If γX is a compactification of X, A(X) can consist of the restrictions to X of functions from C(γX). In particular, if X is locally compact we may consider its one-point compactification αX. • If X is embedded into a Polish space Y as an open subset of Y , A(X) can be {f ∈ Cb(X): lim f(x) = 0 for each z ∈ ∂X}. x→z • If the topopogy of X is induced by a metric d and x0 ∈ X is a fixed point, A(X) can be the set
{f ∈ Cb(X): lim f(x)=0}. d(x,x0)→∞ • If X = (0, 1), we may take
A(X)= {f|(0,1) : f ∈ C([0, 1]),f(1) = 2f(0)}.
Remark 3.4. It is well known and easy to see that any closed subalgebra of Cb(X) is also a sublattice. (This is proved in [10, Lemma 3.2.20] under the additional as- sumption that it contains constant functions. But this assumption is not necessary as the sequence of polynomials (wi) chosen in [10, Lemma 3.2.20] may be easily adapted to satisfy wi(0) = 0 for each i.) So, any subalgebra satisfying (A2) is an acceptable subspace.
Lemma 3.5. Let A(X) be an acceptable subspace of Cb(X). Then the following holds. (i) There is a (in general non-metrizable) compactification K of X and a closed linear sublattice A(K) of C(K) separating points of K such that
A(X)= {f|X : f ∈ A(K)}. (ii) If A(X) contains constant functions and K is as in (i), then A(K)= C(K). (iii) If A(X) is even a subalgebra of Cb(X) and does not contain constant func- tions and K is as in (i), then there is some z ∈ K \ X such that A(K)= {f ∈ C(K): f(z)=0}. (iv) Assume that A(X) does not contain constant functions but there is some h ∈ A(X) with infx h(x) > 0. Then f A (X)= : f ∈ A(X) h h is an acceptable subspace of Cb(X) containing constant functions. Proof. (i) This follows by the standard construction of compactifications, namely K is the closure in RA(X) of the canonical copy of X formed by evaluation functionals. (ii) This follows from the Stone-Weierstrass theorem. (iii) Observe that span(A(K) ∪{1}) is a closed subalgebra of C(K) separating points of K. Hence, by the Stone-Weierstrass theorem we get that span(A(K) ∪ {1})= C(K). It follows that A(K) is a hyperplane in C(K), i.e., it is the kernel of a functional ϕ ∈ C(K)∗. Since 1 ∈/ A(K), we may assume without loss of generality that ϕ(1) = 1. In this case, using the fact that A(K) is also a subalgebra, it is easy to check that ϕ is multiplicative. Hence ϕ is represented by a Dirac measure δz for some z ∈ K (cf. [21, Example 11.13(a)]) . It follows from (A2) that z ∈ K \ X. f (iv) Note that the mapping f 7→ h is an isomorphism of Cb(X) onto itself which is also a lattice homomorphism. It follows that Ah(X) is a closed linear sublattice of Cb(X). Moreover, the validity of condition (A2) is clearly preserved. PLANS ON MEASURES AND AM-MODULUS 7
If A(X) is an acceptable subspace of Cb(X), then, in particular, it separates points of X. Hence M(X) can be considered as a subspace of A(X)∗, so we can equip M(X) with the restricted weak* topology of A(X)∗, which will be denoted by τA. The above-defined topologies τb and τ0 are special cases. We restrict the + topologies τA to the positive cone M (X) of M(X). The duality pairing of a measure µ ∈ M(X) and a function g ∈ A(X) is denoted by hµ,gi. The following proposition summarizes basic properties of these topologies.
Proposition 3.6. Let X be a Polish space and A(X) be an acceptable subspace of Cb(X). Let P (X) be the set of probability measures on X. Then the following assertions hold. + (a) (M (X), τb) is a Polish space. It is σ-compact if and only if X is compact. + (b) If A(X) contains constant functions, then τA = τb on M (X). + (c) If A(X) contains a function h with inf h(X) > 0, then τA = τb on M (X). (d) τA = τb on P (X). + + (e) (M (X), τA) is metrizable if and only if there is a function h ∈ A(X) with inf h(X) > 0. + (f) Any τA-compact subset of M (X) is metrizable. + (g) If X is locally compact, then (M (X), τ0) is σ-compact. Moreover, {µ ∈ + M (X): µ(X) ≤ c} is metrizable in τ0 for each c> 0. + The whole space M (X) is metrizable in τ0 if and only if X is compact. + (h) Any τb-open subset of (M (X), τb) is τA-Fσ. Hence, τb-Borel sets and + τA-Borel sets in M (X) coincide.
Proof. (a) The first statement is well known (see, e.g., [23, Appendix, Theorem 7] or [14, Corollary (a)]). If X is compact, then (M+(X), τb) is σ-compact by Banach-Alaoglu theorem. If X is not compact, then there is a sequence (xn) in X with no cluster point. + Then F = {xn : n ∈ N} is a closed subset of X, thus M (F ) is a closed subset of M+(X) (this follows easily from Lemma 3.1(i)). By Lemma 3.1(ii) the subspace topology on M+(F ) coincide with its τb-topology. Since F is a countable discrete + + set, (M (F ), τb) is homeomorphic to the positive cone ℓ1 of the Banach space ℓ1 N ∗ equipped with the weak topology (note that Cb( ) = ℓ∞ = ℓ1). This cone is not + + σ-compact as, otherwise ℓ1 = ℓ1 − ℓ1 would be also weakly σ-compact and hence ℓ1 would be reflexive, which is not the case. (b) Assume A(X) contains constant functions. It follows from Lemma 3.5(ii) + that there is a compactification K of X such that on M (X) the topology τA coincides with the weak∗ topology inherited from C(K)∗. But this one coincides with τb by Lemma 3.1. (c) It easily follows from Lemma 3.3(ii) that µ 7→ µ(U) is τA-lower semicontinu- ous for each open set U ⊂ X. Further, by the assumption there is h ∈ A(X) with + h ≥ 1. It follows from (A4) that there is a sequence (gn) in A(X) with gn ր h−1. Hence
h ≥ h − gn ց 1. By the monotone convergence theorem we get
hµ,h − gni ց hµ, 1i = µ(X).
Thus the function µ 7→ µ(X) is τA-upper semi-continuous. Since X is also an open set, we deduce that µ 7→ µ(X) is τA-continuous. + Hence, using Lemma 3.1(i) we see that τA is finer than τb on M (X). Since it + is clearly simultaneously weaker that τb, we conclude that τA = τb on M (X). 8VENDULA HONZLOVA´ EXNEROVA,´ ONDREJˇ F.K. KALENDA, JAN MALY,´ AND OLLI MARTIO
(d) It is clear that τA is weaker than τb. Conversely, since µ(X) = 1 for each µ ∈ P (X), it follows from Lemma 3.1(i) that the sets
{µ ∈ P (X): µ(U) >c}, c ∈ R,U ⊂ X open, form a subbase of the topology τb restricted to P (X). But these sets are also τA- open as, in the same way as in the proof of (c), we deduce from Lemma 3.3(ii) that µ 7→ µ(U) is τA-lower semicontinuous for each open set U ⊂ X. This completes the proof. (e) The ‘if part’ follows from assertions (c) and (e). The ‘only if part’ will be proved by contradiction. To this end assume that inf f(X) = 0 for each f ∈ A(X)+ + but (M (X), τA) is metrizable. Then there exist a countable base of neighborhoods of zero. Hence we may find functions 0 ≤ f1 ≤ f2 ≤ ... in A(X) such that
+ Vk := {µ ∈M (X): µ(fk) < 1}, k ∈ N form a base of neighborhoods of the origin. −k Fix k ∈ N. Since inf fk(X) = 0, we find xk ∈ X with fk(xk) < 2 . Applying −k −k condition (A3) to the constant function 2 we find gk ∈ A(X) with 0 ≤ gk ≤ 2 with gk(xk) >fk(xk). Set ∞ g = gk. Xk=1 Then g ∈ A(X)+ and thus
V := {µ ∈M+(X): hµ,gi < 1} is a τA-neighborhood of 0. Given k ∈ N, we may choose ak ∈ R such that fk(xk) < ak < g(xk). Then N δxk /ak ∈ Vk \ V . Hence Vk \ V 6= ∅ for each k ∈ , a contradiction. + (f) Since (M (X), τb) is a Polish space, its topology has a countable base. Since τA is weaker than τb, the base of τb is a network for τA (cf. [10, p. 127]). It is well known that any compact space with countable network has even a countable base (see [10, Theorem 3.1.19]) an hence it is metrizable by [10, Theorem 4.2.8]. (g) The compactness of {µ ∈M+(X): µ(X) ≤ c} follows from Banach-Alaoglu theorem. This set is metrizable by assertion (f) and σ-compactness of the whole space is obvious. + If X is compact, then τ0 = τb, hence (M (X), τ0) is metrizable by assertion (a). If X is not compact, the non-metrizability follows from assertion (e). + (h) A base of (M (X), τb) is formed by sets of type
+ + + Uµ0,g := {µ ∈ M (X): |hµ − µ0, gi| < 1}, µ0 ∈ M (X), g ∈ Cb(X) .
+ Fix µ0 and g as above and use property (A4) to find a sequence gj ∈ A(X) such that gj ր g. Then
∞ 1 U = µ: |hµ − µ , g i| ≤ 1 − , µ0,g 0 j k k[=1 j\≥k n o + so Uµ0,g is τA-Fσ. Since (M (X), τb) is a Polish space, any open set is a countable union of basic open sets, hence it is also τA-Fσ. It follows that any τb-Borel set is τA-Borel. The converse implication follows from the fact that the topology τA is weaker than τb. PLANS ON MEASURES AND AM-MODULUS 9
4. Moduli of families of measures In this section X will be again a Polish space and A(X) a fixed acceptable subspace of Cb(X). In addition, we assume that X is equipped with a reference measure m which is a non-negative Radon measure (not necessarily finite). Definition 4.1. Let E be a subfamily of M+(X). We say that a lsc function ρ: X → [0, ∞] is an admissible function for E if
ρ dµ ≥ 1 for each µ ∈E. ZX Let p ∈ [1, ∞). We define the Lp-modulus of E and its continuous version as
p Mp(Γ) = inf kρkLp(m) : ρ is admissible for E ,
A n p o Mp,c(Γ) = inf kρkLp(m) : ρ ∈ A(X) is admissible for E . n o Definition 4.2 ([18]). Let E be a subfamily of M+(X). We say that a sequence of lsc functions (ρj )j , ρj : X → [0, ∞] is an admissible sequence for E if
lim inf ρj dµ ≥ 1 for each µ ∈E. j ZX Let p ∈ [1, ∞). We define the Lp-approximation modulus of E and its continuous version as p AMp(E) = inf lim inf kρjk p : (ρj )j is admissible for E , j L (m) A n p o AMp,c(E) = inf lim inf kρjk p : (ρj )j ⊂ A(X) is admissible for E . j L (m) n A A o If p = 1, we simplify AM1 to AM and AM1,c to AMc . Definition 4.3. Most frequently, the concept of modulus is used on families of paths. If we speak on paths, the topology on X is induced by a fixed metric, so that the length of a curve ϕ:[a,b] → X makes sense. A path ϕ:[a,b] → X is a non-constant curve of finite length. Then ϕ can be parametrized by arc length, the reparametrization it is of the form γ = ϕ ◦ h, where h: [0,ℓ] → [a,b] is a suitable increasing homeomorphism and ℓ is the total length of the path ϕ. The curvelinear integral of a function ρ: X → R over ϕ is then defined as ℓ ρ ds = ρ(γ(t)) dt Zϕ Z0 whenever the integral on the right makes sense. A path ϕ induces a finite Radon measure µϕ defined as
µϕ(E)= χE ds, E ⊂ X Borel. Zϕ It is the push-forward of the Lebesgue measure under the arc-lenght reparametriza- tion of ϕ. The modulus of a family of paths is then defined as the modulus of the family of induced measures. Note that one Radon measure can be induced by dif- ferent paths even if we insist on arc-length parametrization, as we do not require the paths being injective. By definition, the modulus treats only the resulting measures. Remark 4.4. Due to the motivation, we point out if an example of any phenom- ena can be demonstrated on paths. However, note that even when dealing with + paths we use the τA-topology on M (X), which differs from topologies used usu- ally on families of paths, see e.g. [22]. In a correspondence with [4] we prefer an 10VENDULA HONZLOVA´ EXNEROVA,´ ONDREJˇ F.K. KALENDA, JAN MALY,´ AND OLLI MARTIO approach based on the topology on measures. It would be interesting to develop a parallel research on paths and discuss the dependence of situation on the choice of topologies. Remark 4.5. The AM-modulus on paths is related to BV spaces, see [15]. If u is a precisely represented BV function on Rn, then u ◦ ϕ is BV on AM- almost every path ϕ. This is not true for the M1 modulus. Using the related notion of approximation upper gradient, which is a sequence gj of positive Borel functions on X such that
oscϕ u ≤ lim inf gj ds j Zϕ for AM-almost every path, Martio [18] introduced a version of BV space on a metric measure space X. Very recently, Durand-Cartagena, Eriksson-Bique, Korte and Shanmugalingam [9] proved that, under the assumptions that the reference measure m is doubling and (X, m) supports the 1-Poincar´einequality, the Martio BV space coincides with the BV space introduced by Miranda Jr. [20]
Remark 4.6. It is clear that AMp ≤ Mp. In [15], we show that AMp = Mp for 1 A The situation for AM moduli is different. By the following theorem AMp,c does not depend on A(X) and, moreover, is equal to AMp. A Theorem 4.8. Let p ≥ 1. Then AMp = AMp,c. + A Proof. Choose E ⊂M (X). Obviously AMp(E) ≤ AMp,c(E). For the converse inequality we may assume that AMp(E) < ∞. Let (ρj )j be an admissible sequence for E consisting of lower semicontinuous Lp functions. + By Lemma 3.3(ii) there is, for each j ∈ N, a sequence (uj,i) in A(X) such that uj,i ր ρj . By the dominated convergence theorem we get i→∞ kuj,i − ρj kp −→ 0. Choose ε> 0. Passing to a subsequence, we may assume that for i, j ∈ N we have −j−i kuj,i − ρj kp ≤ 2 ε, so that −j−i kuj,i+1 − uj,ikp ≤kρj − uj,ikp ≤ 2 ε, hence ∞ (1) kuj,i+1 − uj,ikp ≤ ε. i,j=1 X PLANS ON MEASURES AND AM-MODULUS 11 Denote uj = uj,1 and reorder uj,i+1 − uj,i (i, j = 1, 2,... ) into a single sequence (wj )j . Set gj = uj + w1 + ··· + wj . We obtain (2) kgjkp ≤kujkp + ε ≤kρj kp + ε from (1) by the triangle inequality. Next we show that the sequence (gj ) is admissible for E. Choose µ ∈E and set t = lim inf uj dµ. j→∞ ZX Choose δ > 0 and find k ∈ N such that (3) uj dµ>t − δ and ρj dµ > 1 − δ, j ≥ k. ZX ZX Find m ≥ k such that (4) um dµ gj dµ ≥ uj dµ + wi dµ ≥ (t − δ)+(1 − t − 2δ)=1 − 3δ. X X X Z Z Xi≤l Z We have shown that (gj ) is admissible for E and by (2), lim inf kgjkp ≤ lim inf kρj kp + ε. j j A Since gj ∈ A(X) for each j and ε> 0 is arbitrary, we deduce AMp,c(E) ≤ AMp(E). A The previous theorem says, in particular, that the modulus AMp,c does not A depend on the choice of A(X). For Mp,c the situation is different by Example 4.7, but we have at least the following result on compact families of measures. + Theorem 4.9. Let p ≥ 1 and K⊂M (X) be τA-compact. Then A Mp(K)= Mp,c(K). A Proof. Since Mp,c(K) ≥ Mp(K), we need to prove the opposite inequality and for that we may assume that Mp(K) < ∞. Let ε ∈ (0, 1/2). Choose an Mp-admissible function ρ for K such that p ρ dm hµ, ρj i ր hµ, ρi 12VENDULA HONZLOVA´ EXNEROVA,´ ONDREJˇ F.K. KALENDA, JAN MALY,´ AND OLLI MARTIO for every µ ∈ K. Next, let + Gj = {µ ∈M (X): hµ, ρj i > 1 − ε}. Since ρj ∈ A(X), Gj form a τA-open cover of K. By compactness of K and mono- tonicity of (Gj )j there is k ∈ N such that Gk ⊃ K. Then ρk/(1 − ε) is admissible A for Mp,c(K) and we obtain ρp M (K)+ ε M A (K) ≤ k dm < p . p,c (1 − ε)p (1 − ε)p ZX A Letting ε → 0 we conclude that Mp,c(K) ≤ Mp(K) Now, we will study the moduli from the point of view of properties listed in Section 2. Theorem 4.10 ([4], Theorem 5.1). If 1 Remark 4.11. The results of [4] concern the topology τb. Moreover, the notion of a Choquet capacity is defined and studied on Polish spaces, so it is natural to formulate this result for τb. However, even for nonmetrizable topologies it has a sense to study the properties from Definition 2.1. Observe that properties (i) and (ii) are independent of the choice of topology. Concerning (iii), it is shown for τb in [4], but for more general τA it holds as well, see Proposition 6.6 below. By Theorem 4.10, the Mp-modulus, p > 1, has the property (ii) of Definition 2.1, namely (6) lim Mp(Ei)= Mp Ei i→∞ i [ + if Ei ⊂M (X), E1 ⊂E2 ⊂ ... . The situation for p = 1 is more complicated. The property (ii) of Definition 2.1 A holds for neither of the moduli AM, M1 and M1,c, so that these moduli fail to be Choquet capacities. This is shown in Theorems 4.21, 4.17 and 4.16 below. To describe the situation in detail, we start with some positive results. + Lemma 4.12. If E1 ⊂E2 ⊂ ... is a sequence of subsets of M (X) and E = i Ei, then S AM(E) ≤ lim M1(Ei). i→∞ Proof. We may assume that lim M1(Ei) < ∞. Choose admissible functions ρi for i→∞ Ei such that −i ρi dm µ ∈Ek ⊂Ek+1 ⊂ ... for some k and thus lim inf ρi dµ ≥ 1. i→∞ ZX Now we obtain −i AM(E) ≤ lim inf ρi dm ≤ lim inf(M1(Ei)+2 ) = lim M1(Ei). i→∞ i→∞ i→∞ ZX PLANS ON MEASURES AND AM-MODULUS 13 + Corollary 4.13. If E1 ⊂ E2 ⊂ ... are subsets of M (X) and M1(Ei) = AM(Ei) for each i, then (7) lim AM(Ei)= AM Ei i→∞ i [ Proof. By Lemma 4.12, AM Ei ≤ lim M(Ei) = lim AM(Ei). i→∞ i→∞ i [ The reverse inequality follows from monotonicity. The following theorem provides an alternative characterization for the AM- modulus in terms of increasing path families and the M1–modulus. Theorem 4.14. If E⊂M+(X), then AM(E) = inf{ lim M1(Ei): E1 ⊂E2 ⊂ ... and Ei = E} i→∞ i (8) [ A = inf{ lim M1,c(Ei): E1 ⊂E2 ⊂ ... and Ei = E}. i→∞ i [ + A Proof. Fix E ⊂M (X). Since clearly M1 ≤ M1,c, by Lemma 4.12 it suffices to show that for each ε> 0 there is an increasing sequence (Ei) as in (8) and A lim M1,c(Ei) ≤ AM(E)+ ε. i→∞ Assume first that AM(E) < ∞. Fix δ ∈ (0, 1) to be specified later and choose an admissible sequence (ρj ) for E such that lim inf ρj dm ≤ AM(E)+ δ. j→∞ ZX In view of Theorem 4.8, we may assume that ρj in A(X). Let Ei = µ ∈E : ρj dµ ≥ 1 − δ . X j\≥in Z o Then Ei ⊂Ei+1 and E = i Ei. Since ρi/(1 − δ) is admissible for Ei, we obtain A S ρi AM(E)+ δ lim M1,c(Ei) ≤ lim inf dm ≤ < AM(E)+ ε i→∞ i→∞ 1 − δ 1 − δ ZX if δ is small enough. If AM(E)= ∞, then we can choose Ei = E for each i. Theorem 4.15. Let E⊂M+(X). Then the following assertions are equivalent: (i) M1(E)= AM(E), + (ii) M1(E) = lim M1(Ei) for each increasing sequence (Ei) of subsets of M (X) i→∞ with E = i Ei. Proof. (i) =⇒ (ii):S If AM(E)= M1(E) then (ii) follows from Lemma 4.12 as lim M1(Ei) ≤ M1(E). i→∞ (ii) =⇒ (i): Since AM(E) ≤ M1(E), we need only to prove the converse inequality and we may assume that AM(E) < ∞. Choose ε> 0 and use Lemma 4.14 to find an increasing sequence (Ei) of subsets of E with E = i Ei such that lim M1(Ej ) < AM(E)+ ε. i S Then by (ii) M1(E) < AM(E)+ ε 14VENDULA HONZLOVA´ EXNEROVA,´ ONDREJˇ F.K. KALENDA, JAN MALY,´ AND OLLI MARTIO and letting ε → 0 we obtain the desired inequality. We give two examples involving τb-compact families. Since τb is the finest from all τA-topologies, the families in consideration are τA-compact as well. Theorem 4.16. Let X be R2 with the Lebesgue measure. There exists an increasing R2 sequence Γk of τb-compact families of paths in such that, denoting Γ= k Γk, A A M1(Γ) = lim M1(Γk) = lim M1,c(Γk)=0 < ∞ = M1,c(Γ). S k k A Consequently, Γ is Fσ but not M1,c-capacitable (see Definition 2.4). Proof. For any k ∈ N, let Γk be the family of all paths ϕz(t)= tz, t ∈ [0, 1], where 2 z ∈ B(0, 1) \ B(0, 1/k). Then Γk are τb-compact. If ρ: R → R is continuous and zk = (1/k, 0), then inf ρ ds =0, k ϕ Z zk A so that there is no continuous admissible function for Γ and M1,c(Γ) = ∞. On the −1 other hand, for any ε> 0, ε|x| is an admissible function for Γ, so that M1(Γ) = 0. Let ω be a nonnegative continuous function on [0, ∞) with a support in [0, 1] such ∞ that 0 ω(t) dt = 1. Write ρ (x)= jω(j|x|). R j Then, for fixed k, j ≥ k, and z ∈ B(0, 1) \ B(0, 1/k), we have 1/j ρj ds ≥ jω(jr) dr =1. Zϕz Z0 Thus ρj , j ≥ k, are admissible functions for Γk and 1/j inf ρj (x) dx = inf 2πrjω(jr) dr =0. j≥k R2 j≥k Z Z0 A Therefore, M1,c(Γk) = 0, k =1, 2,... . The following theorem is based on a deeper study of an example from [18]. Theorem 4.17. There exists an increasing sequence Γk of τb-compact families of R paths in such that, denoting Γ= k Γk, AM(Γ) ≤ lim M1(Γk) ≤ 1, S k→∞ but M1(Γ) = ∞. Thus, AM(Γ) inf ρ(t) dt =0, r>0 Zϕr so that there is no admissible function for Γ and M1(Γ) = ∞. On the other 1 hand, if η ≥ 0 is a continuous function on [0, ∞) with support on [0, 2 ] such that ∞ 0 η(x) dx = 1, and k k R ρk(x)=2 η(2 x), then ρk is admissible for Γk, k =1, 2,... . It follows lim inf M1(Γk) ≤ 1. k→∞ PLANS ON MEASURES AND AM-MODULUS 15 By Lemma 4.12, AM(Γ) ≤ lim inf M1(Γk) ≤ 1. k→∞ + Remark 4.18. Let E1 ⊂ E2 ⊂ ... be a sequence of subsets of M (X) and E = A j Ej . Then, in general, there is no relation between M1(E) and limj M1,c(Ej ). Indeed, we have S A M1(E) < lim M1,c(Ej ) j A if Ej = E and M1(E) (a) Gm,1, m ∈ N are pairwise disjoint. (b) Gm,1 ⊃ Gm,2 ⊃ Gm,3 ⊃ ... for each m ∈ N. (c) m(Gm,i) > 0 for each m,i ∈ N. (d) limi→∞ m(Gm,i)=0 for each m ∈ N. (e) m( m Gm,1) < ∞. For each m ∈ N and each sequence s = (s ) of integers we consider the set S n n Hm,s = Gn,sn n[≥m + and denote the restriction of m to Hm,s by µm,s. Let E0 ⊂ M (X) contain all measures µm,s. Then there is an increasing sequence (Em)m≥1 of subsets of E0 that AM Em > lim AM(Em). m m [ Proof. Set 1 N gm,i = χGm,i , m,i ∈ . m(Gm,i) Then each gm,i is a lower semicontinuous function satisfying kgm,ik1 = 1. Set Em = µ ∈E0 : lim inf gn,i dµ ≥ 1 . i X n\≥m Z By the very definition the sequence (gm,i)i is admissible for Em, hence AM(Em) ≤ 1. Clearly E1 ⊂E2 ⊂E3 ⊂ ... . Set E = Em. We are going to show that AM(E) ≥ 2. m Assume that AM(E) < 2. Then thereS is a sequence (hk)k admissible for E such that lim inf khkk1 < 2. Since a subsequence of an admissible sequence is again admissible we may assume that there is some ε> 0 such that khkk1 < 2(1 − ε) for N N each k ∈ . For each m ∈ let µm be the restriction of m to n≥m Gn,1. Then µ = µ s for s = (1, 1, 1,... ), therefore µ ∈E . We have m m, m 0 S gn,i dµm = 1 for n ≥ m, i ∈ N, ZX 16VENDULA HONZLOVA´ EXNEROVA,´ ONDREJˇ F.K. KALENDA, JAN MALY,´ AND OLLI MARTIO thus µm ∈En ⊂E for n ≥ m. It follows that lim infk X hk dµm ≥ 1. Thus there is p ∈ N such that for each k ≥ p we have m m R 1 − ε< hk dµm = hk dm. X S Gn,1 Z Zn≥m It follows that for k ≥ pm we have hk dm ≤khkk1 − hk dm < 1 − ε. S Gn,1 S Gn,1 Zn Now, using (d) we find an increasing sequence q = (qm)m of positive integers such that qm ≥ pm and ε h dm < for k ≤ q . k 2i+1 m ZGm,qm Let ν be the the restriction of m to m Gm,qm . Then ν = µ1,q ∈ E0. Moreover, ν ∈E1 ⊂E as S gm,i dν =1, i ≥ qm. ZX Then for each k ∈ N we have hk dν = hk dm ZX ZSm Gm,qm ≤ hk dm + hk dm S Gm,qm S Gm,qm Z{m : qm Remark 4.20. It is clear that the families Em constructed in the proof of the above lemma are Borel subsets of E0. In the following theorem we show that the modulus AM satisfies property (ii) from Definition 2.1 is fulfilled only in trivial cases. We formulate the result as an equivalence to provide a complete picture, but of course, the ‘only if part’ is more important for the theory. Theorem 4.21. Let X be a Polish space equipped with a reference Radon measure m. Then the modulus AM satisfies property (ii) from Definition 2.1 in M+(X) if and only if m = cxδx, xX∈F where F is a closed discrete subset of X (finite or infinite) and (cx)x∈F are positive numbers bounded below by some c> 0. Proof. Let us start by the ‘if part’. Assume that m has the required form. First observe, that for each E⊂M+(X) we have AM(E)= AM({µ ∈E : µ(X \ F )=0}). Indeed, the inequality ≥ is obvious. To prove the converse one fix any sequence (hk) admissible for {µ ∈E : µ(X \F )=0} and observe that the sequence (hk +kχX\F ) is admissible for E and khk + kχX\F k1 = khkk1 for each k ∈ N. Hence the inequality ≤ follows as well. PLANS ON MEASURES AND AM-MODULUS 17 Hence, to complete the proof of the ‘if part’ we may assume that F = X. We give the proof in case X is infinite. The proof in case X is finite is analogous (and easier). Hence we may assume F = X = N. Then M(X) is canonically identified with the Banach space ℓ1 through the mapping κ : M(X) → ℓ1 defined by κ(µ)(k) = µ({k}). (We consider ℓ1 rather as a space of functions than a space of sequences as the comparison with L1(m) is then more accurate.) We will show + that AM(E)= M1(E) for any E⊂M (X). Since the inequality ≤ holds always, it is enough to prove the converse one. If AM(E)= ∞, the inequality is obvious. So, assume that AM(E) < ∞ and fix any ε> 0. Then there is an admissible sequence (ρn)n for E such that lim inf kρnk1 < AM(E)+ ε. Up to passing to a subsequence N we may assume that kρnk1 < AM(E)+ ε for each n ∈ . Since kfk1 ≥ ckfkℓ1 for 1 each f ∈ L (m), we deduce that the sequence (ρn) is bounded in ℓ1. Therefore, up to passing to a subsequence we may assume that the sequence (ρn)n converges ∗ to some ρ ∈ ℓ1 in the weak* topology of c0. In particular, ρn → ρ pointwise, hence 1 kρk1 ≤ AM(E)+ ε by the Fatou lemma. Further, κ(µ) ∈ ℓ ⊂ c0 for any µ ∈ E, thus ρ dµ = hρ, κ(µ)i = limhρn, κ(µ)i = lim ρn dµ ≥ 1, N n n N Z Z so that ρ is admissible for E. It follows that M1(E) ≤ AM(E)+ ε. Since ε > 0 is arbitrary, we get M1(E) ≤ AM(E), thus M1(E)= AM(E). Having proved M1 = AM, the validity of property (ii) of Definition 2.1 follows from Corollary 4.13. We continue by proving the ‘only if part’. Assume that m is not of the given form. Then there are two possibilities: Case 1: The support of m is not discrete. It follows that there is a one-to-one sequence (xn) in the support of m converging to some x ∈ X. Since m is finite N on the compact set {x}∪{xn : n ∈ }, necessarily n m({xn}) < ∞. Now, using outer regularity of m it is easy to find a disjoint sequence (Un) of open sets such N P that m(Un) > 0 for each n ∈ and n m(Un) < ∞. Case 2: The measure m has the above form with cx > 0 but infx cx = 0. Then P one can choose a sequence (xn) in the support with n cxn < ∞. Again, using outer regularity of m it is easy to find a disjoint sequence (Un) of open sets such N P that m(Un) > 0 for each n ∈ and n m(Un) < ∞. From the sequence (Un) constructed in both cases one may easily construct a P family (Gm,i) with properties (a)–(e) from Lemma 4.19. For example, if (pm)m is a sequence of distinct primes, we may set G = U j . m,i pm j[≥i This completes the proof. Lemma 4.19 may be also used to provide a counterexample for families of paths given in the following statement. Example 4.22. There is a compact metric space X equipped with a (doubling) reference Radon measure m such that the modulus AM fails property (ii) of Defi- nition 2.1 on families of paths. Proof. The space X will be constructed as a suitable subset of R2. Denote 2 −m Lm = {x = (x1, x2) ∈ R : x1 ∈ [0, 2 ], x2 = x1/m}, a collection of line segments emerging from 0, and set ∞ X = Lm. m=1 [ 18VENDULA HONZLOVA´ EXNEROVA,´ ONDREJˇ F.K. KALENDA, JAN MALY,´ AND OLLI MARTIO It is clear that X is a compact subset of R2. Let m be the double of the linear measure. (The linear measure itself would serve as well, but with its double we can use Lemma 4.19 directly.) Note that m(X) < ∞ and m is a doubling measure in X, see Remark 4.23. Further, for m,i ∈ N set −m−i+1 Gm,i = {x ∈ Lm : 0 < x1 < 2 }. These sets are open subsets of X and, moreover, they satisfy conditions (a)–(e) from Lemma 4.19. Moreover, each of the measures µm,s described in the lemma is provided by a path (we may take a path which runs twice over {x ∈ Ln : x1 ≤ 2−n−sn+1} for each n ≥ m). Thus we may conclude by using Lemma 4.19. Remark 4.23. It is easy to observe that the measure m from Example is doubling, which is interesting for experts in analysis on metric measure spaces. It is obvious that m(B(0, 2r)) ≤ 2mB(0, r) for any r> 0 and that m(B(x, 2r)) ≤ 2mB(x, r) if x ∈ X and r is sufficiently small. The function m(B(x,2r)) , r> 0, f(x, r)= m(B(x,r)) (2, r =0 is upper semicontinuous on the compact space X × [0, 2] and thus it attains a maximum. Remark 4.24. Let us remark that while failure of property (ii) from Definition 2.1 for families of measures is a rather common feature by Theorem 4.21, the counterex- ample for families of paths is rather special. The reason is the use of Lemma 4.19 where suitable restrictions of the reference measure play a key role. So, the example is designed in such a way that these restrictions are measures generated by paths. This approach is impossible in Rn. So, the question whether property (ii) from Definition 2.1 holds for families of paths in Rn remains open. + Theorem 4.25. AM is not outer regular on (M (X), τA). Proof. Since τb is finer than τA, it is enough to consider τb-openness. Let X = R and m be the Lebesgue measure. We identify any closed intervals I ⊂ R with the one-to one path with locus I (or with the restriction of the Lebesgue measure on I). Let Γ be the family of all paths [0,ℓ], ℓ ∈ (0, 1]. Then AM(Γ) = 1. Indeed, the inequality AM(Γ) ≥ 1 is obvious and irrelevant. For the converse we use the admissible sequence (ρk)k, where 1 k, x ∈ (0, k ), ρk(x)= (0 otherwise. + Choose an τb-open set G⊂M (R) such that Γ ⊂ G. Then there exists δ1 > 0 such that [δ1, 1] belongs to G. Let (gk)k be an admissible sequence for G. Then 1 lim inf gk dx ≥ 1, k Zδ1 but since [0,δ1] ∈ Γ, we have also δ1 lim inf gk dx ≥ 1. k Z0 Altogether, 1 lim inf gk dx ≥ 2, k Z0 which shows that AM(G) ≥ 2. PLANS ON MEASURES AND AM-MODULUS 19 This is enough to disprove the outer regularity. However, we can proceed further to get that in fact AM(G)= ∞. Indeed, by induction we find δj , j =1, 2,... , such that [δj+1,δj ] ∈ G for each j ∈ N. Remark 4.26. We can construct a similar example on paths in R2 equipped with the Lebesgue measure. For x,a,b ∈ R with a Γ= {γx,0,ℓ : x ∈ (0, 1),ℓ ∈ (0, 1]}. Again AM(Γ) ≤ 1 as witnessed by the admissible sequence (ρk) where ρk = kχ 1 . Let G ⊃ Γ be τb-open. (0,1)×(0, k ) For each m ∈ N let Em = {x ∈ (0, 1): γ 1 ∈ G}. x, j ,1 j\≥m Then each Em is a Gδ-subset of (0, 1). Moreover, since for each x ∈ (0, 1) we have γx,0,1 = τb- lim γx, 1 ,1 , n→∞ n we deduce that m Em = (0, 1). N 3 Hence there is m ∈ with |Em| > 4 . Then for each admissible sequence (gk)k for G we have S 1 lim inf gk(x, y) dx dy = lim inf gk(x, y) dy dx k k ZEm×[0,1] ZEm Z0 1 ≥ lim inf gk(x, y) dy dx k ZEm Z0 1/m 1 ≥ lim inf gk(x, y) dy + lim inf gk(x, y) dy dx k k ZEm Z0 Z1/m ! 3 ≥ 2|E |≥ . m 2 3 Hence AM(G) ≥ 2 > 1. A Remark 4.27. The moduli AMp, Mp and Mp,c nevertheless satisfy (v) and thus (iii) of Definition 2.1 for each p ≥ 1 (see Proposition 6.6 below). By Theorem 4.25, the modulus AM is not outer regular. However, a weaker version of outer regularity holds by the following lemma. Lemma 4.28. Let E⊂M+(X) be arbitrary. Then AM(E) = inf{AM(A): A⊃E is τA-Fσ} Proof. The inequality ≤ is obvious. Let us prove the converse one. Choose 0 Then Fk are τA-closed and E ⊂ A := Fk. [k Further, the sequence (ρk/t)k is admissible for A. Therefore 1 AM(A) ≤ (AM(E)+1 − t) t and letting t → 1 we obtain inf{AM(A): A⊃E is τA-Fσ}≤ AM(E), which completes the proof. 5. Plans and a content on M+(X) In this section X continues to be a Polish space equipped with a positive Radon measure m. A(X) is again a fixed acceptable subspace of Cb(X). The space M+(M+(X)) is defined as the family of all nonnegative finite Borel + measures on M (X). Here we do not need to distinguish between τb-Borel subset + and τA Borel subsets of M (X) as they are the same, see Proposition 3.6(h). + + + Moreover, all measures from M (M (X)) are Radon on (M (X), τA). Indeed, for τb this follows from Proposition 3.6(a) and Proposition 2.6. Since τA is a weaker topology, it has more compact sets, so the Radon property remains true for τA. Definition 5.1 ([5, 4]). A finite positive Borel measure η on M+(X) is called a plan. We denote η# : E 7→ µ(E) dη(µ), E ⊂ X Borel. + ZM (X) If η is a probability measure, η# has the interpretation as barycenter, cf. [4]. If η# is absolutely continuous with respect to m, we identify it with its density. q # Under this convention, we can associate the L (m) norm kη kq with the measure η#. Now, if p ∈ [1, ∞), q is the dual exponent to p, E is a subfamily of M+(X), we define the p-content of E as ∗ + + # # (9) Ctp(E) = sup η (E): η ∈M (M (X)), η ≪ m, kη kq ≤ 1 . Recall that η∗ is the outern measure induced by η, see Proposition 2.6, ando the sign ≪ denotes the relation of absolute continuity. We simplify Ct1 to Ct for p = 1. Remark 5.2. In [4], this content is defined for universally measurable sets E. Our novelty is that we use the outer measure, which allows to define the content for all sets. Surprisingly (and in contrast with properties of AM and M1), this set function is a Choquet capacity, see Theorem 6.7. The following theorem is a key step to this observation. + Theorem 5.3. Let Ej ⊂M (X), E1 ⊂E2 ⊂ ... , and E = j Ej. Then Ctp(E) = lim Ctp(Ej ). S j Proof. The inequality ≥ is obvious. To prove the converse one we may assume that Ctp(E) > 0. Choose 0 ≤ t < Ctp(E). Then we can find a measure π ∈ + + # # ∗ ∗ M (M (X)) such that π ≪ m, kπ kq ≤ 1 and π (E) > t. Since π is a ∗ Choquet capacity (Proposition 2.6), there is k ∈ N such that π (Ek) > t. Hence Ctp(Ej ) >t for all j ≥ k. PLANS ON MEASURES AND AM-MODULUS 21 6. Modulus and content The following fundamental theorem is due to Ambrosio, Di Marino and Savar´e. + Theorem 6.1 ([4]). Let p ∈ (1, ∞) and E⊂M (X) be a τb-Suslin set. Then 1/p Ctp(E)= Mp(E) . Since we are deeply interested in M1 and AM-moduli corresponding to p = 1, we were motivated to look what happens for p = 1. Proposition 6.2. Let E⊂M+(X) be a Borel set. Then Ct(E) ≤ AM(E). # # Proof. Let η be a plan with η ≪ m, kη k∞ ≤ 1 and (ρj )j be a sequence admissible for E. Then by the Fatou lemma, η(E) ≤ lim inf ρj dµ dη(µ) ≤ lim inf ρj dµ dη(µ) E j X j E X Z Z Z Z # # = lim inf ρj dη ≤kη k∞ lim inf ρj dm ≤ lim inf ρj dm. j j j ZX ZX ZX Passing to the supremum on the left and to the infimum on the right we obtain the desired inequality Ct(E) ≤ AM(E). + Theorem 6.3. Let K⊂M (X) be a τA-compact family of measures. Then A Ct(K)= AM(K)= M1(K)= M1,c(K). Proof. This can be shown as in the first step of [4, Theorem 5.1]. We simplify a bit the argument, on the other hand, some parts of the proof are longer in order to cover the axiomatic approach. A Since inequalities AM(K) ≤ M1(K) ≤ M1,c(K) are obvious and in view Propo- A sition 6.2, it remains to verify that M1,c(K) ≤ Ct(K). If 0 ∈ K then clearly Ct(K) = ∞, hence we may assume that 0 ∈/ K. Also we A may assume that ξ := M1,c(K) > 0, in particular, K is nonempty. Consider the bounded linear operator J : A(X) → C(K) defined as Jg(µ)= hµ,gi, µ ∈ K,g ∈ A(X). The dual operator J ∗ is η 7→ η#. Define U = {Jρ: ρ ∈ A(X), kρk1 ≤ ξ}, V = {u ∈ C(K): u> 1}. A Then U, V are convex and V is open. Further, U ∩V = ∅ by the definition of M1,c: if A ρ ∈ C0(X) and Jρ ≥ λ> 1, then ρ/λ is admissible and thus kρk1 ≥ λM1,c(K) > ξ. By the Hahn-Banach theorem there exist a ∈ R and η ∈M(K) such that η ≤ a on U and η >a on V . We claim that a> 0. Of course, a ≥ 0 as U contains 0. Further, there is an m-integrable strictly positive function h ∈ A(X). (This is a stronger version of Lemma 3.3(i).) Since X is Polish and m is locally finite, there is a strictly positive continuous m-integrable function h0. The existence of h may be now proved by applying the proof of Lemma 3.3(i) to function h0 in place of 1. We have Jh > 0 on K as K does not contain the null measure. By compactness of K, Jh attains a strictly positive minimum in K, and thus a positive multiple of Jh belongs to V . Therefore hη,Jhi > 0 and since another positive multiple of Jh belongs to U, we have a> 0. Now we can normalize to get a = 1, so that we may assume that η ≤ 1 on U and η > 1 on V . Let u ∈ A(K)+. Then for each ε> 0 we have 2 + u/ε ∈ V and thus u hη,u +2εi = ε η, 2+ > ε. ε D E 22VENDULA HONZLOVA´ EXNEROVA,´ ONDREJˇ F.K. KALENDA, JAN MALY,´ AND OLLI MARTIO Letting ε → 0 we obtain hη,ui≥ 0. By property (A4) and dominated convergence + + theorem we deduce that hη,fi ≥ 0 for each f ∈ Cb(X) , i.e., η ∈ M (X). We claim that η is a probability measure. Indeed, given ε> 0, we have 1+ ε ∈ V =⇒ hη, 1+ εi > 1. On the other hand, we find an admissible g ∈ A(X) such that kgk1 ≤ ξ(1 + ε). Then (1 + ε)−1Jg ∈ U and Jg ≥ 1 on K. Thus hη, 1i ≤ hη, Jgi≤ 1+ ε and we have verified that η is a probability measure. Next, the inequality η ≤ 1 on U implies that for any g ∈ A(X) with kgk1 ≤ 1 we have 1 1 hη#,gi = hJ ∗η,gi = hη,Jgi = = hη,J(ξg)i≤ . ξ ξ It follows that 1 (10) |hη#,gi| ≤ ·kgk , g ∈ A(X). ξ 1 From this inequality we first deduce that η# ≪ m. Let N ⊂ X be an m-null set. Then, given ε> 0 there is an open set G ⊃ X with m(G) <ε. By Lemma 3.3(ii) + there is a sequence (gn) in A(X) with gn ր χG. Then # # 1 1 ε η (G) = limhη ,gni≤ lim sup ·kgnk1 ≤ · m(G) < . n n ξ ξ ξ Thus η#(N) = 0. # # 1 Once we know that η ≪ m, (10) shows that kη k∞ ≤ ξ . Thus, A + M1,c(K)= ξ = ξη(M (X)) ≤ Ct(K). Remark 6.4. Note that the proof of AM(K) = M1(K) is a miracle. One would expect a construction of a single admissible function ρ from an admissible sequence ρj using a suitable covering of the compact set K. However, the sets that would be useful for such a proof are not open. So, instead of this we use a non-constructive proof using the Hahn-Banach theorem. Proposition 6.5. Let (Kk)k be an increasing sequence of τA-compact subsets of M +(X). Then (11) AM Kk = lim AM(Kk). k [k Proof. We obtain (11) from Corollary 4.13 and Theorem 6.3. A Proposition 6.6. The moduli AMp, Mp and Mp,c satisfy condition (v) and thus (iii) of Definition 2.1 for each p ≥ 1. + Proof. Let K⊂M (X) be a τA-compact set, ε > 0 and let ρ ∈ A(X) be an admissible function for K, then the set G = µ ∈M+(X): hµ,ρi > (1 + ε)−1 + A is τA-open in M (X) and (1+ ε)ρ is admissible for Mp,c(G ). Hence A A AMp(G) ≤ Mp(G) ≤ Mp,c(G) ≤ (1 + ε)Mp,c(K)=(1+ ε)Mp(K)=(1+ ε)AMp(K), where the inequalities are obvious, the first equality follows from Theorem 4.9 and the second one from Theorem 6.3 if p = 1 and from Remark 4.6 if p> 1. In view of Theorem 6.1, the following result is a counterpart of Theorem 4.10 for p = 1. PLANS ON MEASURES AND AM-MODULUS 23 + Theorem 6.7. Ct is a Choquet capacity on (M (X), τb). In fact, Ct satisfies + conditions (i)–(iii) from Definition 2.1 also on (M (X), τA) if A is an acceptable subspace of Cb(X). Proof. Property (i) is obvious, (ii) follows from Theorem 5.3 (and does not depend on the topology). It remains to verify (iii). In fact, we can even verify (v). + Let K⊂M (X) be a τA-compact set and c > Ct(K). By Theorem 6.3 we deduce that c>AM(K). Futher, by Proposition 6.6 there is a τA-open set G ⊃ K with AM(G) Remark 6.8. Since Ct is a Choquet capacity and AM not, we cannot expect the equality AM = Ct in general. It is even worse. In view of Theorems 5.3 and 4.21 we see that there may exist a Borel set E⊂M+(X) which fails to be AM-capacitable. Indeed, consider the sets Ej and E as in Lemma 4.19. Then E is a Borel set, but sup{AM(K): K⊂E τ0-compact} = sup{Ct(K): K⊂E τ0-compact} (12) = Ct(E) = lim Ct(Ej ) ≤ lim AM(Ej ) < AM(E). j j 7. Modulus and content on locally compact spaces In this section we assume that X is a locally compact Polish space and A(X)= C0(X). In this seting we obtain a better correspondence between AM-modulus and 1-content; we can express the AM-modulus in terms of the 1-content. The advantage of local compactness is that M(X) is the dual space to C0(X), see Section 3. It follows that, for each c > 0, the set {µ ∈ M+(X): µ(X) ≤ c} is bounded ∗ and weak* closed in C0(X) , hence it is weak* compact, or, in our notation, τ0 + compact. Consequently, M (X) is σ-compact in τ0 and thus each Fσ-subset of + M (X) is a countable union of τ0-compact sets. We obtain the following final result on modulus and content. Theorem 7.1. Let E⊂M+(X) be arbitrary. Then (13) Ct(E) ≤ AM(E) = inf{Ct(G): G⊃E is Fσ in τ0} Proof. By Lemma 4.28, we have AM(E) = inf{AM(G): G⊃E is Fσ}. + However, if G ⊂M (X) is Fσ in τ0, there exists an increasing sequence (Kj )j of τ0-compact sets such that G = j Kj . By Proposition 6.5, Theorem 6.3, and Theorem 5.3, S AM(G) = lim AM(Kj ) = lim Ct(Kj ) = Ct(G) j→∞ j→∞ for each τA-open set G. Hence Ct(E) ≤ inf{Ct(G): G⊃E is Fσ in τ0} = inf{AM(G): G⊃E is Fσ in τ0} = AM(E). Remark 7.2. The equality in (13) resembles usual definitions of outer capacities from “precapacities” on compact set. Unexpectedly, we consider infimum over Fσ sets and not over open sets. The reason is that AM is not outer regular, see Theorem 4.25. 24VENDULA HONZLOVA´ EXNEROVA,´ ONDREJˇ F.K. KALENDA, JAN MALY,´ AND OLLI MARTIO Remark 7.3. Let E be as in Lemma 4.19. Then by (12) and Theorem 7.1, Ct(E) < AM(E) = inf{Ct(G): G⊃E is Fσ in τ0}. So, Ct fails not only outer regularity, but also the weaker form of upper regularity which is satisfied by AM by Lemma 4.28. Thus, from the point of view of conti- nuity on increasing families, Ct is “better” than AM, but Ct has worse regularity properties. References [1] L. Ahlfors and A. Beurling. 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PLANS ON MEASURES AND AM-MODULUS 25 Department of Mathematical Analysis, Faculty of Mathematics and Physics, Charles University in Prague, Sokolovska´ 83, Prague 8, 18675 Czech Republic E-mail address: [email protected] Department of Mathematical Analysis, Faculty of Mathematics and Physics, Charles University in Prague, Sokolovska´ 83, Prague 8, 18675 Czech Republic E-mail address: [email protected] Department of Mathematical Analysis, Faculty of Mathematics and Physics, Charles University in Prague, Sokolovska´ 83, Prague 8, 18675 Czech Republic E-mail address: [email protected] Department of Mathematics and Statistics, FI-00014 University of Helsinki, Finland E-mail address: [email protected]