Carathéodory-type extension theorem with respect to prime end boundaries
aa r X i v : . [ m a t h . M G ] S e p Carath´eodory-type extension theorem with respect toprime end boundaries
Joshua Kline, Jeff Lindquist and Nageswari Shanmugalingam ∗ September 25, 2019
Abstract
We prove a Caratheodory-type extension of BQS homeomorphisms betweentwo domains in proper, locally path-connected metric spaces as homeomor-phisms between their prime end closures. We also give a Caratheodory-typeextension of geometric quasiconformal mappings between two such domains pro-vided the two domains are both Ahlfors Q -regular and support a Q -Poincareinequality when equipped with their respective Mazurkiewicz metrics. We alsoprovide examples to demonstrate the strengths and weaknesses of prime endclosures in this context. The celebrated theorem of Riemann states that every simply connected planar do-main that is not the entire complex plane is conformally equivalent to the unit disk(that is, there is a conformal mapping between the domain and the disk). However,there are plenty of simply connected planar domains with quite complicated bound-aries, and it is not always possible to extend these conformal mappings from suchdomains to their (topological) boundaries. The work of Carath´eodory [C] beau-tifully overcame this obstruction by the use of cross-cuts and the correspondingnotion of prime ends, and showed that every such conformal mapping extends toa homeomorphism between the (Carath´eodory) prime end boundary of the domainand the topological boundary (the unit circle) of the disk. This extension theoremis known in various literature as the Carath´eodory theorem for conformal mappingsand as the prime end theorem (see [C], [Ah1, Section 4.6] and [P, Chapter 2]).In the case of non-simply connected planar domains or domains in higher di-mensional Euclidean spaces (and Riemannian manifolds), the notion of prime endsof Carath´eodory is not as fruitful. Indeed, it is not clear what should play therole of cross-cuts, as not all Jordan arcs in the domain with both end points in theboundary of the domain will separate the domain into exactly two connected com-ponents. There are viable extensions of the Carath´eodory construction for certaintypes of Euclidean domains, called quasiconformally collared domains (see for ex-ample [N, AW]), but this construction is not optimal when the domain of interest is ∗ The third author was partially supported by grant DMS branched quasisymmetrc (BQS) homeomorphisms and geometric quasiconfor-mal homeomorphisms. In Theorem 4.4 we prove that every BQS homeomorphismbetween two bounded domains in locally path-connected metric spaces has a home-omorphic extension to the respective prime end closures. We also show that if thesedomains are finitely connected at their respective boundaries, then the extendedhomeomorphism is also BQS, see Proposition 4.12 and the subsequent remark. InTheorem 5.14 we also prove a Carath´eodory extension theorem for BQS mappingsbetween unbounded domains in locally path-connected metric spaces, where the no-tion of prime ends for unbounded domains is an extension of the one from [ABBS],first formulated in [E]. We point out here that local path-connectedness of the am-bient metric spaces can be weakened to local continuum-connectedness (that is, foreach point in the metric space and each neighborhood of that point, there is a smallerneighborhood of the point such that each pair of points in that smaller neighborhoodcan be connected by a continuum in that smaller neighborhood). However, as localpath-connectedness is a more familiar property, we will phrase our requirements interms of it.In contrast to BQS homeomorphisms, geometric quasiconformal mappings donot always have a homeomorphic extension to the prime end closure for the primeends as constructed in [ABBS]; for example, with Ω the planar unit disk and Ω ′ theharmonic comb given byΩ ′ := (0 , × (0 , \ [ n { n +1 } × [0 , ] ! , (1)we know that Ω and Ω ′ are conformally equivalent, but the prime end closure Ω P of Ω is the closed unit disk which is compact, while Ω ′ P is not even sequentiallycompact. However, we show that if both domains are Ahlfors Q -regular Loewner2omains for some Q > M and d ′ M respectively, then geometric quasiconformal mappings between themextend to homeomorphisms between their prime end closures. Should the twodomains be bounded, then the results in this paper also imply that the geomet-ric quasiconformal mappings extend as homeomorphisms between the respectiveMazurkiewicz boundaries, with the extended map serving as a quasisymmetric mapthanks to [HK, Theorem 4.9] together with [K, Theorem 3.3]. In Theorem 6.9 wegive a Carath´eodory-type extension for geometric quasiconformal maps.Our results are related to those of [A, AW] where extensions, to the prime endclosures, of certain classes of homeomorphisms between domains is demonstrated.In [A, Theorem 5] it is shown that a homeomorphism between two bounded domains,one of which is of bounded turning and the other of which is finitely connected atthe boundary, has a homeomorphic extension to the relevant prime end closures pro-vided that the homeomorphism does not map two connected sets that are a positivedistance apart to two sets that are zero distance apart (i.e., pinched). In general,it is not easy to verify that a given BQS or geometric quasiconformal map satisfiesthis latter non-pinching condition. Moreover, we do not assume that the domainsare finitely connected at the boundary. We therefore prove homeomorphic extensionresults by hand. The paper [AW] deals with domains in the first Heisenberg group H , and the prime ends considered there are not those considered here. The notionof prime ends considered in [AW] are effectively that of Carath´eodory, see [AW, Def-inition 3.3]. In particular, topologically three-dimensional analogs of the harmoniccomb, e.g. Ω ′ × (0 ,
1) with Ω ′ as in (1), will have no prime end considered in [AW]with impression containing any point in { } × { } × [0 , Background notions
In this section we give a definition of prime ends for bounded domains. An extensionof this definition of prime ends (following [E]) for unbounded domain is postponeduntil Section 5 where it is first needed. We follow [ABBS, Section 4] in constructingprime ends for bounded domains.We say that a metric space X is (metrically) doubling if there is a positive integer N such that whenever x ∈ X and r >
0, we can cover the ball B ( x, r ) by at most N number of balls of radius r/ X be a complete, doubling metric space. It is known that a complete dou-bling space is proper, that is, closed and bounded subsets of the space are compact.Let Ω ⊆ X be a domain, that is, Ω is an open, non-empty, connected subset of X . Definition 2.1 (Mazurkiewicz metric) . The Mazurkiewicz metric d M on Ω is givenby d M ( x, y ) := inf E diam( E )for x, y ∈ Ω, where the infimum is over all continua (compact connected sets) E ⊂ Ωthat contain x and y .Note that in general d M can take on the value of ∞ ; however, if X is locallypath-connected (and hence so is Ω), then d M is finite-valued and so is an actualmetric on Ω. Definition 2.2 (Acceptable Sets) . A bounded, connected subset E ⊆ Ω is an acceptable set if E is compact and E ∩ ∂ Ω = ∅ .The requirement that E be compact was not needed to be explicitly statedin [ABBS] because it was assumed there that the metric space X is complete anddoubling, and so closed and bounded subsets are compact; as Ω is bounded and E ⊂ Ω, it follows then that E is compact in the setting considered in [ABBS]. Inour more general setting, we make this an explicit requirement.Given two sets E, F ⊂ Ω, we know from Definition 2.1 thatdist M ( E, F ) = inf γ diam( γ ) , where the infimum is over all continua γ ⊂ Ω with γ ∩ E = ∅ 6 = γ ∩ F . Definition 2.3 (Chains) . A sequence { E i } of acceptable sets is a chain is(a) E i +1 ⊆ E i for all i ≥ M (Ω ∩ ∂E i +1 , Ω ∩ ∂E i ) > i ≥ T ∞ i =1 E i ⊆ ∂ Ω.The second condition above guarantees also that E i +1 is a subset of the interiorint( E k ) of E k . 4 efinition 2.4 (Divisibility of Chains, and resulting ends) . A chain { E i } divides achain { F j } (written { E i }|{ F j } ) if for each positive integer k there exists a positiveinteger i k with E i k ⊆ F k . Two chains { E i } and { F j } are equivalent (written { E i } ∼{ F j } ) if { E i }|{ F j } and { F j }|{ E i } . The equivalence class of a given chain { E i } underthis relation is called an end and is denoted [ E i ]. Notation 2.5.
We write ends using the Euler script font (e.g. E , F , G ). If { E i } isa chain representing the end E , we write { E i } ∈ E .Note that if { E k }|{ G k } and { F k } ∼ { E k } , then { F k }|{ G k } . Moreover, if { E k }|{ G k } and { F k } ∼ { G k } , then { E k }|{ F k } . Therefore the notion of divisibilityis inherited by ends from their constituent chains. Definition 2.6 (Divisibility of Ends) . An end E divides an end F (written E | F ) ifwhenever E i ∈ E and F j ∈ F , then { E i }|{ F j } . Definition 2.7 (Impressions) . The impression of a chain { E i } is T ∞ i =1 E i . If twochains are equivalent, then they have the same impression. Hence, if E is an end, wemay refer to the impression I ( E ) of that end. We may also write I [ E i ] if { E i } ∈ E . Definition 2.8 (Prime Ends and Prime End Boundary) . An end E is called a primeend if F = E whenever F | E . The collection of prime ends of Ω is called the prime endboundary of Ω and is denoted ∂ P Ω. The set Ω together with its prime end boundaryform the prime end closure of Ω, denoted Ω P = Ω ∪ ∂ P Ω.We next describe a topology on Ω P using a notion of convergence. Definition 2.9 (Sequential topology for the prime end boundary) . We say that asequence { x k } of points in Ω converges to a prime end E = [ E k ] if for each positiveinteger k there is a positive integer N k such that x j ∈ E k for all j ≥ N k . It is possiblefor a sequence of points in Ω to converge to more than one prime end, see [ABBS].A sequence of prime ends { E j } is said to converge to a prime end E if, with each(or, equivalently, with some) choice of { E k,j } ∈ E j and { E k } ∈ E , for each positiveinteger k there is a positive integer N k such that whenever j ≥ N k there is a positiveinteger m j,k such that E m j,k ,j ⊂ E k .It is shown in [ABBS] that the above notion of convergence yields a topology onΩ P which may or may not be Hausdorff but satisfies the T1 separation axiom.Recall that a domain Ω is of bounded turning if there is some λ ≥ x, y ∈ Ω we can find a continuum E x,y ⊂ Ω with x, y ∈ E x,y such thatdiam( E x,y ) ≤ λd ( x, y ). Lemma 2.10 (Bounded Turning Boundary) . Let X be a complete metric space.Let Ω ⊆ X be a bounded domain with λ -bounded turning. Then, Ω P is metrizable byan extension of the Mazurkiewicz metric d M and there is a biLipschitz identification Ω = Ω P .Proof. Let x ∈ ∂ Ω. Let x i ∈ Ω be such that d ( x i , x ) < − i . For each i , let γ i bea continuum with x i , x i +1 ∈ γ i and diam( γ i ) ≤ λd ( x i , x i +1 ), and let Γ j = S ∞ i = j γ i .5hen, Γ j is connected for each j and diam(Γ j ) ≤ λ − j +1 . Let E j be the connectedcomponent of B ( x, λ − j +2 ) ∩ Ω containing Γ j . Then x ∈ ∂ Ω ∩ E j , and so E j is anacceptable set.We claim [ E j ] is a prime end with I ( { E j } ) = { x } . We first show { E j } is a chain.Condition (a) is clear as Γ j +1 ⊆ Γ j for each j , and B ( x, λ − j +1 ) ⊆ B ( x, λ − j +2 )for each j . If y ∈ B ( x, λ − j +2 ) ∩ Ω ∩ ∂E j , then there is some ρ > B ( y, ρ ) ⊂ B ( x, λ − j +2 ) ∩ Ω, and there is a point w ∈ B ( y, ρ/ [3 λ ]) ∩ E j . Thuswhenever z ∈ B ( y, ρ/ [3 λ ]), by the bounded turning properety of Ω we know thatthere is a continuum in Ω containing both z and w , with diameter at most 2 ρ/ < ρ ,and so this continuum will lie inside B ( x, ρ ). It follows that B ( y, ρ/ [3 λ ]) ⊂ E j ,violating the assumption that y ∈ ∂E j . Therefore Ω ∩ ∂E j ⊂ ∂B ( x, λ − j +2 ). Hencedist M (Ω ∩ ∂E j , Ω ∩ ∂E j +1 ) ≥ dist(Ω ∩ ∂E j , Ω ∩ ∂E j +1 ) ≥ − j +1 > , that is, Condition (b) is valid. Condition (c) follows as { x } = T j E j . Hence, { E j } is a chain and therefore [ E j ] is an end. That it is a prime end follows from [ABBS]together with the fact that I [ E j ] is a singleton set.We show that if F is a prime end with I ( F ) = { x } , then E = F where E is the oneconstructed above. To prove this it suffices to show that F | E . Let E = [ E k ] as above,and choose { F k } ∈ F . Fix a positive integer k and let w j ∈ Ω ∩ B ( x, − (2+ k ) ) and z j ∈ E k ∩ B ( x, − (2+ k ) ). Then by the bounded turning property, there is a continuum K j containing both w j and z j such that diam( K j ) ≤ λd ( z j , w j ) < λ − k − . It followsthat K j ⊂ B ( x, λ − k +2 ) ∩ Ω, and so K j ⊂ E k . It follows that B ( x, − k − ) ∩ Ω ⊂ E k .On the other hand, T j F j = { x } , and as F is compact with F j +1 ⊂ F j for each j ,it follows that for each k there is some j k ∈ N such that F j k ⊂ B ( x, − k − ) ∩ Ω.It then follows that F j k ⊂ E k . Therefore we have that { F j }|{ E k } , that is, F | E asdesired.We now show that if F is a prime end, then I ( F ) consists of a single point. Let x ∈ I ( F ) and choose { F k } ∈ F . Then for each positive integer k we set τ k := [3 λ ] − min { − k − , dist M (Ω ∩ ∂F j , Ω ∩ ∂F j +1 ) : j = 1 , · · · , k } . Note that τ k +1 ≤ τ k with lim k τ k = 0. We can then find a point x k ∈ B ( x, τ k ) ∩ F k +1 since x ∈ ∂F j for each j ∈ N . Note that as F j +1 ⊂ F j for each j , we necessarily have x k +1 , x k ∈ F k +1 . By the bounded turning property of Ω we can find a continuum K k ⊂ Ω such that x k , x k +1 ∈ K k and diam( K k ) ≤ λd ( x k , x k +1 ) < λτ k . As2 λτ k < dist M (Ω ∩ ∂F k , Ω ∩ ∂F k +1 )and x k , x k +1 ∈ F k +1 , it follows that K k ⊂ F k for each positive integer k . A similarargument now also shows that B ( x, τ k ) ∩ Ω is contained in the connected component G k of B ( x, λτ k ) ∩ Ω containing K k , and hence B ( x, τ k ) ∩ Ω ⊂ G k ⊂ F k . It followsthat { G k } is a chain in Ω with { G k }|{ F k } , and as { F k } is a prime end, it followsthat [ G k ] = F . Thus we have I ( F ) = I ([ G k ]) = { x } .The above argument shows that ∂ P Ω consists solely of singleton prime ends, andhence by the results of [ABBS] we know that Ω P is compact and is metrizable byan extension of the Mazurkiewicz metric d M . Moreover, for each x ∈ ∂ Ω there is6xactly one prime end E x ∈ ∂ P Ω such that I ( E x ) = { x } . To complete the proof ofthe lemma, note that if x, y ∈ Ω, we can find a continuum E x,y ⊂ Ω with x, y ∈ E x,y such that diam( E x,y ) ≤ λd ( x, y ); therefore d ( x, y ) ≤ d M ( x, y ) ≤ λd ( x, y ). Since Ω P is the completion of Ω with respect to the metric d M , this biLipschitz correspondenceextends to Ω → Ω P as wished for.Throughout this paper, Ω and Ω ′ are domains in X . Recall that a set E ⊂ X is a continuum if it is connected and compact. Such a set E is nondegenerate if inaddition it has at least two points. Definition 2.11.
A domain Ω is said to be finitely connected at a point x ∈ ∂ Ω iffor every r > U , · · · U k of B ( x, r ) ∩ Ωwith x ∈ ∂U i for i = 1 , · · · , k ,2 there is some ρ > B ( x, ρ ) ∩ Ω ⊂ S kj =1 U j .It was shown in [ABBS] (see also [BBS1]) that Ω is finitely connected at everypoint of its boundary if and only if every prime end of Ω has a singleton impression(that is, its impression has only one point) and ∂ P Ω is compact.It is an open problem whether, given a bounded domain, every end of thatdomain is divisible by a prime end, see for example [ABBS, AS, ES]. Since thisproperty is of importance in the theory of Dirichlet problem for prime end boundaries(see for instance [AS, ES]), the following useful lemma is also of independent interest.
Lemma 2.12 (Divisibility) . Suppose that Ω is a bounded domain that is finitelyconnected at x ∈ ∂ Ω , and let [ E k ] be an end of Ω with x ∈ I [ E k ] . Then there is aprime end [ G k ] such that [ G k ] | [ E k ] .Proof. For each positive integer j let C j , · · · , C jk j be the connected components of B ( x, − j ) ∩ Ω that contain x in their boundary. For each choice of positive integers j and k , there is at least one choice of m ∈ { , · · · , k j } such that C jm ∩ E k is non-empty.Let V ′ be the collection of all C jm , j ∈ N and m ∈ { , · · · , k j } for which C jm intersects E k for infinitely many positive integers k (and hence, by the nested prop-erty of the chain { E k } , all positive integers k ), and let V := V ′ ∪ { Ω } . As Ω isfinitely connected at the boundary, it follows that for each j there is at least one m ∈ { , · · · , k j } such that C jm has this property. We set C := Ω. For non-negativeinteger j we say that C jm is a neighbor of C j +1 n if and only if C j +1 n ⊂ C jm . Thisneighbor relation creates a tree structure, with V as its vertex set, with each vertexof finite degree.Note that for each positive integer M , there is a path in this tree (starting fromthe root vertex C = Ω) with length at least M . Because each vertex has finitedegree, it follows that there is a path in this tree with infinite length. To see this weargue as follows. Let property (P) be the property, applicable to vertices C jm , thatthe sub-tree with the vertex as its root vertex has arbitrarily long paths starting atthat vertex. For each non-negative integer j the vertex C jm has finite degree, andso if C jm has property (P) then it has at least one descendant neighbor (also known7s a child in graph theory) C j +1 n with property (P). Since C = Ω has property (P)as pointed out above, we know that there is a choice of m ∈ { , · · · , k } such thatthe vertex C m also has property (P). From here we can find m ∈ { , · · · , k } suchthat C m also has property (P) and C m is a neighbor of C m . Inductively we canfind C jm j for each positive integer j to create this path.Denoting this path by Ω ∼ C m ∼ C m ∼ · · · , we see that the collection { C jm j } isa chain for Ω with impression { x } . As a chain with singleton impression, it belongsto a prime end [ G k ]. Recall that X is locally path-connected and Ω ⊂ X is anopen connected set. Therefore, for each positive integer i we see that there is some j i such that C jm j ⊂ E i when j ≥ j i , and so [ G k ] | [ E k ] (for if not, then for each j the open set C jm j contains a point from E i +1 and a point from Ω \ E i , and sodist M (Ω ∩ ∂E i , Ω ∩ ∂E i +1 ) ≤ − j for each j , violating the property of { E k } beinga chain).The following two lemmata regarding ends are useful to us. Lemma 2.13.
Let { E k } be a chain of Ω . Then there is a chain { F k } of Ω that isequivalent to { E k } such that for each k ∈ N the set F k is open.Proof. For each positive integer k we choose F k to be the connected component ofint( E k ) that contains E k +1 . It is clear that F k +1 ⊂ E k ⊂ F k (2)for each k ∈ N and that F k is connected and open. Moreover, as E k +1 ⊂ F k , itfollows that T k E k ⊂ T k F k ⊂ T k E k ⊂ ∂ Ω, and so we have that ∅ 6 = \ k F k . (3)We show now that ∂F k ⊂ ∂E k , from which it will follow thatdist M (Ω ∩ ∂F k , Ω ∩ ∂F k +1 ) > E k , E k +1 also satisfy analogous condition. To see (3), let x ∈ ∂F k . Then,for each ε >
0, the ball B ( x, ε ) intersects F k , and so also intersects E k . Hence x ∈ E k . If x ∂E k , then x ∈ int( E k ), and in this case there is a ball B ( x, τ ) ⊂ E k ;we argue that this is not possible. Indeed, by the local path-connectedness of X there must be a connected open set U ⊂ Ω such that x ∈ U ⊂ B ( x, τ ) ⊂ E k , andas x ∈ ∂F k , necessarily U intersects F k as well. Then F k ∪ U is connected, with F k ∪ U ⊂ E k ; this is not possible as F k is the largest open connected subset of E k .Therefore x int( E k ), and so x ∈ ∂E k .From the above we have that { F k } is a chain of Ω. By (2) we also know thatthis chain is equivalent to the original chain { E k } . Lemma 2.14. If { E k } is a chain of Ω , then its impression I { E k } is a compactconnected subset of X . roof. For each j ∈ N set K j := E j . Then K j is compact, non-empty, and connectedwith K j +1 ⊂ K j . Set K := T j K j . Suppose that K is not connected. Then there areopen sets V, W ⊂ X with K ⊂ V ∪ W , K ∩ V = ∅ , K ∩ W = ∅ , and K ∩ V ∩ W empty.We set K V := K ∩ V and K W := K ∩ W . Then K V = K \ W and K W = K \ V ,and so both K V and K W are compact sets. Moreover, K V ∩ K W = K ∩ V ∩ W = ∅ .Therefore dist( K V , K W ) > , and so there are open sets O V , O W ⊂ X such that K V ⊂ O V , K W ⊂ O W , and O V ∩ O W = ∅ . Set U := O v ∪ O W . As each K j is connected, it follows that wecannot have K j ⊂ U , for then we will have K j ⊂ O V ∪ O W with K j ∩ O V ∩ O W empty but K j ∩ O V ⊃ K ∩ O V = ∅ and K j ∩ O W ⊃ K ∩ O W = ∅ . It follows that foreach j ∈ N there is a point x j ∈ K j \ O . This forms a sequence in the compact set K , and so there is a subsequence x j n that converges to some point x ∈ K .For each k ∈ N we have that K k is compact and { x k } j ≥ k a sequence there; so x ∈ K k for each k ∈ N , that is, x ∈ K . However, K ⊂ O and each x j ∈ X \ O ; thereforewe must have x ∈ X \ O (recall that O is open), leading us to a contradiction.Therfore we conclude that it is not possible for K to be not connected. Definition 2.15 (BQS homeomorphisms) . A homeomorphism f : Ω → Ω ′ is said tobe a branched quasisymmetric homeomorphism (or BQS homeomorphism) if thereis a monotone increasing map η : (0 , ∞ ) → (0 , ∞ ) with lim t → + η ( t ) = 0 such thatwhenever E, F ⊂ Ω are nondegenerate continua with E ∩ F non-empty, we havediam( f ( E ))diam( f ( F )) ≤ η (cid:18) diam( E )diam( F ) (cid:19) . Definition 2.16 (Quasisymmetry) . A homeomorphism f : Ω → Ω ′ is said to be an η -quasisymmetry if η : [0 , ∞ ) → [0 , ∞ ) is a homeomorphism and for each triple ofdistinct points x, y, z ∈ Ω we have d Y ( f ( x ) , f ( y )) d Y ( f ( x ) , f ( z )) ≤ η (cid:18) d X ( x, y ) d X ( x, z ) (cid:19) . Quasisymmetries are necessarily BQS homeomorphisms, but not all BQS home-omorphisms are quasisymmetric as the mapping f in Example 5.5 shows. How-ever, if both Ω and Ω ′ are of bounded turning, then every BQS homeomorphismis necessarily quasisymmetric. BQS homeomorphisms between two locally path-connected metric spaces continue to be BQS homeomorphisms when we replace theoriginal metrics with their respective Mazurkiewicz metrics, see Proposition 4.12.When equipped with the respective Mazurkiewicz metrics, the class of BQS home-omorphisms coincides with the class of quasisymmetric maps. However, a domainequipped with its Mazurkiewicz metric could have a different prime end structurethan the prime end structure obtained with respect to the original metric, see Exam-ple 4.8 for instance. Hence, considering BQS maps as quasisymmetric maps betweentwo metric spaces equipped with a Mazurkiewicz metric would not give a completepicture of the geometry of the domain.There are many notions of quasiconformality in the metric setting, the abovetwo being examples of them. The following is a geometric version.9 efinition 2.17 (Geometric quasiconformality) . Let µ Y be an Ahlfors Q -regularmeasure on Y and µ X be a doubling measure on X . A homeomorphism f : Ω → Ω ′ is a geometrically quasiconformal mapping if there is a constant C ≥ C − Mod Q ( f Γ) ≤ Mod Q (Γ) ≤ C Mod Q ( f Γ) . Here, by Mod p (Γ) we mean the numberMod p (Γ) := inf ρ Z Ω ρ p dµ X with infimum over all non-negative Borel-measurable functions ρ on Ω for which R γ ρ ds ≥ γ ∈ Γ.From the work of Ahlfors and Beurling [Ah1, Ah2, AB] we know that conformalmappings between two planar domains are necessarily geometrically quasiconfor-mal; a nice treatment of quasiconformal mappings on Euclidean domains can befound in [V]. Hence every simply connected planar domain that is not the entirecomplex plane is geometrically quasiconformally equivalent to the unit disk. How-ever, as Example 4.6 shows, not every conformal map is a BQS homeomorphism;hence among homeomorphisms between two Euclidean domains, the class of BQShomeomorphisms is smaller than the class of geometric quasiconformal maps. Fur-thermore, from Example 4.10 we know that while quasisymmetric maps betweentwo Euclidean domains are BQS homeomorphisms, not all BQS homeomorphismsare quasisymmetric. Thus the narrowest classification of Euclidean domains is viaquasisymmetric maps, the next narrowest is via BQS homeomorphisms. Amonglocally Loewner metric measure spaces, geometric quasiconformal maps provide thewidest classification. Therefore, even in classifying simply connected planar do-mains, a deeper understanding of the geometry of the domains is gained by consid-ering the Carath´eodory prime end compactification in tandem with the prime endsconstructed in [ABBS, ES, E].
Not all BQS homeomorphisms are quasisymmetric, as shown by Example 4.10 fromthe previous section, and not all geometrically quasiconformal maps are BQS, asshown by Example 4.6. On the other hand, BQS homeomorphisms need not begeometrically quasiconformal, as the next example shows.
Example 3.1.
Let X = R n be equipped with the Euclidean metric and Lebesguemeasure, and Y = R n be equipped with the metric d Y given by d Y ( x, y ) = p k x − y k .Then X has a positive modulus worth of families of non-constant curves, while Y has none. Therefore the natural identification map f : X → Y is not geometricallyquasiconformal; however, it is not difficult to see that f is a BQS homeomorphismand a quasisymmetric map.In this section we will describe notions of Ahlfors regularity and Poincar´e inequal-ity that are needed in the final section of this paper where geometric quasiconformalmaps are studied. 10 emark 3.2. Note that Ω is a domain in X , and hence it inherits the metric d X from X . However, Ω also has other induced metrics as well, for example theMazurkiewicz metric d M as in Definition 2.1. We denote the balls in Ω with respectto the Mazurkiewicz metric d M , centered at z ∈ Ω and with radius ρ >
0, by B M ( z, ρ ). Since we assume that X is locally path connected, we know that thetopology generated by d X in Ω and the topology generated by d M are the same.To see this, observe that if U ⊂ Ω is open with respect to the metric d X , thenfor each x ∈ U there is some r > B ( x, r ) ⊂ U . Then as d X ≤ d M ,it follows that B M ( x, r ) ⊂ B ( x, r ) ⊂ U ; that is, U is open with respect to d M .Next, suppose that U ⊂ Ω is open with respect to d M , and let x ∈ U . Then thereis some ρ > B M ( x, ρ ) ⊂ U . We can also choose ρ small enough sothat B ( x, ρ ) ⊂ Ω. Then by the local path-connectedness of X , there is an openset W ⊂ B ( x, ρ ) with x ∈ W such that W is path-connected. So for each y ∈ W there is a path (and hence a continuum) K ⊂ W with x, y ∈ K . Note that thend M ( x, y ) ≤ diam( K ) ≤ ρ < ρ , that is, y ∈ B M ( x, ρ ) ⊂ U . Therefore W ⊂ U with x ∈ W and W open with respect to the metric d X . It follows that U is openwith respect to d X as well.From the above remark, given an interval I ⊂ R , a map γ : I → Ω is a pathwith respect to d X if and only if it is a path with respect to d M . Definition 3.3 (Upper gradients) . Let Z be a metric space and d Ω a metric on Ω.Given a function f : Ω → Z , we say that a non-negative Borel function g is an uppergradient of f in Ω with respect to the metric d Ω if whenever γ is a non-constantcompact rectifiable curve in Ω, we have d Z ( f ( x ) , f ( y )) ≤ Z γ g ds, where x, y denote the end points of γ and the integral on the right-hand side is takenwith respect to the arc-length on γ . Lemma 3.4. If γ is a curve in Ω , then γ is rectifiable with respect to d M if andonly if it is rectifiable with respect to d X ; moreover, ℓ d X ( γ ) = ℓ d M ( γ ) . Furthermore,if g is a non-negative Borel measurable function on Ω , then g is an upper gradientof a function u with respect to the metric d X if and only if it is an upper gradientof u with respect to d M . Also, given a family Γ of paths in Ω and ≤ p < ∞ , thequantity Mod p (Γ) is the same with respect to d X and with respect to d M .Proof. It suffices to prove the lemma for non-constant paths γ : I → Ω with I acompact interval in R . Since d X ≤ d M , it follows that ℓ d X ( γ ) ≤ ℓ d M ( γ ) (whether γ isrectifiable or not, this holds). To show that converse inequality, let t < t < · · ·
0, we haveinf c ∈ R µ ( B Ω ) Z B | u − c | dµ ≤ C r (cid:18) µ ( λB Ω ) Z λB g p dµ (cid:19) /p , where B Ω := { y ∈ Ω : d Ω ( x, y ) < r } and λB Ω := { y ∈ Ω : d Ω ( x, y ) < λr } . Definition 3.6 (Ahlfors regularity) . We say that a measure µ is doubling on Ωwith respect to a metric d Ω if µ is Borel regular and there is a constant C ≥ x ∈ Ω and r > µ ( B Ω ( x, r )) ≤ C µ ( B Ω ( x, r )) . We say that µ is Ahlfors Q -regular (with respect to the metric d Ω ) for some Q > C ≥ x ∈ Ω and 0 < r < d Ω ) if r Q C ≤ µ ( B Ω ( x, r )) ≤ C r Q . From the seminal work of Heinonen and Koskela [HK, Theorem 5.7] we knowthat if a complete metric measure space (
X, d, µ ) is Ahlfors Q -regular (with some Q > Q -Poincar´e inequality if and only if it satisfies a strongversion of the Q -Loewner property. Definition 3.7.
We say that (
X, d, µ ) satisfies a Q -Loewner property if there is afunction φ : (0 , ∞ ) → (0 , ∞ ) such that whenever E, F ⊂ X are two disjoint continua(that is, connected compact sets with at least two points),Mod Q (Γ( E, F, X ) ≥ φ (∆( E, F ))where Γ(
E, F, X ) is the collection of all curves in X with one end point in E andthe other in F , and ∆( E, F ) := dist(
E, F )min { diam( E ) , diam( F ) }
12s the relative distance between E and F . The strong version of the Q -Loewnerproperty is that lim inf t → + φ ( t ) = ∞ . Theorem 3.8 ([HK, Theorem 3.6]) . If ( X, d, µ ) is Q -Loewner for some Q > and µ is Ahlfors Q -regular, then we can take φ ( t ) = C max { log(1 /t ) , | log( t ) | − Q } forsome constant C ≥ . The principal focus of this section and the next section is to obtain a Carath´eodory-type extension of BQS homeomorphisms between two domains. In this section wewill focus on bounded domains, and in the next section we will study unboundeddomains. Thus in this section both Ω and Ω ′ are bounded domains in proper, lo-cally connected metric spaces. Theorem 4.4 and Proposition 4.7 are the two mainresults of this section. However, Proposition 4.12 is of independent interest as itdemonstrates that BQS property and geometric quasiconformality can be formu-lated equivalently with respect to d X , d Y or with respect to d M , d ′ M ; moreover, thisproposition holds also for unbounded domains.To understand how BQS homeomorphisms transform prime ends, we first needthe following two lemmata. Lemma 4.1.
Let X and Y be complete, doubling, locally path-connected metricspaces with Ω ⊂ X , Ω ′ ⊂ Y two domains (not necessarily bounded). If there is aBQS homeomorphism f : Ω → Ω ′ , then Ω is bounded if and only if Ω ′ is bounded.Moreover, if A ⊂ Ω with diam M ( A ) < ∞ , then diam ′ M ( f ( A )) < ∞ .Proof. It suffices to show that if Ω is bounded, then Ω ′ is also bounded. We fixa continuum Γ ⊂ Ω such that diam(Γ) ≥ diam(Ω) /
2, and let τ := diam(Γ), s :=diam( f (Γ)). Note that 0 < sτ < ∞ . Let z, w ∈ Ω ′ . Then as Ω is locally path-connected, it follows that there is a path (and hence a continuum) Λ connecting f − ( z ) to f − ( w ). The set f (Λ) is a continuum in Ω ′ with z, w ∈ f (Λ). Let Γ bethe continuum obtained by connecting Γ to Λ in Ω; then diam(Γ ) ≥ τ and so bythe BQS property of f we now havediam( f (Λ))diam( f (Γ )) ≤ η (cid:18) diam(Λ) τ (cid:19) ≤ η (cid:18) diam(Ω) τ (cid:19) . Moreover, as Γ and Γ intersect withdiam(Ω) / ≤ diam(Γ) ≤ diam(Γ ) ≤ diam(Ω) , we know from the BQS property of f thatdiam( f (Γ )) s ≤ η (cid:18) diam(Ω)diam(Ω) / (cid:19) .
13t follows that d Y ( z, w ) ≤ diam( f (Λ)) ≤ diam( f (Γ )) η (cid:18) diam(Ω) τ (cid:19) ≤ s η (2) η (cid:18) diam(Ω) τ (cid:19) < ∞ , and so Ω ′ is also bounded.Now suppose that A ⊂ Ω with diam M ( A ) < ∞ . If A has only one point, thenthere is nothing to prove. Hence assume that 0 < diam M ( A ); then we can find acontinuum Γ ⊂ Ω intersecting A , such thatdiam M ( A ) / ≤ diam(Γ) = diam M (Γ) ≤ M ( A ) . Let z, w ∈ f ( A ); then we can find a path β ⊂ Ω with end points f − ( z ) , f − ( w )such that diam( β ) ≤ M ( A ). We can also find a continuum γ ⊂ Ω with f − ( w ) and a point from A ∩ Γ contained in γ and diam( γ ) ≤ M ( A ) (notethat f − ( w ) ∈ A ∩ β ). It follows from the BQS property of f thatdiam( f ( β ))diam( f ( γ ∪ Γ)) ≤ η (cid:18) diam( β )diam( γ ∪ Γ) (cid:19) ≤ η (cid:18) M (Γ)diam( f (Γ)) (cid:19) . Applying the BQS property of f to the two intersecting continua γ ∪ Γ and Γ givesdiam( f ( γ ∪ Γ))diam( f (Γ)) ≤ η (cid:18) diam( γ ∪ Γ)diam(Γ) (cid:19) ≤ η (8) . It follows that d ′ M ( z, w ) ≤ diam( f ( β )) ≤ η (8) η (cid:18) M (Γ)diam( f (Γ)) (cid:19) . It follows thatdiam ′ M ( f ( A )) ≤ diam( f ( β )) ≤ η (8) η (cid:18) M (Γ)diam( f (Γ)) (cid:19) < ∞ as desired. Lemma 4.2.
Let X and Y be complete, doubling, locally path-connected metricspaces with Ω ⊂ X , Ω ′ ⊂ Y two domains (not necessarily bounded) and let f : Ω → Ω ′ be a BQS homeomorphism. Suppose that E, F ⊂ Ω are two non-empty boundedconnected open sets with E ⊂ F such that dist M (Ω ∩ ∂E, Ω ∩ ∂F ) > . (4) Then dist M ′ (Ω ′ ∩ ∂f ( E ) , Ω ′ ∩ ∂f ( F )) > . Proof.
Suppose γ is a continuum in Ω ′ connecting Ω ′ ∩ ∂f ( E ) , Ω ′ ∩ ∂f ( F ). Thenbecause f is a homeomorphism, f − ( γ ) is a continuum in Ω connecting Ω ∩ ∂E, Ω ∩ ∂F , and hence diam( f − ( γ ) ≥ dist M (Ω ∩ ∂E, Ω ∩ ∂F ) := τ > . A ⊂ Ω be a continuum in F with 2 diam( F ) ≥ diam( A ); as X is locally pathconnected and F is a non-empty bounded connected open set, such a continuumexists. As f − γ intersects F (for by the assumption (4), we know that Ω ∩ ∂E ⊂ F ),we can find a continuum β ⊂ F intersecting both A and f − ( γ ) ∩ F such thatdiam( β ) ≤ F ). By applying the BQS property of f to the two intersectingcontinua A ∪ β and f − ( γ ), we havediam( f ( A ∪ β ))diam( γ ) ≤ η (cid:18) diam( A ∪ β )diam( f − ( γ )) (cid:19) ≤ η (cid:18) F ) τ (cid:19) . As diam( f ( A ∪ β )) ≥ diam( f ( A )) >
0, we see thatdiam( γ ) ≥ diam( f ( A )) η (cid:16) F ) τ (cid:17) > , that is, dist M ′ (Ω ′ ∩ ∂f ( E ) , Ω ′ ∩ ∂f ( F )) ≥ diam( f ( A )) η (cid:16) F ) τ (cid:17) > Lemma 4.3.
Let X and Y be complete, doubling, locally path-connected metricspaces with Ω ⊂ X , Ω ′ ⊂ Y two domains (not necessarily bounded) and let f : Ω → Ω ′ be a BQS homeomorphism. If { E k } is a chain of Ω with singleton impression,then T k f ( E k ) has only one point.Proof. Suppose that and let { E k } is a chain with singleton impression. Thenlim k diam( E k ) = 0. We fix a continuum J ⊂ E with diam( E ) / ≤ diam( J ) ≤ diam( E ), and let k ∈ N . For each pair z, w ∈ f ( E k ) we can find a continuum γ ⊂ E k containing f − ( z ) and f − ( w ) such that diam( γ ) ≤ diam( E k ). Let J k be acontinuum in E that intersects both J and γ ; note that diam( J k ∪ J ) ≤ diam E as well. Then by the BQS property of f , we obtaindiam( f ( γ ))diam( f ( J ∪ J k )) ≤ η (cid:18) diam( γ )diam( J ∪ J k ) (cid:19) ≤ η (cid:18) diam( E k )diam( J ) (cid:19) . On the other hand, by applying the BQS property of f to the two intersectingcontinua J and J ∪ J k , we see thatdiam( f ( J ∪ J k ))diam( f ( J )) ≤ η (cid:18) diam( J ∪ J k )diam( f ( J )) (cid:19) ≤ η (2) . Therefore diam( f ( γ )) ≤ η (2) η (cid:18) diam( E k )diam( J ) (cid:19) . It follows that diam( f ( E k )) ≤ η (2) η (cid:18) diam( E k )diam( J ) (cid:19) → k → ∞ . Therefore T k f ( E k ) has at most one point, and as f ( E ) is compact and ∅ 6 = f ( E k +1 ) ⊂ f ( E k ), it follows that T k f ( E k ) has exactly one point.15 heorem 4.4. Let X and Y be complete, doubling metric spaces that are locallypath connected. Let Ω ⊆ X and Ω ′ ⊆ Y be bounded domains. Let f : Ω → Ω ′ be abranched quasi-symmetric (BQS) homeomorphism. Then, f induces a homeomor-phism f P : Ω P → Ω ′ P . Note that both Ω and Ω ′ are locally path connected because X and Y are;however, we do not assume any further topological properties for these domains.For instance, we do not assume that either Ω or Ω ′ is finitely connected at theirboundaries, and so we do not know that Ω P is compact. Proof.
We must show that f maps prime ends to prime ends.We first show that f maps chains to chains. As f is a homeomorphism, f mapsadmissible sets to admissible sets. Indeed, if E ⊂ Ω is an admissible set, then E is connected and so f ( E ) is also connected. Suppose f ( E ) does not intersect ∂ Ω ′ ;then f ( E ) is a compact subset of Ω ′ , and hence f − ( f ( E )) is a compact subset ofΩ. Therefore E ⊂ f − ( f ( E )) ⊂ Ω, which violates the requirement that E ∩ ∂ Ω benon-empty.Suppose { E i } is a chain in Ω. We check conditions ( a ) − ( c ) of Definition 2.3for the sequence { f E i } . Note that if E ⊆ F ⊆ Ω, then f E ⊆ f F ⊆ Ω ′ as f is ahomeomorphism. Hence, f E i +1 ⊆ f E i for all i , so condition (a) holds. Condition (b)follows from Lemma 4.2 by considering E := E k +1 and F := E k .Lastly, let { E i } ∈ E , and for each i let F i = f E i . Suppose y ∈ Ω ′ . Let x = f − ( y ) ∈ Ω. As T ∞ i =1 E i ⊆ ∂ Ω, there is an i and a ρ > B ( x, ρ ) ⊆ Ωand B ( x, ρ ) ∩ E i = ∅ . Then, f ( B ( x, ρ )) ∩ f ( E i ) is empty. As f ( B ( x, ρ )) open and y ∈ f ( B ( x, ρ )), there is some ρ > B ( y, ρ ) ⊂ Ω ′ ∩ f ( B ( x, ρ )). It followsthat f ( E i ) ∩ B ( y, ρ /
2) is empty, and hence y f ( E i ). Condition (c) follows.The above argument tells us that given a BQS homeomorphism f : Ω → Ω ′ ,each chain { E k } is mapped to a chain f { E k } := { f ( E k ) } . Next we show thatends are mapped to ends. To this end, note that if { G k } is also a chain in Ω with { G k }|{ F k } , then for each k there is a positive integer i k such that G i k ⊂ F k . Hence f ( G i k ) ⊂ f ( E k ), that is, f { G k }| f { E k } . Therefore division relation among chains isrespected by f . Therefore, for each end E we have that f E ⊂ G for some end G ofΩ ′ . Given that f − : Ω ′ → Ω is also a BQS homeomorphism, we see that f E = G .Finally, if E is a prime end, then whenever { G k } is an end for Ω ′ with { G k }| f E , wehave that f − { G k }| E and hence as E is a prime end, we must have f − { G k } ∈ E .Therefore E | f − { G k } , from which it follows that f E |{ G k } , that is, { G k } ∈ f E .Hence f E is also a prime end. Thus we conclude that whenever E ∈ ∂ P Ω, we musthave f E ∈ ∂ P Ω ′ , and we therefore have the natural extension f P : Ω P → Ω ′ P asdesired.The fact that f P is bijective follows from applying the above discussion also to f − (a BQS homeomorphism in its own right) to obtain an inverse of f P .To complete the proof of the theorem it suffices to show that f P is continuous,for then a similar argument applied to f − and f − P yields continuity of f − P . Toshow continuity, it suffices to show that if { ζ k } is a sequence in Ω P and ζ ∈ Ω P such that ζ k → ζ , then f ζ k → f ζ . If ζ ∈ Ω, then the tail-end of the sequencealso must lie in Ω, and thus the claim will follow from the fact that f P | Ω = f is16 homeomorphism of Ω. Therefore it suffices to consider only the case ζ ∈ ∂ P Ω.By separating the sequence into two subsequences if necessary, we can consider twocases, namely, that ζ k ∈ Ω for each k , or ζ k ∈ ∂ P Ω for each k .First, suppose that ζ k ∈ Ω for each k . Let { E k } ∈ ζ . Then the fact that ζ k → ζ tells us that for each k ∈ N we can find a positive integer i k such that whenever i ≥ i k , we have ζ i ∈ E k . It then follows that f ( ζ i ) ∈ f ( E k ). Thus for each k there is apositive integer i k with f ( ζ i ) ∈ f ( E k ) when i ≥ i k , that is, f ( ζ k ) → f ([ E k ]) = f ( ζ ).Finally, suppose that [ F j,k ] = ζ k ∈ ∂ P Ω for each k . We again choose any chain { E k } ∈ ζ . Then for each k there is a positive integer i k such that for each i ≥ i k ,there exists a positive integer j i,k such that whenever j ≥ j i,k we have F j,i ⊂ E k .So we have f ( F j,i ) ⊂ f ( E k ) for j ≥ j i,k , that is, f ([ F j,k ]) = f ( ζ k ) → f ([ E k ]) = f ( ζ ).This completes the proof of the theorem. Remark 4.5.
The only place in the above proof where the BQS property plays akey role is in verifying that if E k , E k +1 are two connected sets in Ω with E k +1 ⊂ E k and dist M (Ω ∩ ∂E k , Ω ∩ ∂E k +1 ) > , (5)then we must have dist M (Ω ′ ∩ ∂f ( E k ) , Ω ′ ∩ ∂f ( E k +1 )) > . (6)This part was explicated into Lemma 4.2. The remaining parts of the proof ofexistence and continuity of the map f P hold for any homeomorphism f : Ω → Ω ′ that satisfies (6) for each pair of connected sets E k +1 ⊂ E k ⊂ Ω that satisfies (5).This is useful to keep in mind in the final section, where we consider quasiconformalmappings that are not necessarily BQS maps.In the situation of the above theorem, if Ω P is compact, then so is Ω ′ P . This isa useful property in determining that two domains are not BQS-homeomorphicallyequivalent, as the following example shows. Example 4.6.
With Ω a planar disk and Ω ′ given byΩ ′ := (0 , × (0 , \ [ n ∈ N { /n } × [0 , / , we know from the Riemann mapping theorem that Ω and Ω ′ are conformally equiv-alent. However, as Ω P is the closed disk, which is compact, and as Ω ′ P is not evensequentially compact, it follows that there is no BQS homeomorphism between thesetwo planar domains.The above theorem is useful in other contexts as well.
Proposition 4.7.
Suppose that X and Y are locally path connected and Ω ⊂ X ,Ω ′ ⊂ Y are bounded open connected sets. Suppose that there is a BQS homeomor-phism f : Ω → Ω ′ .(a) If Ω P is compact, then so is Ω ′ P . 17b) If Ω P is metrizable, then so is Ω ′ P .(c) If Ω is finitely connected at the boundary, then so is Ω ′ . Proof.
By Theorem 4.4, homeomorphically invariant properties of Ω P will be inher-ited by Ω ′ P ; hence the first two claims of the proposition are seen to be true. On theother hand, finite connectivity of a domain at its boundary is not a homeomorphicinvariant. Hence we provide a proof of the last claim of the proposition here.Note by [BBS1] that a bounded domain is finitely connected at the boundary ifand only if all of its prime ends are singleton prime ends and its prime end closureis compact. Thus if Ω is finitely connected at the boundary, then Ω P is compact,and hence so is Ω ′ P . Hence to show that Ω ′ is finitely connected at the boundary,it suffices to show that all of the prime ends of Ω ′ are singleton prime ends. This isdone by invoking Lemma 4.3. Example 4.8.
It is possible to have two homeomorphic equivalent domains withone finitely connected at the boundary and the other, not finitely connected at theboundary. LetΩ := (0 , × (0 , \ [ n ∈ N ([0 , − /n ] × { / n } ) ∪ ([1 /n, × { / (2 n + 1) } )and Ω ′ be the planar unit disk. Then as Ω is simply connected and bounded, it fol-lows from the Riemann mapping theorem that Ω and Ω ′ are conformally, and hencehomeomorphically, equivalent. However, Ω ′ is finitely connected at the boundaryand Ω is not. Example 4.9.
The domain Ω constructed in Example 4.8 has the property thatΩ P is compact and metrizable. This is seen by the fact that Ω P is homeomorphicto U P where U := ϕ (Ω) where ϕ : R → R is given by ϕ ( x, y ) = ( xy, y ). Observethat U is simply connected at its boundary, and so U P is metrizable.On the other hand, the domain Ω ′ given byΩ ′ := (0 , × (0 , \ [ n ∈ N ([0 , / × { / n } ) ∪ ([1 / , × { / (2 n + 1) } )has the property that Ω ′ P is not compact. Hence Ω is not BQS homeomorphicallyequivalent to Ω ′ , even though both are simply connected planar domains that arenot finitely connected at the boundary. Example 4.10.
With Ω the planar domain given byΩ := { z ∈ C : | z | < z ) > } and Ω ′ := { z ∈ C : | z | < }\ [0 ,
1] a slit disk, the map f : Ω → Ω ′ given by f ( z ) = z is a conformal equivalence that is not a quasisymmetric map. However, this map isa BQS homeomorphism, and so has an extension f P : Ω = Ω P → Ω ′ P . This followsfrom [GW2, Theorem 6.50]. 18 xample 4.11. With Ω and Ω ′ as in Example 4.10 above, the map h : Ω → Ω ′ given by h ( re iθ ) = re iθ is also a BQS homeomorphism which is not conformal. Thisagain follows from [GW2, Theorem 6.50].As indicated by the definition of chains in the prime end theory, a domain Ω canalso naturally be equipped with the Mazurkiewicz metric d M . Proposition 4.12.
Let X and Y be locally path-connected metric spaces, and f :Ω → Ω ′ be a homeomorphism. Then f is BQS with respect to the respective metrics d X and d Y on Ω and Ω ′ if and only if it is BQS with respect to the correspondingMazurkiewicz metrics d M , d ′ M . Proof.
To prove the theorem it suffices to show that whenever E ⊂ Ω is a continuum,we have diam( E ) = diam M ( E ). Note that x, y ∈ Ω we have d ( x, y ) ≤ d M ( x, y ). Toprove the reverse of this inequality, let E ⊂ Ω be a continuum, ε >
0, and wechoose x, y ∈ E such that [1 + ε ] d M ( x, y ) ≥ diam M ( E ). By the definition of theMazurkiewicz metric d M (see Definition 2.1), we haved M ( x, y ) ≤ diam( E ) . It follows that diam M ( E ) ≤ [1 + ε ] diam( E ) , The desired claim now follows by letting ε → + . Remark 4.13.
By the above proposition together with [H, Proposition 10.11], ifΩ and Ω ′ are finitely connected at their respective boundaries and f : Ω → Ω ′ is aBQS homeomorphism, then its extension f P : Ω P → Ω ′ P is a BQS homeomorphismwith respect to the respective Mazurkiewicz metrics d M and d ′ M .Given Proposition 4.12 and the above remark, it is natural to ask why we do notconsider prime ends as constructed with respect to the Mazurkiewicz metric inducedby the original metric d X on Ω. However, while the property that a set E ⊂ Ω beconnected is satisfied under both d X and the induced Mazurkiewciz metric d M , thecrucial property that E ∩ ∂ Ω be non-empty depends on the metric (since the closure E depends on the metric), and hence we have fewer acceptable sets and chains, andtherefore prime ends, under the induced metric d M than under the original metric d X . See Lemma 4.14 below for the limitations put on prime ends under d M . Forinstance, the domain Ω as in Example 4.8 will have as prime ends under d M onlythe singleton prime ends, with no prime end with impression containing any point( x, ≤ x ≤
1. Hence under the metric d M the prime end closure of Ω will notbe compact. The issue of considering prime ends with respect to the original metric d X versus the induced metric d M becomes even more of a delicate issue when Ω isnot bounded, see Section 5 below. Lemma 4.14.
Let Ω be a bounded domain in a locally path connected completemetric space X , and let d M be the Mazurkiewicz metric on Ω induced by the metric d X inherited from X . If [ { E k } ] is a prime end in Ω with respect to d M , then [ { E k } ] is a prime end with respect to d X with T k E k a singleton set. roof. Note first that if E ⊂ Ω with E M compact, then E M is sequentially compactand hence so is E ; hence it follows that if E M is compact, then E is also compact.Combining this together with the fact that diam = diam M for connected sets, tellsus that if { E k } is a chain with respect to d M then it is a chain with respect to d .Next, note that if { E k } is a chain with respect to d M and { G k } is a chain withrespect to d such that { G k } | { E k } , then, by passing to a sub-chain if necessary, { G k } is also a chain with respect to d M . Indeed, the only property that needs to beverified here is whether each G kM is compact. By the divisibility assumption, weknow that there is some j ∈ N such that whenever j ≥ j we have G j ⊂ E . Then G j M is a closed subset of the compact set E M , and hence is compact.Since E k is acceptable with respect to d M , we know that E kM , the closure of E k under the metric d M , is compact. It then follows also that T k E kM is non-empty andhas a point ζ ∈ ∂ M Ω. We can then find a sequence x k ∈ E k with lim k d M ( x k , ζ ) = 0.Since d X ≤ d M on Ω, it follows that { x k } is a Cauchy sequence with respect to themetric d X , and as X is complete, there is a point x ∈ Ω such that lim k d X ( x k , x ) = 0.Since { x k } is also a Cauchy sequence with respect to the metric d M , for each positiveinteger k we can find a continuum Γ k in Ω with end points x k and x k +1 such that(by passing to a subsequence if necessary)diam M (Γ k ) = diam(Γ k ) ≤ − ( k +1) . In the above, the first identity is inferred from the proof of Proposition 4.12 above.For each positive ineger k let G k be the connected component of B ( x, − k ) ∩ Ωcontaining the connected set S j ≥ k Γ j ⊂ Ω. Then { G k } is a chain in Ω with respectto the metric d , and { x } = I ( { G k } ). It is clear from the second paragraph abovethat { G k } is a chain also with respect to d M , with { ζ } = T k G kM . As a chain withsingleton impression, it follows that the end (with respect to d M ) that { G k } belongsto is a prime end. Moreover, for each positive integer k we know that infinitelymany of the sets G j , j ∈ N , intersect E k ; it follows from the assumption that [ { E k } ]is a prime end that [ { E k } ] = [ { G k } ] (see the last part of the proof of Lemma 2.12),and so the claim follows. In this section we consider unbounded domains, and so we adopt the modificationto the construction of ends found in [E, Definition 4.12]. However, we introduce aslight modification, namely the boundedness of the separating compacta R k and K wiht respect to the Mazurkiewicz metric d M . Recall that a metric space is proper ifclosed and bounded subsets of the space are compact. Definition 5.1.
A connected set E ⊂ Ω is said to be acceptable if E is proper and E ∩ ∂ Ω is non-empty. A sequence { E k } of acceptable subsets of Ω is a chain if foreach positive integer k we have(a) E k +1 ⊂ E k , 20b) E k is compact, or E k is unbounded and there is a compact set R k ⊂ X suchthat Ω \ R k has two components C ,k and C ,k with Ω ∩ ∂E k ⊂ C ,k andΩ ∩ ∂E k +1 ⊂ C ,k , and diam M ( R k ∩ Ω) is finite,(c) dist M (Ω ∩ ∂E k , Ω ∩ ∂E k +1 ) > ∅ 6 = T k E k ⊂ ∂ Ω.The above notion of ends agrees with the notion of ends from Section 2 when Ωis bounded, for then the properness of E will guarantee that E is compact. WhenΩ is not bounded, the above definition gives us two types of ends, (a) those thatare bounded (i.e. E k is bounded for some positive integer k ) and (b) those that areunbounded (i.e. E k is unbounded for each k ). While the definition above agrees inphilosophy with that from [E], but the requirement that diam M ( R k ∩ Ω) be finiteeliminates some of the candidates for ends from [E] here. However, ends in our senseare necessarily ends in the sense of [E]. From [E, Section 4.2] we know that if X itselfis proper, then ends (and prime ends, see below) as given in [E] are invariant undersphericalization and flattening procedures, and so understanding ends of type (b)above can be accomplished by sphericalizing the domain and understanding endsof the resulting bounded domain with impression containing the new point thatcorresponds to ∞ . Both of these types of ends have non-empty impressions. Thenext definition deals with ends that should philosophically be ends with only ∞ intheir impressions. Definition 5.2.
An unbounded connected set E ⊂ Ω is said to be acceptable at ∞ if E is proper and there is a compact set K ⊂ X such that diam M ( K ∩ Ω) < ∞ and E is a component of Ω \ K . A sequence { E k } of sets that are acceptable at ∞ is said to be a chain at ∞ if for each positive integer k we have(a) E k +1 ⊂ E k ,(b) dist M (Ω ∩ ∂E k , Ω ∩ ∂E k +1 ) > T k E k = ∅ .As with chains and ends for bounded domains in Section 2, we define divisibilityof one chain by another, and say that two chains are equivalent if they divide eachother. An end is an equivalence class of chains, and a prime end is an end that isdivisible only by itself.It was noted in [E, Remark 4.1.14] that every end at ∞ is necessarily a primeend.We now return to some mechanisms that make unbounded ends from Defini-tion 5.1 work. Lemma 5.3.
Let E and E be two open connected subsets of Ω such that E ⊂ E .Suppose in addition that R is a compact subset of X such that no component of Ω \ R contain points from both Ω ∩ ∂E and Ω ∩ ∂E . Let R := R ∩ E \ E . Then E is contained in a component b C of Ω \ R and b C ⊂ E . Moreover, no componentof Ω \ R contains points from both Ω ∩ ∂E and Ω ∩ ∂E . roof. Let C be the union of all the components of Ω \ R that contain points fromΩ ∩ ∂E , and let C be the union of all the components of Ω \ R that contain pointsfrom Ω ∩ ∂E .Since Ω ∩ ∂E ⊂ C and C is open, it follows that C ∪ E is a connected subsetof Ω \ R . So there is a component b C of Ω \ R such that C ∪ E ⊂ b C ⊂ Ω \ R ,and hence C = b C ⊃ E .Let U be a component of Ω \ R such that U ⊂ C . If U ∩ b C is non-empty, then U = b C , and in this case there are two points z ∈ U ∩ ∂E and w ∈ b C ∩ ∂E suchthat z, w ∈ U . As U is an open and connected subset of Ω, we can find a curve γ :[0 , → U with γ (0) = z and γ (1) = w . We can assume, by considering a subcurveof γ and modifying z and w if necessary, that γ ((0 , ∩ ∂E and γ ((0 , ∩ ∂E are empty. As Ω ∩ ∂E ⊂ E , it follows that γ ((0 , ⊂ E \ E . By the propertyof separation that R has, we must have γ ((0 , ∩ R non-empty. Hence γ ((0 , R ∩ E \ E = R . This contradicts the fact that γ ⊂ U with U ∩ R empty. We therefore conclude that C ∩ b C is empty.Finally we are ready to prove that b C ⊂ E . Suppose there is a point z ∈ b C \ E .Since E ⊂ b C and b C is open and connected, there is a point w ∈ E ⊂ E and acurve β ⊂ b C connecting z to w . As z E , it follows that γ intersects Ω ∩ ∂E ,which violates the conclusion reached in the previous paragraph that Ω ∩ ∂E doesnot intersect b C . Hence such z cannot exist, whence it follows that b C ⊂ E .As mentioned above, by [E, Section 4.2], understanding prime ends for un-bounded domains can be done by first sphericalizing them and transforming theminto bounded domains and then studying the prime ends of the transformed domains.There is a metric analog of reversing sphericalizations, called flattening. However,the category of BQS homeomorphisms need not be preserved by the sphericalizationand flattening transformations, see the example below. Therefore sphericalizationand flattening may not provide as useful a tool as we like in the study of BQS maps. Example 5.4 (BQS Sphere ; BQS Plane) . Let f : C ∗ = C \ { } → C ∗ be givenby f ( z ) = 1 /z . The push-forward of f under sphericalization of C ∗ to the twice-punctured sphere gives a BQS homeomorphism, but f is not a BQS homeomorshimfrom C ∗ to itself, for the families of pairs of continua E r := [ r,
1] and F r := S ,0 < r < /
10, have the property thatdiam( E r )diam( F r ) ≈ , diam( f ( E r ))diam( f ( F r )) ≈ r . Example 5.5 (BQS Plane ; BQS Sphere) . For each positive integer k let A k = ( { k } × [1 , k ]) ∪ ([ k +1 , k ]) × { k } ) ∪ ( { k +1 } × [0 , k ]) ∪ ([ k +2 , k +1 ] × { } ) ∪ ( { k +2 } × [0 , , let ε k = k , ℓ ( A k ) = 2 k +1 + k +2 the length of A k , and letΩ := [ j ∈ N [ z ∈ A j Q ( z, ε j ) , A k and its cube neighborhoodwhere, for z ∈ C , the cube Q ( z, t ) = (Re( z ) − t, Re( z ) + t ) × (Im( z ) − t, Im( z ) + t ).For m ≥
1, let a m = P mj =1 ℓ ( A j ), and set a = 0. Let I m = [ a m − , a m ] = [ a m − , a m − + ℓ ( A m ) ] , I = [0 , ∞ ) . Let Ω ′ = S ∞ m =1 I m × ( − ε m , ε m ). Then, there is a BQS map f : Ω ′ → Ω definedas follows. On I , this map is the arclength parametrization of the curve S k A k .On most of Ω ′ , this map is defined similarly; for m ∈ N and ε m +2 ≤ τ < ε m ,the line (0 , a m ) × { τ } is mapped by f as a parametrization to a curve that is an ℓ ∞ -distance τ “above” the curve S k A k (the dashed straight line gets mapped tothe dashed bent line above f ( I ) in Figure 2 below). This parametrization is mostlyan arclength parametrization, although near the corners of Ω we change speed ofthis parametrization by an amount that is determined by τ . We can ensure thatthis speed is at least 1 /
128 and 128. For example, in the picture below, the dashedcurve on the left is mapped to the dashed curve on the right by f at a speed thatis the ratio of the lengths of the dashed curve on the right and the middle curve onthe right. fI f ( I )Figure 2: Fiber map of f near cornersBy using ǫ k much smaller than 2 k (as k ≥
2) and changing the “tube size” at y = 1, this guarantees that near each of these corners the map is biLipschitz andhence BQS there. The arclength parameterization guarantees that the map is BQSat large scales, and hence BQS.After applying the stereographic projection ϕ (projection to the sphere), how-ever, there is some C > i < j we have diam( ϕ ( ∪ jk = i A k )) ≥ C ,23hereas diam( ϕ ( f ( ∪ ∞ k = i A k )) → i → ∞ . Hence the induced map between ϕ (Ω)and ϕ (Ω ′ ) is not BQS.Now we are ready to study boundary behavior of BQS homeomorphisms betweentwo unbounded domains.Let Ω , Ω ′ be two open, connected subsets of metric spaces X, Y respectively, with X and Y both proper and locally path-connected metric spaces. Let f : Ω → Ω ′ bea BQS homeomorphism.In light of Lemma 4.1, we now concentrate on the case of both Ω and Ω ′ be-ing unbounded. So henceforth, we assume that Ω and Ω ′ are unbounded domainsin proper, locally path-connected metric spaces and that f : Ω → Ω ′ is a BQShomeomorphism. Recall that we have now to deal with three types of prime ends. Definition 5.6 ( Prime end types).
We categorize the three types of ends asfollows:(a) Ends [ { E k } ] for which there is some positive integer k with E k compact,(b) Ends [ { E k } ] for which each E k is unbounded and I [ { E k } ] is non-empty,(c) Ends that are ends at ∞ ; these are also unbounded. Remark 5.7.
A few words comparing prime ends with respect to the original metric d X on Ω and prime ends with respect to the induced Mazurkiewicz metric d M arein order here. A prime end of type (c) with respect to d X will continue to be aprime end of type (c) with respect to d M . A prime end of type (b) with respect to d X will be a prime end of type (c) with respect to d M , and so information aboutthe impression of such a prime end is lost. A singleton prime end of type (a) withrespect to d X will continue to be a singleton prime end of type (a) with respect tod M . However, a non-singleton prime end of type (a) with respect to d X will cease tobe a prime end (and indeed, cease to be even an end) with respect to d M . Indeed,if [ { E k } ] is a prime end of type (a) with respect to d X with I [ { E k } ] non-singleton,then no point of I [ { E k } ] can be accessible from inside the chain { E k } , as can beseen from an easy adaptation of the proof of Lemma 2.12. Therefore T k E kM isempty, and so the sequence { E k } fails the definition of an end with respect to d M . Lemma 5.8.
Suppose that Ω M is proper. Fix x ∈ Ω and a strictly increasingsequence { n k } of positive real numbers with lim k n k = ∞ . Then, for each k ∈ N there is only finitely many unbounded components of Ω \ B M ( x , n k ) . If for each k ∈ N we choose an unbounded component F k of Ω \ B M ( x , n k ) such that F k +1 ⊂ F k ,then { F k } is a chain for Ω such that [ { F k } ] is a prime end of Ω .Proof. To see the veracity of the first claim, note that if there are infinitely manyunbounded components of Ω \ B M ( x , n k ), then there are infinitely many componentsof Ω \ B M ( x , n k ) that intersect B M ( x , n k +2 ) ∩ Ω, in which case we can extract outa sequence of points { x j } from B M ( x , n k +1 ) ∩ Ω \ B M ( x , n k +1 ) such that for j = i we have 2 n k +2 ≥ d M ( x j , x i ) ≥ n k +2 − n k +1 >
0, and this sequence will not have anysubsequence converging in Ω M , violating the properness of Ω M .24iven the first claim, it is not difficult to see that we can choose a sequence { F k } as in the second hypothesis of the lemma. Sincedist M (Ω ∩ ∂F k , Ω ∩ ∂F k +1 ) ≥ dist M (Ω ∩ ∂B M ( x , n k ) , Ω ∩ ∂B M ( x , n k +1 )) ≥ n k +1 − n k > , it follows that { F k } is a chain of type (b) if T k F k is non-empty, and is a chainof type (c) (chain at ∞ ) if T k F k is empty. In this latter case, we already havethat [ { F k } ] is prime. Hence suppose that T k F k is non-empty, and suppose that { G k } | { F k } for some chain { G k } of Ω. By passing to a subsequence of { G k } ifnecessary, we can then assume that G k ⊂ F k for all k ∈ N . We need to show that { F k } | { G k } . Suppose this is not the case. Then there is some positive integer k ≥ j ∈ N the set F j \ G k is non-empty. Recall that G k ⊂ F k for each k ∈ N . Therefore, it follows from the nested property of { G k } , that both F j \ G k and F j ∩ G k are non-empty for each j ∈ N . This immediately implies that { G k } cannot be a chain of type (a).Suppose that { G k } is a chain of type (c). As F j is an open connected subset ofΩ, it follows that Ω ∩ ∂G k must intersect F j . Hence, for each j ∈ N we have thatΩ ∩ ∂G k ∩ F j = ∅ , which violates the requirement that diam M (Ω ∩ ∂G k ) < ∞ .Therefore we have that { G k } cannot be of type (c). If { G k } is of type (b), thenas R k separates Ω ∩ ∂G k − from Ω ∩ ∂G k , we must have that for each j ∈ N , R k intersects F j . This again violates the requirement that diam M ( R k ∩ Ω) < ∞ .Therefore we must have that { F k } | { G k } as well. Hence [ { F k } ] is a prime end. Lemma 5.9. If E is a bounded prime end of Ω , then f ( E ) is a bounded prime endof Ω ′ . Moreover, if E is a singleton prime end (that is, its impression contains onlyone point), then f ( E ) is also a singleton prime end.Proof. Let { E k } ∈ E ; by passing to a subsequence if need be, we can also assumethat E is compact. Moreover, by Lemma 2.13 we can also assume that each E k is open. Then the restriction f | E is also a BQS homeomorphism onto f ( E ), andit follows from Lemma 4.1 above that f ( E ) is also bounded. Therefore for eachpositive integer k we have that f ( E k ) is bounded, and by the properness of Y wealso have that f ( E k ) is compact. Moreover, since E k ∩ ∂ Ω is non-empty, we canfind a sequence x m of points in E k converging to some x ∈ E k ∩ ∂ Ω; since f isa homeomorphism, it follows that f ( x m ) cannot converge to any point in Ω ′ , andso by the compactness of f ( E k ) we have a subsequence converging to a point in f ( E k ) ∩ ∂ Ω ′ , that is, f ( E k ) ∩ ∂ Ω ′ is non-empty. A similar argument also shows that T k f ( E k ) is non-empty and is a subset of ∂ Ω ′ .From Lemma 4.2 we know that for each positive integer k ,dist M ′ (Ω ′ ∩ ∂f ( E k ) , Ω ′ ∩ ∂f ( E k +1 )) > . Therefore f ( E ) is an end of Ω ′ .Next, if F is an end that divides f ( E ), then with { F k } ∈ F we must have that F k is bounded (and hence, F k is compact) for sufficiently large k . The above argument,applied to { F k } and the BQS homeomorphism f − tells us that f − ( F ) is an end of25 dividing the prime end E ; it follows that f − ( F ) = E , and so F = f ( E ), that is, f ( E ) is a prime end of Ω ′ .The last claim of the theorem follows from Lemma 4.3. Lemma 5.10.
Let E be a prime end of Ω of type (b). Then f ( E ) is a prime endof type (b) for Ω ′ or there is an end F at ∞ for Ω ′ such that with { E k } ∈ E thereexists { F k } ∈ F with F k +1 ⊂ f ( E k ) ⊂ F k for each positive integer k .Proof. Here we cannot invoke Lemma 4.2 as neither E k nor E k +1 is bounded; butthe idea for the proof is similar, as we now show. Suppose E , E ⊂ Ω be connectedopen sets with E ⊂ E and τ := dist M (Ω ∩ ∂E , Ω ∩ ∂E ) >
0. Suppose alsothat there is a compact set R ⊂ X such that Ω ∩ ∂E and Ω ∩ ∂E belong todifferent components of Ω \ R and that diam M ( R ∩ Ω) < ∞ . We fix a continuum A ⊂ Ω that intersects R with diam( A ) ≤ M ( R ∩ Ω). If γ is a continuumin Ω intersecting both Ω ∩ ∂E and Ω ∩ ∂E , then it must intersect R . Hencewe can find a continuum β ⊂ Ω intersecting both γ ∩ R and A ∩ R such thatdiam M ( β ) = diam( β ) ≤ M ( R ∩ Ω). Now by the BQS property of f , we seethat diam( f ( β ∪ A ))diam( f ( γ )) ≤ η (cid:18) diam( β ∪ A )diam( γ ) (cid:19) ≤ η (cid:18) diam( β ∪ A ) τ (cid:19) . Note that β and A are intersecting continua with diameter at most 2 diam M ( R ∩ Ω).It follows that diam( β ∪ A ) ≤ M ( R ∩ Ω) < ∞ . Hencediam( f ( A ))diam( f ( γ )) ≤ η (cid:18) M ( R ∩ Ω) τ (cid:19) , that is, 0 < diam( f ( A )) η (cid:16) M ( R ∩ Ω) τ (cid:17) ≤ diam( f ( γ )) . It follows that dist M (Ω ′ ∩ ∂f ( E ) , Ω ′ ∩ ∂f ( E )) > . From the second claim of Lemma 4.1, we know that f ( R ∩ Ω) is bounded withrespect to the Mazurkiewicz metric d ′ M and hence with respect to d Y . From thefact that f is a homeomorphism from Ω to Ω ′ , it also follows that Ω ′ ∩ ∂f ( E ) andΩ ′ ∩ ∂f ( E ) belong to different components of Ω ′ \ f ( R ∩ Ω).The properness of Y together with the fact that diam d Y ( f (Ω ∩ R )) ≤ diam ′ M ( f (Ω ∩ R )) < ∞ tells us that f (Ω ∩ R ) is compact in Y .From the above argument, we see that f ( E ) is an end of Ω ′ provided that T k f ( E k ) is non-empty. Similar argument as at the end of the proof of Lemma 5.9then tells us that f ( E ) is a prime end.Suppose now that T k f ( E k ) = ∅ . To see that f ( E ) is of type (b), note thatwith { E k } ∈ E , each E k is unbounded (and without loss of generality, open, seeLemma 2.13). Hence by Lemma 4.1 we know that f ( E k ) is unbounded. Thiscompletes the proof in the case that T k f ( E k ) is non-empty.Finally, suppose that T k f ( E k ) is empty. We construct the end at ∞ , F , asfollows. By Lemma 5.3, we can assume that the separating sets R k have the added26eature that E k +1 is a subset of a component of Ω \ R k and that that componentis a subset of E k . So for each positive integer k we set F ∗ k to be the image of thiscomponent under f . Then T k F k ⊂ T k f ( E k ) = ∅ . Moreover, F ∗ k is a componentof Ω ′ \ K k with K k = f ( R k ∩ Ω), and as R k is compact and diam M ( R k ∩ Ω) < ∞ ,by Lemma 4.1 we know that diam ′ M ( f ( R k ∩ Ω)) < ∞ and so f ( R k ∩ Ω) is compactin Y . We cannot however guarantee that dist M ′ (Ω ′ ∩ ∂F ∗ k , Ω ′ ∩ ∂F ∗ k +1 ) is positive;hence we set F k := F ∗ k − and note thatdist M ′ (Ω ′ ∩ ∂F k , Ω ′ ∩ ∂F k +1 ) ≥ dist M ′ (Ω ′ ∩ ∂E k , Ω ′ ∩ ∂E k +1 ) > . Thus the sequence { F k } is an end at ∞ of Ω ′ , completing the proof.Given that ends at infinity are prime ends (no other ends divide them, see [E]),it follows that should E be associated with an end F at ∞ , there can be only onesuch end at ∞ it can be associated with.We next consider the last of the three types of prime ends of Ω. We say thata chain { E k } in an end E at ∞ of Ω is a standard representative of E if there is apoint x ∈ X and a strictly monotone increasing sequence R k → ∞ such that foreach k the set E k is a connected component of Ω \ B M ( x , R k ). Note that if H ⊂ Ωsuch that diam M ( H ) < ∞ , then diam M ( H ∩ Ω) is also finite.
Lemma 5.11.
Let E be an end at ∞ of Ω . Then there is a standard representative { E k } ∈ E . Moreover, for each x ∈ Ω there is a representative chain { F k } ∈ E suchthat for each k ∈ N , F k is a component of Ω \ B M ( x , k ) .Proof. Let { G k } ∈ E and for each k let K k be a compact subset of X such thatdiam M ( K k ∩ Ω) is finite and G k is an unbounded component of Ω \ K k . Fix x ∈ Ω.Since T E k is empty, we can assume without loss of generality that x E . Then x is in a different component from E k of the set Ω \ K k for each k ∈ N . In whatfollows, by B M ( x , τ ) we mean the ball in Ω centered at x with radius τ withrespect to the Mazurkiewicz metric d M .As τ k := diam M ( K k ) < ∞ , inductively we can find a sequence of strictly mono-tone increasing positive real numbers R k as follows. Let R := max { , diam M ( { x } ∪ K ) } . We claim that there is some L ∈ N such that G L does not intersect B M ( x , R ).If this claim is wrong, then for each positive integer j ≥ x j ∈ G j ∩ B M ( x , R ), in which case, by the compactness of the set B M ( x , R ) there isa subsequence x j n and a point x ∞ ∈ B M ( x , R ) ⊂ X such that x j n → x ∞ ; noteby the nestedness property of the chain { G k } that x ∞ ∈ G k for each k , violatingthe property that T k G k is empty. Therefore the claim holds. We set E to be theconnected component of Ω \ B M ( x , R ) that contains G L . Set n = 1 + L . Wethen set R := max { , R + 1 , diam M ( { x } ∪ K n ) } , and as before, note that there is some n > n such that G n is disjoint from B M ( x , R ). We set E to be the component of Ω \ B M ( x , R ) that contains G n .27hus inductively, once R , · · · , R k have been determined and the correspondingstrictly monotone increasing sequence of integers n , · · · , n k have also been chosen,we set R k +1 := max { k + 1 , R k + 1 , diam M ( { x } ∪ K n k ) } and choose n k +1 such that n k +1 > n k and G n k +1 ∩ B M ( x , R k +1 ) is empty, andset E k +1 to be the component of Ω \ B M ( x , R k +1 ) that contains G n k +1 . Thus weobtain the sequence { E k } ; it is not difficult to see that this sequence is a chain at ∞ according to Definition 5.2. Indeed, as G n k +1 ⊂ E k +1 , and as G n k +1 ⊂ G n k and K n k ∩ E k +1 = ∅ , it follows that E k +1 belongs to the same component as G n k +1 ofΩ \ K n k , and so E k +1 ⊂ G n k . Thus we have G n k +1 ⊂ E k +1 ⊂ G n k ⊂ E k , and so we have both the nested property of the sequence { E k } and the fact that { G k } divides { E k } . Therefore, [ { E k } ] is an end at ∞ of Ω and hence is a prime end,and thus [ { E k } ] = E ; see [E, Remark 4.1.11, Remark 4.1.14] for details of the proof.The last part of the claim follows from knowing that for each k ∈ N there arepositive integers j, l such that R k ≤ j ≤ R k + l . Lemma 5.12.
Let E be an end at ∞ of Ω . Then either T k f ( E k ) = ∅ for each { E k } ∈ E or T k f ( E k ) non-empty for each { E k } ∈ E . In the first case f ( E ) is anend at ∞ of Ω ′ . In the second case f ( E ) is a prime end of type (b) of Ω ′ .Proof. Let { E k } ∈ E be a standard representative; hence, there is some x ∈ Ω andfor each k ∈ N there exists R k > { R k } is strictly monotoneincreasing with lim k R k = ∞ , and E k is a component of Ω \ B M ( x , R k ). Fix acontinuum A ⊂ Ω intersecting Ω ∩ ∂E k and withdiam( A ) ≤ M (Ω ∩ ∂E k ) < ∞ . Let z ∈ Ω ′ ∩ ∂f ( E k ) and w ∈ Ω ′ ∩ ∂f ( E k +1 ), and let γ be a continuum in Ω ′ containing z, w . Then f − ( γ ) is a continuum in Ω containing f − ( z ) ∈ Ω ∩ ∂E k and f − ( w ) ∈ Ω ∩ ∂E k +1 . By the fact that { E k } is a chain of Ω, we know thatdiam( f − ( γ )) ≥ τ := dist M (Ω ∩ ∂E k , Ω ∩ ∂E k +1 ) > . As f − ( γ ) intersects both Ω ∩ ∂E k and Ω ∩ ∂E k +1 , we can find a continuum β connecting A ∩ ∂E k and f − ( γ ) ∩ ∂E k such that diam( β ) ≤ M (Ω ∩ ∂E k ). Bythe BQS property, we now havediam( f ( β ∪ A ))diam( γ ) ≤ η (cid:18) diam( β ∪ A )diam( f − ( γ ) (cid:19) ≤ η (cid:18) M (Ω ∩ ∂E k ) τ (cid:19) , and so 0 < diam( f ( A )) η (cid:16) M (Ω ∩ ∂E k ) τ (cid:17) ≤ diam( γ ) , and so we have thatdist M (Ω ′ ∩ ∂f ( E k ) , Ω ′ ∩ ∂f ( E k +1 )) ≥ diam( f ( A )) η (cid:16) M (Ω ∩ ∂E k ) τ (cid:17) > . (7)28ote that as neither E k nor E k +1 is bounded, we cannot utilize that lemma directly,nor can we directly call upon the relevant part of the proof of Lemma 5.10 as there isno separating set R k ; but the idea of the proof is quite similar. For the convenienceof the reader we gave the complete proof above.Now we have two possibilities, either T k f ( E k ) is empty, or it is not empty. Case 1: T k f ( E k ) = ∅ . For each k ∈ N note that f ( E k ) is a component of Ω ′ \ f ( B M ( x , R k ) ∩ Ω). By Lemma 4.1, we know that dist ′ M ( f ( B M ( x , R k ) ∩ Ω)) < ∞ ,and moreover, as the topology on Ω with respect to d X and with respect to d M arethe same, we also have that f ( B M ( x , R k ) ∩ Ω) ∩ Ω ′ = f ( B M ( x , R k ) ∩ Ω) . Also, f ( B M ( x , R k ) ∩ Ω) is compact as Y is proper and diam( f ( B M ( x , R k ) ∩ Ω)) =diam( f ( B M ( x , R k ) ∩ Ω) ≤ diam M ( f ( B M ( x , R k ) ∩ Ω) < ∞ . Again, as Y is proper, f ( E k ) is also proper. Therefore f ( E k ) is an acceptable set at ∞ , and by (7) we seethat { f ( E k ) } is a chain at ∞ of Ω ′ . Case 2: T k f ( E k ) = ∅ . In this case we need to show that { f ( E k ) } forms a chainof type (b). The only additional condition we need to check is that for each k ∈ N there is some compact set P k ⊂ Y such that Ω ′ ∩ ∂f ( E k ) and Ω ′ ∩ ∂f ( E k +1 ) arein different components of Ω ′ \ P k . In what follows, by S M ( x , τ ) we mean the set { y ∈ Ω : d M ( x , y ) = τ } . We choose P k = f ( S M ( x , R k + R k +1 )) . Since Ω ∩ ∂E k ⊂ Ω ∩ ∂B M ( x , R k ) and Ω ∩ ∂E k +1 ⊂ Ω ∩ ∂B M ( x , R k +1 ), it fol-lows that these two sets cannot intersect the same connected component of Ω \ S M ( x , R k + R k +1 )). Since B M ( x , R k ) ∩ Ω is connected in Ω, it follows that Ω ∩ ∂E k is in one component of Ω \ S M ( x , R k + R k +1 )). As Ω ∩ E k +1 ⊂ Ω \ B M ( x , R k +1 ) is con-nected, it also follows that Ω ∩ ∂E k +1 lies in a component of Ω \ S M ( x , R k + R k +1 )). As f is a homeomorphism, similar statements hold for Ω ′ ∩ ∂f ( E k ), Ω ′ ∩ ∂f ( E k +1 ), thatis, they are contained in two different components of Ω \ P k . As diam M ( P k ∩ Ω ′ ) < ∞ by Lemma 4.1, it follows also that P k is a compact subset of Y . Thus the separationcondition for the sequence { f ( E k ) } is verified, and so this sequence is a chain of Ω ′ of type (b).To see that [ { f ( E k ) } ] is a prime end, note that if { G k } is a chain that divides { f ( E k ) } , then { f − ( G k ) } is a chain of Ω that divides { E k } ; as { E k } is a prime end,it follows that { E k } divides { f − ( G k ) } , and so { f ( E k ) } divides { G k } as well. Thusthe corresponding end is a prime end. Example 5.13.
In Example 5.5, Ω has a prime end of type (b) with impression { } × [0 , ∞ ) (e.g. take E k = (cid:18) (0 , k ) × (0 , ∞ ) (cid:19) ∩ Ω) which gets mapped to an endat ∞ by f . The BQS map f − then also maps an end at infinity to an end of type(b).We now summarize the above study. Recall that we assume X and Y to belocally path-connected proper metric spaces and that Ω ⊂ X , Ω ′ ⊂ Y are domains.29 heorem 5.14. Suppose that Ω is an unbounded domain and let f : Ω → Ω ′ be aBQS homeomorphism. Then there is a homeomorphism f P : Ω P → Ω ′ P such that f ( E ) is a bounded prime end (i.e. a prime end of type (a)) if E is a bounded primeend, and f ( E ) is either an unbounded prime end (i.e. of type (b)) or is an end at ∞ of Ω ′ if E is an unbounded prime end or an end at ∞ of Ω . Moreover, if E is asingleton prime end, then so is f ( E ) . Remark 5.15.
Combining the above theorem with [H, Proposition 10.11] we seethat if Ω is finitely connected at the boundary, then f P | Ω P \ ∂ ∞ Ω is a BQS homeo-morphism with respect to the Mazurkiewicz metric d M , where ∂ ∞ Ω is the collectionof all ends at ∞ of Ω. However, unlike in the case of bounded domains (see Propo-sition 4.7(c)), we cannot conclude here that Ω ′ must also be finitely connected atthe boundary, see Example 5.5; however, this is not an issue as no continuum inΩ P contains points from ∂ ∞ Ω. The obstacle however is to know about f − P , andtherefore it is natural to ask whether there are continua in Ω ′ P that do not lie en-tirely in Ω ′ ∪ ∂ B Ω ′ , where ∂ B Ω ′ is the collection of all bounded prime ends of Ω ′ (which is the same as the collection of all singleton prime ends, thanks to the abovetheorem). Unfortunately it is possible for ∂ P Ω ′ \ [ ∂ B Ω ′ ∪ ∂ ∞ Ω ′ ] to contain manycontinua, as Example 5.16 below shows. We do not have a notion of metric on Ω ′ P in this case, and so it does not make sense to ask that f P is itself a BQS map unlessΩ ′ is also finitely connected at the boundary (in which case, it is indeed a BQShomeomorphism). Example 5.16.
Let Ω be the planar domain { ( x, y ) ∈ R : x > , < y < x }\ [ ≤ n ∈ N [0 , n − × { n } [ n ∈ N [ , n ] × { n +1 } , and set Ω = Ω × (0 , ∞ ). ...Figure 3: Ω Note that Ω is not finitely connected at the boundary. For each t > k ∈ N consider E tk := { ( x, y, z ) ∈ R : y + ( z − t ) < k +1) } . Observe that for each t > { E tk } is a chain of Ω with impression R × { } × { t } . The corresponding end is a prime end, and for each compact interval30 ⊂ (0 , ∞ ), the set { [ { E tk } ] : t ∈ I } is a subset of ∂ P Ω \ [ ∂ B Ω ∪ ∂ ∞ Ω] and is acontinuum. Note that in this example, ∂ ∞ Ω is empty. By opening up the domainalong the slitting planes [0 , n − × { n } and [ , n ] × { n +1 } , we obtain a domainthat is finitely connected at the boundary, and the map from Ω to this domain canbe seen to be a BQS homeomorphism.The following example shows that it is possible to have a prime end in the senseof [E] which makes the prime end closure sequentially compact, but no equivalentprime end in our sense. Example 5.17.
We construct Ω by removing closed subsets of R . The “barrier” G ( h ) centered at (0 ,
0) with height h is given by G ( h ) = S k ∈ Z C k , where C :=(( −∞ , − ∪ [1 , ∞ )) × { } and for h, k > C k = ( { − − k } × [ − h − k , h − k ]) ∪ ([1 − − k , k ] × {− h − k , h − k } ) C − k = ( {− − k } × [ − h − k , h − k ]) ∪ ([ − k , − − k ] × {− h − k , h − k } )Thus C − k is obtained from C k by reflecting it about the y -axis. See Figure 3 belowfor an illustration of the barrier G ( h ). h Figure 4: The gate G ( h ) centered at (0 , A ⊂ R and a point ( a, b ) ∈ R , we denote by A + ( a, b ) the set A + ( a, b ) := { ( x + a, y + b ) : ( x, y ) ∈ A } . We now set Ω := R × (0 , ∞ ) \ [ ≤ j ∈ N (cid:0) G (2 − j ) + (2 j , − j ) (cid:1) . The sets E j := Ω ∩ ( R × (0 , − j )) forms a chain { E j } of Ω in the sense of [E]with [ { E j } ] prime, but this does not form a chain in our sense, nor is it equivalentto a chain that is also a chain in our sense. In this section we consider a Carath´eodory extension theorem for geometric qua-siconformal maps. Unlike with BQS homeomorphisms, geometric quasiconformalmaps can map bounded domains to unbounded domains. In this section we willconsider quasiconformal maps between two domains without assuming boundednessor unboundedness for either domain. As we have less control with geometric quasi-conformal maps than with BQS maps, here we need additional constraints on thetwo domains. The main theorem of this section is Theorem 6.9.In this section, we will assume that X and Y are proper metric spaces that arelocally path-connected, and that Ω ⊂ X , Ω ′ ⊂ Y are domains. We also assume that31oth (Ω , d M ) and (Ω ′ , d ′ M ) are Ahlfors Q -regular and support a Q -Poincar´e inequalityfor some Q > µ is a Radon measure on Ω such that bounded subsets of Ω havefinite measure, and that f : Ω → Ω ′ is a geometric quasiconformal mapping. In lightof the following lemma, if both Ω and Ω ′ are bounded, then their prime end closuresand their Mazurkiewicz metric completions are equivalent, in which case by theresults of [HK] and [H, Proposition 10.11], we know that geometric quasiconformalmaps between them have even a geometric quasiconformal homeomorphic extensionto their prime end closures. Thus the interesting situation to consider here is whenat least one of the domains is unbounded.Before addressing the extension result given in Theorem 6.9 below, we first needto study the properties of prime ends of domains that are Ahlfors regular and supporta Poincar´e inequality as above. Lemma 6.1.
The domain Ω is finitely connected at the boundary if and only if thefollowing three conditions hold:(i) Ω M is proper,(ii) Ω has no type (b) prime ends,(iii) Ω P is sequentially compact.Proof. We first show that Ω M is proper. It suffices to show that bounded sequencesin Ω have a convergent subsequence converging to a point in Ω M . Let { x n } be abounded sequence (with respect to the Mazurkiewicz metric d M ) in Ω. Then it is abounded sequence in X , and as X is proper, it has a subsequence, also denoted { x n } ,and a point x ∈ X such that lim n x n = x , the limit occurring with respect to themetric d X . If x ∈ Ω, then by the local path-connectivity of X , we also have that thelimit occurs with respect to d M as well. If x Ω, then x ∈ ∂ Ω. Since Ω is finitelyconnected at the boundary, it follows that there are only finitely many components U (1) , · · · , U k (1) of B ( x , ∩ Ω with x ∈ ∂U j (1) for j = 1 , · · · , k , and we alsohave that B ( x , ρ ) ∩ Ω ⊂ S k j =1 U j (1) for some ρ >
0. We have a choice of some j ∈ { , · · · , k } such that infinitely many of the terms in the sequence { x n } lie in U j (1). this gives us a subsequence { x n } with d M ( x n , x m ) ≤
1. Similarly, we havefinitely many components U (2) , · · · , U k (2) of B ( x , − ) ∩ Ω, with x ∈ ∂U j (2) for j = 1 , · · · , k , and we also have that B ( x , ρ ) ∩ Ω ⊂ S k j =1 U j (2) for some ρ > { x n } of the sequence { x n } lying entirely in U j (2) for some j ∈ { , · · · , k } . Note that d M ( x n , x m ) ≤ − . Inductively, for each l ∈ N with l ≥ { x ln } contained within a componentof B ( x , − l ) ∩ Ω with that component containing x in its boundary, such that { x ln } is a subsequence of { x l − n } . We then have d M ( x ln , x lm ) ≤ − l . Now a Cantordiagonalization argument gives a subsequence { y l } of the original sequence { x n } such that for each l ∈ N , d M ( y l , y l +1 ) ≤ − l , that is, this subsequence is Cauchywith respect to the Mazurkiewicz metric d M and so is convergent to a point in Ω M .We next verify that there cannot be any prime end of type (b). Suppose E isa prime end of Ω with impression I ( E ) containing more than one point. Then byLemma 2.12 there is a singleton prime end G with G | E , and as the impression of32 is not a singleton set, we have G = E , violating the primality of E . Thus therecannot be a prime end of type (b), nor a non-singleton prime end of type (a).To verify that Ω P is sequentially compact, we first consider sequences { x k } in Ω. If this sequence is bounded in the metric d X , then by the properness of X there is a subsequence { x n j } that converges to a point x ∈ Ω. If x ∈ Ω,then that subsequence converges to x also in Ω P . If x ∈ ∂ Ω, then using thefinite connectivity of Ω at x we can find a singleton prime end E = [ { E k } ] with { x } = I ( E ) such that for each k ∈ N there is an infinite number of positive integers j for which x n j ∈ E k . Now a Cantor-type diagonalization argument gives a furthersubsequence that converges in Ω P to E . If the sequence is not bounded in d X , thenwe argue as follows. Fix z ∈ Ω. Then for each n ∈ N there are only finitely manycomponents of Ω \ B M ( z, n ) that intersect Ω \ B M ( z, n + 1) (for if not, then we canfind a sequence of points y j in the unbounded components of Ω \ B ( z, n ), with notwo belonging to the same unbounded component, such that d M ( z, y j ) = n + andd M ( y j , y l ) ≥ for j = l ; then { y j } would be a bounded sequence with respect to d X ,and the properness of X gives us a subsequence that converges to some y ∈ ∂ Ω,and Ω would then not be finitely connected at y ). Hence there is an unboundedcomponent U of Ω \ B M ( z,
1) that contains x k for infinitely many k ∈ N , that is,there is a subsequence { x j } of { x k } that is contained in U . There is an unboundedcomponent U of Ω \ B M ( z,
2) that contains x j for infinitely many j ∈ N , and so weobtain a further subsequence { x j } lying in U . Since this further subsequence alsolies in U , it follows that U ⊂ U . Thus inductively we find unbounded components U n of Ω \ B M ( z, n ) and subsequence { x nj } of { x n − j } such that x nj ∈ U n ; moreover, U n ⊂ U n − . If T n U n is non-empty, then T n U n ⊂ ∂ Ω (because for every point w ∈ Ωwe know from the local connectivity of X that d M ( z, w ) < ∞ ), and Ω will not befinitely connected at each point in T n U n . It follows that we must have T n U n = ∅ .Therefore { U n } is an end at ∞ for Ω, and by considering the subsequence z n = x nn of the original sequence { x k } , we see that { z n } converges in Ω P to { U n } . Thisconcludes the proof that sequences in Ω have a subsequence that converges in Ω P .Next, if we have a sequence E n of points in ∂ P Ω that are all singleton prime ends(that is, they have a representative in ∂ M Ω), then we can use the Mazurkiewiczmetric to choose { E nk } ∈ E n such that for each n, k ∈ N we have diam M ( E nk ) < − ( k + n ) , and choose y n = x nn ∈ E nn to obtain a sequence in Ω. The above argumentthen gives a subsequence y n j and F ∈ ∂ P Ω such that y n j → F as j → ∞ in the primeend topology. If F is a singleton prime end, then the properness of Ω M implies that E n j → F . If F is not a singleton prime end, then it is an end at ∞ (because we haveshown that there are no ends of type (b)), in which case we can find a sequenceof positive real numbers R k → ∞ and { F k } ∈ F such that F k is a component ofΩ \ B M ( x , R k ) for some fixed x ∈ Ω, see Lemma 5.11. Since we can find L ∈ N such that R k + L − R k >
1, and as y j ∈ F k + L for sufficiently large j ∈ N , it followsthat E n j n j ⊂ F k . Therefore E n j → F . Finally, if E n are all ends at ∞ , then fromthe second part of Lemma 5.11 we can fix x ∈ Ω and choose { E nk } ∈ E n with E nk a component of Ω \ B M ( x , k ). For each k ∈ N there are only finitely manycomponents of Ω \ B M ( x , k ) that are unbounded, and so it follows that there is33ome component F of Ω \ B M ( x ,
1) for which the set I (1) := { n ∈ N : E n = F } has infinitely many terms. Note that necessarily F ⊂ F . It then follows that thereis some component F of Ω \ B M ( x ,
2) such that I (2) := { n ∈ I (1) : n ≥ E n = F } has infinitely many terms. Proceeding inductively, we obtain sets I ( j + 1) ⊂ I ( j ) ofinfinite cardinality and F j such that { F j } is an end at ∞ , and a sequence n j ∈ N with j ≤ n j ∈ I ( j ) such that E n j j = F j . It follows that E n converges in the primeend topology to { F j } .Now we prove that if Ω satisfies conditions (i)–(iii), then it is finitely connectedat the boundary. Recall the definition of finitely connected at the boundary fromDefinition 2.11. Suppose that x ∈ ∂ Ω such that Ω is not finitely connected at x .Then there is some r > B ( x , r ) ∩ Ω with x in their boundaries, or else there are only finitely manysuch components U , · · · , U k such that for all ρ > B ( x , ρ ) \ S kj =1 U j is non-empty. In the first case, B ( x , r/
2) intersects each of those components, and henceinfinitely many components of B ( x , r ) ∩ Ω intersect B ( x , r/ B ( x , r/ \ S kj =1 U j is non-empty; in this case, if there are only finitelymany components of B ( x , r ) ∩ Ω that intersect B ( x , r/ W that are not one of U , · · · , U k , we must have that dist( x , W ) > ρ ≤ r/ x , W ) over allsuch W tells us that B ( x , ρ ) ∩ Ω ⊂ S kj =1 U j , contradicting the fact that Ω is notfinitely connected at x .From the above two cases, we know that there is some r > B ( x , r ) ∩ Ω intersecting B ( x , r/ y j ∈ Ω, with no two belonging to the same componentof B ( x , r ) ∩ Ω, such that d ( x , y j ) = r/
2. Then we have that for each j, l ∈ N with j = l , d M ( y j , y l ) ≥ r/
2. Since by condition (i) we know that Ω M is proper, wemust necessarily have that { y j } has no bounded subsequence with respect to theMazurkiewicz metric d M . By the sequential compactness of Ω P there must then bea prime end E , and a subsequence, also denoted { y j } , such that this subsequenceconverges to E in the topology of Ω P . Since { y j } is a bounded sequence in X , itfollows from the properness of X that there is a subsequence converging to some w ∈ Ω. As w Ω, we must have that w ∈ ∂ Ω, and moreover, w ∈ I ( E ). Thus E is oftype (a) since by condition (ii) we have no type (b) prime ends. It then follows thatwhen { E k } ∈ E , for sufficiently large k the open connected set E k must be boundedin d X and hence in d M . This creates a conflict between the fact that the tail endof the sequence { y j } lies in E k and the fact that { y j } has no bounded subsequencewith respect to the metric d M . Hence Ω must be finitely connected at x . Example 6.2.
Let Ω ⊂ R be given byΩ = (0 , ∞ ) × (0 , \ [ k ∈ N [0 , k ] × { − k } . { x k } in Ω that isbounded in the Mazurkiewicz metric d M will have to lie in a subset (0 , ∞ ) × (2 − k , k ∈ N , and so Ω M is proper. However, Ω is not finitely connected at theboundary. Thus Conditions (i),(ii) on their own do not characterize finite connect-edness at the boundary. Lemma 6.3.
Suppose that Ω M proper. Let A, B ⊂ Ω with dist M ( A M , B M ) = τ > and diam M ( A ) , diam M ( B ) both finite. Then either diam ′ M ( f ( A )) is finite or else diam ′ M ( f ( B )) is finite.Proof. Suppose that diam ′ M ( f ( A )) = ∞ = diam ′ M ( f ( B )). Then we can find twosequences z k ∈ f ( A ) and w k ∈ f ( B ) such that d ′ M ( z k , z k +1 ) ≥ k and d ′ M ( w k , w k +1 ) ≥ k for each k ∈ N . Let x k = f − ( z k ) and y k = f − ( w k ). By the properness of Ω M , wecan find two subsequences, also denoted x k and y k , such that these subsequences areCauchy in d M . We can also ensure that d M ( x k , x k +1 ) < − k τ / M ( y k , y k +1 ) < − k τ /
8. Then for each k there is are continua α k , β k ⊂ Ω with x k , x k +1 ∈ α k , y k , y k +1 ∈ β k , and max { diam M ( α k ) , diam M ( β k ) } < − k τ /
8. Since x k ∈ A , y k ∈ B ,and dist M ( A M , B M ) = τ >
0, we must have that S k α k and S k β k are disjoint withdist M [ k α k , [ k β k ! ≥ τ / . For each n ∈ N set Γ n = S nk =1 α k and Λ n = S nk =1 β k . Then Γ n and Λ n are disjointcontinua with dist M (Λ n , Γ n ) ≥ τ / n ∪ Λ n ⊂ [ z ∈ A ∪ B B ( z, τ + diam( A ) + diam( B )) =: U. Hence Mod Q (Γ(Γ n , Λ n )) ≤ Z Ω ρ Q dµ < ∞ where ρ := τ χ U is necessarily admissible for the family Γ(Γ n , Λ n ) of curves in Ωconnecting points in Γ n to points in Λ n .However, f (Λ n ) is a continuum containing z and z n , and so diam ′ M ( f (Λ n )) → ∞ as n → ∞ ; a similar statement holds for f (Γ n ). Moreover, dist M ′ ( f (Λ n ) , f (Γ n )) ≤ d ′ M ( z , w ) < ∞ . Hence by the Q -Loewner property together with the Ahlfors Q -regularity of (Ω ′ , d ′ M ) gives usMod Q (Γ( f (Γ n ) , f (Λ n ))) ≥ Φ (cid:18) dist M ′ ( f (Γ n ) , f (Λ n ))min { diam ′ M ( f (Λ n )) , diam ′ M ( f (Γ n )) } (cid:19) → ∞ as n → ∞ . this violates the geometric quasiconformality of f . Lemma 6.4.
Suppose that (Ω , d M ) is Ahlfors Q -regular for some Q > and supportsa Q -Poincar´e inequality. Then every type (a) prime end of Ω is a singleton primeend, and Ω M is proper. Moreover, Ω P is sequentially compact. M is proper andΩ P is sequentially compact but Ω is not finitely connected at its boundary. Wedo not know whether this is still possible if we also require (Ω , d M ) to be Ahlfors Q -regular for some Q > Q -Poincar´e inequality. Proof.
The properness of Ω M is a consequence of the fact that H Q (this Hausdorffmeasure taken with respect to the Mazurkiewicz metric d M ) is doubling and thatΩ M is complete, see for example [HKST, Lemma 4.1.14].Let E be a prime end of type (a) for Ω, and let x ∈ I ( E ). Let { E k } ∈ E such that E is bounded (in d X , and hence by the local connectedness of X and the fact that E is open, in the Mazurkiewicz metric d M as well). Let { x j } be a sequence in E with x j ∈ E j such that it converges (in the metric d X ) to x . Then as this sequenceis bounded in d M , it follows from the properness of Ω M that there is a subsequence,also denoted { x j } , such that this subsequence converges in d M to a point ζ ∈ ∂ M Ω.For n ∈ N we claim that B M ( ζ, − n ) ∩ Ω is connected. Indeed, by the definition ofMazurkiewicz closure, we can find a Cauchy sequence x k ∈ Ω with d M ( x k , ζ ) → x ∈ B M ( ζ, − n ) ∩ Ω we have d M ( x, ζ ) < − n , and so for sufficiently large k wealso have that d M ( x, x k ) < − n . Thus we can find a continuum in Ω connecting x to x k with diameter smaller than 2 − n . Note that each point on this continuumnecessarily then lies in B M ( ζ, − n ) ∩ Ω. Now if y ∈ B M ( ζ, − n ) ∩ Ω, then we canconnect both x and y to x k for sufficiently large k and hence obtain a continuumin B M ( ζ, − n ) ∩ Ω connecting x to y . Therefore B M ( ζ, − n ) ∩ Ω is a connected set.Let F n = B M ( ζ, − n ) ∩ Ω that contain x j for sufficiently large j . Then { F n } is achain for Ω with { x } = I ( { F n } ), and so it forms a singleton prime end F = [ { F n } ].We claim now that F | E . For each k ∈ N we know that x j ∈ E k +1 for all j ≥ k + 1;as dist M (Ω ∩ ∂E k , Ω ∩ ∂E k +1 ) >
0, it follows that for large j the connected set F j contains x m for all m ≥ j + 1 withdiam M ( F j ) <
12 dist M (Ω ∩ ∂E k , Ω ∩ ∂E k +1 );hence F j ⊂ E k for sufficiently large j . Now the primality of E tells us that E = F ,that is, E is a singleton prime end.Now we show that Ω P is sequentially compact. If { x j } is a sequence in Ω thatis bounded in the Mazurkiewicz metric d M , then by the properness of Ω M we seethat there is a subsequence that converges to a singleton prime end or to a pointin Ω. Similarly, if we have a sequence of singleton prime ends that is bounded withrespect to d M (and recall that singleton prime ends correspond to points in theMazurkiewicz boundary ∂ M Ω, see [BBS2]), then we have a convergent subsequence.It now only remains to consider sequences { ζ j } from Ω P such that the sequence iseither an unbounded (in d M ) sequence of points from Ω M or is a sequence of primeends that are of types (b) and (c). In this case, we fix x ∈ Ω and construct a primeend [ { F k } ] as follows. With n k = k in Lemma 5.8, we choose F to be the unboundedcomponent of Ω \ B M ( x ,
1) that contain infinitely many points from { ζ j } if this36equence consist of points in Ω M , or tail-end of the chains { E jk } ∈ ζ j for infinitelymany j . We then choose F to be the unbounded component of Ω \ B M ( x ,
2) thatcontain infinitely many of the points { ζ j } that are in F if { ζ j } ⊂ Ω M , or tail-end of the chains { E jk } ∈ ζ j for infinitely many j that also gave a tail-end for F .Proceeding inductively, we obtain a chain { F k } ; by Lemma 5.8 we know that [ { F k } ]is a prime end of Ω, and by the above construction and by a Cantor diagonalization,we have a subsequence of { ζ j } that converges to [ { F k } ].Chains in prime ends of type (b) have corresponding separating sets R k , butno control over the Mazurkiewicz diameter of the boundaries of the acceptable setsthat make up the chains. The following lemma and its corollary give us a goodrepresentative chain in the prime end of type (b) that gives us control over theboundary of the acceptable sets as well as dispenses with the need for the separatingsets R k . Lemma 6.5.
Let E be an end of type (b) of Ω . Let { E k } be a representative chainfor E and let R k be as given in Definition 5.1. Then, there is a representative chain { F k } of E and compact sets S k ⊂ X with the following properties:(i) F k is a component of Ω \ S k ,(ii) diam M ( S k ∩ Ω) < ∞ ,(iii) dist M ( S k ∩ Ω , S k +1 ∩ Ω) > .Furthermore, there are corresponding separating sets R Fk from Definition 5.1 thathave the properties that dist M ( R Fk ∩ Ω , R Fk +1 ∩ Ω) > and diam M ( R Fk ∩ Ω) < ∞ .Proof. Let T k = ( R k ∩ Ω \ E k +1 ) ∩ E k . We see E k +1 belongs to a single component of Ω \ T k because E k +1 is connected and E k +1 ∩ T k = ∅ . This follows as E k +1 is open in X and ( R k ∩ Ω \ E k +1 ) ∩ E k = ∅ . Wealso see that T k ∩ Ω ⊂ E k because R k ∩ ∂E k = ∅ and so T k ∩ Ω = ( R k ∩ E k ) \ E k +1 .By Lemma 5.3, we observe that there are no points x ∈ ∂ Ω ∩ E k and y ∈ Ω ∩ ∂E k +1 that belong to the same component of Ω \ T k .Let S k = T k and let F k be the component of Ω \ S k containing E k +1 . The sets S k are compact as they are closed subsets of the sets R k . We first verify properties(i) - (iii) and then show that { F k } is a chain equivalent to { E k } . Property (i)follows immediately from the definition of F k . Property (ii) follows as S k ⊆ R k anddiam M ( R k ) < ∞ . For property (iii), let x ∈ S k ∩ Ω and y ∈ S k +1 ∩ Ω. Let E ⊂ Ωbe a continuum containing x and y . As x ∈ S k , we have x / ∈ E k +1 . As y ∈ S k +1 , wehave y ∈ E k +1 , so there must exist a point z ∈ E ∩ ∂E k +1 . Similarly, x / ∈ E k +2 and y ∈ E k +2 , so there is a point z ∈ E ∩ ∂E k +2 . Hence,0 < dist M ( ∂E k +1 ∩ Ω , ∂E k +2 ∩ Ω) ≤ d M ( z , z ) . and, as x, y are arbitrary, we have dist M ( S k ∩ Ω , S k +1 ∩ Ω) > F k is a chain of type (b) that is equivalent to { E k } .We first note that F k +1 ⊂ E k +1 ⊂ F k , from which equivalence follows. Indeed,37 k +1 ⊂ F k follows immediately from the definition of F k . To see that F k +1 ⊂ E k +1 ,suppose that there is a point y ∈ F k +1 \ E k +1 . Let x ∈ E k +3 ⊂ F k +1 ; then by theconnectedness of the open set F k +1 together with local connectedness of Ω, there is acontinuum γ ⊂ F k +1 containing the points x, y . Then as x ∈ E k +1 and y E k +1 , γ intersects Ω ∩ ∂E k +1 . Moreover, as x ∈ E k +2 and y E k +2 , we must also havethat γ intersects Ω ∩ ∂E k +2 . Then as T k +1 separates Ω ∩ ∂E k +1 from Ω ∩ ∂E k +2 ,it follows that γ intersects T k +1 , contradicting the fact that γ ⊂ F k +1 .Properties (a) and (d) in Definition 5.1 follow immediately from the above ob-servation. Property (c) follows from (iii) as Ω ∩ ∂F k ⊆ S k ∩ Ω. For property (b),we use the sets R Fk = T k +1 . Compactness of R Fk follows as before: R Fk is a closedsubset of the compact set R k +1 . We see diam M ( R Fk ∩ Ω) ≤ diam M ( R k +1 ∩ Ω) < ∞ .The separation property of R Fk is similar to the separation property for the sets T k :let x ∈ ∂F k and y ∈ ∂F k +1 and suppose γ is a path in Ω \ T k +1 with γ (0) = x and γ (1) = y . Then, y ∈ T k +2 so y ∈ E k +2 and x ∈ T k so x / ∈ E k +1 . Thus, γ contains points in ∂E k +1 and ∂E k +2 . Considering an appropriate subpath of γ asbefore leads to a point in γ ∩ T k +1 , a contradiction.The inequality dist M ( R Fk ∩ Ω , R Fk +1 ∩ Ω) > M ( S k ∩ Ω , S k +1 ∩ Ω) >
0. The inequality diam M ( R Fk ∩ Ω) < ∞ follows as R Fk ∩ Ω ⊆ R k +1 .The following corollary is a direct consequence of the above lemma and its proof. Corollary 6.6. If { E k } is a representative chain of a type (b) prime end of Ω , thenthere is a sequence { F k } of unbounded open connected subsets of Ω such that(a) F k +1 ⊂ F k for each k ∈ N ,(b) diam M (Ω ∩ ∂F k ) < ∞ for each k ∈ N ,(c) dist M (Ω ∩ ∂F k , Ω ∩ ∂F k +1 ) > for each k ∈ N ,(d) ∅ 6 = T k F k ⊂ ∂ Ω ,and in addition, for each k ∈ N we can find j k , i k ∈ N such that F j k ⊂ E k and E i k ⊂ F k . Moreover, if { F k } is a sequence of unbounded open connected subsets of Ω that satisfyConditions (a)–(d) listed above, then { F k } is a representative chain of a type (b)prime end of Ω . Lemma 6.7.
Suppose that both (Ω , d M ) and (Ω ′ , d ′ M ) are Ahlfors Q -regular for some Q > and that they both support a Q -Poincar´e inequality with respect to the respec-tive Ahlfors regular Hausdorff measure. Let E be an end of any type of Ω . Let { E k } ∈ E be a chain with diam M (Ω ∩ ∂E k ) < ∞ for all k and let F k = f ( E k ) .Then, dist M ′ (Ω ′ ∩ ∂F k , Ω ′ ∩ ∂F k +1 ) > . (8)Chains with diam M (Ω ∩ ∂E k ) < ∞ for all k exist for all types of ends by Lemma6.4 (for type (a)), Lemma 6.5 (for type (b)), and Lemma 5.11 (for type (c)). Heredist M ′ is the distance between the two sets with respect to the Mazurkiewicz metricd ′ M . 38 roof. Suppose (8) is not the case for some positive integer k . Then we can findtwo sequences w j ∈ Ω ′ ∩ ∂F k and z j ∈ Ω ′ ∩ ∂F k +1 such that d ′ M ( w j , z j ) → j → ∞ . Let x j ∈ Ω ∩ ∂E k be such that f ( x j ) = w j , and y j ∈ Ω ∩ ∂E k +1 be suchthat f ( y j ) = z j . By passing to a subsequence if necessary, we may assume that w = w and that z = z ; therefore we also have x = x and w = w . Note thatas τ k := dist M (Ω ∩ ∂E k , Ω ∩ ∂E k +1 ) > , we must have that ∞ > diam M ( E ) ≥ d M ( x j , y j ) ≥ τ k for each j .By Lemma 6.4 we know that Ω M is proper, and so sequences that are boundedin Ω have a subsequence that converges in d M (and hence in d X as well) to somepoint in Ω M . By [BBS1, Theorem 1.1] we know that there is a bijective correspon-dence between singleton prime ends and points in ∂ M Ω. Therefore by passing to asubsequence if necessary, we can find ζ, ξ ∈ ∂ P Ω with ζ = ξ such that x j → ζ and y j → ξ . Note that d M ( ξ, ζ ) ≥ τ k . So we can choose chains { G m } ∈ ξ and { H m } ∈ ζ such that for each m the completion in d M of G m and the completion of H m in d M are disjoint (indeed, we can ensure that G m ⊂ B M ( ξ, τ k /
3) and H m ⊂ B M ( ζ, τ k / x j → ζ , by passing to a further subsequence if needed, we can ensure that foreach positive integer j , x j ∈ H ; similarly, we can assume that y j ∈ G . As H and G are open connected subsets of Ω and X is locally path-connected, it follows thatfor each j there is a path γ j in H connecting x j to x j +1 ; similarly there is a path β j in G connecting y j to y j +1 . For each positive integer n let α n be the concatenationof γ j , j = 1 , · · · , n , and let σ n be the concatenation of β j , j = 1 , · · · , n . Then α n and σ n are continua in Ω with dist M ( α n , σ n ) ≥ τ k /
3. Therefore for each n we haveMod Q (Γ( α n , σ n )) ≤ Q τ Qk µ (Ω) < ∞ . On the other hand, with A n = f ( α n ) and Σ n = f ( σ n ), we know that dist M ′ ( A n , Σ n ) → n → ∞ . However,lim inf n →∞ diam ′ M ( A n ) ≥ diam ′ M ( A ) > , lim inf n →∞ diam ′ M (Σ n ) ≥ diam ′ M (Σ ) > . Therefore by the Ahlfors regularity together with the Poincar´e inequality on Ω ′ withrespect to d ′ M , we know thatMod Q (Γ( A n , Σ n )) → ∞ as n → ∞ . Since f (Γ( α n , σ n )) = Γ( A n , Σ n ) , this violates the geometric quasiconformality of f , see Theorem 3.8. Hence (8) musthold true. Lemma 6.8.
Suppose that both (Ω , d M ) and (Ω ′ , d ′ M ) are Ahlfors Q -regular for some Q > and that they both support a Q -Poincar´e inequality with respect to the respec-tive Ahlfors regular Hausdorff measure. Let E be an end of any type of Ω . Let { E k } ∈ E be an appropriate chain with diam M (Ω ∩ ∂E k ) < ∞ for all k and let F k = f ( E k ) . Then, i) If diam( F k ) < ∞ for some k , then { F k } is a chain corresponding to an end oftype (a).(ii) If diam( F k ) = ∞ for all k and T k F k = ∅ , then { F k } is a chain correspondingto an end of type (b).(iii) If diam( F k ) = ∞ for all k and T k F k = ∅ , then { F k } is a chain correspondingto an end of type (c). We use the phrase “appropriate chain” to mean one which has the properties inLemma 6.5 for type (b) or of the form given in Lemma 5.11 for type (c). As f is ahomeomorphism, this means that f induces a map from ends of Ω to ends of Ω ′ . Proof.
Let E be an end and let { E k } be a chain representing E with diam M (Ω ∩ ∂E k ) < ∞ for all k . Let F k = f ( E k ). The proof will be split into several cases. Wenote that in all cases, the containments F k +1 ⊆ F k are immediate and the separationinequality (8) follows from Lemma 6.7, so we only check the other conditions. Case 1 [(a),(b),(c) → (a)]: Suppose diam( F k ) < ∞ for some k . By consideringthe tail end of this sequence, we assume diam( F k ) < ∞ for all k . Then, as f isa homeomorphism, each F k is a bounded, connected, open set. The sets F k arecompact as Y is proper. As Y is complete, it follows that T k F k = ∅ . The factthat T k F k ∩ Ω ′ = ∅ follows as f is a homeomorphism: if x ′ = f ( x ) ∈ T k F k ∩ Ω ′ ,then as Ω ′ is open we can find r ′ > B ( x ′ , r ) ⊂ Ω ′ . Then, there is an r > B ( x, r ) ⊂ f − ( B ( x ′ , r ′ )). For all types of ends, there is an index I with E I ∩ B ( x, r/
2) = ∅ . Hence f ( B ( x, r/ ∩ F I = ∅ , and so x / ∈ F I . In particular, F k ∩ ∂ Ω ′ = ∅ for all k , so the sets F k are admissible, and T k F k ⊂ ∂ Ω, so { F k } is achain. Case 2 [(a),(b),(c) → (b)]: We assume that diam( F k ) = ∞ for all k and that T k F k = ∅ . As in Case 1, we must have T k F k ⊆ ∂ Ω ′ . Moreover, F k is proper for all k as Y is proper. Now the fact that { F k } corresponds to an end of type (b) followsfrom Corollary 6.6 together with Lemma 6.3. Case 3 [(a),(b),(c) → (c)]: We assume that diam( F k ) = ∞ for all k and that T k F k = ∅ . We only need to check that there exists a compact subset K k ⊂ Y withdiam ′ M ( K k ∩ Ω ′ ) < ∞ such that F k is a component of Ω ′ \ K k . Note that for each k we have diam M (Ω ∩ ∂E k ) < ∞ , and dist M (Ω ∩ ∂E k , Ω ∩ ∂E k +1 ) >
0. Thereforefor all except at most one k we have that diam ′ M (Ω ′ ∩ ∂f ( E k )) < ∞ by Lemma 6.3.Set K k = f ( ∂E k ∩ Ω ′ ). To see that F k = f ( E k ) is a component of Ω ′ \ K k , wenote that as F k is connected it is contained in some component of Ω ′ \ K k . If F k is not this entire component, then as open connected sets are path connected, thiscomponent contains a point of ∂F k , but this is impossible as ∂F k ⊆ K k because f is a homeomorphism. Theorem 6.9.
Suppose that both (Ω , d M ) and (Ω ′ , d ′ M ) are Ahlfors Q -regular forsome Q > and that they both support a Q -Poincar´e inequality with respect tothe respective Ahlfors regular Hausdorff measure. Let f : Ω → Ω ′ be a geometricquasiconformal mapping. Then, f induces a homeomorphism f : Ω P → Ω ′ P . roof. By Lemma 6.8, f induces a map from ends of Ω to ends of Ω ′ . As continuityis determined by inclusions of sets in chains and f as a homeomorphism preservesthese inclusions, the extended f is also a homeomorphism.To see that f maps prime ends to prime ends, suppose that E is a prime end ofΩ. Then, by Lemma 6.8, f ( E ) is an end of Ω ′ . If F is an end with F | f ( E ), then byapplying Lemma 6.8 to f − , we see that f − ( F ) is an end with f − ( F ) | E . As E isprime, it follows that E | f − ( F ) as well, and so f ( E ) | F . It follows that f ( E ) = F , so f ( E ) is prime as F was arbitrary.We end this paper by stating the following two open problems:1 If Ω is a domain (open connected set) in a proper locally path-connected metricspace and (Ω , d M ) is Ahlfors Q -regular and supports a Q -Poincar´e inequalityfor some Q >
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