# Convergence of group actions
in metric measure geometry

Hiroki Nakajima  Institute for Excellence in Higher Education, Tohoku
University, Sendai 980-8576, Japan<EMAIL_ADDRESS>and
Takashi Shioya  Mathematical Institute, Tohoku University, Sendai 980-8578,
JAPAN<EMAIL_ADDRESS>
###### Abstract.

We generalize the box and observable distances to those between metric measure
spaces with group actions, and prove some fundamental properties. As an
application, we obtain an example of a sequence of lens spaces with unbounded
dimension converging to the cone of the infinite-dimensional complex
projective space. Our idea is to use the theory of mass-transport.

###### Key words and phrases:

box distance, observable distance, group action

###### 2010 Mathematics Subject Classification:

53C23

This work was supported by JSPS KAKENHI Grant Number 19K03459 and 19J10866.

##  1\. Introduction

Fukaya [Fk:Theory] first introduced the concept of convergence of metric
spaces with isometric group actions, which was applied to the study of the
collapsing theory and fundamental group (see [FY:f-gp]). Recently, it is
further developed in the study of dynamical system (see for example [AR:top-
stab]). In this paper, we study convergence of metric measure spaces (mm-
spaces) with measure preserving and isometric group actions. We generalize the
box and observable distances due to Gromov [Gmv:green] to those between mm-
spaces with group actions, and study their properties. Our main results are
the following.

  1. (1)

We define the equivariant box and observable metric and prove that they are
really metrics (Theorem 4.10 and 4.7).

  2. (2)

We prove that if a sequence of metric measure spaces with group actions
converges with respect to the equivariant box/observable metric, then so does
the sequence of the quotient spaces (Theorem 5.4 and 5.7).

  3. (3)

We prove that the projection from the set of mm-spaces with group actions to
the set of mm-spaces (i.e., the projection dropping group actions) is a proper
map (Theorem 6.5).

  4. (4)

As an application, we have an example of a sequence of lens spaces with
unbounded dimension converging to the cone of the infinite-dimensional complex
projective space in the box metric (Theorem 7.1).

For (1), we remark that the nondegeneracy of the metrics is not so easy to
prove even for mm-spaces without group actions. Our idea is to use the mass-
transport theory to formulate the equivariant box and observable metrics. Such
a formulation enables us to prove the nondegeneracy.

The key point of (2) is to prove the precompactness of groups acting on metric
measure spaces with respect to the Ky Fan metric by using the thick-thin
decomposition of the spaces (see Lemmas 5.2 and 5.3). It is delicate that the
groups do not Gromov-Hausdorff converge to the group acting on the limit space
with respect to the Ky Fan metric.

(1)–(3) was proved in the case of metric spaces without measures in
[Fk:Theory, FY:f-gp]. We emphasize that the proofs in our case are much more
delicate.

(4) is an interesting example of convergence which is only obtained for metric
measure spaces and not for metric spaces.

######  Acknowledgment.

The authors would like to thank Daisuke Kazukawa for pointing out some minor
mistakes in an earlier version of this paper and for valuable comments.

##  2\. Preliminaries

In this paper, a _metric measure space_ , an _mm-space_ for short, means a
complete separable metric space with a Borel probability measure. We assume
that the measure of an mm-space has full-support unless otherwise stated. For
an mm-space $X$, the metric and measure of $X$ are respectively denoted by
$d_{X}$ and $\mu_{X}$. Sometimes, the triple $(X,d_{X},\mu_{X})$ is called an
mm-space. Two mm-spaces $X$ and $Y$ are _mm-isomorphic_ to each other if there
exists an isometry $f:X\to Y$ with $f_{*}\mu_{X}=\mu_{Y}$, where
$f_{*}\mu_{X}$ is the push-forward measure of $\mu_{X}$ by $f$.

For finitely many sets $X_{i}$, $i=1,2,\dots,n$, and for numbers
$i_{1},i_{2},\dots,i_{k}\in\\{1,2,\dots,n\\}$, let $p_{i_{1}i_{2}\dots
i_{k}}:\prod_{i=1}^{n}X_{i}\to\prod_{j=1}^{k}X_{i_{j}}$ be the projection
defined by

$p_{i_{1}i_{2}\dots
i_{k}}(x_{1},x_{2},\dots,x_{n}):=(x_{i_{1}},x_{i_{2}},\dots,x_{i_{k}}),\qquad
x_{i}\in X_{i},\ i=1,2,\dots,n.$

Let $X$, $Y$, and $Z$ be mm-spaces. Denote by $\Pi(X,Y)$ the set of couplings
(or transport plan) between $\mu_{X}$ and $\mu_{Y}$. It is well-known that
$\Pi(X,Y)$ is compact with respect to the Prokhorov metric $\operatorname{{\it
d}_{{\rm P}}}$. For $\sigma\in\Pi(X,Y)$ and $\tau\in\Pi(Y,Z)$, the _gluing_
$\tau\bullet\sigma\in P(X\times Y\times Z)$ is defined by

$\tau\bullet\sigma(A\times B\times
C):=\int_{B}\sigma_{y}(A)\tau_{y}(C)\,d\mu_{Y}(y),$

where $\\{\sigma_{y}\\}_{y\in Y}$ and $\\{\tau_{y}\\}_{y\in Y}$ are the
disintegrations respectively of $\sigma$ and $\tau$ for the projections
$p_{2}:X\times Y\to Y$ and $p_{1}:Y\times Z\to Y$. It holds that
$(p_{12})_{*}(\tau\bullet\sigma)=\sigma$ and
$(p_{23})_{*}(\tau\bullet\sigma)=\tau$. We also define
$\tau\circ\sigma:=(p_{13})_{*}(\tau\bullet\sigma)\in\Pi(X,Z)$. We observe the
associativity $(\upsilon\circ\tau)\circ\sigma=\upsilon\circ(\tau\circ\sigma)$.
For subsets $S\subset X\times Y$ and $T\subset Y\times Z$, we define $T\bullet
S:=(X\times T)\cap(S\times Z)$, $T\circ S:=p_{13}(T\bullet S)$,
$\operatorname{Dom}(S):=p_{1}(S)$, $S^{-1}:=\\{\,(y,x)\mid(x,y)\in S\,\\}$,
and $\operatorname{Im}(S):=\operatorname{Dom}(S^{-1})$. If $S$ and $T$ are
Borel, then $\operatorname{Dom}(S)$, $T\circ S$, and
$\operatorname{Dom}(T\circ S)$ are all analytic sets and measurable with
respect to the completion of any Borel measure. We denote the completion of a
Borel measure by the same notation. Let $\sigma\in\Pi(X\times Y)$ and
$\tau\in\Pi(Y\times Z)$. Since
$\mu_{X}(\operatorname{Dom}(S))=\sigma(\operatorname{Dom}(S)\times
Y)\geq\sigma(S)$, we see

$\displaystyle\mu_{X}(\operatorname{Dom}(T\circ
S))\geq(\tau\circ\sigma)(T\circ S)\geq(\tau\bullet\sigma)(T\bullet S)$
$\displaystyle=1-(\tau\bullet\sigma)((X\times T)^{c}\cup(S\times
Z)^{c})\geq\tau(T)+\sigma(S)-1,$

which is often used in this paper.

##  3\. Equivariant box distance

In this section, we introduce the box distance between mm-spaces with mm-
actions and prove a triangle inequality.

Let $X$ be an mm-space and $G$ a group. An _mm-action_ of $G$ on $X$ is, by
definition, isometric action of $G$ preserving $\mu_{X}$. Denote by
$\operatorname{Aut}(X)$ the group of isometries of $X$ preserving $\mu_{X}$.
If an mm-action of $G$ on $X$ is effective, then $G$ is identified with a
subgroup of $\operatorname{Aut}(X)$.

######  Definition 3.1.

Let $X$ and $Y$ be two mm-spaces and let $G\subset\operatorname{Aut}(X)$ and
$H\subset\operatorname{Aut}(Y)$ be subgroups. We say that $(X,G)$ is
_equivariantly mm-isomorphic_ to $(Y,H)$ and write $(X,G)\cong(Y,H)$ if there
exists an mm-isomorphism from $X$ to $Y$ that is equivariant for some group-
isomorphism from $G$ to $H$. Denote by $\mathcal{X}_{\rm eq}$ the equivariant
mm-isomorphism classes of $(X,G)$, where $X$ is any mm-space and $G$ is any
_closed_ subgroup of $\operatorname{Aut}(X)$.

We sometimes write $(X,G)\in\mathcal{X}_{\rm eq}$, which means that $(X,G)$ is
a representative of a class in $\mathcal{X}_{\rm eq}$.

The following lemma is obvious and the proof is omitted.

######  Lemma 3.2.

Let $X$ be a complete separable metric space and $Y$ a metric space. Let
$\varphi_{i}:X\to Y$, $i=1,2,\dots$, be $C$-Lipschitz maps for a constant
$C\geq 0$, and $X_{0}\subset X$ a dense subset. Then, if $\varphi_{i}$
converges pointwise on $X_{0}$ as $i\to\infty$, then $\varphi_{i}$ converges
pointwise on the whole $X$.

We denote the Prokhorov metric by $\operatorname{{\it d}_{{\rm P}}}$ and the
Ky Fan metric by $\operatorname{{\it d}_{{\rm KF}}}$.

######  Lemma 3.3.

Let $X$ be an mm-space. For a sequence of elements of $\operatorname{Aut}(X)$,
the following (i)—(iii) are equivalent to each other.

  1. (i)

The sequence converges uniformly on compact sets.

  2. (ii)

The sequence converges pointwise.

  3. (iii)

The sequence converges in measure.

Moreover, $\operatorname{Aut}(X)$ is compact with respect to the Ky Fan
metric.

###### Proof.

Let us first prove the relative compactness of the orbit
$\operatorname{Aut}(X)x=\\{\varphi(x)\mid\varphi\in\operatorname{Aut}(X)\\}$
for each $x\in X$. In fact, for any $\varepsilon>0$ and
$\varphi\in\operatorname{Aut}(X)$, since
$\mu_{X}(B_{\varepsilon/2}(x))=\mu_{X}(B_{\varepsilon/2}(\varphi(x)))$, any
$\varepsilon$-discrete net in $\operatorname{Aut}(X)x$ is finite and so
$\operatorname{Aut}(X)x$ is precompact. Since $X$ is complete, the orbit
$\operatorname{Aut}(X)x$ is relatively compact.

The implication (i) $\Rightarrow$ (ii) is clear.

We prove (ii) $\Rightarrow$ (i). Let a sequence
$\\{\varphi_{i}\\}\subset\operatorname{Aut}(X)$ converge pointwise to a map
$\varphi$ and let $K\subset X$ be compact. If $\\{\varphi_{i}\\}$ does not
converge uniformly to $\varphi$ on $K$, then there are a subsequence
$\\{\varphi_{i_{j}}\\}$ and points $x_{j},x\in K$ such that

(3.1)
$\lim_{j\to\infty}x_{j}=x\quad\text{and}\quad\inf_{j}d_{X}(\varphi_{i_{j}}(x_{j}),\varphi(x))>0.$

Since $\operatorname{Aut}(X)$ is equicontinuous and any orbit is relatively
compact, Ascoli-Arzelà theorem implies that $\\{\varphi_{i_{j}}\\}$ has a
subsequence converging uniformly on $K$. However, the limit must be $\varphi$,
which contradicts (3.1). We therefore obtain (i).

The implication (i) $\Rightarrow$ (iii) is clear.

To prove (iii) $\Rightarrow$ (ii), we assume that a sequence
$\\{\varphi_{i}\\}\subset\operatorname{Aut}(X)$ converges in measure to
$\varphi$. For any $\varepsilon>0$, setting $\Phi_{i,\varepsilon}:=\\{\,\xi\in
X\mid d_{X}(\varphi_{i}(\xi),\varphi(\xi))<\varepsilon\,\\}$, we have
$\mu_{X}(\Phi_{i,\varepsilon})\to 1$ as $i\to\infty$. For any $x\in X$ and
$\varepsilon>0$, the ball $B_{\varepsilon}(x)$ intersects
$\Phi_{i,\varepsilon}$ for all sufficiently large $i$. There is a point
$x^{\prime}\in B_{\varepsilon}(x)$ with
$d_{X}(\varphi_{i}(x^{\prime}),\varphi(x^{\prime}))<\varepsilon$. Since
$d_{X}(\varphi(x),\varphi(x^{\prime}))=d_{X}(x,x^{\prime})<\varepsilon$ and
$d_{X}(\varphi_{i}(x),\varphi_{i}(x^{\prime}))=d_{X}(x,x^{\prime})<\varepsilon$,
a triangle inequality implies $d_{X}(\varphi_{i}(x),\varphi(x))<3\varepsilon$.
Thus, $\varphi_{i}(x)$ converges to $\varphi(x)$. (ii) has been obtained.

We have proved that (i)–(iii) are equivalent to each other. The rest is to
prove the compactness of $\operatorname{Aut}(X)$.

Take any sequence $\\{\varphi_{i}\\}\subset\operatorname{Aut}(X)$ and a dense
countable subset $X_{0}\subset X$. Since any $\operatorname{Aut}(X)$-orbit is
relatively compact, $\\{\varphi_{i}(x_{0})\\}$ for any fixed $x_{0}\in X_{0}$
has a convergent subsequence. By a standard diagonal argument, there is a
subsequence $\\{\varphi_{i_{j}}\\}$ that converges pointwise on $X_{0}$. By
Lemma 3.2, $\varphi_{i_{j}}$ converges pointwise (and so, in measure) to a map
$\varphi:X\to X$ on $X$. The map $\varphi$ is an isometry. It follows from
[Sy:mmg]*Lemma 1.26 that $\operatorname{{\it d}_{{\rm
P}}}(\mu_{X},\varphi_{*}\mu_{X})=\operatorname{{\it d}_{{\rm
P}}}((\varphi_{i_{j}})_{*}\mu_{X},\varphi_{*}\mu_{X})\leq\operatorname{{\it
d}_{{\rm KF}}}(\varphi_{i_{j}},\varphi)\to 0$ as $j\to\infty$, which implies
$\varphi_{*}\mu_{X}=\mu_{X}$ and so $\varphi\in\operatorname{Aut}(X)$. This
completes the proof of the lemma. ∎

######  Definition 3.4 (Equivariant box metric).

Let $(X,G),(Y,H)\in\mathcal{X}_{\rm eq}$. For a nonempty subset $S\subset
X\times Y$ and $\gamma_{1},\gamma_{2}\in X^{X}\sqcup Y^{Y}$, we define

$d^{S}(\gamma_{1},\gamma_{2}):=\sup_{S\times
S}|\,f_{\gamma_{1}}-f_{\gamma_{2}}\,|\quad(\leq+\infty),$

where, for $(x_{i},y_{i})\in S$, $i=1,2$,

$f_{\gamma}((x_{1},y_{1}),(x_{2},y_{2})):=\begin{cases}d_{X}(\gamma(x_{1}),x_{2})&\text{for
$\gamma\in X^{X}$,}\\\ d_{Y}(\gamma(y_{1}),y_{2})&\text{for $\gamma\in
Y^{Y}$.}\end{cases}$

We set $d^{\emptyset}(\gamma_{1},\gamma_{2}):=0$. For $\pi\in\Pi(X,Y)$ and
$\gamma_{1},\gamma_{2}\in X^{X}\sqcup Y^{Y}$, we define

$d^{\pi}(\gamma_{1},\gamma_{2}):=\inf_{S}\max\\{1-\pi(S),d^{S}(\gamma_{1},\gamma_{2})\\},$

where $S$ runs over all closed subsets of the support
‘$\operatorname{supp}\pi$’ of $\pi$. Both $d^{S}$ and $d^{\pi}$ are extended
pseudo-metrics, where ‘extended’ means that the metrics take values in
$[\,0,+\infty\,]$. Denoting by $\square^{\pi}((X,G),(Y,H))$ the Hausdorff
distance between $G$ and $H$ with respect to $d^{\pi}$, we define the
_equivariant box distance_ between $(X,G)$ and $(Y,H)$ to be

$\square((X,G),(Y,H)):=\inf_{\pi\in\Pi(X,Y)}\square^{\pi}((X,G),(Y,H)),$

and also define

$\square(X,Y):=\square((X,\\{\operatorname{id}_{X}\\}),(Y,\\{\operatorname{id}_{Y}\\})).$

Note that $\square\leq 1$ because of $d^{\emptyset}=0$.

It follows from [Nk:thesis]*Section 6 (and also [Nk:box-obs]) that
$\square(X,Y)$ coincides with the original box distance between $X$ and $Y$
defined by Gromov [Gmv:green].

In this and the next sections, we prove that $\square$ is a metric on
$\mathcal{X}_{\rm eq}$. The following is obvious by the definition of
$\square^{\pi}$.

######  Lemma 3.5.

Let $(X,G),(Y,H)\in\mathcal{X}_{\rm eq}$ and $\varepsilon>0$. Then, the
inequality
$\square^{\pi}((X,G),(Y,H))<\varepsilon$ is equivalent to both (i) and (ii)
being satisfied.

  1. (i)

For any element $g\in G$ there exist an element $h\in H$ and a closed subset
$S\subset\operatorname{supp}\pi$ with $\pi(S)>1-\varepsilon$ such that

(3.2) $d^{S}(g,h)<\varepsilon.$

  2. (ii)

For any element $h\in H$ there exist an element $g\in G$ and a closed subset
$S\subset\operatorname{supp}\pi$ with $\pi(S)>1-\varepsilon$ such that (3.2)
holds.

######  Lemma 3.6.

$\square$ satisfies a triangle inequality on $\mathcal{X}_{\rm eq}$.

###### Proof.

Let us prove
$\square((X,G),(Z,K))\leq\square((X,G),(Y,H))+\square((Y,H),(Z,K))$ for
$(X,G),(Y,H),(Z,K)\in\mathcal{X}_{\rm eq}$. Take any $\varepsilon_{1}$,
$\varepsilon_{2}$ with $\varepsilon_{1}>\square((X,G),(Y,H))$ and
$\varepsilon_{2}>\square((Y,H),(Z,K))$. There are $\pi_{1}\in\Pi(X,Y)$ and
$\pi_{2}\in\Pi(Y,Z)$ such that
$\square^{\pi_{1}}((X,G),(Y,H))<\varepsilon_{1}$ and
$\square^{\pi_{2}}((Y,H),(Z,K))<\varepsilon_{2}$. We set
$\pi:=\pi_{2}\circ\pi_{1}$. It suffices to prove
$\square^{\pi}((X,G),(Z,K))\leq\varepsilon_{1}+\varepsilon_{2}$. Take any
$g\in G$. Since $\square^{\pi_{1}}((X,G),(Y,H))<\varepsilon_{1}$ and by Lemma
3.5, there are a closed subset $S_{1}\subset\operatorname{supp}\pi_{1}$ with
$\pi_{1}(S_{1})>1-\varepsilon_{1}$ and $h\in H$ such that
$d^{S_{1}}(g,h)<\varepsilon_{1}$. In the same way,
$\square^{\pi_{2}}((Y,H),(Z,K))<\varepsilon_{2}$ implies that there are a
closed subset $S_{2}\subset\operatorname{supp}\pi_{2}$ with
$\pi_{2}(S_{2})>1-\varepsilon_{2}$ and $k\in K$ such that
$d^{S_{2}}(h,k)<\varepsilon_{2}$. Let $S:=\overline{S_{2}\circ S_{1}}$, where
the upper bar means the closure. For any $(x_{i},z_{i})\in S_{2}\circ S_{1}$,
$i=1,2$, there is $y_{i}\in Y$ such that $(x_{i},y_{i})\in S_{1}$ and
$(y_{i},z_{i})\in S_{2}$. We have

$\displaystyle|\,d_{X}(g(x_{1}),x_{2})-d_{Z}(k(z_{1}),z_{2})\,|$
$\displaystyle\leq|\,d_{X}(g(x_{1}),x_{2})-d_{Y}(h(y_{1}),y_{2})\,|+|\,d_{Y}(h(y_{1}),y_{2})-d_{Z}(k(z_{1}),z_{2})\,|$
$\displaystyle\leq\varepsilon_{1}+\varepsilon_{2}.$

Since these inequalities are kept under the limit, these hold for any
$(x_{i},z_{i})\in S$, i.e., we have
$d^{S}(g,k)\leq\varepsilon_{1}+\varepsilon_{2}$. By Lemma 3.5, we obtain
$\square^{\pi}((X,G),(Z,K))\leq\varepsilon_{1}+\varepsilon_{2}$. In the same
way, for any $k\in K$ there is $g\in G$ such that
$d^{\pi}(g,k)\leq\varepsilon_{1}+\varepsilon_{2}$. We thus obtain
$\square((X,G),(Z,K))\leq\square^{\pi}((X,G),(Z,K))\leq\varepsilon_{1}+\varepsilon_{2}$.
This completes the proof. ∎

The nondegeneracy of $\square$ on $\mathcal{X}_{\rm eq}$ is proved in the next
section.

######  Proposition 3.7.

For any $(X,G),(Y,H)\in\mathcal{X}_{\rm eq}$, we have

$\square(X,Y)\leq 2\,\square((X,G),(Y,H)).$

###### Proof.

Let $\varepsilon>\square((X,G),(Y,H))$. We apply Lemma 3.5(ii) for
$h:=\operatorname{id}_{Y}$. Then, there are $g\in G$ and a closed subset
$S\subset\operatorname{supp}\pi$ satisfying $\pi(S)\geq 1-\varepsilon$ and
(3.2). In particular, we have $d_{X}(g(x),x)\leq\varepsilon$ for any $x\in
p_{1}(S)$. A triangle inequality and (3.2) together imply that, for any
$(x_{i},y_{i})\in S$, $i=1,2$,

$|\,d_{X}(x_{1},x_{2})-d_{Y}(y_{1},y_{2})\,|\leq|\,d_{X}(g(x_{1}),x_{2})-d_{Y}(y_{1},y_{2})\,|+\varepsilon\leq
2\varepsilon.$

This completes the proof. ∎

##  4\. Equivariant observable distance

In this section, we introduce the equivariant observable metric on
$\mathcal{X}_{\rm eq}$ and prove that it is a really metric. As an
application, we prove the nondegeneracy of the equivariant box metric.

Denote by $\mathcal{L}{\it ip}_{1}(X)$ the set of real valued $1$-Lipschitz
functions on a metric space $X$.

######  Definition 4.1 (Equivariant observable metric).

Let $(X,G),(Y,H)\in\mathcal{X}_{\rm eq}$, $\pi\in\Pi(X,Y)$, $g\in G$, and
$h\in H$. For $f\in\mathcal{L}{\it ip}_{1}(X)$ and
$f^{\prime}\in\mathcal{L}{\it ip}_{1}(Y)$, we define

$\rho^{\pi}_{g,h}(f,f^{\prime}):=\max\\{\operatorname{{\it d}_{{\rm
KF}}}^{\pi}(f\circ p_{1},f^{\prime}\circ p_{2}),\operatorname{{\it d}_{{\rm
KF}}}^{\pi}(f\circ g\circ p_{1},f^{\prime}\circ h\circ p_{2})\\},$

where $\operatorname{{\it d}_{{\rm KF}}}^{\pi}$ denotes the Ky Fan metric on
the set of Borel measurable functions on $X\times Y$ with respect to $\pi$.
Let $\rho^{\pi}(g,h)$ denote the Hausdorff distance between $\mathcal{L}{\it
ip}_{1}(X)$ and $\mathcal{L}{\it ip}_{1}(Y)$ with respect to
$\rho^{\pi}_{g,h}$, and $\operatorname{{\it d}_{{\rm
conc}}}^{\pi}((X,G),(Y,H))$ the Hausdorff distance between $G$ and $H$ with
respect to $\rho^{\pi}$. Define the _equivariant observable distance_ between
$(X,G)$ and $(Y,H)$ to be

$\operatorname{{\it d}_{{\rm
conc}}}((X,G),(Y,H)):=\inf_{\pi\in\Pi(X,Y)}\operatorname{{\it d}_{{\rm
conc}}}^{\pi}((X,G),(Y,H)),$

and also define

$\operatorname{{\it d}_{{\rm conc}}}(X,Y):=\operatorname{{\it d}_{{\rm
conc}}}((X,\\{\operatorname{id}_{X}\\}),(Y,\\{\operatorname{id}_{Y}\\})).$

Note that, by the same trick in the definition of equivariant box distance, we
are able to extend $\rho^{\pi}$ to a pseudo-metric on $G\sqcup H$, however we
do not need such the extension to obtain the Hausdorff distance and to define
$\operatorname{{\it d}_{{\rm conc}}}^{\pi}$.

It follows from [Nk:thesis]*Section 6 (and also [Nk:box-obs]) that
$\operatorname{{\it d}_{{\rm conc}}}(X,Y)$ coincides with the original
observable distance between $X$ and $Y$ defined by Gromov [Gmv:green]. It is
obvious that $\operatorname{{\it d}_{{\rm conc}}}(X,Y)\leq\operatorname{{\it
d}_{{\rm conc}}}((X,G),(Y,H))$.

######  Lemma 4.2.

Let $(X,G),(Y,H),(Z,K)\in\mathcal{X}_{\rm eq}$, $\pi_{1}\in\Pi(X,Y)$, and
$\pi_{2}\in\Pi(Y,Z)$.

  1. (1)

For any $g\in\operatorname{Aut}(X)$, $h\in\operatorname{Aut}(Y)$, and
$k\in\operatorname{Aut}(Z)$, we have

$\rho^{\pi_{2}\circ\pi_{1}}(g,k)\leq\rho^{\pi_{1}}(g,h)+\rho^{\pi_{2}}(h,k).$

  2. (2)

We have

$\displaystyle\operatorname{{\it d}_{{\rm
conc}}}^{\pi_{2}\circ\pi_{1}}((X,G),(Z,K))\leq\operatorname{{\it d}_{{\rm
conc}}}^{\pi_{1}}((X,G),(Y,H))+\operatorname{{\it d}_{{\rm
conc}}}^{\pi_{2}}((Y,H),(Z,K)).$

  3. (3)

$\operatorname{{\it d}_{{\rm conc}}}$ satisfies a triangle inequality on
$\mathcal{X}_{\rm eq}$.

###### Proof.

We prove (1). Take any $\varepsilon_{i}$ with
$\rho^{\pi_{1}}(g,h)<\varepsilon_{1}$ and
$\rho^{\pi_{2}}(h,k)<\varepsilon_{2}$. For any $f\in\mathcal{L}{\it
ip}_{1}(X)$ there is $f^{\prime}\in\mathcal{L}{\it ip}_{1}(Y)$ such that
$\operatorname{{\it d}_{{\rm KF}}}^{\pi_{2}\bullet\pi_{1}}(f\circ
p_{1},f^{\prime}\circ p_{2})=\operatorname{{\it d}_{{\rm
KF}}}^{\pi_{1}}(f\circ p_{1},f^{\prime}\circ p_{2})<\varepsilon_{1}$ and
$\operatorname{{\it d}_{{\rm KF}}}^{\pi_{2}\bullet\pi_{1}}(f\circ g\circ
p_{1},f^{\prime}\circ h\circ p_{2})=\operatorname{{\it d}_{{\rm
KF}}}^{\pi_{1}}(f\circ g\circ p_{1},f^{\prime}\circ h\circ
p_{2})<\varepsilon_{1}$. For the $f^{\prime}$ there is
$f^{\prime\prime}\in\mathcal{L}{\it ip}_{1}(Z)$ such that $\operatorname{{\it
d}_{{\rm KF}}}^{\pi_{2}\bullet\pi_{1}}(f^{\prime}\circ
p_{2},f^{\prime\prime}\circ p_{3})=\operatorname{{\it d}_{{\rm
KF}}}^{\pi_{2}}(f^{\prime}\circ p_{1},f^{\prime\prime}\circ
p_{2})<\varepsilon_{2}$ and $\operatorname{{\it d}_{{\rm
KF}}}^{\pi_{2}\bullet\pi_{1}}(f^{\prime}\circ h\circ
p_{2},f^{\prime\prime}\circ k\circ p_{3})=\operatorname{{\it d}_{{\rm
KF}}}^{\pi_{2}}(f^{\prime}\circ h\circ p_{1},f^{\prime\prime}\circ k\circ
p_{2})<\varepsilon_{2}$. A triangle inequality implies that
$\operatorname{{\it d}_{{\rm KF}}}^{\pi_{2}\circ\pi_{1}}(f\circ
p_{1},f^{\prime\prime}\circ p_{2})=\operatorname{{\it d}_{{\rm
KF}}}^{\pi_{2}\bullet\pi_{1}}(f\circ p_{1},f^{\prime\prime}\circ
p_{3})<\varepsilon_{1}+\varepsilon_{2}$ and $\operatorname{{\it d}_{{\rm
KF}}}^{\pi_{2}\circ\pi_{1}}(f\circ g\circ p_{1},f^{\prime\prime}\circ k\circ
p_{2})=\operatorname{{\it d}_{{\rm KF}}}^{\pi_{2}\bullet\pi_{1}}(f\circ g\circ
p_{1},f^{\prime\prime}\circ k\circ p_{3})<\varepsilon_{1}+\varepsilon_{2}$. In
the same way, for any $f^{\prime\prime}\in\mathcal{L}{\it ip}_{1}(Z)$ there is
$f\in\mathcal{L}{\it ip}_{1}(X)$ satisfying the last inequality. Therefore we
obtain $\rho^{\pi_{2}\circ\pi_{1}}(g,k)\leq\varepsilon_{1}+\varepsilon_{2}$.
(1) has been proved.

(2) follows from (1) by a straightforward discussion.

(3) follows from (2).

This completes the proof. ∎

We need some lemmas for the proof of the nondegeneracy of $\operatorname{{\it
d}_{{\rm conc}}}$.

######  Lemma 4.3 ([Nk:thesis]*Proposition 6.13).

Let $X$ be a complete separable metric space. For any
$f,f^{\prime}\in\mathcal{L}{\it ip}_{1}(X)$ and for any Borel probability
measures $\mu,\nu$ on $X$, we have

$|\,\operatorname{{\it d}_{{\rm KF}}}^{\mu}(f,f^{\prime})-\operatorname{{\it
d}_{{\rm KF}}}^{\nu}(f,f^{\prime})\,|\leq 2\operatorname{{\it d}_{{\rm
P}}}(\mu,\nu).$

######  Lemma 4.4.

Let $(X,G),(Y,H)\in\mathcal{X}_{\rm eq}$.

  1. (1)

For any $\pi,\pi^{\prime}\in\Pi(X,Y)$ we have

$|\,\operatorname{{\it d}_{{\rm conc}}}^{\pi}((X,G),(Y,H))-\operatorname{{\it
d}_{{\rm conc}}}^{\pi^{\prime}}(X,G),(Y,H))\,|\leq 2\operatorname{{\it
d}_{{\rm P}}}(\pi,\pi^{\prime}).$

  2. (2)

There exists $\pi\in\Pi(X,Y)$ such that

$\operatorname{{\it d}_{{\rm conc}}}((X,G),(Y,H))=\operatorname{{\it d}_{{\rm
conc}}}^{\pi}((X,G),(Y,H)).$

###### Proof.

(1) follows from Lemma 4.3.

(2) follows from (1) and the compactness of $\Pi(X,Y)$. ∎

######  Lemma 4.5.

Let $X$ be an mm-space. Then, for any $f,f^{\prime}\in\mathcal{L}{\it
ip}_{1}(X)$ and $g,g^{\prime}\in\operatorname{Aut}(X)$, we have

$\operatorname{{\it d}_{{\rm KF}}}^{\mu_{X}}(f\circ g,f^{\prime}\circ
g^{\prime})\leq\operatorname{{\it d}_{{\rm
KF}}}^{\mu_{X}}(f,f^{\prime})+\operatorname{{\it d}_{{\rm
KF}}}^{\mu_{X}}(g,g^{\prime}).$

###### Proof.

The lemma follows from

$\displaystyle\operatorname{{\it d}_{{\rm KF}}}^{\mu_{X}}(f\circ
g,f^{\prime}\circ g^{\prime})$ $\displaystyle\leq\operatorname{{\it d}_{{\rm
KF}}}^{\mu_{X}}(f\circ g,f\circ g^{\prime})+\operatorname{{\it d}_{{\rm
KF}}}^{\mu_{X}}(f\circ g^{\prime},f^{\prime}\circ g^{\prime})$
$\displaystyle\leq\operatorname{{\it d}_{{\rm
KF}}}^{\mu_{X}}(f,f^{\prime})+\operatorname{{\it d}_{{\rm
KF}}}^{\mu_{X}}(g,g^{\prime}).$

∎

######  Lemma 4.6.

$\operatorname{{\it d}_{{\rm conc}}}$ is nondegenerate on $\mathcal{X}_{\rm
eq}$.

###### Proof.

Assume $\operatorname{{\it d}_{{\rm conc}}}((X,G),(Y,H))=0$ for two elements
$(X,G),(Y,H)\in\mathcal{X}_{\rm eq}$. By Lemma 4.4, there is $\pi\in\Pi(X,Y)$
satisfying $\operatorname{{\it d}_{{\rm conc}}}^{\pi}((X,G),(Y,H))=0$. Take
any $g\in G$. Then there is a sequence $h_{n}\in H$, $n=1,2,\dots$, with
$\rho^{\pi}(g,h_{n})\to 0$. For any $f\in\mathcal{L}{\it ip}_{1}(X)$ there is
a sequence $f_{n}^{\prime}\in\mathcal{L}{\it ip}_{1}(Y)$, $n=1,2,\dots$, such
that $\operatorname{{\it d}_{{\rm KF}}}^{\pi}(f\circ p_{1},f_{n}^{\prime}\circ
p_{2})\to 0$ and $\operatorname{{\it d}_{{\rm KF}}}^{\pi}(f\circ g\circ
p_{1},f_{n}^{\prime}\circ h_{n}\circ p_{2})\to 0$ as $n\to\infty$. By the
compactness of $H$, we may assume that $h_{n}$ converges in measure to $h$ as
$n\to\infty$ by replacing $\\{h_{n}\\}$ with a subsequence. Also, we may
assume that $f_{n}^{\prime}$ converges in measure to a function
$f^{\prime}\in\mathcal{L}{\it ip}_{1}(Y)$ (see the proof of [Sy:mmg]*Lemma
4.45). Since

$\displaystyle|\operatorname{{\it d}_{{\rm KF}}}^{\pi}(f\circ
p_{1},f_{n}^{\prime}\circ p_{2})-\operatorname{{\it d}_{{\rm
KF}}}^{\pi}(f\circ p_{1},f^{\prime}\circ p_{2})|$
$\displaystyle\leq\operatorname{{\it d}_{{\rm KF}}}^{\pi}(f_{n}^{\prime}\circ
p_{2},f^{\prime}\circ p_{2})=\operatorname{{\it d}_{{\rm
KF}}}^{\mu_{Y}}(f_{n}^{\prime},f^{\prime})\to 0\quad\text{as}\ n\to\infty,$

we see that $\operatorname{{\it d}_{{\rm KF}}}^{\pi}(f\circ
p_{1},f^{\prime}\circ p_{2})=0$, i.e., $f\circ p_{1}=f^{\prime}\circ p_{2}$ on
$\operatorname{supp}\pi$. By using Lemma 4.5, the same discussion yields that
$f\circ g\circ p_{1}=f^{\prime}\circ h\circ p_{2}$ on
$\operatorname{supp}\pi$.

Take any $(x,y)\in\operatorname{supp}\pi$ and fix it. Applying the above for
$f:=d_{X}(x,\cdot)$ yields that there is $f^{\prime}\in\mathcal{L}{\it
ip}_{1}(Y)$ such that $f\circ p_{1}=f^{\prime}\circ p_{2}$ on
$\operatorname{supp}\pi$. For any
$(x^{\prime},y^{\prime})\in\operatorname{supp}\pi$, we have
$d_{X}(x,x^{\prime})=f\circ p_{1}(x^{\prime},y^{\prime})=f^{\prime}\circ
p_{2}(x^{\prime},y^{\prime})=f^{\prime}(y^{\prime})$ and in particular,
$f^{\prime}(y)=0$. Therefore,
$d_{X}(x,x^{\prime})=f^{\prime}(y^{\prime})=f^{\prime}(y^{\prime})-f^{\prime}(y)\leq
d_{Y}(y,y^{\prime})$. Exchanging $x$ and $y$ yields the opposite inequality.
We then obtain $d_{X}(x,x^{\prime})=d_{Y}(y,y^{\prime})$. This determines an
isometry
$\varphi:\operatorname{Dom}(\operatorname{supp}\pi)\to\operatorname{Im}(\operatorname{supp}\pi)$
satisfying $(x,\varphi(x))\in\operatorname{supp}\pi$ for any
$x\in\operatorname{Dom}(\operatorname{supp}\pi)$. Since
$\operatorname{Dom}(\operatorname{supp}\pi)$ is dense in
$\operatorname{supp}\mu_{X}=X$, the map $\varphi$ extends to an isometry from
$X$ to $Y$, which we denote by the same notation $\varphi$. We see that
$\pi=(\operatorname{id}_{X},\varphi)_{*}\mu_{X}$ and
$\varphi_{*}\mu_{X}=\mu_{Y}$, so that $\varphi$ is an mm-isomorphism. We also
see that $f^{\prime}=f\circ\varphi^{-1}$ for any $f\in\mathcal{L}{\it
ip}_{1}(X)$. Since $f\circ g\circ p_{1}=f\circ\varphi^{-1}\circ h\circ p_{2}$
on $\operatorname{supp}\pi$, we have $f\circ g=f\circ\varphi^{-1}\circ
h\circ\varphi$. Take any $x\in X$ and set $f:=d_{X}(g(x),\cdot)$. Then, since
$0=f\circ g(x)=f\circ\varphi^{-1}\circ h\circ\varphi(x)$, we have
$\varphi^{-1}\circ h\circ\varphi(x)=g(x)$ and so $h=\varphi\circ
g\circ\varphi^{-1}$. This implies $\varphi\circ G\circ\varphi^{-1}\subset H$.
In the same way, $\varphi^{-1}\circ H\circ\varphi\subset G$.
$\rho(g):=\varphi\circ g\circ\varphi^{-1}$ is a group-isomorphism from $G$ to
$H$. Thus, $(X,G)$ and $(Y,H)$ are equivariantly mm-isomorphic to each other.
This completes the proof. ∎

######  Theorem 4.7.

$\operatorname{{\it d}_{{\rm conc}}}$ is a metric on $\mathcal{X}_{\rm eq}$.

###### Proof.

The theorem follows from Lemmas 4.2(3) and 4.6. ∎

We next prove the nondegeneracy of $\square$.

For $S\subset X\times Y$ and $f\in\mathcal{L}{\it ip}_{1}(X)$, we define a
function $\tilde{f}_{S}:Y\to\mathbb{R}$ by

$\tilde{f}_{S}(y):=\inf_{(x^{\prime},y^{\prime})\in
S}(d_{Y}(y,y^{\prime})+f(x^{\prime})),\qquad y\in Y,$

which is a variant of the McShane-Whitney extension. It follows from a
standard discussion that $\tilde{f}_{S}$ is $1$-Lipschitz continuous on $Y$.

######  Lemma 4.8.

Let $\pi\in\Pi(X,Y)$ and let
$S,S^{\prime},S^{\prime\prime}\subset\operatorname{supp}\pi$ be three closed
subsets. Then, for any $g,g^{\prime}\in G$ and $h,h^{\prime}\in H$, we have

$\rho^{\pi}(g,h)\leq
2\max\\{1-\pi(S),1-\pi(S^{\prime}),1-\pi(S^{\prime\prime}),d^{S}(g,h),d^{S^{\prime}}(g^{\prime},\operatorname{id}_{Y}),d^{S^{\prime\prime}}(\operatorname{id}_{X},h^{\prime})\\}.$

###### Proof.

We put

$\varepsilon:=\max\\{1-\pi(S),1-\pi(S^{\prime}),1-\pi(S^{\prime\prime}),d^{S}(g,h),d^{S^{\prime}}(g^{\prime},\operatorname{id}_{Y}),d^{S^{\prime\prime}}(\operatorname{id}_{X},h^{\prime})\\}.$

For any $f\in\mathcal{L}{\it ip}_{1}(X)$, we set $f^{\prime}:=\tilde{f}_{S\cap
S^{\prime}}$.

Let us prove

(4.1) $\operatorname{{\it d}_{{\rm KF}}}^{\pi}(f\circ p_{1},f^{\prime}\circ
p_{2})\leq 2\varepsilon.$

Take any $(x,y)\in S\cap S^{\prime}$. It is obvious that $f^{\prime}(y)\leq
f(x)$. From $d^{S^{\prime}}(g^{\prime},\operatorname{id}_{Y})\leq\varepsilon$,
we see that

$\displaystyle f(x)-f^{\prime}(y)$
$\displaystyle=\sup_{(x^{\prime},y^{\prime})\in S\cap
S^{\prime}}(f(x)-f(x^{\prime})-d_{Y}(y,y^{\prime}))$
$\displaystyle\leq\sup_{(x^{\prime},y^{\prime})\in S\cap
S^{\prime}}(d_{X}(x,x^{\prime})-d_{Y}(y,y^{\prime}))\leq 2\varepsilon$

(see the proof of Proposition 3.7 for the last inequality). This together with
$\pi(S\cap S^{\prime})\geq 1-2\varepsilon$ implies (4.1).

We next prove

(4.2) $\operatorname{{\it d}_{{\rm KF}}}^{\pi}(f\circ g\circ
p_{1},f^{\prime}\circ h\circ p_{2})\leq 4\varepsilon.$

Take any $(x,y)\in S\cap S^{\prime}\cap(g^{-1}(\operatorname{Dom}(S\cap
S^{\prime}))\times Y)$. Since $d^{S}(g,h)\leq\varepsilon$, we see

$\displaystyle f^{\prime}\circ h(y)$
$\displaystyle=\inf_{(x^{\prime},y^{\prime})\in S\cap
S^{\prime}}(d_{Y}(h(y),y^{\prime})+f(x^{\prime}))$
$\displaystyle\leq\inf_{(x^{\prime},y^{\prime})\in S\cap
S^{\prime}}(d_{X}(g(x),x^{\prime})+f(x^{\prime}))+\varepsilon\leq f\circ
g(x)+\varepsilon.$

By the $1$-Lipschitz continuity of $f$ and by $d^{S}(g,h)\leq\varepsilon$,

$\displaystyle f\circ g(x)-f^{\prime}\circ h(y)$
$\displaystyle=\sup_{(x^{\prime},y^{\prime})\in S\cap S^{\prime}}(f\circ
g(x)-f(x^{\prime})-d_{Y}(h(y),y^{\prime}))$
$\displaystyle\leq\sup_{(x^{\prime},y^{\prime})\in S\cap
S^{\prime}}(d_{X}(g(x),x^{\prime})-d_{Y}(h(y),y^{\prime}))\leq\varepsilon.$

We thus obtain $|f\circ g(x)-f^{\prime}\circ h(y)|\leq\varepsilon$, which
together with $\pi(S\cap S^{\prime}\cap(g^{-1}(\operatorname{Dom}(S\cap
S^{\prime}))\times Y))\geq 1-4\varepsilon$ implies (4.2).

(4.1) and (4.2) together imply $\rho^{\pi}_{g,h}(f,f^{\prime})\leq
4\varepsilon$.

In the same way, for any $f^{\prime}\in\mathcal{L}{\it ip}_{1}(Y)$ there is
$f\in\mathcal{L}{\it ip}_{1}(X)$ such that $\rho^{\pi}_{g,h}(f,f^{\prime})\leq
4\varepsilon$. We thus obtain $\rho^{\pi}(g,h)\leq 4\varepsilon$. This
completes the proof. ∎

######  Proposition 4.9.

For any $(X,G),(Y,H)\in\mathcal{X}_{\rm eq}$ we have

$\operatorname{{\it d}_{{\rm conc}}}((X,G),(Y,H))\leq
4\,\square((X,G),(Y,H)).$

###### Proof.

Take any $\varepsilon$ with $\varepsilon>\square((X,G),(Y,H))$. There is
$\pi\in\Pi(X,Y)$ such that $\square^{\pi}((X,G),(Y,H))<\varepsilon$. For any
$g\in G$ there are $h\in H$ and a closed subset
$S\subset\operatorname{supp}\pi$ such that $\pi(S)\geq 1-\varepsilon$ and
$d^{S}(g,h)\leq\varepsilon$. This together with Lemma 4.8 implies
$\rho^{\pi}(g,h)\leq 4\varepsilon$. The same is true if we exchange $g$ and
$h$. We thus obtain $\operatorname{{\it d}_{{\rm
conc}}}((X,G),(Y,H))\leq\operatorname{{\it d}_{{\rm
conc}}}^{\pi}((X,G),(Y,H))<4\varepsilon$. This completes the proof. ∎

######  Theorem 4.10.

$\square$ is a metric on $\mathcal{X}_{\rm eq}$.

###### Proof.

Lemma 3.6 says a triangle inequality for $\square$. The symmetry is trivial.
Combining Lemma 4.6 and Proposition 4.9 implies the nondegeneracy of
$\square$. This completes the proof. ∎

Lemma 4.6 and Theorem 4.10 also lead us the nondegeneracy of the original
observable and box metrics without group actions. This gives a new proof of
the nondegeneracy.

##  5\. Convergence of quotient spaces

In this section, we prove that if a sequence of mm-spaces with group actions
converges with respect to $\square/\operatorname{{\it d}_{{\rm conc}}}$, then
so does the sequence of the quotient spaces.

For $(X,G)\in\mathcal{X}_{\rm eq}$, we denote by $d_{X/G}$ the _quotient
metric_ on $X/G$ (which is defined as the distance between orbits) and set
$\mu_{X/G}:=(\pi_{G})_{*}\mu_{X}$, where $\pi_{G}:X\to X/G$ is the quotient
map. Note that $d_{X/G}$ is a pseudo-metric in general.

######  Lemma 5.1.

For any $(X,G)\in\mathcal{X}_{\rm eq}$, any $G$-orbit is closed and in
particular, $d_{X/G}$ is a metric.

###### Proof.

Let $y\in X$ be a point in the closure of the $G$-orbit of a point $x\in X$.
It then suffices to prove that $y$ belongs to the $G$-orbit of $x$. There is a
sequence $\\{g_{n}\\}\subset G$ such that $g_{n}(x)\to y$ as $n\to\infty$. By
Lemma 3.3, replacing $\\{g_{n}\\}$ with a subsequence, we see that $g_{n}$
converges pointwise to an element $g\in\operatorname{Aut}(X)$. Hence,
$g_{n}(x)\to g(x)$ and $y=g(x)$. Since $G$ is closed, $g$ belongs to $G$. This
completes the proof. ∎

We consider the thick-thin decomposition. Assume that a sequence of mm-spaces
$X_{n}$, $n=1,2,\dots$, box-converges to an mm-space $Y$. For $v,r>0$, we put
$Y(v,r):=\\{\,y\in Y\mid\mu_{Y}(B_{r}(y))>v\,\\}$. For any $\varepsilon>0$
there is $v_{\varepsilon}>0$ such that
$\mu_{Y}(Y(v_{\varepsilon},\varepsilon))>1-\varepsilon$ and
$v_{\varepsilon}\to 0$ as $\varepsilon\to 0+$. Let
$Y_{\varepsilon}:=Y(v_{\varepsilon},\varepsilon)$ and
$X_{n,\varepsilon}:=\\{\,x\in
X_{n}\mid\mu_{X_{n}}(B_{2\varepsilon}(x))>v_{\varepsilon}/2\,\\}$, which we
call the _thick parts_ of $Y$ and $X_{n}$ respectively. Then,
$X_{n,\varepsilon}$ is $\operatorname{Aut}(X_{n})$-invariant and
$Y_{\varepsilon}$ is $\operatorname{Aut}(Y)$-invariant.

######  Lemma 5.2.

For any $n$ and $\varepsilon>0$ with $\square(X_{n},Y)<v_{\varepsilon}/2$, we
have

$\mu_{X_{n}}(X_{n,\varepsilon})>1-\varepsilon-\square(X_{n},Y).$

###### Proof.

Let $\varepsilon_{n}:=\square(X_{n},Y)$. There is $\pi_{n}\in\Pi(X_{n},Y)$ for
each $n$ such that $\square^{\pi_{n}}(X_{n},Y)=\varepsilon_{n}$. Take any
$x\in S_{n}^{-1}(Y_{\varepsilon})$. There is $y\in Y_{\varepsilon}$ with
$(x,y)\in S_{n}$. Since $p_{1}((X_{n}\times B_{\varepsilon}(y))\cap
S_{n})=S_{n}^{-1}(B_{\varepsilon}(y))\subset B_{2\varepsilon}(x)$, we see

$\displaystyle\mu_{X_{n}}(B_{2\varepsilon}(x))\geq\mu_{X_{n}}(p_{1}((X_{n}\times
B_{\varepsilon}(y))\cap S_{n}))$
$\displaystyle=\pi_{n}(p_{1}^{-1}(p_{1}((X_{n}\times B_{\varepsilon}(y))\cap
S_{n}))\geq\pi_{n}((X_{n}\times B_{\varepsilon}(y))\cap S_{n})$
$\displaystyle\geq v_{\varepsilon}-\varepsilon_{n}>v_{\varepsilon}/2,$

which implies $x\in X_{n,\varepsilon}$. Therefore,
$S_{n}^{-1}(Y_{\varepsilon})\subset X_{n,\varepsilon}$ and
$\mu_{X_{n}}(X_{n,\varepsilon})\geq\mu_{X_{n}}(S_{n}^{-1}(Y_{\varepsilon}))\\\
\geq\pi_{n}((X_{n}\times Y_{\varepsilon})\cap
S_{n})>1-\varepsilon-\varepsilon_{n}$. This completes the proof. ∎

######  Lemma 5.3.

If a sequence $(X_{n},G_{n})\in\mathcal{X}_{\rm eq}$, $n=1,2,\dots$,
equivariantly box converges to an element $(Y,H)\in\mathcal{X}_{\rm eq}$, then
$\\{(G_{n},\operatorname{{\it d}_{{\rm KF}}})\\}_{n=1}^{\infty}$ is precompact
with respect to the Gromov-Hausdorff topology.

###### Proof.

Take any $\delta>0$. Let $\\{g_{i}\\}_{i=1}^{N}$ be a $\delta$-discrete net of
$G_{n}$ with respect to $\operatorname{{\it d}_{{\rm KF}}}$. It suffices to
prove that $N$ is bounded from above for every $n$. There is $\varepsilon>0$
such that $v_{\varepsilon}/2+4\varepsilon<\delta$. Assume that $n$ is so large
that $\square((X_{n},G_{n}),(Y,H))<\varepsilon/2$.

Let us prove $d^{\pi_{n}}(g_{i},g_{j})\geq v_{\varepsilon}/2$ for all $i\neq
j$. If $d^{\pi_{n}}(g_{i},g_{j})<v_{\varepsilon}/2$ for a pair of distinct $i$
and $j$, then there is a closed subset
$S_{n}\subset\operatorname{supp}\pi_{n}$ such that $\pi_{n}(S_{n})\geq
1-v_{\varepsilon}/2$ and

(5.1) $|\,d_{X_{n}}(g_{i}(x_{1}),x_{2})-d_{X_{n}}(g_{j}(x_{1}),x_{2})\,|\leq
v_{\varepsilon}/2$

for any $x_{1},x_{2}\in p_{1}(S_{n})$. Put
$S_{n,\varepsilon}:=(X_{n,\varepsilon}\times Y)\cap S_{n}$. Since Proposition
3.7 implies $\square(X_{n},Y)<\varepsilon$, we see by Lemma 5.2 that
$\mu_{X_{n}}(X_{n,\varepsilon})\geq 1-2\varepsilon$, which proves
$\pi_{n}(S_{n,\varepsilon})\geq 1-v_{\varepsilon}/2-2\varepsilon$. We also
have
$\mu_{X_{n}}(p_{1}(S_{n}))\geq(p_{1})_{*}\pi_{n}(p_{1}(S_{n}))\geq\pi_{n}(S_{n})\geq
1-v_{\varepsilon}/2$. For any $x\in p_{1}(S_{n,\varepsilon})$, we see
$g_{i}(x)\in X_{n,\varepsilon}$ and so there is $x^{\prime}\in p_{1}(S_{n})$
such that $d_{X_{n}}(g_{i}(x),x^{\prime})\leq 2\varepsilon$. Since (5.1)
implies
$|\,d_{X_{n}}(g_{i}(x),x^{\prime})-d_{X_{n}}(g_{j}(x),x^{\prime})\,|\leq
v_{\varepsilon}/2$, we have $d_{X_{n}}(g_{j}(x),x^{\prime})\leq
v_{\varepsilon}/2+2\varepsilon$. By a triangle inequality,
$d_{X_{n}}(g_{i}(x),g_{j}(x))\leq v_{\varepsilon}/2+4\varepsilon<\delta$. This
contradicts $\operatorname{{\it d}_{{\rm KF}}}(g_{i},g_{j})\geq\delta$. We
have proved $d^{\pi_{n}}(g_{i},g_{j})\geq v_{\varepsilon}/2$ for all $i\neq
j$.

The net $\\{g_{i}\\}_{i=1}^{N}$ is $(v_{\epsilon}/2)$-discrete with respect to
$d^{\pi_{n}}$. It follows from Lemma 3.3 that $G_{n}$ and $H$ are compact with
respect to $d^{\pi_{n}}$, where we remark that $d^{\pi_{n}}$ on $H$ is
independent of $n$. By $\square^{\pi_{n}}((X_{n},G_{n}),(Y,H))\to 0$, we have
the precompactness of $\\{G_{n}\\}$ with respect to $d^{\pi_{n}}$. Thus, $N$
is bounded from above. This completes the proof. ∎

######  Theorem 5.4.

If a sequence $(X_{n},G_{n})\in\mathcal{X}_{\rm eq}$, $n=1,2,\dots$,
equivariantly box converges to an element $(Y,H)\in\mathcal{X}_{\rm eq}$, then
$X_{n}/G_{n}$ box converges to $Y/H$.

###### Proof.

Let $\varepsilon_{n}:=\square((X_{n},G_{n}),(Y,H))$. There is
$\pi_{n}\in\Pi(X_{n},Y)$ for each $n$ such that
$\square^{\pi_{n}}((X_{n},G_{n}),(Y,H))=\varepsilon_{n}$.

Taking any $\varepsilon>0$, we set
$\varepsilon^{\prime}:=\min\\{v_{\varepsilon},\varepsilon\\}/3$. Assume that
$n$ is so large that $\varepsilon_{n}$ is very small compared with
$\varepsilon^{\prime}$. By the precompactness of $\\{G_{n}\\}$ (Lemma 5.3) and
by the compactness of $H$, there are an $\varepsilon^{\prime}$-net
$\\{g_{n,k}\\}_{k=1}^{N}$ of $G_{n}$ and an $\varepsilon^{\prime}$-net
$\\{h_{n,k}\\}_{i=1}^{N}$ of $H$ with respect to the Ky Fan metric such that
$d^{\pi_{n}}(g_{n,k},h_{n,k})\leq\varepsilon_{n}$, where $N$ is independent of
$n$. There is a closed subset $S_{k,n}\subset\operatorname{supp}\pi_{n}$ for
each $k$ such that $\pi_{n}(S_{k,n})\geq 1-\varepsilon_{n}$ and, for any
$(x_{i},y_{i})\in S_{k,n}$,

$|\,d_{X_{n}}(g_{n,k}(x_{1}),x_{2})-d_{Y}(h_{n,k}(y_{1}),y_{2}))\,|\leq\varepsilon_{n}.$

Put $S_{n}:=(X_{n,\varepsilon}\times
Y_{\varepsilon})\cap\bigcap_{k=1}^{N}S_{k,n}$. Define a map $p_{G_{n}}\times
p_{H}:X_{n}\times Y\to X_{n}/G_{n}\times Y/H$ by $p_{G_{n}}\times
p_{H}(x,y):=(p_{G_{n}}(x),p_{H}(y))$, and set $\sigma_{n}:=(p_{G_{n}}\times
p_{H})_{*}\pi_{n}$ and $\mathcal{S}_{n}:=p_{G_{n}}\times p_{H}(S_{n})$. We
have $\sigma_{n}(\mathcal{S}_{n})=\pi_{n}((p_{G_{n}}\times
p_{H})^{-1}(p_{G_{n}}\times p_{H}(S_{n})))\geq\pi_{n}(S_{n})\geq
1-2\varepsilon-N\varepsilon_{n}\geq 1-3\varepsilon$. Take any
$([x_{i}],[y_{i}])\in\mathcal{S}_{n}$, $i=1,2$. We may assume
$(x_{i},y_{i})\in S_{n}$. There is $h\in H$ with
$d_{Y}(h(y_{1}),y_{2})=d_{Y/H}([y_{1}],[y_{2}])$. Also there is $k$ such that
$\operatorname{{\it d}_{{\rm KF}}}(h,h_{n,k})\leq\varepsilon^{\prime}$. It
follows from $y_{1}\in Y_{\varepsilon}$ that
$\mu_{Y}(B_{\varepsilon}(y_{1}))\geq v_{\varepsilon}>\varepsilon^{\prime}$, so
that there is $y_{1}^{\prime}\in B_{\varepsilon}(y_{1})$ with
$d_{Y}(h_{n,k}(y_{1}^{\prime}),h(y_{1}^{\prime}))\leq\varepsilon^{\prime}$. By
the isometric property of $h_{n,k}$, $h$ and by a triangle inequality, we have
$d_{Y}(h_{n,k}(y_{1}),h(y_{1}))\leq 2\varepsilon+\varepsilon^{\prime}\leq
3\varepsilon$ and hence

$\displaystyle d_{X_{n}/G_{n}}([x_{1}],[x_{2}])\leq
d_{X_{n}}(g_{n,k}(x_{1}),x_{2})\leq
d_{Y}(h_{n,k}(y_{1}),y_{2})+\varepsilon_{n}$ $\displaystyle\leq
d_{Y}(h(y_{1}),y_{2})+3\varepsilon+\varepsilon_{n}\leq
d_{Y/H}([y_{1}],[y_{2}])+4\varepsilon.$

In the same way, there is $g\in G_{n}$ with
$d_{Y}(g(x_{1}),x_{2})=d_{X_{n}/G_{n}}([x_{1}],[x_{2}])$. There is $k$ with
$\operatorname{{\it d}_{{\rm KF}}}(g,g_{n,k})\leq\varepsilon^{\prime}$. It
follows from $x_{1}\in X_{n,\varepsilon}$ that
$\mu_{X_{n}}(B_{2\varepsilon}(x_{1}))\geq
v_{\varepsilon}/2>\varepsilon^{\prime}$, so that there is $x_{1}^{\prime}\in
B_{2\varepsilon}(x_{1})$ with
$d_{X_{n}}(g_{n,k}(x_{1}^{\prime}),g(x_{1}^{\prime}))\leq\varepsilon^{\prime}$.
We have $d_{X_{n}}(g_{n,k}(x_{1}),g(x_{1}))\leq
4\varepsilon+\varepsilon^{\prime}\leq 5\varepsilon$ and hence

$d_{Y/H}([y_{1}],[y_{2}])\leq d_{Y}(h_{n,k}(y_{1}),y_{2})\leq
d_{X_{n}/G_{n}}([x_{1}],[x_{2}])+5\varepsilon.$

Thus, for any $([x_{i}],[y_{i}])\in\mathcal{S}_{n}$, $i=1,2$,

$|\,d_{X_{n}/G_{n}}([x_{1}],[x_{2}])-d_{Y/H}([y_{1}],[y_{2}])\,|\leq
5\varepsilon,$

which implies $\square(X_{n}/G_{n},Y/H)\leq 5\varepsilon$. This completes the
proof. ∎

######  Lemma 5.5.

Let $X$ be an mm-space, $f,g:X\to\mathbb{R}$ two $L^{1}$-functions, and let
$D\geq 0$. Then we have the following.

  1. (1)

$\operatorname{{\it d}_{{\rm KF}}}(f,g)^{2}\leq\|f-g\|_{L^{1}}$.

  2. (2)

If $|f|,|g|\leq D$, then $\|f-g\|_{L^{1}}\leq(2D+1)\operatorname{{\it d}_{{\rm
KF}}}(f,g)$.

###### Proof.

Put $\varepsilon:=\operatorname{{\it d}_{{\rm KF}}}(f,g)$.

We prove (1). Since $\mu_{X}(|f-g|\geq\varepsilon)\geq\varepsilon$, we have

$\displaystyle\|f-g\|_{L_{1}}\geq\int_{\\{|f-g|\geq\varepsilon\\}}|f-g|d\mu_{X}\geq\varepsilon\mu_{X}(|f-g|\geq\varepsilon)\geq\varepsilon^{2}$

We prove (2). If $|f|,|g|\leq D$, then

$\displaystyle\|f-g\|_{L_{1}}$
$\displaystyle\leq\int_{\\{|f-g|>\varepsilon\\}}|f-g|\;d\mu_{X}+\int_{\\{|f-g|\leq\varepsilon\\}}|f-g|\;d\mu_{X}$
$\displaystyle\leq
2D\mu_{X}(|f-g|>\varepsilon)+\varepsilon\leq(2D+1)\varepsilon.$

This completes the proof. ∎

######  Definition 5.6 (Observable diameter [Gmv:green]).

For $\kappa>0$, we define the _$\kappa$ -observable diameter_ of an mm-space
$X$ to be

$\operatorname{ObsDiam}(X;-\kappa):=\sup_{f\in\mathcal{L}{\it
ip}_{1}(X)}\operatorname{diam}(f_{*}\mu_{X};-\kappa),$

where $\operatorname{diam}(\nu;-\kappa)$ for a Borel measure $\nu$ on
$\mathbb{R}$ is the infimum of the diameter of Borel subsets
$A\subset\mathbb{R}$ with $\nu(A)\geq 1-\kappa$.

######  Theorem 5.7.

For any $(X,G),(Y,H)\in\mathcal{X}_{\rm eq}$ and $\kappa>0$, setting

$D_{\kappa}:=\max\\{\operatorname{ObsDiam}(X;-\kappa),\operatorname{ObsDiam}(Y;-\kappa)\\},$

we have

$\operatorname{{\it d}_{{\rm
conc}}}(X/G,Y/H)\leq\sqrt{3(D_{\kappa}+1)\operatorname{{\it d}_{{\rm
conc}}}((X,G),(Y,H))}+\kappa.$

In particular, if a sequence $(X_{n},G_{n})\in\mathcal{X}_{\rm eq}$,
$n=1,2,\dots$, $\operatorname{{\it d}_{{\rm conc}}}$-converges to an element
$(Y,H)\in\mathcal{X}_{\rm eq}$, then $X_{n}/G_{n}$ $\operatorname{{\it
d}_{{\rm conc}}}$-converges to $Y/H$.

###### Proof.

We first remark that we have the one to one correspondence between the set of
$G$-invariant functions in $\mathcal{L}{\it ip}_{1}(X)$ and $\mathcal{L}{\it
ip}_{1}(X/G)$, where $f(x)=\bar{f}([x])$ for $G$-invariant
$f\in\mathcal{L}{\it ip}_{1}(X)$ and $\bar{f}\in\mathcal{L}{\it ip}_{1}(X/G)$.

Take any $\varepsilon$ with $\operatorname{{\it d}_{{\rm
conc}}}((X,G),(Y,H))<\varepsilon$. There is $\pi\in\Pi(X,Y)$ such that
$\operatorname{{\it d}_{{\rm conc}}}^{\pi}((X,G),(Y,H))<\varepsilon$. For any
$h\in H$ there is $g\in G$ with $\rho^{\pi}(g,h)<\varepsilon$. We take any
$G$-invariant function $\hat{f}\in\mathcal{L}{\it ip}_{1}(X)$. Since
$\operatorname{diam}(\hat{f}_{*}\mu_{X};-\kappa)\leq D_{\kappa}$, there is a
real number $c$ such that $\mu_{X}(|\hat{f}+c|\leq D_{\kappa}/2)\geq
1-\kappa$. Setting
$f(x):=\min\\{\max\\{\hat{f}(x)+c,-D_{\kappa}/2\\},D_{\kappa}/2\\}$, $x\in X$,
we see that $|f|\leq D_{\kappa}/2$ and $\operatorname{{\it d}_{{\rm
KF}}}(f,\hat{f}+c)\leq\kappa$. By $\rho^{\pi}(g,h)<\varepsilon$, there is
$f_{h}^{\prime}\in\mathcal{L}{\it ip}_{1}(Y)$ with $|f_{h}^{\prime}|\leq
D_{\kappa}/2$ such that $\rho^{\pi}_{g,h}(f,f_{h}^{\prime})<\varepsilon$.
Hence,

$\operatorname{{\it d}_{{\rm KF}}}^{\pi}(f\circ p_{1},f_{h}^{\prime}\circ
p_{2})<\varepsilon\quad\text{and}\quad\operatorname{{\it d}_{{\rm
KF}}}^{\pi}(f\circ p_{1},f_{h}^{\prime}\circ h\circ p_{2})<\varepsilon.$

Setting $f^{\prime}:=f_{\operatorname{id}_{Y}}^{\prime}$ yields
$\operatorname{{\it d}_{{\rm KF}}}^{\pi}(f\circ p_{1},f^{\prime}\circ
p_{2})<\varepsilon$. By a triangle inequality, $\operatorname{{\it d}_{{\rm
KF}}}^{\mu_{Y}}(f^{\prime}\circ h,f_{h}^{\prime}\circ h)=\operatorname{{\it
d}_{{\rm KF}}}^{\mu_{Y}}(f^{\prime},f_{h}^{\prime})=\operatorname{{\it
d}_{{\rm KF}}}^{\pi}(f^{\prime}\circ p_{2},f_{h}^{\prime}\circ
p_{2})<2\varepsilon$. By a triangle inequality again, $\operatorname{{\it
d}_{{\rm KF}}}^{\pi}(f\circ p_{1},f^{\prime}\circ h\circ p_{2})<3\varepsilon$.
Denote by $\mu_{\operatorname{Aut}(Y)}$ the Haar probability measure on
$\operatorname{Aut}(Y)$ and define

$f^{\prime\prime}(y):=\int_{\operatorname{Aut}(Y)}f^{\prime}\circ
h(y)\,d\mu_{\operatorname{Aut}(Y)}(h),\qquad y\in Y.$

The $f^{\prime\prime}$ is an $H$-invariant $1$-Lipschitz function and
satisfies $|f^{\prime\prime}|\leq D_{\kappa}/2$. For any
$(x,y)\in\operatorname{supp}\pi$,

$\displaystyle|f(x)-f^{\prime\prime}(y)|$
$\displaystyle=\left|\int_{\operatorname{Aut}(Y)}(f(x)-f^{\prime}\circ
h(y))\;d\mu_{\operatorname{Aut}(Y)}(h)\right|$
$\displaystyle\leq\int_{\operatorname{Aut}(Y)}|f(x)-f^{\prime}\circ
h(y)|\;d\mu_{\operatorname{Aut}(Y)}(h).$

Integrate this formula with respect to $d\pi(x,y)$ and apply Fubini’s theorem
and Lemma 5.5(2) to obtain

$\displaystyle\|f\circ p_{1}-f^{\prime\prime}\circ p_{2}\|_{L_{1}}$
$\displaystyle\leq\int_{\operatorname{Aut}(Y)}\|f\circ p_{1}-f^{\prime}\circ
h\circ p_{2}\|_{L_{1}}d\mu_{\operatorname{Aut}(Y)}(h)\leq
3(D_{\kappa}+1)\varepsilon,$

which together with Lemma 5.5(1) implies $\operatorname{{\it d}_{{\rm
KF}}}^{\pi}(f\circ p_{1},f^{\prime\prime}\circ
p_{2})\leq\sqrt{3(D_{\kappa}+1)\varepsilon}$. Therefore, $\operatorname{{\it
d}_{{\rm KF}}}^{\pi}(\hat{f}\circ p_{1}+c,f^{\prime\prime}\circ
p_{2})\leq\sqrt{3(D_{\kappa}+1)\varepsilon}+\kappa$. The same is true if we
exchange $X$ and $Y$. We thus obtain the first part of the theorem.

The latter follows from the first because of the boundedness of
$\operatorname{ObsDiam}(X_{n};-\kappa)$ which is implied by
[OzSy:limf]*Theorem 1.1. ∎

##  6\. Convergence of group actions

In this section, we prove that the projection $\mathcal{X}_{\rm
eq}\ni(X,G)\mapsto X\in\mathcal{X}$ is a proper map with respect to $\square$,
which is analogous to [FY:f-gp]*Proposition 3.6, where $\mathcal{X}$ denotes
the set of mm-isomorphism classes of mm-spaces.

We need some definition and lemmas for the proof.

######  Definition 6.1 (Weak Hausdorff convergence).

Let $X$ be a metric space and $S,S_{n}\subset X$, $n=1,2,\dots$, closed
subsets. We say that $S_{n}$ converges to $S$ in the _weak Hausdorff_ sense if
the following (i) and (ii) are satisfied.

  1. (i)

$\lim_{n\to\infty}d_{X}(x,S_{n})=0$ for any $x\in S$.

  2. (ii)

$\liminf_{n\to\infty}d_{X}(x,S_{n})>0$ for any $x\in X\setminus S$.

Note that $S_{n}$ and $S$ may be empty and we set $d_{X}(x,\emptyset)=\infty$.

######  Lemma 6.2 ([Sy:mmg]*Lemma 6.6).

The set of closed subsets of a metric space is sequentially compact with
respect to the weak Hausdorff convergence.

######  Lemma 6.3.

Let $X$ be a metric space and $S,S_{n}\subset X$, $n=1,2,\dots$, closed
subsets. If $S_{n}$ converges to $S$ in the weak Hausdorff sense, then for any
compact subset $K\subset X$ and for any $\varepsilon>0$, there is a number
$n_{0}$ such that $S_{n}\cap K\subset U_{\varepsilon}(S\cap K)$ for $n\geq
n_{0}$.

Lemma 6.3 follows from an easy discussion using the compactness of $K$ and the
proof is omitted.

######  Lemma 6.4.

Let $X$ be a complete separable metric space, $S,S_{n}\subset X$ closed
subsets, and $\mu,\mu_{n}$ Borel probability measures, $n=1,2,\dots$. If
$\mu_{n}$ converges weakly to $\mu$ and if $S_{n}$ converges to $S$ in the
weak Hausdorff, then

$\mu(S)\geq\limsup_{n\to\infty}\mu_{n}(S_{n}).$

###### Proof.

Since Prokhorov’s theorem implies the tightness of $\\{\mu_{n}\\}$, there is a
sequence of compact subsets $K_{m}\subset X$, $m=1,2,\dots$, such that
$\mu_{n}(K_{m})\geq 1-1/m$ and $\mu(K_{m})\geq 1-1/m$ for every $m$. It
follows from Lemma 6.3 that, for any $\varepsilon>0$,

$\mu(U_{\varepsilon}(S\cap K_{m}))\geq\limsup_{n\to\infty}\mu(S_{n}\cap
K_{m})\geq\limsup_{n\to\infty}\mu(S_{n})-1/m,$

which implies

$\mu(S)\geq\mu(S\cap K_{m})\geq\limsup_{n\to\infty}\mu(S_{n})-1/m.$

This proves the lemma. ∎

The following is a main theorem of this section.

######  Theorem 6.5.

The projection $\mathcal{X}_{\rm eq}\ni(X,G)\mapsto X\in\mathcal{X}$ is a
proper map with respect to $\square$, i.e., for given
$(X_{n},G_{n})\in\mathcal{X}_{\rm eq}$, $n=1,2,\dots$ and $Y\in\mathcal{X}$,
if $X_{n}$ box converges to $Y$ as $n\to\infty$, then there exists a closed
subgroup $H\subset\operatorname{Aut}(X)$ such that $(X_{n},G_{n})$
equivariantly box converges to $(Y,H)$.

###### Proof.

By $\square(X_{n},Y)\to 0$, there are $\pi_{n}\in\Pi(X_{n},Y)$, closed subsets
$S_{n}\subset X_{n}\times Y$, and $\varepsilon_{n}\to 0+$ such that
$\pi_{n}(S_{n})\geq 1-\varepsilon_{n}$ and

$\sup_{(x_{i},y_{i})\in
S_{n}}|\,d_{X_{n}}(x_{1},x_{2})-d_{Y}(y_{1},y_{2})\,|\leq\varepsilon_{n}.$

For any $g\in G_{n}$ we put $\rho_{n}(g):=S_{n}\circ g\circ S_{n}^{-1}\subset
Y\times Y$ and
$\pi_{n}^{g}:=\pi_{n}\circ(\operatorname{id}_{X_{n}},g)_{*}\mu_{X_{n}}\circ\pi_{n}^{-1}\in\Pi(Y,Y)$,
where we define $tr(y,y^{\prime}):=(y^{\prime},y)$ for $y,y^{\prime}\in Y$ and
$\pi_{n}^{-1}:=tr_{*}\pi_{n}$. It follows from the $\operatorname{{\it
d}_{{\rm P}}}$-compactness of $\Pi(Y,Y)$ that if we take a subsequence of
$\\{n\\}$, then $\\{\,\pi_{n}^{g}\mid g\in G_{n}\,\\}$ converges to a closed
subset $\Pi\subset\Pi(Y,Y)$ in the Hausdorff sense. Take any $\pi\in\Pi$.
There is a sequence $g_{n}\in G_{n}$ with $\pi_{n}^{g_{n}}\to\pi$. Taking a
subsequence, we see that $\rho_{n}(g_{n})$ converges to a closed subset
$\hat{h}\subset Y\times Y$ in the weak Hausdorff sense. By Lemma 6.4, we have
$\pi(\hat{h})\geq\lim_{n\to\infty}\pi_{n}^{g_{n}}(\rho_{n}(g_{n}))=1$ and so
$\hat{h}\supset\operatorname{supp}\pi$. For any
$y_{i}\in\operatorname{Dom}(\rho_{n}(g_{n}))$ and
$y_{i}^{\prime}\in\rho_{n}(g_{n})(y_{i})$, $i=1,2$, there are $x_{1},x_{2}\in
X_{n}$ such that $(x_{i},y_{i})\in S_{n}$ and
$(g_{n}(x_{i}),y_{i}^{\prime})\in S_{n}$ for $i=1,2$. Therefore,

$\displaystyle|\,d_{X_{n}}(x_{1},x_{2})-d_{Y}(y_{1},y_{2})\,|\leq\varepsilon_{n}\
\text{and}\
|\,d_{X_{n}}(g_{n}(x_{1}),g_{n}(x_{2}))-d_{Y}(y_{1}^{\prime},y_{2}^{\prime})\,|\leq\varepsilon_{n}.$

Since $g_{n}$ is an isometry, these inequalities together imply

(6.1) $|\,d_{Y}(y_{1},y_{2})-d_{Y}(y_{1}^{\prime},y_{2}^{\prime})\,|\leq
2\varepsilon_{n}.$

Taking a subsequence we assume that $\varepsilon_{n}\leq 1/2^{n}$ for every
$n$.
Since $\mu_{Y}(\operatorname{Dom}(\rho_{n}(g_{n})))\geq
1-\pi_{n}(S_{n})-\pi_{n}^{-1}(S_{n}^{-1})\geq 1-2\varepsilon_{n}$, setting

$D:=\bigcup_{m=1}^{\infty}\bigcap_{n=m}^{\infty}\operatorname{Dom}(\rho_{n}(g_{n}))$

we see that

$\mu_{Y}(Y\setminus
D)\leq\mu_{Y}\left(\bigcup_{n=m}^{\infty}(Y\setminus\operatorname{Dom}(\rho_{n}(g_{n}))\right)\leq
2\sum_{n=m}^{\infty}\varepsilon_{n}=\frac{4}{2^{m}}\to 0\ \text{as}\
m\to\infty,$

which implies $\mu_{Y}(D)=1$. For any fixed $y_{1},y_{2}\in D$, since (6.1)
holds for infinitely many $n$, we have
$d_{Y}(y_{1},y_{2})=d_{Y}(y_{1}^{\prime},y_{2}^{\prime})$. Moreover, by the
closedness of $\hat{h}$, the set $\hat{h}$ is the graph of an isometry, say
$h$, and satisfies $\operatorname{supp}\pi=\hat{h}$. we have
$(\operatorname{id}_{Y},h)_{*}\mu_{Y}=\pi$, which implies
$h_{*}\mu_{Y}=\mu_{Y}$ and then $h\in\operatorname{Aut}(Y)$. Thus, the support
of any measure in $\Pi$ is the graph of an element of $\operatorname{Aut}(Y)$.
Let $H$ be the set of all such maps of $\operatorname{Aut}(Y)$.

Let us prove the closedness of $H$ with respect to $\operatorname{{\it
d}_{{\rm KF}}}$. Assume that a sequence $h_{n}\in H$ converges in measure to a
map $h\in\operatorname{Aut}(Y)$. It holds that $\operatorname{{\it d}_{{\rm
P}}}((\operatorname{id}_{Y},h_{n})_{*}\mu_{Y},(\operatorname{id}_{Y},h)_{*}\mu_{Y})\leq\operatorname{{\it
d}_{{\rm
KF}}}((\operatorname{id}_{Y},h_{n}),(\operatorname{id}_{Y},h))=\operatorname{{\it
d}_{{\rm KF}}}(h_{n},h)\to 0$. Since
$(\operatorname{id}_{Y},h_{n})_{*}\mu_{Y}\in\Pi$ and by the compactness of
$\Pi$, we have $(\operatorname{id}_{Y},h)_{*}\mu_{Y}\in\Pi$ and so $h\in H$.
This proves the closedness of $H$.

Let us next prove that $H$ is a subgroup of $\operatorname{Aut}(Y)$. Taking
any $h,h^{\prime}\in H$ we first prove $h^{\prime}\circ h\in H$. By taking a
subsequence of $\\{n\\}$, there are $g_{n},g_{n}^{\prime}\in G_{n}$ such that
$\rho_{n}(g_{n})$ and $\rho_{n}(g_{n}^{\prime})$ respectively converge to the
graphs of $h$ and $h^{\prime}$ in the weak Hausdorff sense. Taking a
subsequence again, we see that $\rho_{n}(g_{n}^{\prime}\circ g_{n})$ converges
to the graph of a map $k\in H$ in the weak Hausdorff sense. It suffices to
prove $h^{\prime}\circ h=k$. We put
$D_{n}:=\operatorname{Dom}(\rho_{n}(g_{n})\circ\rho_{n}(g_{n}^{\prime}))\cap\operatorname{Dom}(\rho_{n}(g_{n}^{\prime}\circ
g_{n}))$. For any $x\in D_{n}$, there are $y,z\in Y$ such that
$(x,y)\in\rho_{n}(g_{n})$ and $(y,z)\in\rho_{n}(g_{n}^{\prime})$. Also, there
are $x^{\prime},y^{\prime}\in X_{n}$ such that
$(x^{\prime},x),(g_{n}(x^{\prime}),y),(y^{\prime},y),(g_{n}^{\prime}(y^{\prime}),z)\in
S_{n}$. There is $z^{\prime}\in Y$ such that
$(x,z^{\prime})\in\rho_{n}(g_{n}^{\prime}\circ g_{n})$, and there is
$x^{\prime\prime}\in X_{n}$ such that
$(x^{\prime\prime},x),(g_{n}^{\prime}\circ
g_{n}(x^{\prime\prime}),z^{\prime})\in S_{n}$. Since
$d_{X_{n}}(g_{n}^{\prime}\circ
g_{n}(x^{\prime}),g_{n}^{\prime}(y^{\prime}))=d_{X_{n}}(g_{n}(x^{\prime}),y^{\prime})\leq\varepsilon_{n}$
and $d_{X_{n}}(g_{n}^{\prime}\circ g_{n}(x^{\prime}),g_{n}^{\prime}\circ
g_{n}(x^{\prime\prime}))=d_{X_{n}}(x^{\prime},x^{\prime\prime})\leq\varepsilon_{n}$,
we see, by a triangle inequality, $d_{X_{n}}(g_{n}^{\prime}\circ
g_{n}(x^{\prime\prime}),g_{n}^{\prime}(y^{\prime}))\leq 2\varepsilon_{n}$,
which proves

(6.2) $d_{Y}(z,z^{\prime})\leq 3\varepsilon_{n}.$

In the same way as before, taking a subsequence of $\\{n\\}$, we may assume
that $D:=\bigcup_{m=1}^{\infty}\bigcap_{n=m}^{\infty}D_{n}$ has
$\mu_{Y}$-measure equal to one. By (6.2), we have $h^{\prime}\circ h(x)=k(x)$
for any $x\in D$, which implies $h^{\prime}\circ h=k\in H$. A similar
(simpler) discussion leads us to $h^{-1}\in H$. Thus, $H$ is a subgroup of
$\operatorname{Aut}(Y)$.

Let us prove that $\square((X_{n},G_{n}),(Y,H))\to 0$. We set
$\delta_{n}:=\operatorname{{\it d}_{{\rm H}}}(\\{\,\pi_{n}^{g}\mid g\in
G_{n}\,\\},\Pi)$, which is the Hausdorff distance with respect to
$\operatorname{{\it d}_{{\rm P}}}$. Let $g\in G_{n}$ and $h\in H$ be such that
$\operatorname{{\it d}_{{\rm P}}}(\pi_{n}^{g},\pi^{h})\leq\delta_{n}$, where
$\pi^{h}:=(\operatorname{id}_{Y},h)_{*}\mu_{Y}\in\Pi$. It follows from
Strassen’s theorem that there are a coupling $\omega$ between $\pi_{n}^{g}$
and $\pi^{h}$ and a closed subset $T^{g,h}\subset\operatorname{supp}\omega$
with $\omega(T^{g,h})\geq 1-\delta_{n}$ such that $d_{Y\times
Y}((y,y^{\prime}),(\eta,\eta^{\prime}))\leq\delta_{n}$ for any
$(y,y^{\prime},\eta,\eta^{\prime})\in T^{g,h}$, where $d_{Y\times Y}$ denotes
the $l^{2}$-product metric of $d_{Y}$. Note that $\eta^{\prime}=h(\eta)$
holds. We put

$\displaystyle S_{n}^{g,h}:=\\{\,(x,y)\in S_{n}\mid(g(x),y^{\prime})\in
S_{n},\ (y,y^{\prime},\eta,h(\eta))\in T^{g,h}\ \text{for some
$y^{\prime},\eta\in Y$}\,\\}.$

For any $(x_{i},y_{i})\in S_{n}^{g,h}$, $i=1,2$, there are
$y_{1}^{\prime},\eta\in Y$ such that $(g(x_{1}),y_{1}^{\prime})\in S_{n}$,
$(y_{1},y_{1}^{\prime},\eta,h(\eta))\in T^{g,h}$. Since

$\displaystyle|\,d_{X_{n}}(g(x_{1}),x_{2})-d_{Y}(y_{1}^{\prime},y_{2})\,|\leq\varepsilon_{n},$
$\displaystyle d_{Y}(y_{1}^{\prime},h(y_{1}))\leq
d_{Y}(y_{1}^{\prime},h(\eta))+d_{Y}(h(\eta),h(y_{1}))\leq 2\delta_{n},$

we have

(6.3)
$|\,d_{X_{n}}(g(x_{1}),x_{2})-d_{Y}(h(y_{1}),y_{2})\,|\leq\varepsilon_{n}+2\delta_{n}.$

We are going to estimate $\pi_{n}(S_{n}^{g,h})$. Note that

$\displaystyle S_{n}^{g,h}=\\{\,(x,y)\in S_{n}\mid(x,y^{\prime})\in S_{n}\circ
g,\ (y,y^{\prime})\in p_{12}(T^{g,h})\ \text{for some $y^{\prime}\in
Y$}\,\\}.$

Let

$\displaystyle\tilde{S}_{n}^{g,h}$ $\displaystyle:=\\{\,(y,x,y^{\prime})\in
Y\times X_{n}\times Y\mid(x,y)\in S_{n},\ (x,y^{\prime})\in S_{n}\circ g,\
(y,y^{\prime})\in p_{12}(T^{g,h})\,\\}$ $\displaystyle=(S_{n}^{-1}\times
Y)\cap(Y\times S_{n}\circ g)\cap[p_{12}(T^{g,h})\times X_{n}]_{23},$
$\displaystyle\tilde{\pi}_{n}^{g}$
$\displaystyle:=(\pi_{n}\circ(\operatorname{id}_{X_{n}},g)_{*}\mu_{X_{n}})\bullet\pi_{n}^{-1},$

where $[...]_{23}$ means to exchange the second and third components. Since
$S_{n}^{g,h}=p_{21}(\tilde{S}_{n}^{g,h})$, $\pi_{n}(S_{n})\geq
1-\varepsilon_{n}$, $\pi_{n}^{g}(p_{12}(T^{g,h}))\geq\omega(T^{g,h})\geq
1-\delta_{n}$, $(p_{13})_{*}\tilde{\pi}_{n}^{g}=\pi_{n}^{g}$,
$(p_{12})_{*}\tilde{\pi}_{n}^{g}=\pi_{n}^{-1}$, and
$(p_{23})_{*}\tilde{\pi}_{n}^{g}=\pi_{n}\circ(\operatorname{id}_{X_{n}},g)_{*}\mu_{X}$,
we have

(6.4)
$\displaystyle\pi_{n}(S_{n}^{g,h})=(p_{21})_{*}\tilde{\pi}_{n}^{g}(p_{21}(\tilde{S}_{n}^{g,h}))=\tilde{\pi}_{n}^{g}(p_{21}^{-1}(p_{21}(\tilde{S}_{n}^{g,h})))\geq\tilde{\pi}_{n}^{g}(\tilde{S}_{n}^{g,h})$
$\displaystyle=1-\tilde{\pi}_{n}^{g}((S_{n}^{-1}\times Y)^{c}\cup(Y\times
S_{n}\circ g)^{c}\cup[p_{12}(T^{g,h})\times X_{n}]_{23}^{c})$
$\displaystyle\geq 1-(1-\tilde{\pi}_{n}^{g}(S_{n}^{-1}\times
Y))-(1-\tilde{\pi}_{n}^{g}(Y\times S_{n}\circ g))$
$\displaystyle\qquad-(1-\tilde{\pi}_{n}^{g}([p_{12}(T^{g,h})\times
X_{n}]_{23}))$
$\displaystyle=\pi_{n}^{-1}(S_{n}^{-1})+\pi_{n}\circ(\operatorname{id}_{X_{n}},g)_{*}\mu_{X}(S_{n}\circ
g)+\pi_{n}^{g}(p_{12}(T^{g,h}))-2$ $\displaystyle\geq
2\pi_{n}(S_{n})+(\operatorname{id}_{X_{n}},g)_{*}\mu_{X}(g)+\pi_{n}^{g}(p_{12}(T^{g,h}))-3$
$\displaystyle\geq 1-2\varepsilon_{n}-\delta_{n}.$

By remarking the definition of $\delta_{n}$, for any $g\in G_{n}$ there is
$h\in H$ such that (6.3) and (6.4) hold. Also, for any $h\in H$ there is $g\in
G_{n}$ such that (6.3) and (6.4) hold. This completes the proof of the
theorem. ∎

As a consequence of Theorem 6.5, Proposition 3.7, and the completeness of
$(\mathcal{X},\square)$ (see [Sy:mmg]*Theorem 4.14), we have the following.

######  Corollary 6.6.

$\mathcal{X}_{\rm eq}$ is complete with respect to the equivariant box metric.

##  7\. Application

In this section, we prove that a sequence of lens spaces with unbounded
dimension box converges to the cone of an infinite-dimensional complex
projective space, as an application of Theorem 5.4.

Consider the sequence of ellipsoids:

$\displaystyle
E_{j}:=\\{\;z\in\mathbb{C}^{n(j)}\mid\sum_{i=1}^{n(j)}\frac{|z_{i}|^{2}}{\alpha_{ij}^{2}}=1\;\\},$
$\displaystyle\alpha_{ij}>0,\ i=1,2,\dots,n(j),\ j=1,2,\dots.$

We equip each $E_{j}$ with the push-forward measure of the normalized volume
measure on the unit sphere $S^{n(j)}(1)$ by the linear transformation and also
with the restriction of the Euclidean distance function. Then, $E_{j}$ is an
mm-space.

Assume that $n(j)\to\infty$ as $j\to\infty$ and set
$a_{ij}:=\alpha_{ij}/\sqrt{n(j)}$. We also assume
$\sum_{i=1}^{\infty}a_{i}^{2}<\infty$. Then, the Gaussian space
$\Gamma^{\infty}_{\\{a_{i}\\}}:=(H,\|\cdot\|_{2},\gamma^{\infty}_{\\{a_{i}\\}})$
is an mm-space, where
$H:=\\{\,z\in\mathbb{C}^{\infty}\mid\|z\|_{2}<+\infty\,\\}$ is a Hilbert space
and
$\gamma^{\infty}_{\\{a_{i}^{2}\\}}:=\bigotimes_{i=1}^{\infty}(\gamma^{1}_{a_{i}^{2}}\otimes\gamma^{1}_{a_{i}^{2}})$
is the infinite product of the one-dimensional Gaussian measure
$\gamma^{1}_{a_{i}^{2}}$ of variance $a_{i}^{2}$. The $U(1)$-Hopf action on
$H$ (which is defined as scalar multiplication of unit complex number) defines
an mm-action on $\Gamma^{\infty}_{\\{a_{i}\\}}$ and also on $E_{j}$. By the
homomorphism $\mathbb{Z}_{j}:=\mathbb{Z}/j\mathbb{Z}\ni[k]\mapsto
e^{\sqrt{-1}k/j}\in U(1)$, we define the mm-action of $\mathbb{Z}_{j}$ on
$E_{j}$. Then the quotient space $E_{j}/\mathbb{Z}_{j}$ is diffeomorphic to a
lens space. The quotient space $\Gamma^{\infty}_{\\{a_{i}^{2}\\}}/U(1)$ is
homeomorphic to the cone of the infinite-dimensional complex projective space.

######  Theorem 7.1.

If $\lim_{j\to\infty}\sum_{i=1}^{n(j)}(a_{ij}-a_{i})^{2}=0$, then, as
$j\to\infty$,

  1. (1)

$(E_{j},\mathbb{Z}_{j})$ equivariantly box converges to
$(\Gamma^{\infty}_{\\{a_{i}\\}},U(1))$.

  2. (2)

$E_{j}/\mathbb{Z}_{j}$ box converges to $\Gamma^{\infty}_{\\{a_{i}\\}}/U(1)$.

###### Proof.

We prove (1). By [KS:ellipsoid]*Proposition 4.2, the measure of $E_{j}$
converges to $\gamma^{\infty}_{\\{a_{i}^{2}\\}}$ on $H$ with respect to the
$2$-Wasserstein metric. Therefore, for any fixed $\varepsilon>0$, if $j$ is
large enough, then there are
$\pi\in\Pi(E_{j},\Gamma^{\infty}_{\\{a_{i}^{2}\\}})$ and a closed subset
$S\subset\operatorname{supp}\pi$ such that $\pi(S)\geq 1-\varepsilon$ and
$\|x-x^{\prime}\|_{2}\leq\varepsilon$ for any $(x,x^{\prime})\in S$. We
observe that $d^{S}(g,g)\leq\varepsilon$ for any $g\in U(1)$, which together
with an easy discussion proves (1).

(2) follows from (1) and Theorem 5.4. ∎

######  Problem 7.2.

If $\lim_{j\to\infty}a_{ij}=a_{i}$, does $\operatorname{{\it d}_{{\rm
conc}}}((E_{j},\mathbb{Z}_{j}),(\Gamma^{\infty}_{\\{a_{i}\\}},U(1)))\to 0$
hold?

##  8\. Further problem

Gromov [Gmv:green, Sy:mmg] constructs a natural compactification of
$\mathcal{X}$ by using pyramids. It is a natural problem if we have the same
construction for $\mathcal{X}_{\rm eq}$. If that is obtained, then we are able
to solve Problem 7.2 affirmatively. For that, it is possible to define the
Lipschitz order relation on $\mathcal{X}_{\rm eq}$ and a pyramid in a suitable
way. However, we have many obstacles. For instance, we do not know if the
limit of pyramids is a pyramid. Also it seems to be difficult to formulate the
concept of measurement with group action for $\mathcal{X}_{\rm eq}$.

## References