# Some stability results of positive mass theorem for uniformly asymptotically
flat $3$-manifolds

Conghan Dong  Mathematics Department, Stony Brook University, NY 11794, United
States<EMAIL_ADDRESS>
(Date: August 27, 2024)

###### Abstract.

In this paper, we show that for a sequence of orientable complete pointed
uniformly asymptotically flat $3$-manifolds $(M_{i},g_{i},p_{i})$ with
nonnegative integrable scalar curvature $R_{g_{i}}\geq 0$, if their mass
$m(g_{i})\to 0$, then by subtracting some subsets $Z_{i}$ whose boundary area
$|\partial Z_{i}|\leq Cm(g_{i})^{\frac{1}{2}}$, $(M_{i}\setminus
Z_{i},g_{i},p_{i})$ converge to the Euclidean space $\mathbb{R}^{3}$ in the
sense of pointed weak volume convergence. Moreover, if we assume the Ricci
curvature bounded from below uniformly by $\text{Ric}_{g_{i}}\geq-2\Lambda$,
then $(M_{i},g_{i},p_{i})$ converge to $\mathbb{R}^{3}$ in the pointed Gromov-
Hausdorff topology.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Asymptotically flat $3$-manifolds
    2. 2.2 Space with Ricci curvature bounded below
  3. 3 Effective estimates on outer region
  4. 4 Regular subregion and the weak volume convergence
  5. 5 Removable singularity under Ricci lower bound

##  1\. Introduction

A pointed smooth orientable connected complete Riemannian $3$-manifold
$(M^{3},g,p)$ with one end is called $(A,B,\sigma)$-asymptotically flat (or
$(A,B,\sigma)$-AF), for some given $A,B>0,\sigma>\frac{1}{2}$, if there exists
a compact subset $K\subset M$ and a $C^{\infty}$-diffeomorphism
$\Phi:M\setminus K\to\mathbb{R}^{3}\setminus B(0,A)$ such that under this
identification,

$|\partial^{k}(g_{uv}-\delta_{uv})(x)|\leq B|x|^{-\sigma-|k|},$

for all multi-indices $|k|=0,1,2$, and $p=\Phi^{-1}(p^{*})$ for some fixed
$p^{*}\in\mathbb{R}^{3}\setminus B(0,A)$. Furthermore, we always assume the
scalar curvature $R_{g}$ is integrable.

In general, a $3$-manifold $(M^{3},g)$ with more than one end is called an AF
$3$-manifold if there exists a compact set $K\subset M$ such that $M\setminus
K=\sqcup_{k=1}^{N}M_{end}^{k}$, where the ends $M_{end}^{k}$ are pairwise
disjoint and for some $A_{k},B_{k}>0,\sigma_{k}>\frac{1}{2}$, each
$M_{end}^{k}$ is $(A_{k},B_{k},\sigma_{k})$-AF defined as above.

In the following, we will say $(M^{3},g,p)$ is pointed $(A,B,\sigma)$-AF if
it’s an AF $3$-manifold with one or more ends, and one of these ends contains
the base point $p$ and is $(A,B,\sigma)$-AF in the sense of above definition.

The ADM mass from general relativity [ADM61] of an AF $3$-manifold $(M,g)$ is
defined as

$m(g)=\lim_{r\to\infty}\frac{1}{16\pi}\int_{S_{r}}\sum_{j,k=1}^{3}(g_{jk,j}-g_{jj,k})v^{k}dA,$

where $v$ is the unit normal vector to the standard sphere
$S_{r}\subset\mathbb{R}^{3}$. It follows from [Bar86] that the mass is finite
and independent of AF coordinates.

The positive mass theorem in dimension three is the following.

######  Theorem 1.1.

Let $(M^{3},g)$ be an AF $3$-manifold with nonnegative integrable scalar
curvature. Then $m(g)\geq 0$ with equality if and only if
$(M,g)=(\mathbb{R}^{3},g_{E})$.

This theorem was first proved by Schoen-Yau [SY79a] in 1979 by constructing
stable minimal surface. Later Schoen-Yau extended the theorem to the case when
the dimension was less than eight [Sch89, SY79b]. In 1981, Witten [Wit81]
later proved the positive mass theorem for spin manifolds of any dimension. In
2001, as a byproduct of the proof of the Penrose inequality, Huisken-Ilmanen
[HI01] proved the three dimensional positive mass theorem by using the inverse
mean curvature flow. Recently, for three dimensional positive mass theorem, Li
[Li18] gave a proof using Ricci flow, Bray-Kazaras-Khuri-Stern [BKKS22] gave a
proof using a mass inequality.

The positive mass theorem gives a rigidity statement. Then a natural question
to ask is the stability problem.

######  Question 1.2.

Would the smallness of the mass imply that the manifold is close to the
Euclidean space in some topology?

It’s not known which topology should be a good choice and suitable for
nonnegative scalar curvature. One conjecture about this question in terms of
Gromov-Hausdorff topology was stated in 2001 by Huisken-Ilmanen in [HI01].

######  Conjecture 1.3.

Suppose $M_{i}$ is a sequence of asymptotically flat $3$-manifolds with
nonnegative scalar curvature and ADM mass tending to zero. Then there is a set
$Z_{i}\subset M_{i}$ such that $|\partial Z_{i}|\to 0$ and $M_{i}\setminus
Z_{i}$ converges to $\mathbb{R}^{3}$ in the Gromov-Hausdorff topology.

######  Remark 1.1.

Huisken-Ilmanen’s motivation of this conjecture comes from the Riemannian
Penrose inequality (see Section 2.1 for more details).

In 2014 Lee-Sormani [LS14] gave a conjecture in terms of intrinsic flat metric
topology (see also [S+21], Conjecture 10.1). Denote the exterior region by
$M_{ext}$ (see Section 2.1 for more details).

######  Conjecture 1.4.

Suppose $M_{i}$ is a sequence of asymptotically flat $3$-manifolds with
nonnegative scalar curvature and ADM mass tending to zero. Suppose
$\Omega_{i}\subset M_{i}$ are regions inside $M_{i,ext}$ containing $\partial
M_{i,ext}$, and $\Sigma_{i}$ is the outer boundary of $\Omega_{i}$ such that
$\text{diam}(\Omega_{i})\leq D$ and $|\Sigma_{i}|=A_{0}$, then
$\Omega_{i}\to\Omega_{0}$ in the intrinsic flat metric topology, where
$\Omega_{0}$ is a ball in the Euclidean space such that
$|\partial\Omega_{0}|=A_{0}.$

Recently Lee-Naber-Neumayer [LNN20] defined a $d_{p}$-convergence, and one can
also formulate a conjecture with respect to this topology.

Assuming additional conditions, there have been several progress toward above
question and conjectures, see among others [BF99, FK02, Cor05, Lee09, LS14,
HL15, KKL21, ABK22].

In this paper, motivated by Conjectures 1.3, 1.4 and the recent work of
Kazaras-Khuri-Lee [KKL21], we study the stability of the positive mass theorem
without imposing additional conditions.

Inspired by Cheeger-Gromov’s smooth convergence, we modify the flat metric
topology in geometric measure theory and give the following definition.

######  Definition 1.1.

Given a sequence of pointed smooth complete Riemannian manifolds
$(M_{i},g_{i})$ and pointed connected submanifolds $(\Omega_{i},p_{i})\subset
M_{i}$ with or without boundary, we say $(\Omega_{i},g_{i},p_{i})$ converge to
a smooth manifold $(M,g,p)$ in the sense of pointed weak volume convergence,
if there exist a sequence of positive numbers $\tau_{i}\to 0$, uniform numbers
$r_{0},c_{0}>0$ and embeddings $\varphi_{i}:\Omega_{i}\to
Y_{i}:=\varphi_{i}(\Omega_{i})\subset M$ with $\varphi_{i}(p_{i})=p$ such that
for any $r>r_{0}$,

$B(p,r)\setminus B(p,r_{0})\subset\varphi_{i}(\Omega_{i}\cap
B(p_{i},(1+\tau_{i})r+c_{0})),$

and as $2$-tensors

$|\varphi_{i,\sharp}g_{i}-g|(y)\leq\tau_{i},\forall y\in Y_{i},$

and for any $r>0$, for the measures
$\mu_{i}:=\chi_{Y_{i}}d\text{Vol}_{\varphi_{i,\sharp}g_{i}}$, the restricted
measures $\mu_{i}|_{B(p,r)}$ weakly converge to $d\text{Vol}_{g}|_{B(p,r)}$.
In particular, this convergence implies that for any $r>0$,
$\text{Vol}_{g_{i}}(\varphi_{i}^{-1}(B_{g}(p,r))\cap\Omega_{i})\to\text{Vol}_{g}B_{g}(p,r)$.

Our main result gives one answer to Question 1.2 in terms of this pointed weak
volume convergence.

To make the statement precisely, we give some notations. For any $r>0$ and a
uniformly AF $3$-manifold with diffeomorphism $\Phi$ on the uniform end,
define $M_{r}$ to be the interior component of $M\setminus\Phi^{-1}(S_{r})$,
where $S_{r}$ is the standard sphere with radius $r$ in $\mathbb{R}^{3}$.
Similarly define $M_{ext,r}$ to be the bounded region in
$M_{ext}\setminus\Phi^{-1}(S_{r})$. If $r>0$ is uniform, then for a uniformly
pointed AF $3$-manifold $(M,g,p)$, without loss of generality, we can assume
$p\in M\setminus M_{r}$ by a uniform perturbation.

######  Theorem 1.5.

Fix $A,B>0$, $\sigma>\frac{1}{2}$. Let $(M_{i}^{3},g_{i},p_{i})$ be a sequence
of orientable complete pointed $(A,B,\sigma)$-AF $3$-manifolds with
nonnegative integrable scalar curvature $R_{g_{i}}\geq 0$. If the ADM masses
$m(g_{i})$ of $(M_{i},g_{i})$ converge to zero, then up to a subsequence,
there exist $Z_{i}\subset M_{i,r_{0}}$ for some uniform $r_{0}>0$ with
boundary area $|\partial Z_{i}|\leq Cm(g_{i})^{\frac{1}{2}}$ for a uniform
constant $C>0$, such that $(M_{i}\setminus Z_{i},g_{i},p_{i})$ converge to
$\mathbb{R}^{3}$ in the sense of pointed weak volume convergence as defined
above.

######  Remark 1.2.

Ilmanen conjectured that for an asymptotically flat complete Riemannian
$3$-manifold $(M,g)$ with nonnegative scalar curvature, there exists a surface
$\Sigma$ bounding a region $\Omega$ such that area $|\Sigma|\leq 16\pi
m(g)^{2}$ and $M\setminus\Omega$ is ”flat-like”. Our result implies that we
can find such surface with area smaller that $Cm(g)^{\frac{1}{2}}$ such that
$M\setminus\Omega$ is ”flat-like” in the sense of pointed weak volume. This
conjecture was told to the author by Prof. Marcus Khuri, and it comes from a
private conversation between Prof. Hubert Bray and Prof. Tom Ilmanen.

######  Remark 1.3.

If we weaken the smallness on area of boundary, then for any uniform small
$0<\varepsilon\ll 1$, there exists $Z_{i}\subset\tilde{Z}_{i}\subset
M_{i,r_{0}}$ such that $|\partial\tilde{Z}_{i}|\leq
Cm(g_{i})^{\frac{1}{2}-\varepsilon}$ and
$(M_{i}\setminus\tilde{Z}_{i},g_{i},p_{i})\to\mathbb{R}^{3}$ in the pointed
flat metric topology. Also, for any $\delta>0$, $(M_{i}\setminus
B_{\delta}(\tilde{Z}_{i}),g_{i},p_{i})$ converges to a subset in
$\mathbb{R}^{3}$ locally in the $C^{0}$-topology, where
$B_{\delta}(\tilde{Z}_{i})$ is the $\delta$-neighborhood of $\tilde{Z}_{i}$.

######  Remark 1.4.

Recently in a paper joint with Antoine Song [DS23], we will prove a stronger
version of this theorem. Namely, we will remove the uniformly asymptotic
flatness assumption, improve the bound of boundary area to be almost optimal
and prove the pointed measured Gromov-Hausdorff convergence for reduced length
metrics. As a result, we will confirm Huisken-Ilmanen’s original conjecture.

If we assume stronger conditions on curvature, we can improve the pointed weak
volume convergence to pointed Gromov-Hausdorff convergence as following for
original manifolds without removing any subsets.

######  Theorem 1.6.

Under the same conditions as in Theorem 1.5, for a fixed $\Lambda>0$, if the
Ricci curvature are bounded uniformly from below by
$\text{Ric}_{g_{i}}\geq-2\Lambda$, then up to a subsequence,
$(M_{i},g_{i},p_{i})$ converge to $\mathbb{R}^{3}$ in the pointed Gromov-
Hausdorff topology.

Now we explain the ideas in the proof of the stability of positive mass
theorem and give an outline of this paper. Let’s firstly recall Kazaras-Khuri-
Lee’s result in [KKL21], where they proved the Gromov-Hausdorff convergence
under additional conditions that $H_{2}(M^{3},\mathbb{Z})=0$ and Ricci
curvature bounded uniformly from below by $\text{Ric}_{g}\geq-2\Lambda$. Their
basic ingredient is the mass inequality proved by Bray-Kazaras-Khuri-Stern in
[BKKS22], which is the following (see Section 2.1 for more details)

$m(g)\geq\frac{1}{16\pi}\int_{M_{ext}}\left(\frac{|\nabla^{2}u^{j}|^{2}}{|\nabla
u^{j}|}+R_{g}|\nabla u^{j}|\right)dV,$

where $\\{u^{j}\\}$, $j\in\\{1,2,3\\}$, are harmonic functions defined on
$M_{ext}$ and asymptotic to each AF coordinate function. When
$H_{2}(M,\mathbb{Z})=0$, this mass inequality holds on $M$. Then under the
Ricci lower bound condition, they used Cheng-Yau’s gradient estimate, Cheeger-
Colding’s techniques, like segment inequality, to derive almost pointwise
estimate for $u^{j}$ from the integral inequality, and take such $u^{j}$ as
splitting functions to prove the limit space is $\mathbb{R}^{3}$ when mass
tends to zero.

There are at least two difficulties when generalizing their arguments. If we
drop the condition that $H_{2}(M,\mathbb{Z})=0$, then $u^{j}$ is only defined
on $M_{ext}$. Even we only restrict on $M_{ext}$, we have no control on
$|\nabla u^{j}|$ around $\partial M_{ext}$. So we can not directly take
$\\{u^{j}\\}$ as splitting functions. If we drop the lower bound on Ricci
curvature, then almost nothing could be controlled, and we can not expect the
Gromov-Hausdorff convergence. The metric could either concentrate or
degenerate inside $M_{ext}$, so a different argument should be involved.

Our idea is inspired by Huisken-Ilmanen’s conjecture. We want to remove a bad
subset and consider the convergence on the remaining set, which we call it the
regular subregion. The regular subregion $E^{\tau}$ is defined by a connected
subregion of

$\\{x\in M_{ext}:\sum_{j,k=1}^{3}|\left<\nabla u^{j},\nabla
u^{k}\right>-\delta_{jk}|^{2}(x)\leq\tau\\}$

containing the outer region. In Proposition 4.4, we show that such regular
subregion is well defined and has smooth boundary with small area. We need
uniform effective estimates on $u^{j}$, which are proved in Section 3, to
ensure $E^{\tau}$ is non-empty and contains the outer region. To show
$\partial E^{\tau}$ has small area, we need to use the co-area formula, the
mass inequality and choose some suitable $\tau$ depending on the mass $m(g)$.

When restricted on the regular subregion, the harmonic map

$\mathcal{U}=(u^{1},u^{2},u^{3}):E^{\tau}\to Y^{\tau}\subset\mathbb{R}^{3}$

is a diffeomorphism onto its image, under which the metric $g$ is
$C^{0}$-close to the Euclidean metric. Applying integral by parts to the
integral of $|\nabla u^{j}|^{2}$, together with the fact that $\partial
E^{\tau}$ has small area, we can show that $E^{\tau}$ has almost Euclidean
volume. Finally, in Section 4, we can prove our main result that up to
diffeomorphism $\mathcal{U}$, $E^{\tau}$ converge to $\mathbb{R}^{3}$ in the
sense of pointed weak volume convergence.

In the case when Ricci curvature has a uniform lower bound, we always have a
subsequence converge to some limit space in the pointed Gromov-Hausdorff
topology, and let’s say $(M_{i},g_{i},p_{i})\to(X,d_{X},p_{X})$. Since
$\partial E^{\tau_{i}}$ has small area, as an application of Bishop-Gromov
volume comparison theorem, we can perturb the geodesic segment between points
in $E^{\tau_{i}}$ to avoid $\partial E^{\tau_{i}}$. This can be used to show
that $(E^{\tau_{i}},d_{i},p_{i})$ equipped with the restricted metric converge
to $(X,d_{X},p_{X})$. Also from effective estimate on outer region, we know
the outer region in $X$ is isometric to the outer region in $\mathbb{R}^{3}$.
Then we can fill the interior region by geodesic segments between points in
the outer region. To make sure all points in the interior region of $X$ can be
obtained in this way, we need to use some structure theory on three
dimensional Ricci limit space, particularly that $X$ is a topological manifold
and geodesics are non-branching. This filling process gives an isometric
embedding

$\Xi:X\to\mathbb{R}^{3}.$

Intuitively, one may think this map $\Xi$ as the limit of harmonic maps
$\mathcal{U}_{i}:E^{\tau_{i}}\to\mathbb{R}^{3}$. But since $E^{\tau_{i}}$ may
not be a length space and it’s not clear whether $\mathcal{U}_{i}$ is
Lipschitz on $E^{\tau_{i}}$, we can not directly take the limit of
$\mathcal{U}_{i}$. So we need above perturbation and filling process to
overcome this difficulty. All these arguments are included in Section 5.

### Notations

We use $C$ to denote a uniform constant which may be different from line to
line; $\Psi(\varepsilon)$ to denote a small uniform number satisfying
$\lim_{\varepsilon\to 0}\Psi(\varepsilon)=0$, and $\Psi(\varepsilon|a)$
satisfying $\lim_{\varepsilon\to 0}\Psi(\varepsilon|a)=0$ for each fixed $a$.
We denote the Euclidean metric and the induced distance by $g_{E},d_{E}$
respectively; and the geodesic segment and distance between
$x,y\in\mathbb{R}^{3}$ with respect to $g_{E}$ by $[xy],|xy|$. The geodesic
ball around $p\in(M,g)$ is denoted by $B(p,r)$, and $B(0,r)$ is a geodesic
ball in $\mathbb{R}^{3}$. Also for a surface $\Sigma\subset\mathbb{R}^{3}$,
$|\Sigma|$ is the area under metric $d_{E}$ ; for a domain
$\Omega\subset\mathbb{R}^{3}$, $|\Omega|$ is the volume under metric $d_{E}$.

### Acknowledgements

The author would like to thank his advisor Prof. Xiuxiong Chen for his
encouragements. The author thanks Prof. Marcus Khuri for his interests in this
work, helpful discussions and telling the author Ilmanen’s conjecture. The
author thanks Prof. Hubert Bray’s interests in this work. The author thanks
Hanbing Fang for introducing him the reference [Hei10] and helpful
discussions. The author thanks Antoine Song’s helpful discussions and
suggestions.

##  2\. Preliminaries

###  2.1. Asymptotically flat $3$-manifolds

In this section, we recall some basic facts and estimates of asymptotically
flat $3$-manifold. Let $(M^{3},g)$ be a complete AF $3$-manifold. By Lemma 4.1
in [HI01], we know inside $M$ there is a trapped compact region $T$ whose
topological boundary consists of smooth embedded minimal $2$-spheres. The
exterior region $M_{ext}$ is defined as the metric completion of a connected
component of $M\setminus T$ associated to one end. Then $M_{ext}$ is connected
and AF, has a compact minimal boundary, and contains no other compact minimal
surfaces (even immersed).

We have the following Riemannian Penrose inequality proved in [HI01, Bra01].

######  Proposition 2.1.

Let $(M^{3},g)$ be a complete AF $3$-manifold with nonnegative scalar
curvature, with total mass $m(g)$. For an exterior region $M_{ext}$ associated
to one end, the total area of the boundary is bounded by

$|\partial M_{ext}|\leq 16\pi m(g)^{2}.$

Given $(M_{ext},g)$ associated to one end, let $x^{j},j=1,2,3$ denote the
given asymptotically flat coordinate system in the end. Then there exist
functions $u^{j}\in C^{\infty}(M_{ext})$ satisfying

$\Delta_{g}u^{j}=0\text{ on }M_{ext},\ \frac{\partial u^{j}}{\partial
v}=0\text{ on }\partial M_{ext},\ |u^{j}-x^{j}|=o(|x|^{1-\sigma})\text{ as
}|x|\to\infty,$

where the Neumann boundary condition can be taken over components in $\partial
M_{ext}$ which do not bound a compact region. We call $u^{j}$ to be harmonic
function asymptotic to an AF coordinate function $x^{j}$. See [BKKS22, KKL21]
for more details.

The following mass inequality was proved in [BKKS22], whose proof was based on
the techniques introduced by Stern in [Ste22], where he proved similar mass
inequality for $S^{1}$-harmonic maps over closed $3$-manifolds.

######  Proposition 2.2 (Theorem 1.2 in [BKKS22]).

Let $(M_{ext},g)$ be an exterior region of a complete AF Riemannian
$3$-manifold $(M,g)$ with mass $m(g)$. Let $u$ be a harmonic function on
$(M_{ext},g)$ asymptotic to one of the AF-coordinate functions of the
associated end. Then

$m(g)\geq\frac{1}{16\pi}\int_{{M_{ext}}}\left(\frac{|\nabla^{2}u|^{2}}{|\nabla
u|}+R_{g}|\nabla u|\right)dV_{g}.$

By Proposition 3.1 and Lemma 3.2 in [KKL21], we have the following estimates
for harmonic functions.

######  Proposition 2.3.

Let $(M,g)$ be an orientable complete $(A,B,\sigma)$-AF $3$-manifold with
$R_{g}\geq 0$ and mass $m(g)\leq m_{0}$. Assume $u$ is a harmonic function
defined on $M_{ext}$ asymptotic to a AF coordinate function $x^{j}$ and
satisfies the Neumann boundary condition on $\partial M_{ext}$. Then there
exists a uniform big $r_{0}=r_{0}(A,B,\sigma)>0$ such that for any $r>r_{0}$
and small $\delta>0$, after the normalization such that the average of $u$
over $M_{r}\setminus M_{r_{0}}$ is zero, we have

$\sup_{M_{ext,r}}|u|\leq C(r),$

where $C(r)$ is unifrom and depends on $A,B,\sigma,m_{0}$. And on
$M_{ext}\setminus M_{ext,r_{0}}$,

$|\nabla u-\partial_{x^{j}}|(x)\leq C(r_{0})|x|^{-\sigma}.$

###  2.2. Space with Ricci curvature bounded below

For two pointed metric spaces $(X,x),(Y,y)$, define the pointed Gromov-
Hausdorff distance by

$d_{GH}((X,x),(Y,y))=\inf\\{d_{H}(X,Y)+|xy|\\},$

where the infimum is taken over all admissible distance on $X\sqcup Y$.

Alternatively, we can use Gromov-Hausdorff approximation (GHA) to define the
GH-distance. A map $f:X\to Y$ is called an $\varepsilon$-GHA if it satisfies
the following conditions:

  * (1)

$Y\subset B_{\varepsilon}(f(X))$;

  * (2)

$|d(x_{1},x_{2})-d(f(x_{1}),f(x_{2}))|<\varepsilon$.

Define

$\hat{d}_{GH}((X,x),(Y,y))=\inf\\{\varepsilon:\exists\varepsilon\text{-GHA
}f:X\to Y,f(x)=y\\}.$

Then $\frac{1}{2}d_{GH}\leq\hat{d}_{GH}\leq 2d_{GH}$ (see for example [Ron06]
for a proof).

For a compact metric space $X$, define the capacity and covering functions by

$Cap_{X}(\varepsilon)=\text{ maximum number of disjoint
}\frac{\varepsilon}{2}\text{-balls in }X,$ $Cov_{X}(\varepsilon)=\text{
minimum number of }\varepsilon\text{-balls it takes to cover }X.$

######  Proposition 2.4 (Gromov’s compactness).

For a class $\mathcal{C}$ of compact metric spaces with uniformly bounded
diameter, $\mathcal{C}$ is precompact is equivalent to
$Cap_{X}(\varepsilon)\leq N_{1}(\varepsilon)$ for all $X\in\mathcal{C}$, which
is also equivalent to $Cov_{X}(\varepsilon)\leq N_{2}(\varepsilon)$ for all
$X\in\mathcal{C}$.

See [Pet06] for a proof and more details. As corollaries, we have the
following lemmas.

######  Lemma 2.5.

If $(X_{i},d_{i})$ are metric spaces with diameter bounded uniformly by
$D<\infty$ and $(X_{i},d_{i})\to(X,d)$ in the Gromov-Hausdorff topology, for
any compact subset $\Omega_{i}\subset X_{i}$ with restricted metric, up to a
subsequence, $(\Omega_{i},d_{i})\to(\Omega,d)\subset(X,d)$.

######  Lemma 2.6.

The collection of complete pointed Riemannian $n$-manifolds with Ricci
curvature uniformly bounded below by $\text{Ric}\geq-(n-1)\Lambda$ is
precompact in the pointed Gromov-Hausdorff topology.

For manifolds with Ricci curvature bounded from below, we have the Bishop-
Gromov’s volume comparison (see [Pet06] for a proof).

######  Proposition 2.7.

Let $(M,g,p)$ be a complete Riemannian $n$-manifold with
$\text{Ric}_{g}\geq-(n-1)\Lambda g$. Then for any $0<r_{1}<r_{2}$, we have

$\frac{\text{Vol}B(p,r_{1})}{|B_{-\Lambda}(r_{1})|}\geq\frac{\text{Vol}B(p,r_{2})}{|B_{-\Lambda}(r_{2})|},$

where $|B_{-\Lambda}(r)|$ is the volume of a geodesic ball with radius $r$ in
the space form $S_{-\Lambda}^{n}$ with constant curvature $-\Lambda$.
Moreover, for $v\in T_{p}M$ a unit tangent vector and
$\gamma_{v}(t),t\in[0,l]$ a minimal geodesic segment with
$\gamma_{v}(0)=p,\dot{\gamma}(0)=v$, for any $0<t_{1}<t_{2}<l$, we have

$\frac{\mathcal{A}(t_{1}v)}{\underline{\mathcal{A}}(t_{1})}\geq\frac{\mathcal{A}(t_{2}v)}{\underline{\mathcal{A}}(t_{2})},$

where $\mathcal{A}(tv)$ is the area element at $\gamma_{v}(t)$ and
$\underline{\mathcal{A}}(t)$ is the corresponding area element in
$S_{-\Lambda}^{n}$.

Also by the Bishop-Gromov’s volume comparison, we have the following
perturbation proposition. Our argument follows from Lemma 3.7 in [Hei10]. We
give full details for reader’s convenience.

######  Lemma 2.8.

Assume $(M^{n},g)$ is a complete Riemannian manifold with Ricci curvature
bounded from below by $\text{Ric}_{g}\geq-(n-1)\Lambda$, and
$\Sigma^{n-1}\subset M$ is a smooth hypersurface with the area
$|\Sigma^{2D}|\leq\varepsilon$, where $\Sigma^{D}:=\Sigma\cap B(p,D)$. Assume
$\delta_{1},v_{0}>0$, and $\text{Vol}(B(p,1))\geq v_{0}$.

Then there exists $\delta_{0}=\Psi(\varepsilon|n,v_{0},\Lambda,D,\delta_{1})$
such that for any $x,y\in B(p,D)$ and $d(x,\Sigma)>2\delta_{1}$, there exists
$z\in B(y,\delta_{0})$ and a geodesic segment $\gamma_{xz}$ connecting $x,z$
such that $\gamma_{xz}\subset M\setminus\Sigma.$

###### Proof.

Otherwise, for any $z\in B(y,\delta_{0})$ and any segment $\gamma_{xz}$, there
is a $y\in\Sigma\cap\gamma_{xz}$. Note $\gamma_{xz}\subset B(p,2D)$, so
$y\in\Sigma^{2D}$.

Denote $B=B(y,\delta_{0})$ and $B^{\prime}\subset B$ the complement of cut
locus of $x$, so $B^{\prime}$ has the same volume as $B$. Let $\Sigma^{1}$ be
the set of all $y\in\Sigma$ which occur as the first intersection with
$\Sigma$ of $\gamma_{xz}$ for $z\in B^{\prime}$. Define
$d_{1},d_{2}:\Sigma^{1}\to\mathbb{R}^{+}$ by $d_{1}(y)=d(x,y)$ and
$d_{2}(y)=\sup\\{t>0:\gamma_{xy}(d_{1}(y)+t)\in B^{\prime}\\}$,
$U:=\\{(y,t)\in\Sigma^{1}\times\mathbb{R}:d_{1}(y)<t\leq
d_{1}(y)+d_{2}(y)\\}$, and $\Phi:U\to M$ by $\Phi(y,t)=\gamma_{xy}(t)$. Then
$\delta_{1}\leq d_{1}(y)\leq d_{1}(y)+d_{2}(y)\leq 2D$ and

$\Phi^{*}\left(d\text{Vol}_{g}\right)|_{(y,t)}=\frac{\mathcal{A}(tv_{y})}{\mathcal{A}(d_{1}(y)v_{y})}\cos\alpha_{y}dA_{\Sigma}\wedge
dt,$

where $v_{y}=\dot{\gamma}_{xy}(0)$, $\alpha_{y}$ is the angle between
$\dot{\gamma}_{xy}(d_{1}(y))$ and exterior normal to $\Sigma$ at $y$. Then by
Bishop-Gromov’s comparison, integrating over $U$ gives

$\displaystyle\text{Vol}(B^{\prime})$
$\displaystyle\leq\int_{\Sigma^{1}}\int_{d_{1}(y)}^{d_{1}(y)+d_{2}(y)}\frac{\mathcal{A}(tv_{y})}{\mathcal{A}(d_{1}(y)v_{y})}dtdA(y)$
$\displaystyle\leq\int_{\Sigma^{1}}\int_{d_{1}(y)}^{d_{1}(y)+d_{2}(y)}\frac{\underline{\mathcal{A}}(t)}{\underline{\mathcal{A}}(d_{1}(y))}dtdA(y)$
$\displaystyle\leq C(D,\delta_{1},\Lambda,n)|\Sigma^{2D}|.$

By Bishop-Gromov’s volume comparison again,

$\text{Vol}B(y,\delta_{0})\geq\frac{\text{Vol}B_{-\Lambda}(\delta_{0})}{\text{Vol}B_{-\Lambda}(3D)}\cdot\text{Vol}B(p,D)\geq\frac{\text{Vol}B_{-\Lambda}(\delta_{0})}{\text{Vol}B_{-\Lambda}(3D)}\cdot
v_{0}.$

Choosing $\delta_{0}$ such that
$\frac{\text{Vol}B_{-\Lambda}(\delta_{0})}{\text{Vol}B_{-\Lambda}(3D)}\cdot
v_{0}=2C(D,\delta_{1},\Lambda,n)\varepsilon$ gives a contradiction. ∎

######  Remark 2.1.

In this proof, we actually proved that if for a set $B\subset B(p,D)$ and for
all points $y\in B$, there is a geodesic segment between $x,y$ intersecting
$\Sigma$, then $\text{Vol}(B)\leq C(D,\delta_{1},\Lambda,n)|\Sigma|$.

######  Remark 2.2.

If we denote the subset in $B(y,\delta_{0})$ by $\hat{B}$ consisting of points
which could be connected to $x$ by segments in $M\setminus\Sigma$, then by the
same argument, we can prove that
$\text{Vol}(\hat{B})\geq(1-\sqrt{\varepsilon})\text{Vol}(B(y,\delta_{0}))$. So
for any uniform positive number $N$ and points $x_{1},\cdots,x_{N}$ with
$d(x_{j},\Sigma)>2\delta_{1}$, if $\varepsilon$ is small enough depending on
$N$, there exists $z\in B(y,\delta_{0})$ such that each geodesic segment
connecting $z,x_{j}$ lies in $M\setminus\Sigma$.

We also need some structure theory on Ricci limit space. The first is the non-
branching property. The following proposition is a special case of Theorem 1.3
in [Den20].

######  Proposition 2.9.

If $(M_{i},g_{i},p_{i})$ is a sequence of Riemannian manifolds with Ricci
curvature bounded uniformly from below, and $(M_{i},g_{i},p_{i})$ converge to
$(X,d,p)$ in the pointed Gromov-Hausdorff topology, then $(X,d)$ is non-
branching. In other words, if there are two geodesics $\gamma_{1}$ and
$\gamma_{2}$ in $X$, and for some $t>0$, $\gamma_{1}(s)=\gamma_{2}(s)$ for
$s\in[0,t]$, then $\gamma_{1}=\gamma_{2}$.

In dimension three, we know Ricci limit space is a manifold. The following
proposition is Corollary 1.5 in [ST21].

######  Proposition 2.10.

If $(M_{i}^{3},g_{i},p_{i})$ is a sequence of Riemannian $3$-manifolds with
Ricci curvature bounded uniformly from below and noncollapsing in the sense
that $\text{Vol}B(p_{i},1)\geq v_{0}>0$, and $(M_{i},g_{i},p_{i})$ converge to
$(X,d,p)$ in the pointed Gromov-Hausdorff topology, then there exists a
topological $3$-manifold $M$ such that $M=X$ and the topology generated by $d$
is the same as $M$. Moreover, the charts of $M$ can be taken to be bi-Hölder
with respect to $d$.

##  3\. Effective estimates on outer region

In this section, we always assume $(M,g,p)$ is an orientable complete pointed
$(A,B,\sigma)$-AF $3$-manifold with nonnegative integrable scalar curvature
$R_{g}\geq 0$, and the mass $m(g)<1$. Let $\Phi:M\setminus
K\to\mathbb{R}^{3}\setminus B(0,A)$ be the corresponding
$C^{\infty}$-diffeomorphism. For this end defined by $\Phi$, let $M_{ext}$ be
the associated exterior region, and $\\{u^{j}\\},j=1,2,3$ be harmonic
functions asymptotic to AF-coordinate functions $x^{j}\circ\Phi$, which are
defined over $M_{ext}$ with Neumann boundary condition on $\partial M_{ext}$
(see Section 2.1 for more details on rigorous definitions and properties).

Recall that $M_{r}$ is defined to be the interior component of
$M\setminus\Phi^{-1}(S_{r})$, where $S_{r}\subset\mathbb{R}^{3}$ is the
standard sphere with radius $r$.

######  Lemma 3.1.

For a sufficiently large uniform $r_{0}>0$, and any $x\in M\setminus
M_{r_{0}}$,

$|\nabla^{2}u^{j}|(x)\leq Cm(g)^{\frac{5}{96}}=\Psi(m(g)).$

###### Proof.

Fix a $j\in\\{1,2,3\\}$ and denote $u^{j}$ by $u$ for simplicity. For large
$r_{0}$ and $x\in M\setminus M_{r_{0}}$, we know

$B(x,2)\subset\Phi^{-1}(\mathbb{R}^{3}\setminus B(0,A)),$

where the metric $g$ is $C^{2}$-close to the Euclidean metric, i.e. $\forall
y\in B(x,2)$, and for all multi-index $|l|=0,1,2$,

$|\partial^{l}(g_{jk}-\delta_{jk})|(y)\leq B|y|^{-\sigma-|l|}\leq
2Br_{0}^{-\sigma}\ll 1.$

By Proposition 2.3, we have estimate

$\sup_{B(x,1)}|\partial u|\leq C.$

From the harmonic equation and Schauder estimate, $|\partial^{2}u|(y)\leq C$
over $B(x,1)$, and thus

$|\nabla u|(y)+|\nabla^{2}u|(y)\leq C,\ \forall y\in B(x,1).$

From the Bochner formula, we have

$\displaystyle\frac{1}{2}\Delta|\nabla^{2}u|^{2}$
$\displaystyle=|\nabla^{3}u|^{2}+\nabla_{j}\nabla_{i}u\cdot\left(\nabla_{k}(R(\partial_{x^{k}},\partial_{x^{j}})(\nabla_{i}u))+\nabla_{j}(R(\partial_{x^{k}},\partial_{x^{i}})(\nabla_{k}u))\right)$
$\displaystyle\ \
+R_{lkij}\cdot\nabla_{j}\nabla_{i}u\cdot\nabla_{k}\nabla_{l}u.$

Now we can apply the Moser iteration to control $|\nabla^{2}u|$ by its
$L^{2}$-integral as following. For any cut-off function $\varphi$ with support
in $B(x,1)$, integral by parts gives

$-\frac{1}{2}\int\nabla\varphi\cdot\nabla|\nabla^{2}u|^{2}\geq-\int\left(\varphi|\nabla
u|^{2}+|\nabla\varphi||\nabla^{2}u||\nabla
u|+\varphi|\nabla^{2}u|^{2}\right),$

where we have used the fact that $|Rm|\ll 1$. Take a cutoff function $\eta$
with support in $B(x,1)$ and a constant $\beta\geq 2$ to be determined below.
Choosing $\varphi=\eta^{2}|\nabla^{2}u|^{2(\beta-1)}$, we have

$\displaystyle(\beta$
$\displaystyle-1)\int\eta^{2}|\nabla^{2}u|^{2(\beta-2)}|\nabla|\nabla^{2}u|^{2}|^{2}+\int
2\eta|\nabla^{2}u|^{2(\beta-1)}\nabla\eta\cdot\nabla|\nabla^{2}u|^{2}$
$\displaystyle\leq\int\eta^{2}|\nabla^{2}u|^{2(\beta-1)}|\nabla
u|^{2}+\int\eta|\nabla\eta||\nabla^{2}u|^{2(\beta-1)}\cdot|\nabla^{2}u||\nabla
u|$ $\displaystyle\ \ \ \
+\int(\beta-1)\eta^{2}|\nabla^{2}u|^{2(\beta-2)}|\nabla|\nabla^{2}u|^{2}|\cdot|\nabla^{2}u||\nabla
u|+\int\eta^{2}|\nabla^{2}u|^{2\beta}.$

By Hölder inequality and $|\nabla u|\leq C$, we have

$\displaystyle(\beta-1)\int\eta^{2}|\nabla^{2}u|^{2(\beta-2)}|\nabla|\nabla^{2}u|^{2}|^{2}$
$\displaystyle\leq\frac{C}{\beta-1}\int|\nabla\eta|^{2}|\nabla^{2}u|^{2\beta}+\int\eta^{2}|\nabla^{2}u|^{2\beta}$
$\displaystyle\ \
+C\int(\beta\eta^{2}|\nabla^{2}u|^{2(\beta-1)}+\eta|\nabla\eta||\nabla^{2}u|^{2\beta-1}).$

Set $v=|\nabla^{2}u|^{2}$. Then

$\displaystyle\int|\nabla(\eta v^{\frac{\beta}{2}})|^{2}$ $\displaystyle\leq
C\int|\nabla\eta|^{2}v^{\beta}+C\beta\int(\eta
v^{\frac{\beta}{2}})^{2}+C\beta^{2}\int\eta^{2}v^{\beta-1}+C\beta\int\eta|\nabla\eta|v^{\beta-\frac{1}{2}}$
$\displaystyle\leq
C\beta^{2}\int(\eta^{2}+\eta|\nabla\eta|+|\nabla\eta|^{2})v^{\beta-1},$

where we have used $v\leq C$. By Sobolev inequality,

$\displaystyle\|\eta v^{\frac{\beta}{2}}\|_{L^{6}}^{2}\leq C\cdot\|\eta
v^{\frac{\beta}{2}}\|_{W^{1,2}}^{2}\leq
C\beta^{2}\int(\eta^{2}+\eta|\nabla\eta|+|\nabla\eta|^{2})v^{\beta-1}.$

Let $\eta\geq 0$ be a cutoff function satisfying $\eta(s)=1,s\leq 0$,
$\eta(s)=0,s\geq 1$, and $|\eta^{\prime}|\leq 2$. Set
$r_{k}=\frac{1}{2}+\frac{1}{2^{k+1}}$, $\Omega_{k}=B(x,r_{k})$, and

$\eta_{k}(y)=\eta\left(\frac{d(x,y)-r_{k}}{r_{k-1}-r_{k}}\right).$

Then

$\displaystyle\|v\|_{L^{3\beta}(\Omega_{k})}$
$\displaystyle\leq\|\eta_{k}v^{\frac{\beta}{2}}\|_{L^{6}(\Omega_{k-1})}^{\frac{2}{\beta}}\leq(C\beta^{2}4^{k})^{\frac{1}{\beta}}\left(\int_{\Omega_{k-1}}v^{\beta-1}\right)^{\frac{1}{\beta}}$
$\displaystyle\leq(C|B(x,1)|^{\frac{1}{\beta}}\beta^{2}4^{k})^{\frac{1}{\beta}}\|v\|_{L^{\beta}(\Omega_{k-1})}^{1-\frac{1}{\beta}}$
$\displaystyle\leq(C\beta^{2}4^{k})^{\frac{1}{\beta}}\|v\|_{L^{\beta}(\Omega_{k-1})}^{1-\frac{1}{\beta}}.$

Taking $\beta=2\cdot 3^{k-1}$ gives

$\displaystyle\|v\|_{L^{2\cdot 3^{k}}(\Omega_{k})}$ $\displaystyle\leq
C^{\frac{k}{3^{k}}}\|v\|_{L^{2\cdot 3^{k-1}}(\Omega_{k-1})}^{1-\frac{1}{2\cdot
3^{k-1}}}$ $\displaystyle\leq
C^{\frac{k}{3^{k}}+\frac{k-1}{3^{k-1}}(1-\frac{1}{2\cdot
3^{k-1}})}\|v\|_{L^{2\cdot 3^{k-2}}(\Omega_{k-2})}^{(1-\frac{1}{2\cdot
3^{k-1}})(1-\frac{1}{2\cdot 3^{k-2}})}$ $\displaystyle\leq
C^{\sum_{j=1}^{k}\frac{j}{3^{j}}}\|v\|_{L^{2}(B(x,1))}^{\prod_{j=0}^{k-1}(1-\frac{1}{2\cdot
3^{j}})}.$

Notice that

$\sum_{j=1}^{\infty}\frac{j}{3^{j}}<\sum_{j=1}^{\infty}\frac{2^{j}}{3^{j}}<1,$
$\prod_{j=0}^{\infty}\left(1-\frac{1}{2\cdot
3^{j}}\right)>\frac{1}{2}\cdot\frac{5}{6}\cdot\prod_{j=2}^{\infty}\left(1-\frac{1}{j^{2}}\right)>\frac{1}{2}\cdot\frac{5}{6}\cdot\frac{1}{2},$

and

$\displaystyle\int_{B(x,1)}|\nabla^{2}u|^{4}$ $\displaystyle\leq
C\int_{B(x,1)}|\nabla^{2}u|$ $\displaystyle\leq
C\left(\int_{B(x,1)}\frac{|\nabla^{2}u|^{2}}{|\nabla
u|}\right)^{\frac{1}{2}}\left(\int_{B(x,1)}|\nabla u|\right)^{\frac{1}{2}}$
$\displaystyle\leq C\cdot m(g)^{\frac{1}{2}}.$

So

$\|v\|_{L^{\infty}(B(x,\frac{1}{2}))}\leq
C\|v\|_{L^{2}(B(x,1))}^{\frac{5}{24}}=Cm(g)^{\frac{5}{96}}=\Psi(m(g)).$

∎

######  Lemma 3.2.

For a sufficiently large uniform $r_{0}>0$, and any $x\in M\setminus
M_{r_{0}}$,

$|\left<\nabla u^{j},\nabla u^{k}\right>-\delta_{jk}|(x)\leq
Cm(g)^{\frac{1}{192}}=\Psi(m(g)).$

###### Proof.

By Proposition 2.3, we know on $M\setminus M_{r_{0}}$,

$|\nabla u^{j}-\partial x^{j}|(x)\leq C|x|^{-\sigma},$

so

$\displaystyle|\left<\nabla u^{j},\nabla u^{k}\right>-\delta_{jk}|$
$\displaystyle\leq|\left<\nabla u^{j}-\partial x^{j},\nabla
u^{k}\right>+\left<\partial x^{j},\nabla u^{k}-\partial
x^{k}\right>+\left<\partial x^{j},\partial x^{k}\right>-\delta_{jk}|$
$\displaystyle\leq C|x|^{-\sigma}.$

Choose $L>0$ big enough, depending on $m(g)$, to be determined below. If $x\in
M\setminus M_{L}$, then

$|\left<\nabla u^{j},\nabla u^{k}\right>-\delta_{jk}|(x)\leq CL^{-\sigma}.$

If $x\in M_{L}\setminus M_{r_{0}}$, choose $y\in\partial M_{L}$ such that
$|xy|=L-r_{0}$ and $[xy]\subset M_{L}\setminus M_{r_{0}}$.

Then

$\displaystyle|\left<\nabla u^{j},\nabla u^{k}\right>-\delta_{jk}|^{2}(x)$
$\displaystyle\leq|(\left<\nabla u^{j},\nabla
u^{k}\right>-\delta_{jk})^{2}(x)-|\left<\nabla u^{j},\nabla
u^{k}\right>-\delta_{jk}|^{2}(y)|$ $\displaystyle\ \ +(\left<\nabla
u^{j},\nabla u^{k}\right>-\delta_{jk})^{2}(y)$ $\displaystyle\leq L\cdot
Cm(g)^{\frac{5}{96}}+CL^{-2\sigma}.$

Taking $L=m(g)^{-\frac{1}{96}}$, then $\frac{1}{L}=\Psi(m(g))$ and
$Lm(g)^{\frac{5}{96}}=m(g)^{\frac{1}{24}}=\Psi(m(g))$ finishes the proof. ∎

Set

$\mathcal{U}:=(u^{1},u^{2},u^{3}):M_{ext}\to\mathbb{R}^{3}.$

For $C(r_{0})$ in Proposition 2.3, we can choose $r_{1}$ such that
$C(r_{0})r_{1}^{-\sigma}<\frac{1}{2}$. For $j\in\\{1,2,3\\}$, we know
$\sup_{M_{r_{1}}}|u^{j}|\leq C(r_{1})$. So if we choose $r_{2}=2C(r_{1})$,
then

$\mathcal{U}^{-1}(\mathbb{R}^{3}\setminus B(0,r_{2}))\subset M\setminus
M_{r_{1}}.$

We prove that around outer region, $\mathcal{U}$ is one to one.

######  Lemma 3.3.

For all $L\gg r_{2}$ big enough, and any $w\in B(0,L)\setminus B(0,L-1)$,
$\mathcal{U}^{-1}\\{w\\}$ consists of at most one point.

###### Proof.

Assume that $\mathcal{U}(y)=\mathcal{U}(z)=w$. Then $y,z\in M\setminus
M_{r_{1}}$. Under the diffeomorphism $\Phi:M\setminus
M_{r_{1}}\to\mathbb{R}^{3}$, we can assume $y,z\in\mathbb{R}^{3}\setminus
B(0,r_{1})$, where the metric $g$ is $C^{2}$-close to the Euclidean metric.
Then $\mathcal{U}:(\mathbb{R}^{3}\setminus B(0,r_{1}),g)\to\mathbb{R}^{3}$.
Consider the straight line segment $[yz]$ between $y,z$ and assume
$y^{\prime},z^{\prime}\in\partial B(0,r_{1})$ are points on $[yz]\cap\partial
B(0,r_{1})$ closest to $y,z$ respectively. Denote the unit direction vector of
$[yz]$ by $a\in S^{2}\subset\mathbb{R}^{3}$. Set
$u^{a}=\sum_{j=1}^{3}a_{j}u^{j}$, and $l=|yz|$. Then there are two cases.

Case 1): if no such points $y^{\prime},z^{\prime}$ exist, then
$[yz]\subset\mathbb{R}^{3}\setminus B(0,r_{1})$. So

$\displaystyle u^{a}(z)-u^{a}(y)$
$\displaystyle=\int_{0}^{l}(u^{a}(y+at))^{\prime}dt$
$\displaystyle=\int_{0}^{l}\sum_{j,k=1}^{3}\left<a_{j}\nabla
u^{j},a_{k}\partial y^{k}\right>$
$\displaystyle\geq(1-C(r_{0})r_{1}^{-\sigma})l$
$\displaystyle\geq\frac{1}{2}l,$

which implies that $l=0$.

Case 2): if such $y^{\prime},z^{\prime}$ exist, then by choosing $L\gg r_{2}$,
we can assume $l\gg r_{1}$, since otherwise no such $y^{\prime},z^{\prime}$
exist. We can construct a piecewise straight line segment
$\gamma_{y^{\prime}z^{\prime}}$ between $y^{\prime},z^{\prime}$ satisfying
$\gamma_{y^{\prime}z^{\prime}}\subset\mathbb{R}^{3}\setminus B(0,r_{1})$ and
$|\gamma_{y^{\prime}z^{\prime}}|\leq 10r_{1}$. Consider the new piecewise
straight line segment
$\gamma:=yy^{\prime}\cup\gamma_{y^{\prime}z^{\prime}}\cup[z^{\prime}z]$, then
$\gamma\subset\mathbb{R}^{3}\setminus B(0,r_{1})$. Similarly,

$u^{a}(z)-u^{a}(z^{\prime})\geq\frac{1}{2}|zz^{\prime}|,\
u^{a}(y^{\prime})-u^{a}(y)\geq\frac{1}{2}|yy^{\prime}|,$

and

$|u^{a}(z^{\prime})-u^{a}(y^{\prime})|\leq C(r_{1}).$

So

$u^{a}(z)-u^{a}(y)\geq\frac{1}{2}(|zz^{\prime}|+|yy^{\prime}|)-C(r_{1})\geq\frac{1}{2}l-C(r_{1})>0,$

a contradiction. ∎

######  Lemma 3.4.

For any $L_{0}>0$, there exists $L_{1}$ big enough such that for all
$L>L_{1}$, $\mathcal{U}^{-1}(\mathbb{R}^{3}\setminus B(0,L))\subset M\setminus
M_{L_{0}}$, and $\mathcal{U}(M\setminus M_{L})\subset\mathbb{R}^{3}\setminus
B(0,L_{0})$.

###### Proof.

By Proposition 2.3, we know $\sup_{M_{L_{0}}}|u^{j}|\leq C(L_{0})$, so for any
$L>C(L_{0})$, $\mathcal{U}^{-1}(\mathbb{R}^{3}\setminus B(0,L))\subset
M\setminus M_{L_{0}}$. For any $y\in M\setminus M_{L}$, choose a straight line
segment $[xy]$ with $x\in\partial M_{L_{0}}$ being the closest point to $y$.
Denote the unit direction vector of $xy$ by $a$, then similar to above
calculation, we know

$u^{a}(y)\geq u^{a}(x)+\frac{1}{2}|xy|\geq\frac{1}{2}(L-L_{0})-C(L_{0}),$

which is bigger than $L_{0}$ if $L$ is chosen big enough. ∎

In the following, for simplicity, we will use $L_{0}$ to denote a uniform
positive number such that conclusions in Lemma 3.3 and Lemma 3.4 hold.

##  4\. Regular subregion and the weak volume convergence

In this section, we define the regular subregion, where the geometry is
locally uniformly controlled. Then we will use this regular subregion to prove
our convergence theorem.

For any small $\tau>0$, we can find $\frac{\tau}{2}<\tau^{j}_{1}<\tau$ such
that

$\\{x\in M_{ext}:|\nabla u^{j}|^{2}=1+\tau^{j}_{1}\\}$

is a smooth surface.

Define

$E_{1}^{\tau}:=M_{ext}\cap\cap_{j=1}^{3}\\{|\nabla u^{j}|^{2}\leq
1+\tau^{j}_{1}\\}.$

By the co-area formula, we have

$\displaystyle\int_{0}^{\infty}\mathcal{H}_{g}^{2}($ $\displaystyle
M_{r_{0}}\cap E^{\tau}_{1}\cap\\{\sum_{j,k=1}^{3}|\left<\nabla u^{j},\nabla
u^{k}\right>-\delta_{jk}|^{2}=t\\})dt$ $\displaystyle=\int_{M_{r_{0}}\cap
E_{1}^{\tau}}\left|\nabla\sum_{j,k=1}^{3}|\left<\nabla u^{j},\nabla
u^{k}\right>-\delta_{jk}|^{2}\right|$ $\displaystyle\leq
2((1+\tau)^{2}+1)\int_{M_{r_{0}}\cap
E_{1}^{\tau}}\sum_{j,k=1}^{3}(|\nabla^{2}u^{j}||\nabla
u^{k}|+|\nabla^{2}u^{k}||\nabla u^{j}|)$ $\displaystyle\leq
C\sum_{j,k=1}^{3}\left(\int_{M_{r_{0}}\cap
E_{1}^{\tau}}\frac{|\nabla^{2}u^{j}|^{2}}{|\nabla
u^{j}|}\right)^{\frac{1}{2}}\left(\int_{M_{r_{0}}\cap E_{1}^{\tau}}|\nabla
u^{j}|^{3}\right)^{\frac{1}{6}}\left(\int_{M_{r_{0}}\cap E_{1}^{\tau}}|\nabla
u^{k}|^{3}\right)^{\frac{1}{3}}$ $\displaystyle\leq
C(r_{0})m(g)^{\frac{1}{2}},$

where in the last inequality, we used the mass inequality Proposition 2.2 and
the fact that

$\int_{M_{r_{0}}\cap E_{1}^{\tau}}|\nabla
u^{j}|^{3}\leq(1+\tau)\int_{M_{r_{0}}}|\nabla u^{j}|^{2}\leq C(r_{0}),$

where the last inequality can be proved by integration by parts (see also the
proof of Lemma 4.5).

So for $\tau_{1}:=\min\\{\tau_{1}^{j}\\}$, there exists
$\frac{\tau_{1}}{2}<\tau_{2}<\tau_{1}$ such that

$\\{x\in M_{ext}:\sum_{j,k=1}^{3}|\left<\nabla u^{j},\nabla
u^{k}\right>-\delta_{jk}|^{2}=\tau_{2}\\}$

is a smooth surface and

$\mathcal{H}^{2}_{g}(M_{r_{0}}\cap
E_{1}^{\tau}\cap\\{\sum_{j,k=1}^{3}|\left<\nabla u^{j},\nabla
u^{k}\right>-\delta_{jk}|^{2}=\tau_{2}\\})\leq\frac{C(r_{0})m(g)^{\frac{1}{2}}}{\tau}.$

Define

$E_{2}^{\tau}:=M_{ext}\cap\\{\sum_{j,k=1}^{3}|\left<\nabla u^{j},\nabla
u^{k}\right>-\delta_{jk}|^{2}\leq\tau_{2}\\}.$

Then $E_{2}^{\tau}\subset E_{1}^{\tau}$, and $\partial M_{ext}\cap
E_{2}^{\tau}=\emptyset$. This is because on $\partial M_{ext}$, $\left<\nabla
u^{j},n\right>=0$ for the normal vector $n$ of $\partial M_{ext}$ and all
$j\in\\{1,2,3\\}$, but $\\{\nabla u^{j}\\}_{j=1}^{3}$ is an almost orthonormal
basis at any point in $E^{\tau}_{2}$. So

$\partial E_{2}^{\tau}=\\{\sum_{j,k=1}^{3}|\left<\nabla u^{j},\nabla
u^{k}\right>-\delta_{jk}|^{2}=\tau_{2}\\},$

which is a smooth surface.

######  Lemma 4.1.

For any $0<\varepsilon\ll 1$, there exists $\tau=\Psi(m(g))$ such that

$M\setminus M_{r_{0}}\subset E_{2}^{\tau},$

and

$\mathcal{H}^{2}_{g}(\partial E_{2}^{\tau})\leq
Cm(g)^{\frac{1}{2}-\varepsilon}=\Psi(m(g)).$

###### Proof.

We can take small $\tau=m(g)^{\varepsilon}$ with
$0<\varepsilon<\frac{1}{192}$, which is very big compared with $\Psi(m(g))$ in
Lemma 3.2, satisfying $\tau\to 0$ as $m(g)\to 0$. Then

$M\setminus M_{r_{0}}\subset E_{2}^{\tau},\ \ \partial E_{2}^{\tau}\subset
M_{r_{0}}.$

By above arguments,

$\displaystyle\mathcal{H}^{2}_{g}(\partial
E_{2}^{\tau})=\mathcal{H}^{2}_{g}(M_{r_{0}}\cap\partial E_{2}^{\tau})\leq
C(r_{0})m(g)^{\frac{1}{2}-\varepsilon}=\Psi(m(g)).$

∎

For any fixed small $\varepsilon>0$, we define the $E^{\tau}_{3}$ to be the
connected component of such $E_{2}^{\tau}$ containing $M\setminus M_{r_{0}}$.
So $E^{\tau}_{3}$ also satisfies the properties in above lemma. Consider the
harmonic map $\mathcal{U}$ restricted to $E^{\tau}_{3}$,

$\mathcal{U}=(u^{1},u^{2},u^{3}):E^{\tau}_{3}\to\mathbb{R}^{3}.$

Since $E^{\tau}_{3}\cap B(p,3r_{0})\neq\emptyset$, we can perturb the base
point and without loss of generality we assume that $p\in E^{\tau}_{3}$, and
$\mathcal{U}(p)=p^{*}\in\mathbb{R}^{3}\setminus B(0,L_{0})$.

By the definition, $\mathcal{U}$ is nondegenerate and thus a local
diffeomorphism over $E^{\tau}_{3}$. The regular subregion is defined as the
subregion of $E^{\tau}_{3}$ where $\mathcal{U}$ is a diffeomorphism as
following.

Set $Y^{\tau}_{3}:=\mathcal{U}(E^{\tau}_{3})$, and

$Y^{\tau}_{4}:=\\{y\in
Y^{\tau}_{3}:\mathcal{H}^{0}(\mathcal{U}^{-1}\\{y\\}\cap E^{\tau}_{3})=1\\},\
Y^{\tau}_{s}:=Y^{\tau}_{3}\setminus Y^{\tau}_{4}.$

######  Lemma 4.2.

$Y^{\tau}_{4}$ is open in $Y^{\tau}_{3}$, and $\partial
Y^{\tau}_{4}\subset\mathcal{U}(\partial E^{\tau}_{3})$.

###### Proof.

Otherwise, assume $Y^{\tau}_{4}$ is not open in $Y^{\tau}_{3}$, then for any
$y\in Y^{\tau}_{4}\setminus\partial Y^{\tau}_{3}$, there exists a sequence of
$y_{k}\to y$ and two different sequences of $x_{k}^{1}\neq x_{k}^{2}$ such
that $\mathcal{U}(x_{k}^{1})=\mathcal{U}(x_{k}^{2})=y_{k}$. Assume
$\mathcal{U}(x)=y$, then $x_{k}^{1}\to x,x_{k}^{2}\to x$, which contradicts
with the fact that $\mathcal{U}$ is a local diffeomorphism.

To see $\partial Y^{\tau}_{4}\subset\mathcal{U}(\partial E^{\tau}_{3})$,
assume otherwise there exists $y\in\partial Y^{\tau}_{4}$ but
$y\notin\mathcal{U}(\partial E^{\tau}_{3})$. Notice that $\partial
Y^{\tau}_{3}\subset\mathcal{U}(\partial E^{\tau}_{3})$, so $y\notin\partial
Y^{\tau}_{3}$. If $y\in Y_{4}^{\tau}$, then since $Y_{4}^{\tau}$ is open in
$Y^{\tau}_{3}$ from previous arguments, $y\notin\partial Y_{4}^{\tau}$. So
$y\in Y^{\tau}_{s}$. Assume $x^{1}\neq x^{2}\in\mathcal{U}^{-1}\\{y\\}\cap
E^{\tau}_{3}$ and $x^{1},x^{2}\notin\partial E^{\tau}_{3}$. We can then choose
small neighborhoods $x^{i}\in U_{i}$ and $y\in U$ such that the restrictions
$\mathcal{U}:U_{i}\to U$ are diffeomorphisms for $i=1,2$. This is a
contradiction since $Y_{4}^{\tau}\cap U\neq\emptyset$, but there are at least
two preimages of $Y_{4}^{\tau}\cap U$ in $U_{i}$ for $i=1,2$. ∎

######  Lemma 4.3.

There exists a uniform $L_{0}>0$ such that $\mathbb{R}^{3}\setminus
B(0,L_{0})\subset Y^{\tau}_{4}$.

###### Proof.

From Lemma 3.4, we know $\mathcal{U}$ maps the infinity of $E^{\tau}_{3}$ to
infinity of $\mathbb{R}^{3}$. Also $\mathcal{U}(\partial E^{\tau}_{3})\subset
B(0,L_{0})$. From Lemma 3.3, we know the mod $2$ degree of
$\mathcal{U}:(E^{\tau}_{3},\partial
E^{\tau}_{3})\to(\mathbb{R}^{3},B(0,L_{0}))$ is one. So from the degree
theory, we know $\mathbb{R}^{3}\setminus B(0,L_{0})\subset Y^{\tau}_{3}$. By
Lemma 3.3 again, $Y^{\tau}_{4}\cap\mathbb{R}^{3}\setminus
B(0,L_{0})=Y^{\tau}_{3}\cap\mathbb{R}^{3}\setminus B(0,L_{0})$.

∎

######  Proposition 4.4.

There exists a connected subregion $E^{\tau}\subset E^{\tau}_{3}$ such that
for a uniform $L_{0}>0$,

$M\setminus M_{L_{0}}\subset E^{\tau},\ \mathcal{H}^{2}_{g}(\partial
E^{\tau})\leq 4\mathcal{H}^{2}_{g}(\partial E^{\tau}_{2}),$

and the restriction map $\mathcal{U}:E^{\tau}\to Y^{\tau}$ is a diffeomorphism
onto $Y^{\tau}:=\mathcal{U}(E^{\tau})\subset\mathbb{R}^{3}$. Moreover,
$\mathbb{R}^{3}\setminus B(0,L_{0})\subset Y^{\tau}$. Such $E^{\tau}$ is
called the regular subregion.

###### Proof.

Notice that $\mathcal{U}(\partial E^{\tau}_{3})$ is a smooth surface in
$\mathbb{R}^{3}$, and from above lemma $\partial
Y^{\tau}_{4}\subset\mathcal{U}(\partial E^{\tau}_{3})$, so we can choose a
smaller approximated region $Y^{\tau}_{5}\subset Y^{\tau}_{4}$ such that
$\partial Y^{\tau}_{5}$ is smooth, $Y^{\tau}_{5}\cup\partial
Y^{\tau}_{5}\subset Y^{\tau}_{4}$, $Y^{\tau}_{4}\setminus Y^{\tau}_{5}$ lies
in a small neighborhood of $\mathcal{U}(\partial E^{\tau}_{3})$ and $|\partial
Y^{\tau}_{5}|\leq 2|\mathcal{U}(\partial E^{\tau}_{3})|$. Define $Y^{\tau}$ be
the connected component of $Y^{\tau}_{5}$ such that $\mathbb{R}^{3}\setminus
B(0,L_{0})\subset Y^{\tau}$, and $E^{\tau}:=\mathcal{U}^{-1}(Y^{\tau})$.

By our definition, we know $\mathcal{U}:E^{\tau}\to Y^{\tau}$ is a
diffeomorphism and $M\setminus M_{L_{0}}\subset E^{\tau}$. From the area
formula, we know

$\mathcal{H}^{2}_{g}(\partial E^{\tau})=\int_{\partial
Y^{\tau}}|J\mathcal{U}^{-1}|\leq(1+4\tau)|\partial Y^{\tau}|\leq
3|\mathcal{U}(\partial E^{\tau}_{2})|\leq 3\int_{\partial
E^{\tau}_{2}}|J\mathcal{U}|\leq 4\mathcal{H}^{2}_{g}(\partial E^{\tau}_{2}).$

∎

For computation simplicity, we will usually take integration over coordinate
cylinder. For any fixed unit vector $(a_{1},a_{2},a_{3})\in
S^{2}\subset\mathbb{R}^{3}$, let $u:=\sum_{j=1}^{3}a_{j}u^{j}$. Define the
coordinate cylinder $C_{L}:=D_{L}^{\pm}\cup T_{L}$, where

$D_{L}^{\pm}:=\\{x\in M_{ext}:u(x)=\pm L,|\mathcal{U}(x)|^{2}-u(x)^{2}\leq
L^{2}\\},$ $T_{L}:=\\{x\in M_{ext}:|u(x)|\leq
L,|\mathcal{U}(x)|^{2}-u(x)^{2}=L^{2}\\}.$

Set $\Omega_{L}\subset M_{ext}$ be the closure of the bounded component of
$M_{ext}\setminus C_{L}$. In the following, we will also use $\Omega_{L}^{a}$
to emphasize that $\Omega_{L}$ is associated to the harmonic function along
$a=(a_{1},a_{2},a_{3})$ direction.

######  Lemma 4.5.

For a fixed $j\in\\{1,2,3\\}$ and $u=u^{j}$,

$\text{Vol}_{g}(\Omega_{L}\cap E^{\tau})\geq(1-4\tau)(1-\Psi(m(g)))2\pi
L^{3}-\Psi(m(g)).$

###### Proof.

Since $\Delta u=0$, integration by parts gives

$\displaystyle\int_{\Omega_{L}}|\nabla u|^{2}$ $\displaystyle=\int_{T_{L}\cup
D_{L}^{\pm}}u\left<\nabla u,n\right>+\int_{\partial M_{ext}}u\left<\nabla
u,n\right>$ $\displaystyle=(1\pm\Psi(m(g)))2\pi L^{3},$

where we used Lemma 3.2 in the last equality.

Note that $M_{ext}\setminus E^{\tau}$ is a bounded domain inside $M_{r_{0}}$
with boundary $\partial M_{ext}\cup\partial E^{\tau}$, where $u$ is uniformly
bounded. Integration by parts gives

$\displaystyle\int_{M_{ext}\setminus E^{\tau}}|\nabla u|^{2}$
$\displaystyle=\int_{\partial E^{\tau}}u\left<\nabla u,n\right>+\int_{\partial
M_{ext}}u\left<\nabla u,n\right>$ $\displaystyle=\int_{\partial
E^{\tau}}u\left<\nabla u,n\right>$ $\displaystyle\leq
C(r_{0})(1+\tau)\mathcal{H}_{g}^{2}(\partial E^{\tau})$
$\displaystyle\leq\Psi(m(g)).$

So

$\displaystyle(1+\Psi(m(g)))2\pi L^{3}$ $\displaystyle\geq\int_{\Omega_{L}\cap
E^{\tau}}|\nabla u|^{2}$ $\displaystyle\geq(1-\Psi(m(g)))2\pi
L^{3}-\Psi(m(g)).$

The conclusion follows from the fact that inside $\Omega_{L}\cap E^{\tau}$,
$||\nabla u|^{2}-1|\leq 2\tau.$ ∎

######  Lemma 4.6.

For any $D>L_{0}$,

$|B(p^{*},D)\setminus Y^{\tau}|\leq\Psi(m(g)|D).$

###### Proof.

Otherwise, assume $|B(p^{*},D)\setminus Y^{\tau}|>c>0$. Choose $L=2D$, then we
have

$\displaystyle|\Omega_{L}\cap Y^{\tau}|$
$\displaystyle\leq|\Omega_{L}\setminus B(p^{*},D)|+|B(p^{*},D)\cap Y^{\tau}|$
$\displaystyle\leq|\Omega_{L}\setminus B(p^{*},D)|+|B(p^{*},D)|-c$
$\displaystyle\leq|\Omega_{L}|-c$ $\displaystyle=2\pi L^{3}-c.$

On the other hand, from above lemma, we know

$|\Omega_{L}\cap Y^{\tau}|\geq(1-4\tau)\text{Vol}_{g}(\Omega_{L}\cap
Y^{\tau})\geq(1-4\tau)^{2}(1-\Psi(m(g)))2\pi L^{3}-\Psi(m(g)).$

Letting $m(g)\to 0$ gives a contradiction. ∎

######  Remark 4.1.

Alternatively, one can prove this lemma by using the remark below Lemma 2.8
and the fact that $|\partial Y^{\tau}|\leq\Psi(m(g))$. Such argument will also
be used later.

Now we prove the convergence theorem in the pointed flat metric topology.

###### Proof of Theorem 1.5.

Assume that $(M_{i},g_{i},p_{i})$ is a sequence of orientable complete
$(A,B,\sigma)$-AF $3$-manifolds with nonnegative integrable scalar curvature
$R_{g_{i}}\geq 0$, and mass $m(g_{i})\to 0$.

By Proposition 4.4, there exists $\tau_{i}\to 0$ such that the regular
subregion $E^{\tau_{i}}$ is well defined. And we have diffeomorphisms
$\mathcal{U}_{i}:E^{\tau_{i}}\to Y^{\tau_{i}}\subset\mathbb{R}^{3}$, under
which $g_{i}^{jk}(y)=\left<\nabla u_{i}^{j},\nabla u_{i}^{k}\right>$ and for
any $y\in Y^{\tau_{i}}$,

$|g_{i}^{jk}-\delta_{jk}|(y)\leq\tau_{i}.$

So for any $\delta>0$, we know $(E^{\tau_{i}}\setminus B_{\delta}(\partial
E^{\tau_{i}}),g_{i},p_{i})$ converges to the Euclidean metric locally in
$C^{0}$-topology. Since $\mathbb{R}^{3}\setminus
B(0,L_{0})\subset\mathcal{U}_{i}(E^{\tau_{i}}\setminus B_{\delta}(\partial
E^{\tau_{i}}))$, for any Euclidean ball $B(q,1)\subset\mathbb{R}^{3}\setminus
B(0,2L_{0})$, $\forall q^{1},q^{2}\in B(q,1)$, if we set
$q^{1}_{i}=\mathcal{U}_{i}^{-1}(q^{1}),q^{2}_{i}=\mathcal{U}^{-1}_{i}(q^{2})\in
E^{\tau_{i}}$, then geodesics between $q^{1}_{i},q^{2}_{i}$ are inside
$E^{\tau_{i}}$. So $d_{g_{i}}(q_{i}^{1},q_{i}^{2})\to d_{E}(q^{1},q^{2})$, in
other words,

$\mathcal{U}_{i,\sharp}d_{g_{i}}\to d_{E}\text{ over }B(q,1).$

The associated current of $Y^{\tau_{i}}$ is

$T_{i}(f):=\int_{Y^{\tau_{i}}}f(y)d\text{Vol}_{g_{i}}(y)=\int_{\mathbb{R}^{3}}f(y)\chi_{Y^{\tau_{i}}}(y)\sqrt{\det
g_{i}(y)}dy,\ \forall f\in C_{c}^{\infty}(\mathbb{R}^{3}).$

We claim that $T_{i}\to\mathbb{R}^{3}$ in the pointed flat metric topology. By
definition, we need to show that for any $D\gg 1$,

$\int_{B(p^{*},D)}|\chi_{Y_{i}}(y)\sqrt{\det g_{i}(y)}-1|dy\to 0.$

By Lemma 4.6, $|B(p^{*},D)\setminus Y^{\tau_{i}}|\leq\Psi(m(g_{i})|D)\to 0$.
So

$\displaystyle\int_{B(p^{*},D)}|\chi_{Y_{i}}(y)\sqrt{\det g_{i}(y)}-1|dy$
$\displaystyle=\int_{B(p^{*},D)\cap Y^{\tau_{i}}}|\sqrt{\det
g_{i}(y)}-1|+|B(p^{*},D)\setminus Y^{\tau_{i}}|$
$\displaystyle\leq\Psi(m(g_{i})|D)\to 0,$

as $i\to\infty$.

So we proved that for $\tilde{Z}_{i}:=M_{i}\setminus E^{\tau_{i}}$,
$(M_{i}\setminus\tilde{Z}_{i},g_{i},p_{i})\to\mathbb{R}^{3}$ in the sense of
pointed weak volume convergence, $|\partial\tilde{Z}_{i}|\leq C\cdot
m(g_{i})^{\frac{1}{2}-\varepsilon}$, and $(M_{i}\setminus
B_{\delta}(\tilde{Z}_{i}),g_{i},p_{i})$ converge to a subset in
$\mathbb{R}^{3}$ locally in $C^{0}$-topology.

To get better control on the area of boundary, we can choose a fixed small
$\tau_{0}\ll 1$ and define $E_{i}^{\tau_{0}}$ to be the connected regular
subregion inside

$\\{x\in M_{i,ext}:\sum_{j,k=1}^{3}|\left<\nabla
u^{j}_{i},u^{k}_{i}\right>-\delta_{jk}|^{2}\leq\tau_{0}\\}$

containing $E^{\tau_{i}}$. The only difference from $E^{\tau_{i}}$ is that in
the definition of $E^{\tau_{0}}_{i}$, we take $\tau_{0}$ to be a fixed small
number such that over $E_{i}^{\tau_{0}}$, $\mathcal{U}_{i}$ is also
nondegenerate. So previous arguments also work for $E_{i}^{\tau_{0}}$ and in
particular, $|\partial E_{i}^{\tau_{0}}|\leq
C(r_{0},\tau_{0})m(g_{i})^{\frac{1}{2}}$, and $\mathcal{U}_{i}$ is a
diffeomorphism onto its image when restricted on $E_{i}^{\tau_{0}}$. The
missing property is that we may not have locally $C^{0}$-convergence of
$g_{i}$ away from the boundary $\partial E_{i}^{\tau_{0}}$.

If we take $Y^{\tau_{0}}_{i}=\mathcal{U}_{i}(E_{i}^{\tau_{0}})$, then since
inside $Y^{\tau_{0}}_{i}$, $|\sqrt{\det g_{i}}-1|\leq 4\tau_{0}$, we know that
for any $D\gg 1$,

$\displaystyle\int_{B(p^{*},D)}$
$\displaystyle|\chi_{Y^{\tau_{0}}_{i}}(y)\sqrt{\det
g_{i}(y)}-1|dy=\int_{B(p^{*},D)\cap Y^{\tau_{i}}}|\sqrt{\det g_{i}(y)}-1|$
$\displaystyle\ \ \ +\int_{B(p^{*},D)\cap(Y^{\tau_{0}}_{i}\setminus
Y^{\tau_{i}})}|\sqrt{\det g_{i}(y)}-1|+|B(p^{*},D)\setminus Y^{\tau_{0}}_{i}|$
$\displaystyle\leq\Psi(m(g_{i})|D)+(4\tau_{0}+1)|B(p^{*},D)\setminus
Y^{\tau_{i}}|$ $\displaystyle\leq\Psi(m(g_{i})|D)\to 0$

as $i\to\infty$.

Define $Z_{i}:=M_{i}\setminus E_{i}^{\tau_{0}}$. So $(M_{i}\setminus
Z_{i},g_{i},p_{i})\to\mathbb{R}^{3}$ in the sense of pointed weak volume
convergence, and $|\partial Z_{i}|\leq C\cdot m(g_{i})^{\frac{1}{2}}$. ∎

##  5\. Removable singularity under Ricci lower bound

In this section, we prove the Gromov-Hausdorff convergence when Ricci
curvature is uniformly bounded below. Let’s assume $(M_{i},g_{i},p_{i})$ is a
sequence of orientable complete $(A,B,\sigma)$-AF $3$-manifolds with
nonnegative integrable scalar curvature $R_{g_{i}}\geq 0$, Ricci curvature
bounded from below by $\text{Ric}_{g_{i}}\geq-2\Lambda$ and the mass
$m(g_{i})\to 0$.

By Gromov’s compactness theorem, up to a subsequence,

$(M_{i},g_{i},p_{i})\to(X,d_{X},p_{X})$

in the pointed Gromov-Hausdorff topology, where $(X,d_{X})$ is a complete
length space.

From Proposition 4.4, there exist regular subregions $E^{\tau_{i}}\subset
M_{i,ext}$, and without loss of generality, assume $p_{i}\in E^{\tau_{i}}$.
Consider the restricted metric space $(E^{\tau_{i}},d_{i},p_{i})$, where
$d_{i}$ is the restriction of the distance induced by $g_{i}$. Up to a
subsequence, assume

$(E^{\tau_{i}},d_{i},p_{i})\to(X^{\prime},d_{X},p_{X})\subset(X,d_{X}).$

Similarly, consider the restricted metric and assume $(\partial
E^{\tau_{i}},d_{i})\to(\Sigma,d_{X})\subset(X^{\prime},d_{X})$. By Proposition
4.4, $X^{\prime}\setminus\Sigma\neq\emptyset$.

In general, $(X^{\prime},d_{X})$ is not a length space and is different from
$(X,d_{X})$. In our case, by using the fact that $|\partial E^{\tau_{i}}|\to
0$ and the conditon on lower bound of Ricci curvature, we can prove the
following.

######  Proposition 5.1.

$X^{\prime}=X$.

###### Proof.

We argue by contradiction. Otherwise, take $y_{0}\in X\setminus X^{\prime}$.
Choose $\delta_{0}>0$ small enough such that
$d(y_{0},X^{\prime})>2\delta_{0}>0$. Choose $x_{0}\in
X^{\prime}\setminus\Sigma$ and by choosing $\delta_{0}$ smaller, assume
$d(x_{0},\Sigma)>2\delta_{0}$. Choose $D>0$ such that $x_{0},y_{0}\in
B(p_{X},\frac{D}{4})$. Assume $d(x_{0},y_{0})=L>2\delta_{0}$ by choosing
$\delta_{0}$ smaller. Take $x_{i},y_{i}\in B(p_{i},D)$ such that $x_{i}\to
x_{0},y_{i}\to y_{0}$, and $x_{i}\in E^{\tau_{i}}$, $d_{i}(x_{i},\partial
E^{\tau})>2\delta_{0},d_{i}(y_{i},\partial E^{\tau_{i}})>2\delta_{0}$.

By Proposition 4.4 and Lemma 2.8, there is a $z_{i}\in B(y_{i},\delta_{0})$
and a segment $\gamma_{x_{i}z_{i}}\subset(M_{i}\setminus\partial
E^{\tau_{i}},g_{i,\varepsilon})$. Note that $M_{i,ext}$ is a connected
component of $M_{i}\setminus\partial M_{i,ext}$ and
$\gamma_{x_{i}z_{i}}\subset M_{i}\setminus\partial M_{i,ext}$, so
$\gamma_{x_{i}z_{i}}\subset M_{i,ext}$ and $z_{i}\in M_{i,ext}$.

Also $\sum_{j,k=1}^{3}|\left<\nabla u_{i}^{j},\nabla
u_{i}^{k}\right>-\delta_{jk}|^{2}(z_{i})<\tau_{i}$, otherwise since
$\sum_{j,k=1}^{3}|\left<\nabla u_{i}^{j},\nabla
u_{i}^{k}\right>-\delta_{jk}|^{2}(x_{i})<\tau_{i}$, there will be an
intersection point of $\gamma_{x_{i}z_{i}}$ and $\partial E^{\tau_{i}}$. So
$z_{i}\in E^{\tau_{i}}$.

Taking $i\to\infty$, $z_{i}\to z_{0}\in B(y_{0},\delta_{0})\cap X^{\prime}$, a
contradiction. ∎

Notice that under diffeomorphisms $\mathcal{U}_{i}$, for metrics
$(\mathcal{U}_{i}^{-1})^{*}g_{i}$ over $\mathbb{R}^{3}\setminus
B(0,L_{0})\subset Y^{\tau_{i}},$ which we still denote by $g_{i}$,

$(\mathbb{R}^{3}\setminus B(0,L_{0}),g_{i})\to(\mathbb{R}^{3}\setminus
B(0,L_{0}),g_{E})$

locally in $C^{0}$-topology. We can assume
$\mathcal{U}_{i}(p_{i})=p^{*}\in\mathbb{R}^{3}\setminus B(0,L_{0})$ as before.
Moreover, we claim that this convergence also holds in the pointed Gromov-
Hausdorff topology, i.e.

$(\mathbb{R}^{3}\setminus B(0,L_{0}),d_{i})\to(\mathbb{R}^{3}\setminus
B(0,L_{0}),d_{E})\subset(X^{\prime},d_{X}).$

This claim follows from the following lemma.

######  Lemma 5.2.

Fix $D>0,\delta>0$. For any $y,z\in E^{\tau_{i}}\cap B(p_{i},D)$ with
$d_{i}(y,\partial E^{\tau_{i}})\geq\delta$, $d_{i}(z,\partial
E^{\tau_{i}})\geq\delta$,

$||\mathcal{U}_{i}(y)-\mathcal{U}_{i}(z)|-d_{i}(y,z)|\leq\Psi(m(g_{i})|D,\delta).$

###### Proof.

Take $\delta_{0}=\Psi(m(g_{i})|D,\delta)$ as in Lemma 2.8. For all $i$ large
enough, $\delta_{0}<\frac{1}{2}\delta$, and by Proposition 4.4 and Lemma 2.8,
there exist $z^{\prime}$ and a geodesic segment $\gamma_{yz^{\prime}}$
connecting $y,z^{\prime}$ such that $z^{\prime}\in B(z,\delta_{0})$ and
$\gamma_{yz^{\prime}}\cap\partial E^{\tau_{i}}=\emptyset$. So
$\gamma_{yz^{\prime}}\subset E^{\tau_{i}}$.

Set $l=d_{i}(y,z^{\prime})$. Note that

$|\mathcal{U}_{i}(y)-\mathcal{U}_{i}(z^{\prime})|^{2}=\sum_{j=1}^{3}(u_{i}^{j}(\gamma_{yz^{\prime}}(l))-u_{i}^{j}(\gamma_{yz^{\prime}}(0)))^{2}=\sum_{j=1}^{3}\left(\int_{0}^{l}\left<\nabla
u_{i}^{j},\gamma_{yz^{\prime}}^{\prime}\right>\right)^{2},$

and

$\left|1-\sum_{j=1}^{3}\left<\nabla
u_{i}^{j},\gamma_{yz^{\prime}}^{\prime}\right>^{2}\right|<4\tau_{i}.$

So

$\displaystyle|\mathcal{U}_{i}(y)-\mathcal{U}_{i}(z^{\prime})|$
$\displaystyle\leq\left(\sum_{j=1}^{3}l\int_{0}^{l}\left<\nabla
u_{i}^{j},\gamma_{yz^{\prime}}^{\prime}\right>^{2}\right)^{\frac{1}{2}}$
$\displaystyle\leq(1+4\tau_{i})l.$

Similarly, since $d_{i}(z,z^{\prime})<\frac{1}{2}\delta$ and $d_{i}(z,\partial
E^{\tau_{i}})>\delta$, there exists a geodesic segment connecting
$z,z^{\prime}$ inside $E^{\tau_{i}}$, so we have

$|\mathcal{U}_{i}(z)-\mathcal{U}_{i}(z^{\prime})|\leq(1+4\tau_{i})d_{i}(z,z^{\prime}).$

Thus

$|\mathcal{U}_{i}(y)-\mathcal{U}_{i}(z)|\leq(1+4\tau_{i})(d_{i}(y,z^{\prime})+d_{i}(z,z^{\prime}))\leq(1+4\tau_{i})d_{i}(y,z)+4\delta_{0},$

i.e.

$|\mathcal{U}_{i}(y)-\mathcal{U}_{i}(z)|-d_{i}(y,z)\leq\Psi(m(g_{i})|D,\delta).$

On the other hand, denote $y_{i}=\mathcal{U}_{i}(y),z_{i}=\mathcal{U}_{i}(z)$.
Then choose a line segment $[y_{i}w]$ with $w\in\partial Y^{\tau_{i}}$ such
that $|y_{i}w|=d_{E}(y_{i},\partial Y^{\tau_{i}})$. Integration along
$[y_{i}w]$ as above implies that

$d_{i}(y,\mathcal{U}^{-1}_{i}(w))\leq(1+4\tau_{i})|y_{i}w|=(1+4\tau_{i})d_{E}(y_{i},\partial
Y^{\tau_{i}}).$

So

$d_{i}(y,\partial E^{\tau_{i}})\leq(1+4\tau_{i})d_{E}(y_{i},\partial
Y^{\tau_{i}}),$

and particularly,

$d_{E}(y_{i},\partial Y^{\tau_{i}}),d_{E}(z_{i},\partial
Y^{\tau_{i}})>\frac{1}{2}\delta.$

Also, from above arguments,

$|y_{i}z_{i}|\leq d_{i}(y,z)+\Psi(m(g_{i})|D,\delta)\leq
2D+\Psi(m(g_{i})|D,\delta).$

Similarly, there exist $z_{i}^{\prime}\in B(z_{i},\delta_{0})$ and a straight
line segment $[y_{i}z_{i}^{\prime}]\subset Y^{\tau_{i}}$. Denote
$z^{\prime}=\mathcal{U}_{i}^{-1}(z_{i}^{\prime})\in E^{\tau_{i}}$. Then
$\gamma_{1}:=\mathcal{U}_{i}^{-1}([y_{i}z_{i}^{\prime}])\subset E^{\tau_{i}}$
is a smooth curve between $y,z^{\prime}$,
$\gamma_{2}:=\mathcal{U}_{i}^{-1}([z_{i}z_{i}^{\prime}])\subset E^{\tau_{i}}$
is a smooth curve between $z,z^{\prime}$. So

$d_{i}(y,z)\leq d_{i}(y,z^{\prime})+d_{i}(z,z^{\prime})\leq
L_{g_{i}}(\gamma_{1})+L_{g_{i}}(\gamma_{2})\leq(1+\Psi(m(g_{i})))(|y_{i}z_{i}|+\delta_{0}),$

i.e.

$d_{i}(y,z)-|\mathcal{U}_{i}(y)-\mathcal{U}_{i}(z)|\leq\Psi(m(g_{i})|D,\delta).$

∎

From this lemma, we know

$(\mathbb{R}^{3}\setminus B(0,L_{0}),d_{X},p_{X})=(\mathbb{R}^{3}\setminus
B(0,L_{0}),d_{E},p^{*})\subset(X,d_{X}).$

Let $\\{y^{1},y^{2},y^{3}\\}$ be the coordinate function of $\mathbb{R}^{3}$
and $e_{j}$ the unit vector along $y^{j}$-axis. Let $q_{\pm}^{j}=\pm Le_{j}$
for $L>L_{0}$. By elementary geometry, we know that any $y\in
B(0,L)\subset\mathbb{R}^{3}$ is uniquely determined by
$\\{|yq_{\pm}^{j}|\\}_{j=1}^{3}$, and particularly there exists a continuous
function $\xi:\mathbb{R}^{6}\to\mathbb{R}$ with $\xi(0)=0$ such that for any
$y,z\in B(0,L)$,

$|yz|=\xi(|yq_{\pm}^{j}|-|zq_{\pm}^{j}|).$

We also choose an $\frac{1}{N}$-net $\\{q^{a_{k}}\\}_{k=1}^{N}$ of
$A(0,L_{0},L)$, where $A(0,L_{0},L)=B(0,L)\setminus
B(0,L_{0})\subset\mathbb{R}^{3}$ is the standard annulus. We always assume
$\\{q_{\pm}^{j}\\}_{j=1}^{3}\subset\\{q^{a_{k}}\\}_{k=1}^{N}$.

######  Lemma 5.3.

For any fixed $N>10$, there exists $\delta_{0}=\Psi(m(g_{i})|N,L)$ such that
for any $y\in Y^{\tau_{i}}\cap B(p^{*},L)$, there exists $y^{\prime}\in
Y^{\tau_{i}}$ with $d_{i}(y,y^{\prime})<\delta_{0}$ and

$|d_{i}(y^{\prime},q^{a_{k}})-|y^{\prime}q^{a_{k}}||\leq\Psi(m(g_{i})),\forall
1\leq k\leq N.$

###### Proof.

By the remark below Lemma 2.8, there exists $\hat{B}\subset
B_{i}(y,\delta_{0})\subset M_{i}$ with
$\text{Vol}_{i}(\hat{B})\geq(1-\Psi(m(g_{i})|N,L))\text{Vol}_{i}(B(y,\delta_{0}))$
such that for any $y^{\prime}\in\hat{B}$, the geodesic segment
$\gamma_{q^{a_{k}}y^{\prime}}$ lies in $Y^{\tau_{i}}$ for all $1\leq k\leq N$.
By volume comparison theorem, we know

$\frac{\text{Vol}(B_{i}(y,\delta_{0}))}{|B_{-\Lambda}(\delta_{0})|}\geq\frac{\text{Vol}(B_{i}(y,2L))}{|B_{-\Lambda}(2L)|}\geq\frac{\text{Vol}(B_{i}(p_{i},L))}{|B_{-\Lambda}(2L)|}\geq\frac{|B(1)|}{|B_{-\Lambda}(2L)|}.$

So

$\frac{\text{Vol}(B_{i}(y,\delta_{0}))}{|B(\delta_{0})|}\geq\frac{|B(1)|}{|B_{-\Lambda}(2L)|}\cdot\frac{|B_{-\Lambda}(\delta_{0})|}{|B(\delta_{0})|}\geq
c(L)>0.$

Note that $\hat{B}\subset Y^{\tau_{i}}$ and
$|\hat{B}|\geq(1-\Psi)\text{Vol}_{i}(\hat{B})$. So

$|\hat{B}|\geq(1-\Psi)c(L)|B(\delta_{0})|.$

We claim that there exists $y^{\prime}\in\hat{B}$ such that each straight line
segment $[q^{a_{k}}y^{\prime}]$ lies in $Y^{\tau_{i}}$. Otherwise, by the same
argument as the proof of Lemma 2.8, we have

$|\hat{B}|\leq C(N,L)\cdot|\partial Y^{\tau_{i}}|,$

which is a contradiction if we choose $\delta_{0}=\Psi(m(g_{i})|N,L)$ bigger.

So for such $y^{\prime}\in\hat{B}$, these geodesic segments
$\gamma_{q^{a_{k}}y^{\prime}}$ and $[q^{a_{k}}y^{\prime}]$ all lie in
$Y^{\tau_{i}}$. Similar to the proof of Lemma 5.2, by taking integral along
these segments, we can get the conclusion. ∎

Now we can define a map from $X^{\prime}$ to $\mathbb{R}^{3}$. For any
$L>2L_{0}$ and $y\in X^{\prime}\cap B(p_{X},\frac{L}{2})$, assume $x_{i}\in
E^{\tau_{i}}\cap B(p_{i},\frac{L}{2})$ and $x_{i}\to y$ in the GH-topology.
Denote $y_{i}=\mathcal{U}(x_{i})$. Then by Lemma 5.2, $y_{i}\in
Y^{\tau_{i}}\cap B(p^{*},L)$. For any fixed $N>10$, by Lemma 5.3, there exists
$y_{N,i}^{\prime}\in Y^{\tau_{i}}$ with
$d_{i}(y_{i},y_{N,i}^{\prime})<\Psi(m(g_{i}))$ and

$|d_{i}(y_{N,i}^{\prime},q^{a_{k}})-|y_{N,i}^{\prime}q^{a_{k}}||\leq\Psi(m(g_{i})),\forall
1\leq k\leq N.$

Up to a subsequence, we can assume $y_{N,i}^{\prime}\to y_{N}^{\prime}\in
B(p^{*},L)\subset\mathbb{R}^{3}$ in the $d_{E}$-topology. By taking a further
subsequence, assume $y_{N_{i},i}^{\prime}\to y^{\prime}$ in the
$d_{E}$-topology. Define

$\Xi_{L}:X^{\prime}\cap B(p_{X},\frac{L}{2})\to
B(p^{*},L)\subset\mathbb{R}^{3},$

by $\Xi_{L}(y)=y^{\prime}.$

######  Lemma 5.4.

$\Xi_{L}$ is well defined.

###### Proof.

It’s enough to show that for $y\in X^{\prime}$ with
$y_{i}^{\prime},y_{i}^{\prime\prime}\in Y^{\tau_{i}}\cap B(p^{*},L)$
satisfying $y_{i}^{\prime},y_{i}^{\prime\prime}\to y$ in the GH-topology and

$|d_{i}(y_{i}^{\prime},q_{\pm}^{j})-|y_{i}^{\prime}q_{\pm}^{j}||\leq\Psi(m(g_{i})),\
|d_{i}(y_{i}^{\prime\prime},q_{\pm}^{j})-|y_{i}^{\prime\prime}q_{\pm}^{j}||\leq\Psi(m(g_{i})),\forall
j\in\\{1,2,3\\},$

if $y_{i}^{\prime}\to y^{\prime}$ and $y_{i}^{\prime\prime}\to
y^{\prime\prime}$ in the $d_{E}$-topology, then $y^{\prime}=y^{\prime\prime}$.
To see this, by the assumptions, we know

$d_{X}(y,q_{\pm}^{j})=|y^{\prime}q_{\pm}^{j}|=|y^{\prime\prime}q_{\pm}^{j}|,\forall
j\in\\{1,2,3\\}.$

So
$|y^{\prime}y^{\prime\prime}|=\xi(|y^{\prime}q_{\pm}^{j}|-|y^{\prime\prime}q_{\pm}^{j}|)=\xi(0)=0$.

∎

######  Proposition 5.5.

$\Xi_{L}:(B(p_{X},\frac{L}{2}),d_{X})\to(\mathbb{R}^{3},d_{E})$ is an isometry
onto $B(p^{*},\frac{L}{2})$.

###### Proof.

For any $y,z\in X^{\prime}\cap B(p_{X},\frac{L}{2})$ and assume $y_{i}\to
y,z_{i}\to z$ in the GH-topology with $y_{i},z_{i}\in Y^{\tau_{i}}\cap
B(p^{*},L)$. Take $y_{N_{i},i}^{\prime},z_{N_{i},i}^{\prime}$ as above and
assume $y_{N_{i},i}^{\prime}\to y^{\prime},z_{N_{i},i}^{\prime}\to z^{\prime}$
in the $d_{E}$-topology. So $\Xi(y)=y^{\prime}$ and $\Xi(z)=z^{\prime}$.

Note that $d_{X}(y,z)=\lim_{i\to\infty}d_{i}(y_{i},z_{i})$ and

$\displaystyle|d_{i}(y_{N_{i},i}^{\prime},z_{N_{i},i}^{\prime})-d_{i}(y_{i},z_{i})|\leq
d_{i}(y_{i},y_{N_{i},i}^{\prime})+d_{i}(z_{i},z_{N_{i},i}^{\prime})\leq\Psi(m(g_{i})),$

so
$d_{X}(y,z)=\lim_{i\to\infty}d_{i}(y_{N_{i},i}^{\prime},z_{N_{i},i}^{\prime})$.

If we take $z=q^{a_{k}}$, $\forall 1\leq k<\infty$, then

$d_{X}(y,q^{a_{k}})=|y^{\prime}q^{a_{k}}|.$

Since $\\{q^{a_{k}}\\}_{k=1}^{\infty}$ is dense in $A(0,L_{0},L)$, for any
$q^{a}\in A(0,L_{0},L)$, we know

$d_{X}(y,q^{a})=|y^{\prime}q^{a}|.$

In particular, $d_{X}(p_{X},y)=|p^{*}y^{\prime}|$. Also $\Xi_{L}$ is the
identity map when restricted on $A(0,L_{0},L)$, and
$\Xi_{L}^{-1}(A(0,L_{0},L))=A(0,L_{0},L)$.

For any $y,z\in X^{\prime}\cap B(p_{X},L)$, since $y^{\prime},z^{\prime}\in
B(p^{*},L)$, we can extend $[y^{\prime}z^{\prime}]$ to a straight line segment
in $B(p^{*},L)\subset\mathbb{R}^{3}$ such that $q^{a}$ is one end point for
some $q^{a}\in A(0,L_{0},L)$, say $z^{\prime}\in[q^{a}y^{\prime}]$. Then

$|y^{\prime}z^{\prime}|=|q^{a}y^{\prime}|-|q^{a}z^{\prime}|=d_{X}(q^{a},y)-d_{X}(q^{a},z)\leq
d_{X}(y,z).$

So $\Xi_{L}$ is an $1$-Lipschitz map, and particularly continuous.

We claim that $\Xi_{L}$ is injective on $B(p_{X},\frac{L}{2})$. Otherwise,
assume $\Xi_{L}(y_{1})=\Xi_{L}(y_{2})=y^{\prime}$ with $y_{1},y_{2}\in
B(p_{X},\frac{L}{2})$. Choose geodesic segments $\gamma_{1},\gamma_{2}$
between $y_{1},y_{2}$ and $q^{a}$ respectively for some $q^{a}\in
A(0,L_{0},2L_{0})$. Then $\gamma_{1},\gamma_{2}\subset B(p_{X},L)$. Let
$\sigma_{j}=\Xi_{L}(\gamma_{j})$ for $j=1,2$. Since $\Xi_{L}$ is
$1$-Lipschitz, the length of $\sigma_{j}$ in $\mathbb{R}^{3}$ is smaller than
the length of $\gamma_{j}$ in $X$, i.e.

$|y^{\prime}q^{a}|\leq|\sigma_{j}|\leq d_{X}(y_{j},q^{a})=|y^{\prime}q^{a}|.$

So $\sigma_{1}=\sigma_{2}=[y^{\prime}q^{a}]$ is the unique line segment
between $y^{\prime},q^{a}$. Since $\Xi_{L}$ is one to one around $q^{a}$, we
know $\gamma_{1}$ coincides with $\gamma_{2}$ around $q^{a}$. If $y_{1}\neq
y_{2}$, then we have branching geodesic segments in $X$. This is a
contradiction with Proposition 2.9.

By Proposition 2.10 and Lemma 5.1, $X^{\prime}\cap
B(p_{X},\frac{L}{2})=B(p_{X},\frac{L}{2})$ is a topological manifold with
boundary $\\{x\in X:d(p_{X},x)=\frac{L}{2}\\}$. Since
$d_{X}(p_{X},x)=|p^{*}\Xi_{L}(x)|$, we know $\Xi_{L}$ maps the boundary of
$B(p_{X},\frac{L}{2})$ to boundary of
$B(p^{*},\frac{L}{2})\subset\mathbb{R}^{3}$. We have proved that $\Xi_{L}$ is
one to one on $A(0,L_{0},L)$, so for the induced map
$\Xi_{L}:(B(p_{X},\frac{L}{2}),\partial
B(p_{X},\frac{L}{2}))\to(B(p^{*},\frac{L}{2}),\partial B(p^{*},\frac{L}{2}))$,
the mod $2$ degree is $1$. From the degree theory,

$\Xi_{L}(B(p_{X},\frac{L}{2}))=B(p^{*},\frac{L}{2}).$

So $\Xi_{L}:B(p_{X},\frac{L}{2})\to B(p^{*},\frac{L}{2})$ is a homeomorphism.
To show it’s also isometric, for any $y,z\in B(p_{X},\frac{L}{2})$, assume
$y^{\prime}=\Xi_{L}(y),z^{\prime}=\Xi_{L}(z)$. Choose a line segment
$[q^{a_{-}}q^{a_{+}}]$ with $q^{a_{-}},q^{a_{+}}\in A(0,L_{0},\frac{L}{2})$,
length $l$ and $[y^{\prime}z^{\prime}]\subset[q^{a_{-}}q^{a_{+}}]\subset
B(p^{*},\frac{L}{2})$. Let $\gamma:=\Xi_{L}^{-1}([q^{a_{-}}q^{a_{+}}])$. Then
$d_{X}(\gamma(t),\gamma(s))\geq|t-s|$ since $\Xi_{L}$ is $1$-Lipschitz, and
$d_{X}(\gamma(0),\gamma(l))=d_{X}(q^{a_{-}},q^{a_{+}})=|q^{a_{-}}q^{a_{+}}|=l$.
So it must hold that

$d_{X}(\gamma(t),\gamma(s))=|t-s|.$

In particular, $d_{X}(y,z)=|y^{\prime}z^{\prime}|$, i.e. $\Xi_{L}$ is an
isometry.

∎

Since $L$ could be any large positive number, $(X,d_{X})=\mathbb{R}^{3}$. This
finishes the proof of Theorem 1.6.

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