are independent conditionally on $\mathscr{C}$, and

  2. ii)

for each $b$, $(\beta_{b}^{\prime},\beta_{b}^{\prime\prime})$ are independent
random variables with distribution
$\nu(\,\cdot\,\cap\mathscr{C}^{\prime}_{b})/\nu(\mathscr{C}_{b}^{\prime})\otimes\nu(\,\cdot\,\cap\mathscr{C}_{b}^{\prime\prime})/\nu(\mathscr{C}^{\prime\prime}_{b})$.

The points $(\beta^{\prime}_{b},\beta^{\prime\prime}_{b})$, $b\in\mathscr{C}$,
now being defined, we are ready to verify that
$\Phi(\mathfrak{C},\boldsymbol{\beta})$ and
$\operatorname{CMT}({\textup{e}},{\mathbf{U}})$ are almost surely isometric.
The arguments above should already make this pretty clear: indeed, for each
$b\in\operatorname{Br}(\mathscr{C})$, the set of marks
$\\{\beta_{b}^{\prime},\beta_{b}^{\prime\prime}\\}$ is precisely the image in
$\mathscr{C}$ the set of points $\\{{{\text{\bsifamily j}}}(a_{ui}),a_{ui}\\}$
which are identified at time $\tau_{ui}$. To make this formal, fix any
$i,j\in\mathbb{N}$, and consider $Y_{n}(i,j)$ and
$\hat{d}_{n}(\zeta_{i},\zeta_{j})$. The choice of the marks
$(\beta_{b}^{\prime},\beta^{\prime\prime}_{b})$,
$b\in\operatorname{Br}(\mathscr{C})$, is precisely made so that, for every
$n\geq 1$, sorting the sets $\\{\nu(C_{u}),|u|=n\\}$ and
$\\{|\Pi_{u}|,|u|=n\\}$ in decreasing order yields the same sequence, and
therefore

$Y_{n}(i,j)=\sqrt{\frac{\pi}{2}}\sum_{|u|=n}\nu(C_{u})^{1/2}=\sqrt{\frac{\pi}{2}}\sum_{|u|=n}|\Pi_{u}|^{1/2}=\hat{d}_{n}(\zeta_{i},\zeta_{j})=d_{n}(\zeta_{i},\zeta_{j})\,,$

where the last step follows from Proposition 4.5. Taking the limit as
$n\to\infty$, this implies that, for each $k\geq 1$ the metric spaces
$(\\{\zeta_{i},1\leq i\leq k\\},d)$ and $([k],Y)$ are isometric (with the
correspondence $(i,\zeta_{i})$, $i\in[k]$). Since $(\\{\zeta_{i},1\leq i\leq
k\\},d)$ increases to $\operatorname{CMT}({\textup{e}},{\mathbf{U}})$
(Proposition 3.22), the claim follows by taking the limit as $k\to\infty$.

Finally, we are ready to prove Theorem 1.5 which shows that the convex
minorant tree provides a coupling between the two classical constructions of
the additive coalescent by Aldous & Pitman [15] on the one hand, and Bertoin
[20] on the other. Let ${\mathcal{P}}=\\{(\pi(x),{{\text{\bsifamily
s}}}(x)):x\in\mathscr{L}(e)\\}$.

###### Proof of Theorem 1.5.

Observe that, by Lemma 4.8, with probability one, all the local minima of e
are of the form $\eta_{u}\curlywedge\eta_{ui}$ defined in Lemma 4.3, and
therefore, almost surely,
${\mathcal{P}}=\\{(\pi(a_{ui}),\tau_{ui}):u\in{\mathcal{U}},i\in\mathbb{N}\\}$.
Since, a.s. for all $u\in{\mathcal{U}}$, we have $\ell_{u}>0$, this can
equivalently be put as
${\mathcal{P}}=\\{(\pi(\kappa(\zeta_{i},\zeta_{j})),\tau(\zeta_{i},\zeta_{j})):i\neq
j\in\mathbb{N}\\}$. From there, the claim is an easy consequence of Theorem
16, and Corollaries 17-18 of [5] (it is even simpler since we do not need to
infer the $\tau(\zeta_{i},\zeta_{j})$ from the cut tree, they can be read
directly from the fragmentation). ∎

##  5 Compactness of the Brownian parabolic tree
$\operatorname{CMT}(X,{\mathbf{U}})$

In this section, we prove the compactness of
$\operatorname{CMT}(X,{\mathbf{U}})$ constructed in Section 3.7. The
completeness is plain from the definition and we only need to verify that
$(\mathscr{M},d)$ is totally bounded.

We will proceed by controlling the growth of a well-chosen sequence of
subspaces that increase to $\mathscr{M}$ using a chaining argument. This leads
us to a process that is reminiscent to a certain version of Prim’s exploration
at the discrete level, and that also turns out to be crucial in the
calculation of the Hausdorff dimension (see Section 6). The general strategy
is inspired from the arguments of Curien and Haas [30] for the compactness of
trees constructed by aggregation of segments.

###  5.1 The growth process

In the entire section, we consider the process $X$, and the random variables
$Z^{\lambda}=Z^{\lambda}(X)$ refer to this case. We may see the metric space
$(\mathscr{M},d)$ as obtained from the coalescent process induced by
$Z^{\lambda}$ on $\mathbb{R}_{+}$, which turns out to be the standard
multiplicative coalescent [24, 17]. In this process, fragments only merge by
pairs, but obtaining fine quantitative estimates is delicate since for any
$\lambda\in\mathbb{R}$ and $h>0$, $Z^{\lambda}\setminus Z^{\lambda+h}$ is a.s.
not contained in any compact interval. We shall thus track a single connected
component as $\lambda$ increases.

######  Remark 5.1.

The most natural choice of a connected component to track is the largest one,
but this leads to some inconvenient conditioning. One could also track a
connected component containing a fixed point (at the discrete level), but,
without any additional structure, such a node must be uniformly random, and
thus the corresponding component would be too small to lead to anything
interesting. Here, the structure imposed by the representation on
$\mathbb{R}_{+}$ allows us to track any fixed point _among the ones that do
matter_ even though they are a negligible for the mass measure (that is, the
nodes $v_{i}$ for $\epsilon n^{2/3}\leq i\leq Cn^{2/3}$ for constants
$0<\epsilon<C$).

For each $\lambda\in\mathbb{R}$, let $L_{\lambda}=\sup Z^{\lambda}\cap[0,1)$
and $R_{\lambda}=\inf Z^{\lambda}\cap(1,\infty)$, and define
$H_{\lambda}=[L_{\lambda},R_{\lambda})$. So, up to inclusion of the left-most
point, $H_{\lambda}\subseteq\mathbb{R}_{+}$ is the interval of
$\mathbb{R}_{+}\setminus Z^{\lambda}$ which contains the point $1$; since
$1\not\in\bigcup_{\lambda}Z^{\lambda}$ with probability one, this is well-
defined for all $\lambda\in\mathbb{R}$. We have the following asymptotics,
whose proofs are found in Section 5.2.

######  Lemma 5.2.

There exist constant $c>0$ and $x_{0},\lambda_{0}$ such that, for all
$x>x_{0}$ and $\lambda\geq\lambda_{0}$ we have

${\mathbf{P}}(L_{\lambda}\geq\tfrac{x}{\lambda^{2}})\leq
e^{-cx}\qquad\text{and}\qquad{\mathbf{P}}(|R_{\lambda}-2\lambda|>1)\leq
e^{-c\lambda}\,.$

It follows that $H_{\lambda}\uparrow(0,\infty)$ as $\lambda\to\infty$, so that
$H_{\lambda}$ provides a suitable increasing family of subspaces of
$\mathscr{M}$. So for any $x\in\mathbb{R}_{+}\setminus\\{0,1\\}$, we let
$\lambda(x)=\inf\\{\lambda:x\in H_{\lambda}\\}$ be the time at which $x$ joins
the connected component containing $1$. The intervals that join $H_{\lambda}$
play a different role depending on whether they lie to the left or to the
right, and we define
$\Lambda^{\\!\scalebox{0.6}{$R$}}=\\{\lambda\in\mathbb{R}:R_{\lambda}>R_{\lambda-}\\}$
and
$\Lambda^{\\!\scalebox{0.6}{$L$}}:=\\{\lambda\in\mathbb{R}:L_{\lambda}<L_{\lambda-}\\}$;
then $\Lambda^{\\!\scalebox{0.6}{$R$}}$ and $\Lambda^{\\!\scalebox{0.6}{$L$}}$
are both countable. Furthermore, they are almost surely disjoint; this is
because the standard multiplicative coalescent is binary [14], and could also
be proved directly from the representation with $X$ (see the proof of
Proposition 7.7). We let
$\Lambda=\Lambda^{\\!\scalebox{0.6}{$R$}}\cup\Lambda^{\\!\scalebox{0.6}{$L$}}$.

For $x,y\in\mathbb{R}_{+}$, we let $x\leftrightarrow y$ if
$\lambda(x)=\lambda(y)$, and let $q_{\lambda}$, $\lambda\in\Lambda$, be the
equivalence classes of this relation. For
$\lambda\in\Lambda^{\\!\scalebox{0.6}{$L$}}$, we have
$q_{\lambda}=[L_{\lambda},L_{\lambda-})$, while
$q_{\lambda}=[R_{\lambda-},R_{\lambda})$ for
$\lambda\in\Lambda^{\\!\scalebox{0.6}{$R$}}$. Because of this,
$(q_{\lambda})_{\lambda\in\Lambda^{\\!\scalebox{0.6}{$L$}}}$ and
$(q_{\lambda})_{\lambda\in\Lambda^{\\!\scalebox{0.6}{$R$}}}$ respectively
define partitions of $(0,1)$ and $(1,\infty)$ into countably many disjoint
intervals.

For an interval $I\subset\mathbb{R}_{+}$, we let $\mathscr{M}|_{I}$ be the
_intrinsic_ metric space induced by $d$ on $I$: this is the metric space
$(I,d_{I})$ where $d_{I}(x,y)=d(x,y)$ if $\llbracket x,y\rrbracket\subseteq I$
and $d_{I}(x,y)=+\infty$ otherwise. So in general, $\mathscr{M}|_{I}$ might be
disconnected. Let $\mathscr{M}_{\lambda}:=\mathscr{M}|_{H_{\lambda}}$; then
$\mathscr{M}_{\lambda}$ is connected for all $\lambda\in\mathbb{R}$. For
$\lambda\in\Lambda^{\\!\scalebox{0.6}{$R$}}$ (resp.
$\Lambda^{\\!\scalebox{0.6}{$L$}}$), we also let
$\mathscr{T}^{\scalebox{0.6}{$R$}}_{\lambda}:=\mathscr{M}|_{q_{\lambda}}$
(resp.
$\mathscr{T}_{\lambda}^{\scalebox{0.6}{$L$}}:=\mathscr{M}_{q_{\lambda}}$).
Quite plainly, and up to the metric completion, the metric space $\mathscr{M}$
is obtained by combining the $\mathscr{T}^{\scalebox{0.6}{$R$}}_{\lambda}$,
$\lambda\in\Lambda^{\\!\scalebox{0.6}{$R$}}$, and
$\mathscr{T}^{\scalebox{0.6}{$L$}}_{\lambda},\lambda\in\Lambda^{\\!\scalebox{0.6}{$L$}}$,
using the identifications performed during the construction (using the random
points constructed from ${\mathbf{U}}$). This process actually turns out
rather agreeable: the metric spaces
$\mathscr{T}^{\scalebox{0.6}{$R$}}_{\lambda}$ and
$\mathscr{T}^{\scalebox{0.6}{$L$}}_{\lambda}$ are easy to understand because
they are small as $\lambda\to\infty$, and the way they are put together is
also easy to control. Informally, the dynamics as $\lambda$ increases are as
follows:

  * •

at time $\lambda\in\Lambda^{\\!\scalebox{0.6}{$R$}}$,
$\mathscr{T}^{\scalebox{0.6}{$R$}}_{\lambda}$ merges with
$\mathscr{M}_{\lambda-}$ by identifying $\inf q_{\lambda}$ with a uniform
point in $H_{\lambda-}$;

  * •

at time $\lambda\in\Lambda^{\\!\scalebox{0.6}{$L$}}$, $\mathscr{M}_{\lambda-}$
connects with $\mathscr{T}^{\scalebox{0.6}{$L$}}_{\lambda}$ by identifying
$L_{\lambda-}$ with a uniform point in
$\mathscr{T}^{\scalebox{0.6}{$L$}}_{\lambda}$.

The $(\mathscr{M}_{k^{3}})_{k\geq 1}$ is the convenient sequence of subspaces
of $\mathscr{M}$ that we mentioned before. The following decomposition which
takes advantage of these dynamics will be useful. What matters for now is the
global picture, we will fill out the details later on.

  * •

_The annuli of forests to the right._ For any $k\geq 1$ define
$\Lambda_{k}:=\\{\lambda\in\Lambda^{\\!\scalebox{0.6}{$R$}}:k^{3}<\lambda\leq(k+1)^{3}\\}$,
and let ${\mathscr{F}}_{k}$ be the intrinsic metric space induced by
$\mathscr{M}$ on $(R_{k^{3}},R_{(k+1)^{3}}]$. For each $k\geq 1$,
${\mathscr{F}}_{k}$ is a forest consisting of infinitely many connected
components obtained when only the identifications within
$(R_{k^{3}},R_{(k+1)^{3}}]$ are performed. Our aim is to bound the maximum
diameter of the connected component ${\mathscr{F}}_{k}$ in order to control
the worst case accumulation of length when putting all the ${\mathscr{F}}_{k}$
together. Formally, for $\lambda,\lambda^{\prime}\in\Lambda_{k}$ we write
$\lambda\equiv_{k}\lambda^{\prime}$ if $R_{\lambda-}\wedge
R_{\lambda^{\prime}-}>R_{k^{3}}$, which implies that
$\mathscr{T}_{\lambda}^{\scalebox{0.6}{$R$}}$ and
$\mathscr{T}_{\lambda^{\prime}}^{\scalebox{0.6}{$R$}}$ are connected within
${\mathscr{F}}_{k}$. We will prove that, almost surely, every equivalence
class of this relation is finite. So for each $\lambda\in\Lambda_{k}$, we may
define
$\rho_{\lambda}:=\min\\{R_{\lambda^{\prime}-}:\lambda^{\prime}\equiv_{k}\lambda\\}$
as the leftmost point of the connected component containing
$\mathscr{T}^{\scalebox{0.6}{$R$}}_{\lambda}$ within ${\mathscr{F}}_{k}$. More
generally, the equivalence relation $\equiv_{k}$ naturally extends as follows:
for $x,y\in(R_{k^{3}},R_{(k+1)^{3}}]$ we let $x\sim_{k}y$ if $x\in
q_{\lambda}$, $y\in q_{\lambda^{\prime}}$ and
$\lambda\equiv_{k}\lambda^{\prime}$. Defining the diameter of a potentially
disconnected metric space as the supremum of the diameters of its connected
components, we therefore have
$\operatorname{diam}({\mathscr{F}}_{k})=\sup\\{d(x,y):x,y\in(R_{k^{3}},R_{(k+1)^{3}}],x\equiv_{k}y\\}$.

  * •

_The chain of beads to the origin._ For any $\lambda\in\mathbb{R}$, we let
${\mathscr{P}}_{\lambda}$ denote the intrinsic metric space induced by
$\mathscr{M}$ on $[0,L_{\lambda})$. This metric is almost surely connected and
has the structure of a “string of beads” that we now describe. For
$a\in\mathbb{R}$, let
$\Lambda_{a}^{\\!\scalebox{0.6}{$L$}}:=\Lambda^{\\!\scalebox{0.6}{$L$}}\cap(a,\infty)$.
Almost surely for any $a\in\mathbb{R}$, $\Lambda^{\\!\scalebox{0.6}{$L$}}_{a}$
only contains finitely many points in any compact interval of $(0,+\infty)$,
so that we may enumerate its elements in increasing order as
$(\lambda_{i})_{i\geq 1}$ (Lemma 3.3). The metric space
${\mathscr{P}}_{\lambda}$ is obtained by putting together the metric spaces
$\mathscr{T}^{\scalebox{0.6}{$L$}}_{\lambda_{j}}$, $j\geq 1$, into a chain by
connecting $\mathscr{T}^{\scalebox{0.6}{$L$}}_{\lambda_{i}}$ to a uniform
random point in $\mathscr{T}^{\scalebox{0.6}{$L$}}_{\lambda_{i+1}}$ for each
$i\geq 1$.

By giving an estimate of the extent of $H_{\lambda}$, Lemma 5.2 provides an
effective way to control the locations of the “gluing points” which lies at
the core of the proofs of the compactness and of the computation of the
Hausdorff dimensions. The contribution of ${\mathscr{P}}_{\lambda}$ is easily
treated separately, and the crucial steps consists in controlling the
diameters of the ${\mathscr{F}}_{k}$, $k\geq 1$.

######  Proposition 5.3 (Diameter of annuli forests).

There exists $k_{0}\in\mathbb{N}$ such that for all $k\geq k_{0}$ we have

${\mathbf{P}}\left(\operatorname{diam}({\mathscr{F}}_{k})>k^{-3/2}\right)\leq
11k^{-5/4}\,.$

######  Proposition 5.4 (Diameter of the string of beads).

Almost surely,

$\lim_{\lambda\to\infty}\operatorname{diam}({\mathscr{P}}_{\lambda})=0\,.$

Taking Propositions 5.3 and 5.4 for granted for now, the proof of compactness
is then straightforward.

######  Proposition 5.5 (Compactness of $\mathscr{M}$).

The metric space $(\mathscr{M},d)$ is almost surely compact.

###### Proof.

By Proposition 5.3 and the Borel–Cantelli lemma, with probability one, there
exists an almost surely finite random variable $K$ such that
$\operatorname{diam}({\mathscr{F}}_{k})\leq k^{-3/2}$ for all $k\geq K$. In
particular, for all $k\geq K$,

$\operatorname{d}_{\text{{h}}}(\mathscr{M}_{k^{3}},\mathscr{M})\leq
2k^{-1/2}+\operatorname{diam}({\mathscr{P}}_{k^{3}})\,.$

Fix any $\epsilon>0$. By Proposition 5.4,
$\operatorname{diam}({\mathscr{P}}_{k^{3}})<\epsilon/3$ for all $k$ large
enough; it follows that there exists $k\geq K$ large enough such that
$\operatorname{d}_{\text{{h}}}(\mathscr{M}_{k^{3}},\mathscr{M})<\epsilon/2$.
Recall that, for any fixed $\lambda\in\mathbb{R}$, the restriction
$\mathscr{M}_{\lambda}$ of $\mathscr{M}$ to $H_{\lambda}$ is almost surely
compact by absolute continuity with the Brownian continuum random tree. So
there exists a cover of $\mathscr{M}_{k^{3}}$ by finitely many balls of radius
$\epsilon/2$; increasing the radius of each ball to $\epsilon$ yields a finite
cover of $\mathscr{M}$. We have thus proved that $\mathscr{M}$ is totally
bounded. Since it is complete by definition, it is compact. ∎

The remainder of the section is devoted to the proof of Propositions 5.3 and
5.4. We first prove Lemma 5.2 in Section 5.2. The forests ${\mathscr{F}}_{k}$
are made of the trees $\mathscr{T}^{\scalebox{0.6}{$R$}}_{\lambda}$, for
$\lambda\in\Lambda_{k}$. For all $k\geq 1$, $\Lambda_{k}$ is infinite, which
causes some difficulties. Still, we expect that, for large $k$, the components
$\mathscr{T}_{\lambda}^{\scalebox{0.6}{$R$}}$ with $\lambda\in\Lambda_{k}$
should be rather small; Section 5.3 deals with the question of uniform bounds
on distances in the $\mathscr{T}^{\scalebox{0.6}{$R$}}_{\lambda}$ in terms of
the lengths $|q_{\lambda}|$. We then obtain in Section 5.4 the relevant
statistics about the connected components
$\mathscr{T}_{\lambda}^{\scalebox{0.6}{$R$}}$ for $\lambda\in\Lambda_{k}$
which includes information about the lengths $|q_{\lambda}|$ and their
diameters. In Section 5.5 we put together all the pieces and prove Proposition
5.3 which essentially says that the diameter of ${\mathscr{F}}_{k}$ is
comparable to the maximum diameter of the
$\mathscr{T}_{\lambda}^{\scalebox{0.6}{$R$}}$, $\lambda\in\Lambda_{k}$.
Finally, we prove Propostition 5.4 in Section 5.6.

###  5.2 The position of the component containing $1$: Proof of Lemma 5.2

Recall that for a continuous process $\omega=(\omega_{t})_{t\geq 0}$, we let
$\underline{\omega}$ and $\overline{\omega}$ denote respectively the running
infimum and supremum processes:
$\underline{\omega}_{t}:=\inf\\{\omega_{s}:0\leq s\leq t\\}$ and
$\overline{\omega}_{t}=\sup\\{\omega_{t}:0\leq s\leq t\\}$.

Recall that $(W_{t})_{t\geq 0}$ denotes a standard Brownian motion. We will
use repeatedly the following simple fact (see, e.g., [41] page 96, consequence
of the fact that $\overline{W}_{t}$ has same law as $|W_{t}|$, for a fixed
$t$): for all $x\geq 0$, we have

${\mathbf{P}}(\overline{W}_{t}\geq
x)={\mathbf{P}}(\underline{W\\!}_{t}\leq-x)\leq e^{-x^{2}/(2t)}\,.$ (17)

Let $\lambda\geq 0$. Let us first focus on the upper bound. For any
$t\in\mathbb{R}_{+}$, we have $R_{\lambda}>t$ if only if there is an excursion
of $X^{\lambda}$ above its running minimum that straddles both $1$ and $t$,
that is if $1\sim_{\lambda}t$. Thus

$\displaystyle{\mathbf{P}}(R_{\lambda}>2\lambda+1)$
$\displaystyle={\mathbf{P}}(\underline{X}_{1}^{\lambda}=\underline{X}_{2\lambda+1}^{\lambda})$
$\displaystyle={\mathbf{P}}(\underline{X}_{1}^{\lambda}\leq\underline{X}_{2\lambda+1}^{\lambda})$
$\displaystyle\leq\inf_{x}\\{{\mathbf{P}}(\underline{X}_{1}^{\lambda}\leq
x)+{\mathbf{P}}(\underline{X}_{2\lambda+1}^{\lambda}>x)\\}\,.$ (18)

A quick inspection of the expected values leads to the choice
$x=-\tfrac{\lambda}{2}$. On the one hand
$\underline{X}_{1}^{\lambda}\geq-\frac{1}{2}+\underline{W\\!}_{1}$ so that, by
(17), we obtain

$\displaystyle{\mathbf{P}}(\underline{X\\!}^{\lambda}_{1}\leq-\tfrac{\lambda}{2})\leq{\mathbf{P}}(-\tfrac{1}{2}+\underline{W}_{1}\leq-\tfrac{\lambda}{2})\leq
e^{-\lambda^{2}/9}\,,$ (19)

for all $\lambda$ large enough. On the other hand,
$\underline{X}_{2\lambda+1}^{\lambda}\leq
X^{\lambda}_{2\lambda+1}=-\lambda-\tfrac{1}{2}+W_{2\lambda+1}$ which implies
that

$\displaystyle{\mathbf{P}}(\underline{X}_{2\lambda+1}^{\lambda}>-\tfrac{\lambda}{2})\leq{\mathbf{P}}(W_{2\lambda+1}>\tfrac{\lambda}{2})\leq
e^{-\lambda/20}\,,$ (20)

for all $\lambda$ large enough. Putting together (5.2)–(20) completes the
proof of the upper bound.

For the lower bound, observe that

$\displaystyle{\mathbf{P}}\left(R_{\lambda}<2\lambda-1\right)$
$\displaystyle={\mathbf{P}}(\underline{X}^{\lambda}_{1}>\underline{X}^{\lambda}_{2\lambda-1})$
$\displaystyle\leq\inf_{x}\big{\\{}{\mathbf{P}}(X_{1}^{\lambda}<x)+{\mathbf{P}}(\underline{X}_{1}^{\lambda}>\underline{X}^{\lambda}_{2\lambda-1},X_{1}^{\lambda}\geq
x)\big{\\}}\,.$ (21)

Considering the fact that $X_{1}^{\lambda}=-\tfrac{1}{2}+\lambda+W_{1}$, we
are lead to choosing $x=\tfrac{\lambda}{2}$. We have
${\mathbf{P}}(X_{1}^{\lambda}<\frac{\lambda}{2})\leq e^{-\lambda^{2}/9}$ for
all $\lambda$ large enough. To deal with the second part of the right-hand
side of (5.2), we use Markov’s property at time $1$ and observe that for
$s\geq 1$, $X_{s}^{\lambda}-X^{\lambda}_{1}$ is distributed like
$X^{\lambda-1}_{s-1}$:

$\displaystyle{\mathbf{P}}(\underline{X}_{1}^{\lambda}>\underline{X}_{2\lambda-1}^{\lambda}~{}|~{}X_{1}^{\lambda}\geq\tfrac{\lambda}{2})$
$\displaystyle\leq{\mathbf{P}}(\underline{X}^{\lambda-1}_{2\lambda-2}<-\tfrac{\lambda}{2})\,.$
(22)

However, since $s\mapsto-\tfrac{s^{2}}{2}+(\lambda-1)s$ is non-negative on
$[0,2(\lambda-1)]$, we have

$\displaystyle\underline{X}^{\lambda-1}_{2\lambda-2}=\inf\\{-\tfrac{s^{2}}{2}+(\lambda-1)s+W_{s}:s\leq
2\lambda-2\\}\geq\underline{W}_{2\lambda-2}\,,$

so that, by (17),

$\displaystyle{\mathbf{P}}(\underline{X}^{\lambda-1}_{2\lambda-2}<-\tfrac{\lambda}{2})\leq{\mathbf{P}}(\underline{W}_{2\lambda-2}<-\tfrac{\lambda}{2})\leq
e^{-\lambda/20}\,,$

for all $\lambda$ large enough. Putting this together with (5.2) and (22)
yields the lower bound on $R_{\lambda}$.

Finally, we deal with the lower bound on $L_{\lambda}$. Observe that
$\mathbf{E}[X^{\lambda}_{s}]=-\tfrac{s^{2}}{2}+\lambda s\geq s\lambda/2$ for
all $s\in[0,1]$ provided that $\lambda$ is large enough. So writing
$t=a/\lambda^{2}$, we have, since $\underline{X}^{\lambda}_{t}\leq 0$,

$\displaystyle{\mathbf{P}}(L_{\lambda}\geq
t)={\mathbf{P}}(\underline{X}_{t}^{\lambda}>\underline{X}_{1}^{\lambda})$
$\displaystyle\leq{\mathbf{P}}(\exists s\in[t,1]:X^{\lambda}_{s}\leq 0)$
$\displaystyle\leq{\mathbf{P}}(\exists s\in[t,1]:W_{s}\leq-\lambda s/2)\,.$

Observe that $\\{(s,y):s\in[t,1],y\leq-\lambda s/2\\}\subseteq\bigcup_{i\geq
1}\\{(s,y):s\leq(i+1)t,y\leq-\lambda it/2\\}$. So writing
$\tau(-x):=\inf\\{s\geq 0:W_{s}<-x\\}$, we have

$\displaystyle{\mathbf{P}}(\exists s\in[t,1]:W_{s}\leq-\lambda s/2)$
$\displaystyle\leq\sum_{i\geq 1}{\mathbf{P}}(\tau(-\lambda it/2)\leq(i+1)t)$
$\displaystyle\leq\sum_{i\geq
1}{\mathbf{P}}(\underline{W}_{(i+1)t}\leq-\lambda i\tfrac{t}{2})$

which, using (17) is at most $e^{-a/20}$ for all $a$ large enough. This
completes the proof of Lemma 5.2.

###  5.3 Distances in small aggregated components

In this section, we are interested in the intrinsic metric space induced by
$d$ on $q_{\lambda}$. Write $D_{\lambda}=\sup\\{d(x,y):x,y\in q_{\lambda}\\}$
for the diameter of this metric space, and for $\zeta_{\lambda}$ a random
variable with uniform distribution in $q_{\lambda}$, independent of everything
else, let $Y_{\lambda}:=d(l_{\lambda},\zeta_{\lambda})$ where
$l_{\lambda}=\inf q_{\lambda}$.

Recall that $X^{\lambda}_{t}=W_{t}-\tfrac{t^{2}}{2}+\lambda t$ and
$B^{\lambda}_{t}=X^{\lambda}_{t}-\underline{X}^{\lambda}_{t}$ is the process
reflected in the running infimum. We define $m_{\lambda}=|q_{\lambda}|$ and
$\varepsilon_{\lambda}:\mathbb{R}_{+}\to\mathbb{R}_{+}$ by

$\varepsilon_{\lambda}(s):=B^{\lambda-h}(l_{\lambda}+s)\mathbf{1}_{{0\leq
s\leq m_{\lambda}}}\,.$ (23)

The excursion $\varepsilon_{\lambda}$ has duration $m_{\lambda}$ and encodes
the metric space $\mathscr{T}_{\lambda}^{\scalebox{0.6}{$R$}}$ or
$\mathscr{T}_{\lambda}^{\scalebox{0.6}{$L$}}$ supported by $q_{\lambda}$. For
any fixed $\lambda^{\prime}\in\mathbb{R}$, and $q$ an interval of
$\mathbb{R}_{+}\setminus Z^{\lambda^{\prime}}$, conditionally on $|q|=\sigma$,
the distribution of the excursion of $B^{\lambda^{\prime}}$ straddling $q$ is
given by

$\tilde{\mathbf{n}}_{\sigma}(A)=\frac{\int\mathbf{1}_{{\omega\in
A}}\exp(\int_{0}^{\sigma}\omega(r)dr){\mathbf{n}}_{\sigma}(d\omega)}{\int\exp(\int_{0}^{\sigma}\omega(r)dr){\mathbf{n}}_{\sigma}(d\omega)}\,$
(24)

where ${\mathbf{n}}_{\sigma}$ is the law of a Brownian excursion of duration
$\sigma>0$. Consider now, $\lambda\in\Lambda$, which is random. For any
$\lambda^{\prime}<\lambda$, $l_{\lambda}\in Z^{\lambda^{\prime}}$;
furthermore, as $\lambda^{\prime}\uparrow\lambda$, we have
$\inf\\{s>0:B^{\lambda^{\prime}}(l_{\lambda}+s)=0\\}\uparrow m_{\lambda}$ and
the excursion of $B^{\lambda^{\prime}}$ starting at $l_{\lambda}$ converges to
$\varepsilon_{\lambda}$. It follows that, for a bounded continuous functional
$\phi$,

$\mathbf{E}[\phi(\varepsilon_{\lambda})~{}|~{}m_{\lambda}=\sigma]=\int\phi(\omega)\tilde{\mathbf{n}}_{\sigma}(d\omega)\,.$
(25)

We note further that, by the strong Markov property, the excursions
$\varepsilon_{\lambda}$, $\lambda\in\Lambda$ are mutually independent
conditionally on $m_{\lambda}$, $\lambda\in\Lambda$.

######  Proposition 5.6 (Distances in small components).

For any $\epsilon>0$, there exist

  1. i)

a sub-Gaussian random variable $D^{\star}$ such that, if
$m_{\lambda}\in(0,\epsilon]$ then
$m_{\lambda}^{-1/2}D_{\lambda}\leq_{st}D^{\star}$, and

  2. ii)

a random variable $Y^{\star}$ with $\mathbf{E}[1/Y^{\star}]<\infty$ and such
that if $m_{\lambda}\in(0,\epsilon]$ then
$Y^{\star}\leq_{st}m_{\lambda}^{-1/2}Y_{\lambda}$.

###### Proof.

_i)_ Let $(\omega,{\mathbf{u}})\mapsto d_{\omega,{\mathbf{u}}}$ a
deterministic measurable function which gives the metric of
$\operatorname{CMT}(\omega,{\mathbf{u}})$ for ${\mathbf{n}}\otimes
d{\mathbf{u}}$-a.e. $(\omega,{\mathbf{u}})$, and let
$\bar{d}_{\omega,{\mathbf{u}}}=\sup_{x,y}d_{\omega,{\mathbf{u}}}(x,y)$ be the
corresponding diameter. From the law of $\varepsilon_{\lambda}$ in (24)–(25)
and the Cauchy–Schwarz inequality, for any $r\in\mathbb{R}$,

$\displaystyle\mathbf{E}\left[e^{rm_{\lambda}^{-1/2}D_{\lambda}}~{}\Big{|}~{}m_{\lambda}=\sigma\right]$
$\displaystyle=\int\exp(r\sigma^{-1/2}\bar{d}_{\omega,{\mathbf{u}}})\tilde{\mathbf{n}}_{\sigma}(d\omega)d{\mathbf{u}}$
$\displaystyle=\int\exp(r\sigma^{-1/2}\bar{d}_{\omega,{\mathbf{u}}})\frac{\exp(\int_{0}^{\sigma}\omega(s)ds){\mathbf{n}}_{\sigma}(d\omega)}{\int\exp(\int_{0}^{\sigma}\omega(s)ds){\mathbf{n}}_{\sigma}(d\omega)}d{\mathbf{u}}$
$\displaystyle\leq\frac{\sqrt{\int\exp(2r\sigma^{-1/2}\bar{d}_{\omega,{\mathbf{u}}}){\mathbf{n}}_{\sigma}(d\omega)d{\mathbf{u}}}\cdot\sqrt{\int\exp(2\int_{0}^{\sigma}\omega(s)ds){\mathbf{n}}_{\sigma}(d\omega)}}{\int\exp(\int_{0}^{\sigma}\omega(s)ds){\mathbf{n}}_{\sigma}(d\omega)}.$
(26)

In the first factor above, the metric space is encoded by a Brownian
excursion, and is therefore exactly distributed like a Brownian continuum
random tree of mass $\sigma$ (Theorem 1.3 with $s=0$). It follows by Brownian
scaling that, under ${\mathbf{n}}_{\sigma}\otimes d{\mathbf{u}}$,
$\sigma^{-1/2}\bar{d}_{\omega,{\mathbf{u}}}$ is dominated by $2\|e\|$, twice
the supremum of a standard Brownian excursion, which is well-known to be sub-
Gaussian (see, for instance, [31, 42]). On the other hand, observe that
$\psi(z):=\int\exp(z\int_{0}^{1}\omega(s)ds){\mathbf{n}}_{1}(d\omega)=\mathbf{E}[\exp(z\int_{0}^{1}e(s)ds)]$,
the Laplace transform of Brownian excursion area [40], is continuous and
positive on any interval $[0,\epsilon]$. It follows that there exist constants
$A=A_{\epsilon}$ and $v>0$ such that, for any $\sigma\in[0,\epsilon]$,

$\mathbf{E}\left[\exp(rm_{\lambda}^{-1/2}D_{\lambda})~{}\Big{|}~{}m_{\lambda}=\sigma\right]\leq
e^{r^{2}/(2v)}\cdot\sup_{\sigma\in[0,\epsilon]}\frac{\sqrt{\psi(2\sigma^{3/2})}}{\psi(\sigma^{3/2})}\leq
Ae^{r^{2}/(2v)}\,.$ (27)

One then easily constructs (the law of a variable) $D^{\star}$ by inverse
transform. For $\sigma>0$ let $F_{\sigma}(x):={\mathbf{P}}(D_{\lambda}\leq
x~{}|~{}m_{\lambda}=\sigma)$, and
$F_{\star}(x):=\inf\\{F_{\sigma}(x):\sigma\leq\epsilon\\}$. Then for $U$ a
$[0,1]$-uniform random variable, $F_{\star}$ is the distribution function of a
random variable $D^{\star}:=F_{\star}^{-1}(U)$ that dominates all the
$m_{\lambda}^{-1/2}D_{\lambda}\mathbf{1}_{{m_{\lambda}\leq\epsilon}}$. The
random variables $F^{-1}_{\sigma}(U)$ are uniformly sub-Gaussian by (27), and
so is $D^{\star}$. This completes the proof of _i)_.

The proof of _ii)_ about the distance to a random point is similar: we only
discuss the adaptation of the arguments in _i)_ to bound
$\mathbf{E}[m_{\lambda}^{1/2}/Y_{\lambda}~{}|~{}m_{\lambda}=\sigma]$. Instead
of Cauchy–Schwarz, using Hölder’s inequality (with exponents $3/2$ and $3$)
provides an upper bound similar to (5.3) where the main term involves the
$d_{\omega,{\mathbf{u}}}(0,\xi)$ for $\xi$ independent and uniform in
$[0,\sigma]$ under ${\mathbf{n}}(d\omega)$: this is the distance between two
random points in a unit mass Brownian CRT, which is a random variable $Y$ with
density $xe^{-x^{2}/2}dx$ on $\mathbb{R}_{+}$ [11], so that
$\mathbf{E}[Y^{-3/2}]<\infty$. The multiplicative error term is bounded just
as above, and the random variable $Y^{\star}$ is constructed similarly, using
the supremum of the distribution functions instead of the infimum. We omit the
details. ∎

###  5.4 Statistics of the aggregated components
$\mathscr{T}_{\lambda}^{\scalebox{0.6}{$R$}}$

From the results of the previous section, especially the law of the
$\varepsilon_{\lambda}$, it is crucial to understand the distribution of the
$m_{\lambda}$. In the following we let ${\mathscr{E}}$ be the space of
continuous excursions, that is, the functions
$f\in\mathcal{C}(\mathbb{R}_{+},\mathbb{R}_{+})$ with $f(0)=0$ such that there
exists $\sigma\in[0,\infty)$ such that $f(s)>0$ for $s\in(0,\sigma)$ and
$f(s)=0$ for $s\geq\sigma$.

Fix any $\lambda\in\mathbb{R}$ and consider the excursions of $X^{\lambda}$
away from its running infimum $\underline{X}^{\lambda}$. For
$y\in\mathbb{R}_{+}$, let
$\tau^{\lambda}_{y}:=\inf\\{s>0:-\underline{X}_{s}^{\lambda}>y\\}$. Then,
$\\{(y,\tau^{\lambda}_{y}-\tau^{\lambda}_{y-}):\tau_{y}^{\lambda}>\tau_{y-}^{\lambda}\\}$
is a Poisson point process on $\mathbb{R}_{+}\times\mathbb{R}_{+}$ of
intensity $dy\tilde{\varrho}^{\lambda}_{\tau_{y-}}$, where the inhomogeneous
measure $\tilde{\varrho}_{x}^{\lambda}$ is defined by (see, e.g., [14],
Section 5.2)

$\tilde{\varrho}^{\lambda}_{x}(l,\infty):=\lim_{\epsilon\to
0}\frac{1}{\epsilon}{\mathbf{P}}\left(\inf\\{s>0:X^{\lambda}_{x+s}\leq
0\\}>l~{}\Big{|}~{}X_{x}^{\lambda}=\epsilon\right)\,.$ (28)

######  Proposition 5.7.

The process $\\{(\lambda,\varepsilon_{\lambda}):R_{\lambda}>R_{\lambda-}\\}$
is a Poisson point process on $\mathbb{R}\times{\mathscr{E}}$ of intensity
$R_{\lambda-}d\lambda\tilde{\varrho}^{\lambda}_{R_{\lambda}-}(d\sigma)\tilde{\mathbf{n}}_{\sigma}(d\omega)$.

###### Proof.

The distribution of the excursions $\varepsilon_{\lambda}$ conditionally on
their duration is known from the previous section, and we only need to deal
with the sizes of the jumps of $(R_{\lambda})_{\lambda\in\mathbb{R}}$. These
are formed by agglomeration of some of the excursion lengths of $B^{\lambda}$,
for $\lambda\in\mathbb{R}$, which are described by the excursion length
measure in (28): As $\lambda$ increases, the excursions of $B^{\lambda}$ away
from $0$ merge together until they eventually join the connected component
containing $1$.

We proceed geometrically using the process $X^{0}$ only. The excursions
$\varepsilon_{\lambda}$ are simply read from $X^{0}$: for any $\lambda$ such
that $R_{\lambda}>R_{\lambda-}$, $\varepsilon_{\lambda}$ is obtained as

$\varepsilon_{\lambda}(s)=(X^{0}(R_{\lambda-}+s)-X^{0}(R_{\lambda-})+s\lambda)\mathbf{1}_{{R_{\lambda-}+s\leq
R_{\lambda}}}\qquad\text{for }s\geq 0.$

Here, notice that $\varepsilon_{\lambda}$ is indeed obtained from the
agglomeration of countably many excursions of
$X^{\lambda-h}-\underline{X}^{\lambda-h}$, which might be described using the
process involving straight lines with slopes $\lambda-h$ (at time $\lambda$,
the excursions of interest are those of $X^{0}$ above the process
$s\mapsto\underline{X}_{s}^{\lambda}-\lambda s$). Now, knowing the intensity
of jumps of $\tau^{\lambda}$ for each $\lambda\in\mathbb{R}$, it is routine to
deduce the intensity of excursions of jumps of
$(R_{\lambda})_{\lambda\in\mathbb{R}}$: at time $\lambda$, we always have an
excursion, the increase in local time is $d\lambda R_{\lambda-}$, and this
gives rise to excursions whose durations are governed by
$\tilde{\rho}_{R_{\lambda}-}^{\lambda}$. ∎

For each $k\geq 1$ we say $\lambda\in\Lambda_{k}$ is of level $i\geq 0$ and
write $\lambda\in\Lambda_{k,i}$ if its duration satisfies
$m_{\lambda}\in[k^{-6-i},k^{-5-i})$. We define the $M_{k,i}$ total duration
(mass) of excursions of level at least $i$

$M_{k,i}=\sum_{j\geq i}\sum_{\lambda\in\Lambda_{k,j}}m_{\lambda}\,.$

######  Lemma 5.8 (Statistics for fragments
$\mathscr{T}^{\scalebox{0.6}{$R$}}_{\lambda}$,
$\lambda\in\Lambda^{\\!\scalebox{0.6}{$R$}}$).

For any $k\geq 1$ and $i\geq 0$, there exists an event $B_{k,i}$ of
probability at most $8\min\\{k^{-1-i/4},k^{-5/4}\\}$ such that outside of
$B_{k,i}$ we have

  1. i)

the longest excursion: $\sup\\{m_{\lambda}:\lambda\in\Lambda_{k}\\}\leq
k^{-5}$

  2. ii)

total duration of excursions of level at least $i$:
$M_{k,i}\leq\min\\{k^{7/2-i/4},7k^{2}\\},$

  3. iii)

number of excursions of level $i$: $\\#\Lambda_{k,i}\leq 7k^{8+i}$,

  4. iv)

maximum diameter of an excursion of level $i$:
$D_{k,i}=\sup\\{D_{\lambda}:\lambda\in\Lambda_{k,i}\\}\leq k^{-2-i/4}$.

In particular, $\bigcup_{i\geq 0}B_{k,i}$ occurs with probability at most
$10k^{-5/4}$.

The proof of Lemma 5.8 is based on upper bounds on the durations
$|q_{\lambda}|$, $\lambda\in\Lambda_{k}$. The relevant calculations are
simplified if we upper bound the Brownian with parabolic drift by a Brownian
with a suitable linear drift. This is why the following is especially useful.
In the following, we let $\varrho^{\mu}$ be the excursion length measure for
Brownian motion with linear drift $\mu$, $W^{\mu}:=s\mapsto W_{s}+\mu s$. We
are mostly interested in what happens at large positions, for which the drift
is negative: we note that, for $\mu\geq 0$,

$\varrho^{-\mu}(x,\infty)=\lim_{\epsilon\to
0}\frac{1}{\epsilon}{\mathbf{P}}\left(\inf\\{s>0:W^{-\mu}_{s}\leq
0\\}>x\Big{|}~{}W^{-\mu}_{0}=\epsilon\right)=\int_{x}^{\infty}\frac{e^{-\mu^{2}r/2}}{\sqrt{2\pi
r^{3}}}dr\,.$

The following proof relies this and the fact that $X^{-\mu}_{s}\leq
W^{-\mu}_{s}$ for all $s\geq 0$ in the natural coupling.

###### Proof of Lemma 5.8.

Let $B_{k}$ be the event that $R_{k^{3}}>2k^{3}-1$ or
$R_{(k+1)^{3}}<2(k+1)^{3}+1$. Then, by Lemma 5.2,
${\mathbf{P}}(B_{k})\leq\exp(-ck^{3})$ for some constant $c>0$. Now, on the
one hand, the quantity of local time $LT_{k}$ corresponding to $\Lambda_{k}$
is

$LT_{k}:=\int_{k^{3}}^{(k+1)^{3}}R_{\lambda-}d\lambda=\int_{k^{3}}^{(k+1)^{3}}R_{\lambda}d\lambda\leq
R_{(k+1)^{3}}[(k+1)^{3}-k^{3}]\,,$

so that $LT_{k}\leq 7k^{5}$ for all $k$ large enough on the complement event
$B_{k}^{c}$. On the other hand, $\sup\\{\lambda-\inf
q_{\lambda}:\lambda\in\Lambda_{k}\\}\leq(k+1)^{3}-R_{k^{3}}\leq-k^{3}/2$ for
large $k$ on $B_{k}^{c}$.

By Proposition 5.7, it follows that, on the event $B_{k}^{c}$, for every
$x>0$, $\\#\\{\lambda\in\Lambda_{k}:m_{\lambda}>x\\}$ is stochastically
dominated by a Poisson random variable with parameter
$\Delta_{k}\varrho^{-\mu_{k}}(x,\infty)$ with $\Delta_{k}=7k^{5}$ and
$\mu_{k}=k^{3}/2$. The properties _i)_ to _iv)_ in the statement then follow
easily.

_i) Longest excursion._ We have

$\displaystyle{\mathbf{E}}\Bigg{[}\sum_{\lambda\in\Lambda_{k}}\mathbf{1}_{{m_{\lambda}>k^{-5}}}\mathbf{1}_{{B_{k}^{c}}}\Bigg{]}\leq
7k^{5}\varrho^{-\mu_{k}}(k^{-5},\infty)=7k^{5}\int_{k^{-5}}^{\infty}\frac{e^{-k^{6}t/8}}{\sqrt{2\pi
t^{3}}}dt\leq e^{-ck}\,,$

for some constant $c$ and all $k$ large enough. Markov’s inequality then
implies the claim.

_ii) Total length of excursions of level $i$._ We proceed similarly for the
upper bound on $M_{k,i}$. We have

$\displaystyle{\mathbf{E}}\Bigg{[}\sum_{\lambda\in\Lambda_{k}}m_{\lambda}\mathbf{1}_{{m_{\lambda}\leq
k^{-5-i}}}\mathbf{1}_{{B_{k}^{c}}}\Bigg{]}\leq
7k^{5}\int_{0}^{k^{-5-i}}\\!\\!\\!\\!\\!\\!x\varrho^{-\mu_{k}}(dx)\leq
7k^{5}\int_{0}^{k^{-5-i}}\frac{xe^{-k^{6}x/8}}{\sqrt{2\pi x^{3}}}dx\leq
7k^{5/2-i/2}\,.$

Moreover, we also always have the bound
$\sum_{\lambda\in\Lambda_{k}}m_{\lambda}\leq R_{(k+1)^{3}}-R_{k^{3}}$ which is
at most $7k^{2}$ on $B_{k}^{c}$. Markov’s inequality then yields

${\mathbf{P}}(M_{k,i}\geq
k^{7/2-i/4},B_{k}^{c})\leq\min\\{7k^{-1-i/4},7k^{-3/2+i/4}\\}\,.$

_iii) Cardinality of $\Lambda_{k,i}$._ This is a deterministic bound on
$B_{k}^{c}$: we have $\\#\Lambda_{k,i}\leq 7k^{2}\cdot k^{6+i}=7k^{8+i}$.

_iv) Maximum diameter in level $i$._ By Proposition 5.6, when the event of
_i)_ occurs, $D_{k,i}$ is stochastically dominated by

$k^{-5/2-i/2}\cdot\max\\{D^{\star}_{j}:1\leq j\leq\\#\Lambda_{k,i}\\}\,,$

where $D^{\star}_{j}$ are iid copies of the random variable $D^{\star}$ that
is sub-Gaussian (choose $\epsilon=1$). Using the bound for $\\#\Lambda_{k,i}$
in _iii)_ above, it follows easily that

$\displaystyle{\mathbf{P}}(D_{k,i}\geq k^{-2-i/4},B_{k}^{c})$
$\displaystyle\leq 7k^{8+i}\cdot{\mathbf{P}}(D^{\star}\geq k^{1/2+i/4})$
$\displaystyle\leq 7k^{8+i}\cdot\exp(-\tfrac{k^{1+i/2}}{2v})\,,$

for some constant $v>0$.

Finally, writing $B_{k,i}$ for the event that either $B_{k}$ or any of the bad
events in _i)_ – _iv)_ occur, we have ${\mathbf{P}}(B_{k,i})\leq
8\min\\{7k^{-1-i/4},k^{-5/4}\\}$ for large $k$ (this is essentially limited by
the event in _ii)_). The union bound yields the last claim. ∎

###  5.5 The accumulation of length in an annulus: Proof of Proposition 5.3

Each component ${\mathscr{F}}_{\lambda}$ gets connected to some point to its
left (in $\mathbb{R}_{+}$), which falls within some
${\mathscr{F}}_{\lambda^{\prime}}$, for some $\lambda^{\prime}<\lambda$, and
so on. The proof of Proposition 5.3 consists in bounding the accumulation of
these lengths before a connection to $\mathscr{M}_{k^{3}}$ eventually occurs.
We are only interested here in the points of $(1,\infty)$.

Recalling the notation ${{\text{\bsifamily j}}}(\cdot)$ from Remark 3.11, by
construction, for each $\lambda\in\Lambda^{\\!\scalebox{0.6}{$R$}}$, the point
$l_{\lambda}=\inf q_{\lambda}$ is identified with ${{\text{\bsifamily
j}}}(l_{\lambda})$ in the metric space $(\mathscr{M},d)$. Furthermore,
${{\text{\bsifamily j}}}(l_{\lambda})$ is uniform in $H_{\lambda-}$. For any
$x\in q_{\lambda}$, with $\lambda\in\Lambda^{\scalebox{0.6}{$R$}}$, the
segment between $x$ and $1$ must contain the points ${{\text{\bsifamily
j}}}(l_{\lambda})$ that we may see as a projection of $q_{\lambda}$ on
$H_{\lambda-}$. With this in mind, we let $p(x)={{\text{\bsifamily
j}}}(l_{\lambda})$ if $x\in q_{\lambda}$,
$\lambda\in\Lambda^{\\!\scalebox{0.6}{$R$}}$. Then, for any point $x>1$, we
consider the sequence of successive projections defined by $p^{0}(x)=x$, and
provided that $p^{n}(x)>1$, $p^{n+1}(x)=p(p^{n}(x))$, until we eventually find
a point in $[0,1]$.

Fix now some natural number $k\geq 1$ and let
$A_{k}:=(R_{k^{3}},R_{(k+1)^{3}}]$ denote the set of points of the annulus of
fragments ${\mathscr{F}}_{\lambda}$, for $\lambda\in\Lambda_{k}$. For each
$\lambda\in\Lambda_{k}$, and $n\geq 0$, let $\lambda_{n}$ be such that
$p^{n}(l_{\lambda})\in q_{\lambda_{n}}$. Recall that
$D_{\lambda}=\sup\\{d(x,y):x,y\in q_{\lambda}\\}$. Clearly, the distance from
any point $x\in A_{k}$ to $\mathscr{M}_{k^{3}}$ is at most

$\sup_{\lambda\in\Lambda_{k}}\sup_{x\in
q_{\lambda}}d(x,\mathscr{M}_{k^{3}})\leq\sup_{\lambda\in\Lambda_{k}}\sum_{1\leq
n\leq N_{\lambda}}D_{\lambda_{n}}\,,$

where $N_{\lambda}=\inf\\{n\geq 1:p^{n}(l_{\lambda})<R_{k^{3}}\\}$. However,
since $\Lambda_{k}$ is infinite, we shall refine the analysis and rely on the
decomposition into different levels introduced in the previous section.

Recall that we say that $q_{\lambda}$ is an interval of level $i$, and write
$\lambda\in\Lambda_{k,i}$ if $m_{\lambda}\in[k^{-6-i},k^{-5-i})$; let
$A_{k,i}=\bigcup_{\lambda\in\Lambda_{k,i}}q_{\lambda}$ be the subset of
$A_{k}$ consisting of the intervals of level $i$. For $i\geq 0$, and
$\lambda\in\Lambda_{k,i}$, let $N_{i}(\lambda):=\inf\\{n\geq
1:p^{n}(l_{\lambda})\not\in A_{k,i}\\}$ be the number of hops until hitting an
interval of level lower than $i$, or exiting $A_{k}$ altogether from the left.
We then have

$\displaystyle\sup_{\lambda\in\Lambda_{k}}\sum_{1\leq n\leq
N_{\lambda}}D_{\lambda_{n}}$ $\displaystyle\leq\sum_{i\geq
0}\sup_{\lambda\in\Lambda_{k,i}}\sum_{1\leq n\leq
N_{i}(\lambda)}D_{\lambda_{n}}$ $\displaystyle\leq\sum_{i\geq
0}\sup_{\lambda\in\Lambda_{k,i}}N_{i}(\lambda)\cdot\sup_{j\geq
i}\sup_{\lambda\in\Lambda_{k,j}}D_{\lambda}$ $\displaystyle\leq\sum_{i\geq
0}\sup_{\lambda\in\Lambda_{k,i}}N_{i}(\lambda)\cdot k^{-2-i/4}\,,$ (29)

provided that the event $\bigcap_{i\geq 0}B^{c}_{k,i}$ from Lemma 5.4 occurs.
So it remains only to upper bound
$N_{k,i}=\sup\\{N_{i}(\lambda):\lambda\in\Lambda_{k,i}\\}$. We do this using
the properties of the sequence of the projections.

######  Lemma 5.9.

For any $k\geq 1$ large enough, with probability at least $1-11k^{-5/4}$, we
have for every $i\geq 0$,

$N_{k,i}:=\sup_{\lambda\in\Lambda_{k,i}}N_{\lambda}<20\,.$

###### Proof.

Let ${\mathcal{G}}$ be the sigma-algebra generated by $(X_{s})_{s\geq 0}$.
Then $(q_{\lambda})$, $\lambda\in\Lambda^{\\!\scalebox{0.6}{$R$}}$ is
${\mathcal{G}}$-measurable while, conditionally on ${\mathcal{G}}$, the random
variables $\xi(l_{\lambda})$ are independent and uniform in $H_{\lambda-}$.
Let $B_{k}$ be the event that $R_{k^{3}}<2k^{3}-1$ or
$R_{(k+1)^{3}}>2(k+1)^{3}+1$. For any $\lambda\in\Lambda_{k}$,
$(1,R_{k^{3}}]\subseteq H_{\lambda-}$; therefore, on the event $B_{k}^{c}$ for
any Borel set $A$, ${\mathbf{P}}(\xi_{\lambda}\in
A~{}|~{}B_{k}^{c},{\mathcal{G}})\leq\operatorname{Leb}(A)/k^{3}$. Furthermore,
by Lemma 5.4, on $B^{c}_{k,i}$ we have
$M_{k,i}\leq\max\\{k^{7/2-i/4},7k^{2}\\}$ and $\\#\Lambda_{k,i}\leq 7k^{8+i}$.
It follows by the union bound, that for any natural number $m\geq 1$, we have

$\displaystyle{\mathbf{P}}(N_{k,i}\geq
m~{}|~{}B^{c}_{k,i},B^{c}_{k},{\mathcal{G}})$
$\displaystyle\leq\\#\Lambda_{k,i}\cdot\sup_{\lambda\in\Lambda_{k,i}}{\mathbf{P}}(N_{\lambda}\geq
m~{}|~{}M_{k,i}\leq k^{7/2-i/4})$ $\displaystyle\leq
7k^{8+i}\cdot(k^{7/2-i/4}/k^{3})^{m}$ $\displaystyle\leq 7k^{8+i}\cdot
k^{-m(i-2)/4}\,.$

As a consequence, for $m=20$, we obtain

${\mathbf{P}}(N_{k,i}\geq 20~{}|~{}B^{c}_{k,i},B^{c}_{k},{\mathcal{G}})\leq
7k^{18-4i}\,,$

which will be good enough for $i\geq 5$. On the other hand, for $0\leq i\leq
4$, the alternative bound $M_{k,i}\leq 7k^{2}$ yields a bound of
$7k^{8+i-m}\leq 7k^{-8}$ for $m=20$. Putting everything together, we have
$N_{k,i}\geq 20$ for some $i\geq 0$ with probability at most
$10k^{-5/4}+28k^{-8}+8k^{-2}\leq 11k^{-5/4}$ for all $k$ large enough. ∎

Going back to (5.5), Lemma 5.9 implies that

$\sup_{\lambda\in\Lambda_{k}}\sum_{1\leq n\leq N_{\lambda}}D_{\lambda_{n}}\leq
20\sum_{i\geq 0}k^{-2-i/4}\leq k^{-3/2}\,$

with probability at least $1-11k^{-5/4}$ which completes the proof of
Proposition 5.3.

###  5.6 The diameter of the string of beads: Proof of Proposition 5.4

By construction, for any $\lambda\in\mathbb{R}$, the diameter of
${\mathscr{P}}_{\lambda}$ is no greater than

$\displaystyle\sum_{\lambda^{\prime}>\lambda}\operatorname{diam}(\mathscr{T}^{-}_{\lambda^{\prime}})\mathbf{1}_{{\lambda^{\prime}\in\Lambda^{\\!\scalebox{0.6}{$L$}}}}$
$\displaystyle\leq\sum_{\lambda^{\prime}>\lambda}\sqrt{m_{\lambda^{\prime}}}\cdot
D_{\lambda}^{\star}\mathbf{1}_{{\lambda^{\prime}\in\Lambda^{\\!\scalebox{0.6}{$L$}}}}\,,$
(30)

where $D^{\star}_{\lambda}$, $\lambda\in\Lambda^{\\!\scalebox{0.6}{$L$}}$, are
i.i.d. copies of the sub-Gaussian random variable whose existence is
guaranteed by Proposition 5.6 with $\epsilon=1$. We have already bounded a
similar sum in Section 3.7; in particular, the arguments there show that
almost surely

$\sum_{\lambda^{\prime}\in\Lambda^{\\!\scalebox{0.6}{$L$}}}\sqrt{m_{\lambda^{\prime}}}<\infty\,.$
(31)

Finally, consider the process $M_{\lambda}(s)$ defined for $s\geq 0$ by

$M_{\lambda}(s):=\sum_{\lambda<\lambda^{\prime}\leq\lambda+s}\sqrt{m_{\lambda^{\prime}}}\cdot(D^{\star}_{\lambda^{\prime}}-\mathbf{E}[D^{\star}_{\lambda^{\prime}}])\,.$

Conditionally on the $m_{\lambda}$, $\lambda\in\Lambda^{-}$,
$(M_{\lambda}(s))_{s\geq 0}$ is a martingale. Since $D^{\star}$ is sub-
Gaussian, $M_{\lambda}(s)$ is bounded in $L^{2}$ and thus converges almost
surely to a finite limit as $s\to\infty$. Putting this together with (31)
shows that the right-hand side of (30) and hence
$\operatorname{diam}({\mathscr{P}}_{\lambda})$ tends to zero as
$\lambda\to\infty$, which completes the proof of Proposition 5.4.

##  6 The mass measure and Hausdorff dimension of
$\operatorname{CMT}(X,{\mathbf{U}})$

In this section we prove the lower bound on the Hausdorff dimension of
$\mathscr{M}$. We use the mass distribution principle using the mass measure
$\mu$ that is defined in Section 6.1. The asymptotics for the $\mu$-mass of
small balls are provided in Section 6.2 and relies heavily on the growth
process defined in Section 5.

###  6.1 The mass measure

We start with the construction of the mass measure $\mu$ on $\mathscr{M}$. The
measures in this section will always be seen as Borel measures on
$\mathscr{M}$, the completion of $\mathbb{R}_{+}$ with respect to $d$. For
$t\in\mathbb{R}_{+}$ we let $\mu_{t}$ be the rescaled Lebesgue measure on
$[0,t]$: $\mu_{t}(A)=t^{-1}\operatorname{Leb}(A\cap[0,t])$. For each $t$,
$\mu_{t}$ is a probability measure on $\mathscr{M}$ which charges only a
subtree containing the root $0$ (it is easy to see that $[0,t]$ is connected).
Let ${\mathcal{L}}$ be the set of leaves of $\mathscr{M}$, that is the set of
points $x$ such that $\mathscr{M}\setminus\\{x\\}$ is connected. Our aim in
this section is the following

######  Proposition 6.1.

With probability one, as $t\to\infty$, $\mu_{t}$ converges weakly to a limit
probability measure that we denote by $\mu$ and call the mass measure on
$\mathscr{M}$. Furthermore $\mu(\mathbb{R}_{+})=0$ so $\mu({\mathcal{L}})=1$.

Recall the notation in Section 5.1. For a subset $S$ of the tree $\mathscr{M}$
and $x\in\mathbb{R}_{+}$, we define the projection of $x$ onto $S$ as the
point of $S$ that is closest to $x$. Fix $\lambda_{0}\in\mathbb{R}$. We are
interested in the projection onto the subset of $\mathscr{M}$ consisting of
the points $[0,R_{\lambda_{0}}]$. For $x\in\mathscr{M}$, we let
$[x]_{\lambda_{0}}$ denote the corresponding point. Observe that, with the
notation of the previous section, a.s.
$[x]_{\lambda_{0}}=\sup\\{p^{n}(x)\cap[0,R_{\lambda_{0}}],n\geq 0\\}$.

Even though $[0,R_{\lambda_{0}}]$ is not closed in $\mathscr{M}$, we will
always have $[x]_{\lambda_{0}}\in[0,R_{\lambda_{0}})$ for the points $x$ we
consider. Rather than working with the measures $\mu_{t}$,
$t\in\mathbb{R}_{+}$, it will be more convenient to work with
$\bar{\mu}_{\lambda}:=\mu_{R_{\lambda}}$ for $\lambda\in\mathbb{R}$; Lemma 5.2
which says that $R_{\lambda}\to\infty$ guarantees that taking the limits as
$t\to\infty$ or $\lambda\to\infty$ is equivalent.

We define the following process: for a Borel set
$S\subseteq[0,R_{\lambda_{0}}]$ and $\lambda\geq\lambda_{0}$,

$M_{\lambda}=M_{\lambda}(S):=\bar{\mu}_{\lambda}(\\{x\leq
R_{\lambda}:[x]_{\lambda_{0}}\in S\\})\,.$ (32)

We will consider only the randomness coming from ${\mathbf{U}}$ and study
$M_{\lambda}$ conditionally on $\sigma(R_{\lambda}:\lambda\geq\lambda_{0})$.
We let ${\mathbf{P}}^{\shortdownarrow}$ and ${\mathbf{E}}^{\shortdownarrow}$
be the corresponding probability and expectation.

######  Lemma 6.2.

The process $(M_{\lambda},\lambda\geq\lambda_{0})$ is almost surely a
supermartingale under ${\mathbf{P}}^{\shortdownarrow}$.

###### Proof.

For each $\lambda\in\Lambda^{\\!\scalebox{0.6}{$R$}}$, $\lambda>\lambda_{0}$,
all the points $x\in q_{\lambda}$ have the same projection on
$[0,R_{\lambda_{0}}]$ since $p(x)=\xi(l_{\lambda})$. Furthermore, in order to
determine where an interval $q_{\lambda}$ projects onto $H_{\lambda_{0}}$ it
suffices to follow the sequence of random projections/jumps $p^{n}(x)$, $n\geq
1$. Almost surely, $\inf\\{n\geq 0:p^{n}(l_{\lambda})\in
H_{\lambda_{0}}\\}<\infty$ and the point $[l_{\lambda}]_{\lambda_{0}}$ is
uniformly random in $H_{\lambda_{0}}$. In the following, $\lambda_{0}$ being
fixed, we use $\phi(\lambda)$ as a short-hand for
$[l_{\lambda}]_{\lambda_{0}}$. The points $\phi(\lambda)$,
$\lambda>\lambda_{0}$, are of course not independent because of the
coalescence of the trajectories. Then,

$M_{\lambda}=\frac{1}{R_{\lambda}}\left[{M_{\lambda_{0}}\cdot
R_{\lambda_{0}}}+\sum_{\lambda_{0}<\lambda^{\prime}\leq\lambda}m_{\lambda^{\prime}}\mathbf{1}_{{\phi(\lambda^{\prime})\in
S}}\right].$ (33)

It follows that, writing $\mathcal{F}_{\lambda}$ for the sigma-algebra
generated by $({{\text{\bsifamily
j}}}(l_{\lambda^{\prime}}):\lambda^{\prime}\leq\lambda)$, the random variable
$M_{\lambda}$ is bounded and $\mathcal{F}_{\lambda}$-measurable. The
expression in (33) is amenable to a simple evaluation of the conditional
expectations: for $h\geq 0$,

$\displaystyle{\mathbf{E}}^{\shortdownarrow}[M_{\lambda+h}~{}|~{}\mathcal{F}_{\lambda}]$
$\displaystyle=\frac{1}{R_{\lambda+h}}\cdot{\mathbf{E}}^{\shortdownarrow}\Bigg{[}M_{\lambda_{0}}\cdot
R_{\lambda_{0}}+\sum_{\lambda_{0}<\lambda^{\prime}\leq\lambda+h}m_{\lambda^{\prime}}\mathbf{1}_{{\phi(\lambda^{\prime})\in
S}}~{}\Bigg{|}~{}\mathcal{F}_{\lambda}\Bigg{]}$
$\displaystyle=\frac{1}{R_{\lambda+h}}\Bigg{(}R_{\lambda}\cdot
M_{\lambda}+\sum_{\lambda<\lambda^{\prime}\leq\lambda+h}m_{\lambda^{\prime}}{\mathbf{P}}^{\shortdownarrow}(\phi(\lambda^{\prime})\in
S~{}|~{}\mathcal{F}_{\lambda})\Bigg{)}\,.$

However, almost surely conditionaly on $\mathcal{F}_{\lambda}$,
$\phi(\lambda^{\prime})\in S$ if and only if the first point of the sequence
$(p^{i}(l_{\lambda^{\prime}}))_{i\geq 1}$ that falls in $H_{\lambda}$ lies in
some interval $q_{\lambda^{\prime\prime}}$ itself such that
$\phi(\lambda^{\prime\prime})\in S$. By definition, such a point is the
projection of $l_{\lambda^{\prime}}$ on $H_{\lambda}$, and is uniform in
$H_{\lambda}\subset[0,R_{\lambda}]$ and therefore
${\mathbf{P}}^{\shortdownarrow}(\phi(\lambda^{\prime})\in
S~{}|~{}\mathcal{F}_{\lambda})\leq M_{\lambda}$. It follows that

$\displaystyle{\mathbf{E}}^{\shortdownarrow}[M_{\lambda+h}~{}|~{}\mathcal{F}_{\lambda}]$
$\displaystyle\leq\frac{M_{\lambda}}{R_{\lambda+h}}\cdot\Bigg{(}R_{\lambda}+\sum_{\lambda<\lambda^{\prime}\leq\lambda+h}m_{\lambda^{\prime}}\Bigg{)}=M_{\lambda}\,,$

which completes the proof. ∎

###### Proof of Proposition 6.1.

Since $\mathscr{M}$ is compact by Proposition 5.5, the collection of measures
$(\mu_{t})_{t>0}$ is tight. We prove that it is Cauchy for the Prohorov metric
using the super-martingales $M_{\lambda}$ we have just introduced. Recall
that, for two Borel measures $\nu$ and $\nu^{\prime}$ on $(\mathscr{M},d)$,
the Prohorov distance is given by

$\operatorname{d}_{\text{{p}}}(\nu,\nu^{\prime})=\inf\\{\epsilon>0:\nu(A)\leq\nu^{\prime}(A^{\epsilon})+\epsilon,\nu^{\prime}(A)\leq\nu(A^{\epsilon})+\epsilon\text{
for all Borel sets }A\\}\,$

where $A^{\epsilon}=\\{x:d(x,A)<\epsilon\\}$.

The arguments for compactness in Section 5 show that for any $\epsilon>0$,
there exists a $\lambda_{0}\in\mathbb{R}_{+}$ such that
$\sup_{x}d(x,[0,R_{\lambda}])<\epsilon$ for all $\lambda\geq\lambda_{0}$. With
this choice for $\lambda_{0}$, it follows that, for any $\lambda>\lambda_{0}$,

$\operatorname{d}_{\text{{p}}}(\mu_{\lambda},[\mu_{\lambda}]_{\lambda_{0}})<\epsilon\,,$

where $[\mu_{\lambda}]_{\lambda_{0}}$ denotes the image of $\mu_{\lambda}$ by
the projection onto $[0,R_{\lambda_{0}}]$. Therefore, for any
$\lambda,\lambda^{\prime}\geq\lambda_{0}$,

$\operatorname{d}_{\text{{p}}}(\mu_{\lambda},\mu_{\lambda^{\prime}})\leq\operatorname{d}_{\text{{p}}}([\mu_{\lambda}]_{\lambda_{0}},[\mu_{\lambda^{\prime}}]_{\lambda_{0}})+2\epsilon\,.$

To complete the proof, cover $[0,R_{\lambda_{0}}]$ with finitely many balls of
diameter $\epsilon$, say $B_{1},B_{2},\dots,B_{k}$. Then, by definition of
$M_{\lambda}(S)$, we can construct a coupling $(X,Y)$ with
$X\sim[\mu_{\lambda}]_{\lambda_{0}}$ and
$Y\sim[\mu_{\lambda^{\prime}}]_{\lambda_{0}}$ such that
$(X,Y)\not\in\cup_{i}B_{i}\times B_{i}$ with probability at most
$\sum_{i}|M_{\lambda}(B_{i})-M_{\lambda^{\prime}}(B_{i})|$. Since the diameter
of $B_{i}$ is at most $\epsilon$, the cost of the coupling on $B_{i}\times
B_{i}$ is at most $\epsilon$. It follows that

$\operatorname{d}_{\text{{p}}}([\mu_{\lambda}]_{\lambda_{0}},[\mu_{\lambda^{\prime}}]_{\lambda_{0}})\leq\sum_{i=1}^{k}|M_{\lambda}(B_{i})-M_{\lambda^{\prime}}(B_{i})|+\epsilon\,,$

which is at most $2\epsilon$ for all $\lambda,\lambda^{\prime}$ large enough
because of the convergence of the mass super-martingales of Lemma 6.2. This
completes the proof of convergence.

The two additional properties are straightforward from the definition. First,
for any interval $[i,i+1)$,
$\mu([i,i+1))=\lim_{\lambda}\mu_{\lambda}([i,i+1])=0$ since
$R_{\lambda}\to\infty$, and thus $\mu(\mathbb{R}_{+})=0$. The completion of
$\mathbb{R}_{+}$ with respect to $d$ only adds leaves, so
$\mathscr{M}\setminus\mathbb{R}_{+}\subseteq{\mathcal{L}}$, and therefore
$\mu({\mathcal{L}})=1$. ∎

###  6.2 The mass of balls around zero

It is proved in [4] that the Minkowski dimension of $\mathscr{M}$ is almost
surely equal to 3, and we thus only need to find a lower bound. For this, we
aim at using the mass distribution principle with the mass measure. In this
direction, one needs to upper bound the $\mu$-mass of balls centered at points
with distribution $\mu$. In general, this might be delicate since we need to
identify the balls around these points, and they are almost surely not in
$\mathbb{R}_{+}$ (Proposition 6.1). This is why the following result is
crucial; the intuition should be intuitively clear from the discrete setting,
where the point $1$ can be replaced in Prim’s algorithm by a uniformly random
point in $[n]$ without altering the distributions. For a point
$x\in\mathscr{M}$ and $r>0$ we let $B_{x}(r)$ denote the open ball of radius
$r$ centered at $x$ (for the metric $d$).

######  Lemma 6.3.

Let $\zeta$ be a point of $\mathscr{M}$ with distribution $\mu$. Then, the
processes $(\mu(B_{0}(r)))_{r\geq 0}$ has the same distribution as
$(\mu(B_{\zeta}(r)))_{r\geq 0}$.

###### Proof.

For $n\geq 1$, recall from Section 1.2 that $v_{1},v_{2},\dots,v_{n}$ denote
the Prim order on $[n]$ on the complete graph with edge weights $(w_{e})$,
$e\in E^{n}$; in what follows, we will occasionally write $v(i)$ instead of
$v_{i}$. For $n\geq 1$, let $V^{n}_{\lambda}$ be the collection of vertices
connected to $v(\lfloor n^{2/3}\rfloor)$ in the random graph with edge weights
at most $p_{n}(\lambda)$, and let $H^{n}_{\lambda}$ denote the collection of
their Prim ranks, and let $L^{n}_{\lambda}=\min H^{n}_{\lambda}$. Let
$\hat{\mu}^{n}_{\lambda}$ denote the uniform probability distribution on
$H^{n}_{\lambda}$. By Lemma 7.4, conditionally on $V^{n}_{\lambda}$, the
vertex $v(L^{n}_{\lambda})$ is uniformly random in $V^{n}_{\lambda}$, and
independent of the random variables $w_{e}$ associated to the edges with end
points in $V^{n}_{\lambda}$. It follows in particular that, for any $r\geq 0$,

$\displaystyle\mu^{n}_{\lambda}(\\{u\in
H^{n}_{\lambda}:d^{n}_{\lambda}(L^{n}_{\lambda},u)\leq
rn^{1/3}\\})\stackrel{{\scriptstyle d}}{{=}}\mu^{n}_{\lambda}(\\{u\in
H^{n}_{\lambda}:d^{n}_{\lambda}(\zeta^{n}_{\lambda},u)\leq rn^{1/3}\\})\,,$
(34)

where $\zeta^{n}_{\lambda}$ denotes an independent point with distribution
$\hat{\mu}^{n}_{\lambda}$ (uniform in $H^{n}_{\lambda}$). By Proposition 7.13,
$(H^{n}_{\lambda},d^{n}_{\lambda},\mu^{n}_{\lambda},L^{n}_{\lambda},\zeta^{n}_{\lambda})$
converges in distribution in the sense of Gromov–Prokhorov to
$(H_{\lambda},d,\hat{\mu}_{\lambda},L_{\lambda},\zeta_{\lambda})$, where
$\zeta_{\lambda}$ is an independent point with distribution $\mu_{\lambda}$.
This implies the convergence of the random variables in (34) as $n\to\infty$
towards

$\hat{\mu}_{\lambda}(\\{u\in H_{\lambda}:d(L_{\lambda},u)\leq
r\\})\stackrel{{\scriptstyle d}}{{=}}\hat{\mu}_{\lambda}(\\{u\in
H_{\lambda}:d(\zeta_{\lambda},u)\leq r\\})\,.$

Now, $d(0,L_{\lambda})\leq\operatorname{diam}({\mathscr{P}}_{\lambda})\to 0$
as $\lambda\to\infty$ by Proposition 5.4. Note also that a straightforward
coupling yields
$\operatorname{d}_{\text{{p}}}(\hat{\mu}_{\lambda},\mu_{\lambda})\leq
L_{\lambda}/R_{\lambda}\leq L_{\lambda}$ (independently of the metric since we
can match the points exactly on a set of probability
$1-L_{\lambda}/R_{\lambda}$). Taking the limit as $\lambda\to\infty$, Lemma
5.2 and Proposition 6.1 yield the claim for every fixed $r\geq 0$. This is
easily extended to the joint convergence for finitely many values
$r_{1}<r_{2}<\dots<r_{k}$, which completes the proof. ∎

In order to upper bound $\mu(B_{0}(r))$ we will proceed in two steps: we will
first upper bound $\bar{\mu}_{\lambda}(B_{0}(r))$ showing that it is of the
correct order of magnitude, that is roughly $r^{3}$ for some well-chosen
$\lambda$ depending on $r$ ($\lambda$ of order $1/r$); we will then rely on
the concentration for the mass supermartingales of the previous section, which
controls the evolution of the mass as $\lambda$ increases, to show that
$\mu(B_{0}(r))=\lim_{\lambda}\bar{\mu}_{\lambda}(B_{0}(r))$ remains of order
$r^{3}$. Once we have the relevant upper bound for a fixed $r$, the proof is
easily completed using routine arguments (taking a suitable subsequence and
the Borel–Cantelli lemma).

######  Proposition 6.4.

There exists a constant $c>0$ such that, for any $\epsilon\in(0,1)$ and every
$r>0$ small enough,

${\mathbf{P}}(\bar{\mu}_{r^{\epsilon-1}}(B_{0}(r))>r^{3-\epsilon})\leq
r^{c\epsilon}\,.$

###### Proof.

Observe that if $d(0,H_{\lambda})>r$ then no point of $H_{\lambda}$ lies
within $B_{0}(r)$, so that $\bar{\mu}_{\lambda}(B_{0}(r))$ is at most
$L_{\lambda}/R_{\lambda}$. Using this with $\lambda=r^{\epsilon-1}$, it
follows that

$\displaystyle{\mathbf{P}}(\bar{\mu}_{r^{\epsilon-1}}(B_{0}(r))>r^{3-6\epsilon})$
$\displaystyle\leq{\mathbf{P}}\left(\bar{\mu}_{r^{\epsilon-1}}(B_{0}(r))>\frac{L_{r^{\epsilon-1}}}{R_{r^{\epsilon-1}}}\right)+{\mathbf{P}}\left(\frac{L_{r^{\epsilon-1}}}{R_{r^{\epsilon-1}}}>r^{3-6\epsilon}\right)$
$\displaystyle\leq{\mathbf{P}}(d(0,H_{r^{\epsilon-1}})\leq
r)+{\mathbf{P}}(L_{r^{\epsilon-1}}>r^{2-4\epsilon})+{\mathbf{P}}(R_{r^{\epsilon-1}}\leq
r^{-1+2\epsilon})$ $\displaystyle\leq{\mathbf{P}}(d(0,H_{r^{\epsilon-1}})\leq
r)+\exp(-r^{-\epsilon})\,,$ (35)

where the last line follows, for all $r>0$ small enough, from the bounds in
Lemma 5.2.

Most of the work now consists in bounding the first term in (6.2) above.
Observe that the geodesic from $H_{r^{\epsilon-1}}$ to $0$ must cross every
single one of the metric spaces induced by $\mathscr{M}$ on the intervals
$q_{\lambda}\subseteq[0,1]$ with $\lambda>r^{\epsilon-1}$. Furthermore, with
the notation of Section 5.3, $d(0,H_{r^{\epsilon-1}})$ decomposes as follows:
since here the portion of path in $q_{\lambda}$ is precisely between
${{\text{\bsifamily j}}}(\sup q_{\lambda})\in q_{\lambda}$ a.s. and $\inf
q_{\lambda}$; we have

$\displaystyle
d(0,H_{r^{\epsilon-1}})=\sum_{\lambda\in\Lambda^{\\!\scalebox{0.6}{$L$}}}Y_{\lambda}\mathbf{1}_{{\lambda>r^{\epsilon-1}}}\geq_{st}\sum_{\lambda\in\Lambda^{\\!\scalebox{0.6}{$L$}}}Y^{\star}_{\lambda}\cdot
m_{\lambda}^{1/2}\cdot\mathbf{1}_{{\lambda>r^{\epsilon-1}}}\,,$

where the last inequality is a stochastic minoration that relies on
Proposition 5.6: the $m_{\lambda}$,
$\lambda\in\Lambda^{\\!\scalebox{0.6}{$L$}}$, are the sizes of the jumps of
$L_{\lambda}$, and the $Y^{\star}_{\lambda}$ are conditionally independent
from the entire collection
$(m_{\lambda}:\lambda\in\Lambda^{\\!\scalebox{0.6}{$L$}})$. In order to lower
bound $d(0,H_{r^{\epsilon-1}})$ it suffices to focus on a single term of the
sum in the right-hand side: if any of those terms is greater than $r$, then
$d(0,H_{r^{\epsilon-1}})>r$ as well, thus

$\displaystyle{\mathbf{P}}(d(0,H_{r^{\epsilon-1}})\leq r)$
$\displaystyle\leq{\mathbf{P}}\left(\\#\\{\lambda>r^{1-\epsilon}:\lambda\in\Lambda^{\\!\scalebox{0.6}{$L$}},Y^{\star}_{\lambda}m_{\lambda}^{1/2}>r\\}=0\right)$
$\displaystyle\leq{\mathbf{P}}\left(\sup\\{m_{\lambda}:\lambda\in\Lambda^{\\!\scalebox{0.6}{$L$}},\lambda>r^{1-\epsilon}\\}\leq
r^{2-\epsilon}\right)+{\mathbf{P}}(Y^{\star}\leq r^{\epsilon/2})\,.$ (36)

The second term is at most $r^{c\epsilon}$ by Proposition 5.6 _ii)_ and
Markov’s inequality. The $(m_{\lambda})$ are also the lengths of the faces of
the convex minorant of $X$ on the interval $[0,1]$ by Lemma 4.2, and to deal
with the first term, we relate it to the convex minorant of a standard
Brownian motion $W$.

Let $(Q_{t})_{t\geq 0}$ be defined by

$\displaystyle Q_{t}$
$\displaystyle=\exp\left(\int_{0}^{t}sdW_{s}-\frac{1}{2}\int_{0}^{t}s^{2}ds\right)\,,$

Then by the Cameron–Martin–Girsanov formula (Theorem 38.5 of [61]), the laws
of $X$ and $W$ are related by a change of measure whose density is given by
the martingale $Q_{t}$. For a function
$\omega\in\mathcal{C}(\mathbb{R}_{+},\mathbb{R})$, we consider the convex
minorant of $\omega$ on $[0,1]$ and we let $\chi_{r}(\omega)$ denote the
indicator that the longest face with slope (strictly) smaller than
$-r^{\epsilon-1}$ has length at most $r^{2-\epsilon/2}$. Then, by Lemma 4.2 we
have

$\displaystyle{\mathbf{P}}\left(\sup\\{m_{\lambda}:\lambda\in\Lambda^{\\!\scalebox{0.6}{$L$}},\lambda>r^{1-\epsilon}\\}\leq
r^{2-\epsilon}\right)$ $\displaystyle=\mathbf{E}[\chi_{r}(X^{0})]$
$\displaystyle=\mathbf{E}[\chi_{r}(W)\cdot Q_{1}]$
$\displaystyle\leq\mathbf{E}[\chi_{r}(W)^{2}]^{1/2}\cdot\mathbf{E}[Q_{1}^{2}]^{1/2}\,,$
(37)

by the Cauchy–Schwarz inequality. Observe that in the right-hand side above,
$\mathbf{E}[Q_{1}^{2}]\leq\mathbf{E}[\exp(2\overline{W}_{1})]$ is finite and
independent of $r$ thanks to the Gaussian tails of $\overline{W}_{1}$.

The first factor in (6.2) can be estimated using the results of Pitman and
Ross [58, Theorem 1] and Brownian scaling. Let $(x_{i},s_{i})_{i}$ be the
points of a Poisson point process with intensity $(2\pi
x)^{-1/2}\cdot\exp(-(2+s^{2})x/2)dxds$ on $\mathbb{R}_{+}\times\mathbb{R}$.
Then $(x_{i},s_{i})$ are the lengths and slopes of the faces of the convex
minorant of $W$ on the interval $[0,E]$ where $E$ is an independent
exponential random variable with mean one, and here $E=\sum x_{j}$. Therefore,
by Brownian scaling,

$\displaystyle\mathbf{E}[\chi_{r}(W)]$
$\displaystyle={\mathbf{P}}\left(\sup\\{x_{i}:s_{i}\sqrt{E}<-r^{\epsilon-1}\\}\leq
E\cdot r^{2-\epsilon}\right)$
$\displaystyle\leq{\mathbf{P}}\left(A^{c}\right)+{\mathbf{P}}\left(\sup\\{x_{i}:s_{i}r^{1-5\epsilon/6}<-1\\}\leq
r^{2-4\epsilon/3}\right)\,,$

where $A$ denotes the event that $E=\sum
x_{j}\in[r^{\epsilon/3},r^{-\epsilon/3}]$. Since $E$ is exponential with mean
one, we have ${\mathbf{P}}(A^{c})\leq 2r^{\epsilon/3}$ for all $r>0$ small
enough. We claim that there exists a constant $c>0$ such that the second term
above is no larger than $r^{c\epsilon}$, and in order the complete the proof,
it suffices to justify that claim. We slightly change the scaling and write
$\delta=r^{1-5\epsilon/6}$ and $\gamma=\epsilon/3$ to lighten the notation.
Then $r^{2-4\epsilon/3}\leq\delta^{2+\gamma}$, and we focus on

$\displaystyle{\mathbf{P}}\left(\sup\\{x_{i}:s_{i}\delta<-1\\}\leq\delta^{2+\gamma}\right)$
$\displaystyle=\exp\left(-\int\\!\\!\\!\\!\int\frac{e^{-(2+s^{2})x/2}}{\sqrt{2\pi
x}}\mathbf{1}_{{x\geq\delta^{2+\gamma},s\delta<-1}}dsdx\right)\,.$ (38)

We just need to lower bound the integral in the right-hand side: consider the
subregion $\Sigma$ of $[\delta^{2+\gamma},\infty)\times(-\infty,-1/\delta]$
where $u=xs^{2}\leq 1$, that is $\Sigma:=\\{(x,s):\delta^{2+\gamma}\leq
x\leq\delta^{2},-x^{-1/2}\leq s\leq-1/\delta\\}$:

$\displaystyle\int\\!\\!\\!\\!\int\frac{e^{-(2+s^{2})x/2}}{\sqrt{2\pi
x}}\mathbf{1}_{{x\geq\delta^{2+\gamma},s\delta<-1}}dsdx$
$\displaystyle\geq\int\\!\\!\\!\\!\int_{\Sigma}\frac{e^{-(s^{2}+2)x/2}}{\sqrt{2\pi
x}}dsdx$ $\displaystyle\geq
e^{-\delta^{2}}\int\\!\\!\\!\\!\int_{\Sigma}\frac{e^{-s^{2}x/2}}{\sqrt{2\pi
x}}dsdx$
$\displaystyle\geq\frac{e^{-\delta^{2}-1/2}}{\sqrt{2\pi}}\int_{\delta^{2+\gamma}}^{\delta^{2}}\left[\frac{1}{x}-\frac{1}{\delta\sqrt{x}}\right]dx$
$\displaystyle\geq\frac{-\gamma\log\delta
e^{-\delta^{2}-1/2}}{\sqrt{2\pi}}\,,$

for all $\delta>0$ small enough. It follows easily that, there exists a
constant $c>0$ such that for all $\delta>0$ small enough the right-hand side
of (38) is at most $\delta^{c\gamma}$, which translated into the original
parameters yields a bound of $r^{c^{\prime}\epsilon}$ for the right-hand side
of (6.2), and in turn for (6.2) and (6.2). This completes the proof. ∎

Finally, using Lemma 6.3, the following proposition completes the proof of
Theorem 1.2

######  Proposition 6.5.

Let $\zeta$ be a point of $\mathscr{M}$ with distribution $\mu$. Then for
every $\epsilon\in(0,1)$, almost surely, for all $r>0$ small enough we have

$\mu(B_{\zeta}(r))\leq r^{3-\epsilon}\,.$

As a consequence $\operatorname{\dim_{\text{{h}}}}(\mathscr{M})\geq 3$.

###### Proof.

By Lemma 6.3, it suffices to prove the bound for $\mu(B_{0}(r))$. Fix
$\lambda_{0}=1/r$, set $S=B_{0}(r)$ and recall the process
$M_{\lambda}=\bar{\mu}_{\lambda}(\\{x\leq
R_{\lambda}:d(0,[x]_{\lambda_{0}})\leq r\\})$ of (32). Then, for any
$\lambda\geq\lambda_{0}$, $\bar{\mu}_{\lambda}(B_{0}(r))$ is no larger than
the $\bar{\mu}_{\lambda}$-mass of the excursions which are grafted within
distance $r$ or the origin: we have

$\bar{\mu}_{\lambda}(B_{0}(r))\leq M_{\lambda}\,.$

Since $M_{\lambda}$ is bounded, Lemma 6.2 implies that $M_{\lambda}$ converges
almost surely as $\lambda\to\infty$, but it also implies some concentration
results since the increments of $M_{\lambda}$ are bounded by the
$m_{\lambda}$, $\lambda\in\Lambda^{\\!\scalebox{0.6}{$R$}}$. By the
Azuma–Hoeffding inequality [19, 39, 23], for any $x>0$, we have, conditionally
on $(m_{\lambda},\lambda\in\Lambda)$,

$\displaystyle{\mathbf{P}}\left(M_{\lambda}-M_{\lambda_{0}}>x~{}|~{}m_{\lambda},\lambda>\lambda_{0}\right)$
$\displaystyle\leq\exp\bigg{(}-\frac{x^{2}}{2\sum_{\lambda_{0}<\lambda^{\prime}}m_{\lambda^{\prime}}^{2}}\bigg{)}\,.$
(39)

Bounding the $\ell^{2}$-norm of $(m_{\lambda})_{\lambda>\lambda_{0}}$ is
routine using Lemma 5.8. Indeed, for any $k\geq 1$,

$\displaystyle\mathbf{E}\left[\sum_{\lambda}m_{\lambda}^{2}\cdot\mathbf{1}_{{\lambda\in\Lambda_{k}}}\right]$
$\displaystyle\leq 8k^{5}\int_{0}^{\infty}\frac{t^{2}}{\sqrt{2\pi
t^{3}}}e^{-k^{6}t/8}dt=64\cdot k^{-4}\,,$

and Markov’s inequality then yields, for some constant $K$,

${\mathbf{P}}\left(\sum_{\lambda:\lambda r>1}m_{\lambda}^{2}\geq
r^{3-\epsilon}\right)\leq
r^{\epsilon-3}\cdot\mathbf{E}\left[\sum_{\lambda}m_{\lambda}^{2}\cdot\mathbf{1}_{{\lambda>1/r}}\right]\leq
Kr^{\epsilon}\,.$ (40)

Since $M_{\lambda_{0}}=\bar{\mu}_{\lambda_{0}}(B_{0}(r))$, it follows from
Proposition 6.4 and (39)–(40) that

$\displaystyle{\mathbf{P}}(\mu(B_{0}(r)\geq 2r^{3-\epsilon}))$
$\displaystyle\leq{\mathbf{P}}(\mu_{\lambda_{0}}(B_{0}(r))\geq
r^{3-\epsilon})+{\mathbf{P}}(M_{\infty}-M_{\lambda_{0}}\geq r^{3-\epsilon})$
$\displaystyle\leq r^{c\epsilon}+\exp(-cr^{-\epsilon})+Kr^{\epsilon}\,.$

From there, completing the proof is standard: take a subsequence
$r_{i}=2^{-i}$, $i\geq 1$; the Borel–Cantelli implies that for all but
finitely many values of $i\geq 1$, we have $\mu(B_{0}(r_{i}))\leq
2r_{i}^{3-\epsilon}$, and thus $\mu(B_{0}(r))\leq 16r^{3-\epsilon}$ for all
$r>0$ small enough. As a consequence, the mass distribution principle (see,
e.g., Proposition 4.9 of [34]) implies that
$\operatorname{\dim_{\text{{h}}}}(\mathscr{M})\geq 3-\epsilon$, which
completes the proof since $\epsilon>0$ was arbitrary. ∎

##  7 Distances in the Brownian parabolic tree

All the proofs of convergence will be based on couplings with discrete
objects. It would be possible to identify the distribution of
$\operatorname{CMT}(X,{\mathbf{U}})$ as that of the scaling limit of the
minimum spanning tree constructed in [4] directly in the continuum using the
dynamics as $\lambda$ evolves and the tools developed in [6]. However, since
we need comparisons with discrete objects anyway for Theorems 1.3 and 1.4, we
do not pursue this here. All the limit theorems essentially boil down to
proving that, in a suitable coupling, and for every $\lambda\in\mathbb{R}$,
the restriction of the metric space $\operatorname{CMT}(X,{\mathbf{U}})$ to
any interval $(a,b)$ of $\mathbb{R}_{+}\setminus Z^{\lambda}$ is the limit (in
probability) of the minimum spanning tree of a connected component induced by
a vertex set whose node have Prim ranks in an interval
$\\{a_{n},a_{n}+1,\dots,b_{n}-1\\}$ where $a_{n}\sim an^{2/3}$ and $b_{n}\sim
bn^{2/3}$. Our coupling will be “global” in the sense that it allows a
transparent application to any collection of times
$\lambda_{1}<\lambda_{2}<\dots<\lambda_{k}$ and any finite collection of
intervals at these times.

###  7.1 Discrete preliminaries

In this section, we provide the discrete representation that we will use to
prove our limit theorems. They all heavily rely on the Prim order introduced
in [24] and its properties. We will in particular give a representation of the
minimum spanning tree $M_{n}$, and of the random graph $G(n,p)$ that we will
see as the union of a portion of the minimum spanning tree, the Kruskal forest
denoted by $K(n,p)$, together with additional cyclic edges.

Recall the Prim algorithm and the Prim order $v_{1},v_{2},\dots,v_{n}$
discussed in Section 1.3. Recall also that $V_{k}=\\{v_{1},\dots,v_{k}\\}$.
For $k\in[n]$, let $N_{k}^{n,p}$ be the number of nodes in $[n]\setminus
V_{k}$ which have a neighbour in $V_{k}$ in the graph $G(n,p)$ whose edge set
is $\\{e:w_{e}\leq p\\}$. For $\lambda\in\mathbb{R}$, set
$p_{n}(\lambda)=1/n+\lambda n^{-4/3}$. Then define, for $t\geq 0$,

$X^{n,\lambda}_{t}:=n^{-1/3}\left(N_{\lfloor
tn^{2/3}\rfloor}^{n,p_{n}(\lambda)}-\\#\\{i\leq
tn^{2/3}:N_{i}^{n,p_{n}(\lambda)}=0\\}\right)\,.$ (41)

Let $Z^{n,\lambda}$ be $n^{2/3}$ times the collection of instants when
$X^{n,\lambda}$ reaches a new minimum. Then, the points of $Z^{n,\lambda}$ are
the Prim ranks of the first vertices of the connected components of
$G(n,p_{n}(\lambda))$. Furthermore, recalling the definitions in Section 1.3,
the collection of the edges of the minimum spanning tree $M_{n}$ are precisely
$\\{e_{i}=(u_{i},v_{i}):2\leq i\leq n\\}$. The identities of the nodes can of
course not be recovered from $(X^{n,\lambda})_{\lambda\in\mathbb{R}}$, but one
may use the Prim ranks to construct a graph on $[n]$ that is isomorphic to
$M_{n}$ using $(X^{n,\lambda})_{\lambda\in\mathbb{R}}$ only. However, the
information about the location of the $u_{i}$ vanishes in the limit, and we
shall construct a graph that has the correct distribution of the left-end
points $u_{i}$, conditionally on $(Z^{n,\lambda})_{\lambda\in\mathbb{R}}$.

We start with an encoding of the merges. Note that there are precisely $n-1$
jumps to the process $(Z^{n,\lambda})_{\lambda\in\mathbb{R}}$, each one
corresponding to the appearance of one of the edges $e_{i}=(u_{i},v_{i})$ for
some $2\leq i\leq n$. Let ${{\text{\bsifamily l}}}_{n}(i)$ and
${{\text{\bsifamily r}}}_{n}(i)$ be respectively the Prim ranks of the left-
most and right-most vertices of the connected component of $v_{i}$ at time
$w_{e_{i}}$; let also ${{\text{\bsifamily s}}}_{n}(i)=(1-nw_{e_{i}})n^{1/3}$
be the discrete analog of the slope of a point $t\in\mathscr{L}$ in the
continuous setting. Then, the set

$\operatorname{Merge}_{n}(X^{n}):=\\{(n^{-2/3}{{\text{\bsifamily
l}}}_{n}(i),n^{-2/3}i,n^{-2/3}{{\text{\bsifamily
r}}}_{n}(i),-{{\text{\bsifamily s}}}_{n}(i)):2\leq i\leq n\\}$ (42)

contains all the information about the merges of connected components. We can
rephrase the fact that the extremities $u_{i}$ of the edges
$e_{i}=(u_{i},v_{i})$ are uniform in the connected component containing
$v_{i-1}$ as follows. Let $(U_{i})_{i\geq 1}$ be i.i.d. uniform on $[0,1]$,
also independent of $\operatorname{Merge}(X^{n})$. For each $i$, let
${{\text{\bsifamily j}}}_{n}(i)={{\text{\bsifamily l}}}_{n}(i)+\lfloor
U_{i}(i-{{\text{\bsifamily l}}}_{n}(i))\rfloor$. Then, ${{\text{\bsifamily
j}}}_{n}(i)$ is uniform in $\\{{{\text{\bsifamily
l}}}_{n}(i),{{\text{\bsifamily l}}}_{n}(i)+1,\dots,i-1\\}$. The following
lemma is a simple reformulation of Lemma 1.6.

######  Proposition 7.1 (A representation of the minimum spanning forest).

Conditionally on $\operatorname{Merge}_{n}(X^{n})$, the collection of Prim
ranks of the nodes $(u_{i}:2\leq i\leq n)$ has the same distribution as
$({{\text{\bsifamily j}}}_{n}(i):2\leq i\leq n)$. In particular, up to a
relabelling of the nodes of $M_{n}$ using the Prim ranks:

  1. i)

the graph on $[n]$ with edges $\\{{{\text{\bsifamily j}}}_{n}(i),i\\}$, $2\leq
i\leq n$, is distributed like $M_{n}$;

  2. ii)

the graph on $[n]$ with edges $\\{{{\text{\bsifamily j}}}_{n}(i),i\\}$, $2\leq
i\leq n$ with $-{{\text{\bsifamily s}}}_{n}(i)\leq\lambda$ is distributed like
the Kruskal forest $K(n,p_{n}(\lambda))$.

We now move on the representation of the random graphs. We say that an edge is
_cyclic_ if it is the maximum weight edge of some cycle. For each $p\in[0,1]$,
the graph $G(n,p)$ is formed of the portion of the minimum spanning tree
consisting of the edges of weight at most $p$, together with the cyclic edges
of weight at most $p$. Observe that while the edges of the minimum spanning
tree are all a.s. a function of $(X^{n,\lambda})_{\lambda\in\mathbb{R}}$, this
is not the case for the cyclic edges (with positive probability some
information is lost, even at the discrete level). Again, rather than
collecting the information from the random graph, it is more instructive to
construct this information with the correct distribution conditionally on
$(Z^{n,\lambda})_{\lambda\in\mathbb{R}}$.

Let $\\{Y_{ij},1\leq i<j\leq n\\}$ be i.i.d. random variables uniform on
$[0,1]$ and independent of everything else (namely
$\operatorname{Merge}_{n}(X^{n})$ and $(U_{i})_{2\leq i\leq n}$). For each
$1\leq i<j\leq n$, let $\lambda^{n}_{ij}=(nY_{ij}-1)n^{1/3}$. We store the
information concerning cyclic edges in a point process. Define

$\Xi_{n}:=\Big{\\{}(in^{-2/3},jn^{-2/3},\lambda^{n}_{ij}):1\leq i<j\leq
n,Z^{n,\lambda^{n}_{ij}}\cap\\{i+1,i+2,\dots,j\\}=\varnothing\Big{\\}}\,,$
(43)

so that $\Xi_{n}$ is the collection of triples
$(i/n^{2/3},j/n^{2/3},\lambda^{n}_{ij})$ for which $v_{i}$ and $v_{j}$ are in
the same connected component of $G(n,p_{n}(\lambda^{n}_{ij})-\delta)$ for
$\delta>0$ small enough. The total number of cyclic edges, sometimes called
the surplus, of a connected component is also a quantity of interest, and can
be expressed in terms of $\Xi_{n}$. Recall that $C^{n,\lambda}_{i}$, $i\geq
1$, denote the collection of vertex sets of the connected components of the
random graph $G(n,p_{n}(\lambda))$, sorted in decreasing order of their sizes.
For a discrete connected component $C^{n,\lambda}_{j}$, the number of surplus
edges in $G(n,p_{n}(\lambda))$ is given by

$\operatorname{surp}^{n,\lambda}_{j}=\\#\big{\\{}(x,y,\lambda^{\prime})\in\Xi_{n}:xn^{2/3},yn^{2/3}\in
C^{n,\lambda}_{j},\lambda^{\prime}\leq\lambda\big{\\}}\,.$ (44)

######  Proposition 7.2 (A representation of the random graph).

Up to a relabelling of the nodes with the Prim ranks, the graph on $[n]$ with
edge set consisting of the union of

  * •

the edges $\\{{{\text{\bsifamily j}}}_{n}(i),i\\}$, $2\leq i\leq n$ such that
${{\text{\bsifamily s}}}_{n}(i)\geq-\lambda$, and

  * •

the edges $\\{i,j\\}$, $1\leq i<j\leq n$ such that
$\lambda^{n}_{ij}\leq\lambda$ and
$Z^{n,\lambda^{n}_{ij}}\cap\\{i+1,i+2,\dots,j\\}=\varnothing$

has the same distribution as $G(n,p_{n}(\lambda))$.

###### Proof.

Note first that we only care about the distribution of the edges of the second
set that are not already in the first one. From Kruskal’s algorithm, it is
clear that, conditionally on the minimum spanning tree $M_{n}$, the weights of
the edges in the complement are independent. Furthermore, for any fixed pair
of nodes $u,v\in[n]$ which are not adjacent in the minimum spanning tree, the
weight of the edge between $u$ and $v$ is uniform, conditioned on being larger
than the value $p_{n}(\lambda)$ at which $u$ and $v$ first become part of the
same connected component. This is precisely what the second condition says
when expressed in terms of the Prim ranks. ∎

Recall that $E^{n}_{p}=\\{e\in E^{n}:w_{e}\leq p\\}$ denotes edge set of the
random graph $G(n,p)$. Define similarly $F^{n}_{p}\subseteq E^{n}_{p}$ the
edge set of the minimum spanning forest $K(n,p)$, that is the collection of
edges of the minimum spanning tree which have weight at most $p$.

######  Lemma 7.3.

Let $p\in[0,1]$.

  1. i)

Conditionally on $E^{n}_{p}$, $(w_{e}:e\in E^{n}_{p})$ is a family of i.i.d.
uniform random variables (r.v.) on $[0,p]$;

  2. ii)

Conditionally on $F^{n}_{p}$, $(w_{e}:e\in F^{n}_{p})$ is dominated by a
family of i.i.d. uniform r.v. on $[0,p]$.

###### Proof.

_i)_ The first assertion is immediate since $\\{w_{e}:e\in E^{n}_{p}\\}$ is
simply a collection of i.i.d. uniform r.v. on $[0,1]$ conditioned on being at
most $p$. _ii)_ The second claim is a consequence of Kruskal’s algorithm: The
set $F^{n}_{p}\subseteq E^{n}_{p}$ is obtained from $E^{n}_{p}$ by iteratively
removing the edge with maximum weight that belongs to a cycle, until there are
no more cycles. The remaining edge have thus been selected for not being the
maximum edge of any cycle; by _i)_ the initial weights in $E^{n}_{p}$ are
i.i.d. uniform random variables on $[0,p]$, and the weights in $F^{n}_{p}$ are
therefore dominated by a collection of i.i.d. uniform r.v. on $[0,p]$. ∎

######  Lemma 7.4.

Fix any $\lambda\in\mathbb{R}$. Conditionally on $a,b\in[n]$, $a<b$, being two
successive points of $Z^{n,\lambda}$,

  1. i)

the vertices whose Prim ranks are in $\\{a,a+1,\dots,b-1\\}$ form a connected
component of $G(n,p_{n}(\lambda))$;

  2. ii)

conditionally on $S=\\{v_{a},\dots,v_{b-1}\\}$, the vertex $v_{a}$ with Prim
rank $a$ is uniformly random in $S$, and independent of $G(n,p_{n}(\lambda))$.

###### Proof.

The first claim is immediate from the definition of $X^{n,\lambda}$; see
Section 4.1 of [24]. The second point is a consequence of the definition of
the Prim order. Consider the time in Prim’s algorithm when we decide who gets
to have rank $a$: conditionally on the event in _i)_ , this depends on an edge
with weight (strictly) larger than $p_{n}(\lambda)$; conditionally on having
its extremity in the set of vertices with Prim ranks $a,a+a,\dots,b-1$, the
end point is uniformly random, and declared to have Prim rank $a$. This
completes the proof. ∎

###  7.2 Asymptotic properties of random graphs for $\lambda\to-\infty$

Recall that we identify the nodes with their Prim ranks, so $v_{i}$ is simply
denoted by $i$. For points $s_{1},\dots,s_{k}\in(0,\infty)$, let
$\operatorname{Span}_{n}(s_{1},\dots,s_{k})$ denote the collection of vertices
that belong to one of the paths in the minimum spanning tree $M_{n}$ between
some $s_{i}^{n}=\lfloor s_{i}n^{2/3}\rfloor$ and $s_{j}^{n}=\lfloor
s_{j}n^{2/3}\rfloor$. For $\lambda\in\mathbb{R}$ and
$s_{1},s_{2},\dots,s_{k}\in(0,\infty)$, let
$J^{n}_{\lambda}(s_{1},s_{2},\dots,s_{k})$ be the collection of indices $j\geq
1$ such that $C^{n,\lambda}_{j}$ intersects
$\operatorname{Span}_{n}(s_{1},\dots,s_{k})$.

The following lemma shows that all the connected components containing part of
the path in the minimum spanning tree between a collection of random points
have a size of order $n^{2/3}$.

######  Proposition 7.5.

Let $I\subset(0,\infty)$ be any compact interval, and let
$s_{1},s_{2},\dots,s_{k}\in(0,\infty)$ be i.i.d. uniform in $I$. Then, for any
$\epsilon>0$ there exists $\lambda\in\mathbb{R}$, and $\delta>0$, such that,
with probability at least $1-\epsilon$, all the connected components of
$G(n,p_{n}(\lambda))$ containing nodes of
$\operatorname{Span}_{n}(s_{1},\dots,s_{k})$ are trees and have size at least
$\delta n^{2/3}$.

###### Proof.

_i)_ We abbreviate $J^{n}_{\lambda}(s_{1},\dots,s_{k})$ as $J^{n}_{\lambda}$.
For any fixed $\lambda$, the collection of connected components containing any
of the $s_{i}^{n}$, $1\leq i\leq k$, have Prim ranks at most $n^{2/3}\sup
I+|C^{n,\lambda}_{1}|$. With high probability, this is at most $tn^{2/3}$ for
some fixed $t$ for all $\lambda\leq 0$ (say). However, by the representations
in [2] or [24], the number of surplus edges involving pairs of nodes with Prim
rank at most $tn^{2/3}$ converges to a Poisson random variable whose parameter
the $\int_{0}^{t}(X^{\lambda}_{s}-\underline{X}^{\lambda}_{s})ds$, which tends
to zero almost surely as $\lambda\to-\infty$. Thus, for any $\epsilon>0$, we
can indeed choose $\lambda$ small enough for all the $C^{n,\lambda}_{j}$,
$j\in J^{n}_{\lambda}$ to be trees with probability at least $1-\epsilon$.

_ii)_ We shall prove that the family of random variables
$\max\\{n^{2/3}/|C^{n,\lambda}_{j}|:j\in J^{n}_{\lambda}\\}$, $n\geq 1$, is
tight. The arguments are all routine, and we only provide the main structure
of the proof. Fix $\epsilon>0$. First, let $\overline{\lambda}$ be large
enough that $s_{1}^{n},\dots,s_{k}^{n}$ are all in the same connected
component $H^{n}$ which also contains the point $\lfloor n^{2/3}\rfloor$ with
probability at least $1-\epsilon$ (see for instance, Lemma 5.2).

By Lemma 7.3 _ii)_ , when decreasing $p$ from $p_{n}(\overline{\lambda})$ to
$p_{n}(\lambda)$, each edge is removed with probability at most
$(p_{n}(\overline{\lambda})-p_{n}(\lambda))/p_{n}(\overline{\lambda})\sim(\overline{\lambda}-\lambda)n^{-1/3}$
independently of the others. So, the number of edges removed on a prescribed
path of length at most $Cn^{1/3}$ is dominated by a binomial random variable
with parameters $Cn^{1/3}$ and $(\overline{\lambda}-\lambda)n^{-1/3}$ and is
thus tight. Since $n^{-1/3}\operatorname{diam}(H^{n})\leq
n^{-1/3}\operatorname{diam}(M_{n})$ which is tight ([1]), the same holds for
the length of the path between any of two of the
$\\{s_{1}^{n},\dots,s_{k}^{n}\\}$. This implies the tightness of
$(|J^{n}_{\lambda}|)_{n\geq 1}$, for any $\lambda\in\mathbb{R}$. This also
readily implies that $n^{1/3}$ divided by the smallest distance in the minimum
spanning tree between any two removed edges is tight. On the other hand, the
minimum distance between any two of the $\\{s_{1}^{n},\dots,s_{k}^{n}\\}$ is
itself of order $n^{1/3}$ (this is lower bounded by the distance in the
corresponding graph $G(n,p_{n}(\overline{\lambda}))$, and thus follows from
the results in [2, 4]). It follows that the smallest portion of a path
connecting the $\\{s_{1}^{n},\dots,s_{k}^{n}\\}$ in the Kruskal forest
$K(n,p_{n}(\lambda))$ is also of order at least $n^{1/3}$.

Now, by _i)_ , let $\lambda$ be small enough that all the involved connected
components are trees at time $p_{n}(\lambda)$ with probability at least
$1-\epsilon$. Conditionally on the number of its nodes being $m$, the diameter
any such connected component is of order $m^{1/2}$ ([11, 12]). Putting this
together with the facts that, for this value of $\lambda$, the number of
portions of paths is tight $J^{n}_{\lambda}$, that each of the portions has
length of order $n^{1/3}$, this implies that each of the portions is contained
in a connected component whose size is indeed of order $n^{2/3}$ (and no
smaller). ∎

The following folklore global asymptotic properties for the connected
components will be useful.

######  Lemma 7.6.

For any $\epsilon,\delta,\delta^{\prime}>0$, there exists
$\lambda\in\mathbb{R}$ such that, with probability at least $1-\epsilon$,

  1. i)

the largest connected component of $G(n,p_{n}(\lambda))$ contains at most
$\delta n^{2/3}$ nodes;

  2. ii)

the maximum diameter of a connected component of $G(n,p_{n}(\lambda))$ is at
most $\delta^{\prime}n^{1/3}$.

###### Proof.

For any $\alpha\leq\delta$, the probability that either _i)_ or _ii)_ fails is
at most

$\displaystyle{\mathbf{P}}(|C^{n,\lambda}_{1}|\geq\alpha
n^{2/3})+{\mathbf{P}}\left(\max_{j\geq
1}\operatorname{diam}(C^{n,\lambda}_{j})\geq\delta^{\prime}n^{1/3},|C^{n,\lambda}_{1}|\leq\alpha
n^{2/3}\right)\,.$

By Theorem 1.3 of [56], there exists $\alpha>0$ small enough such that the
second term is at most $\epsilon/2$. Then, choose $\lambda$ small enough that
the first term is also at most $\epsilon/2$. The fact that such a $\lambda$
exists follows for instance from the results of [16] on the entrance boundary
for the standard multiplicative (Theorem 4 there), and the relation between
the random graph and the multiplicative coalescent in [14] (Proposition 4). ∎

###  7.3 A global coupling argument

Before actually proving the convergence of the trees or graphs seen as metric
spaces, we verify that the main objects, on which the representations of the
previous section rely, do converge. The objective is to eventually construct a
rich enough probability space on which enough parameters converge almost
surely, in order the make the final proof of convergence of the metric as easy
as possible. The starting point is the process
$(X^{n,\lambda})_{\lambda\in\mathbb{R}}$ introduced in (41). By Theorem 7 of
[24], we have

$(X^{n,\lambda})_{\lambda\in\mathbb{R}}\xrightarrow[n\to\infty]{}(X^{\lambda})_{\lambda\in
R}\,,$ (45)

in distribution in $\mathbb{D}(\mathbb{R},\mathbb{C}([0,\infty),\mathbb{R}))$.
The first essential ingredient consists in proving that this implies that the
macroscopic merges restricted to any compact region of time and space also
converge. Define

$\operatorname{Merge}(X)=\\{({{\text{\bsifamily l}}}(t),t,{{\text{\bsifamily
r}}}(t),-{{\text{\bsifamily s}}}(t)):t\in\mathscr{L}(X)\\}\,.$

We say that
$\operatorname{Merge}_{n}((X^{n,\lambda})_{\lambda\in\mathbb{R})}\to\operatorname{Merge}(X)$
if for any compact intervals $I\subset(0,\infty)$ and
$\Lambda\subset\mathbb{R}$, and any threshold $\epsilon>0$, the subset of
$\operatorname{Merge}_{n}((X^{n,\lambda})_{\lambda\in\mathbb{R}}))$ consisting
of points $(l,t,r,-s)$ such that $r-t,t-l>\epsilon$, $l,t,r\in I$ and
$-s\in\Lambda$ converges to the corresponding subset of
$\operatorname{Merge}(X)$.

######  Proposition 7.7 (Convergence of large merges).

Consider a probability space in which
$(X^{n,\lambda})_{\lambda\in\mathbb{R}}\to(X^{\lambda})_{\lambda\in\mathbb{R}}$
almost surely. Then, in probability,

$\operatorname{Merge}_{n}((X^{n,\lambda})_{\lambda\in\mathbb{R}})\xrightarrow[n\to\infty]{}\operatorname{Merge}(X)\,.$

###### Proof.

We will use the following fact: a.s., there does not exist three local minima
of $t,t^{\prime},t^{\prime\prime}\in\mathscr{L}(X)$ such that the points
$(t,X_{t})$, $(t^{\prime},X_{t^{\prime}})$ and
$(t^{\prime\prime},X_{t^{\prime\prime}})$ all lie on the same line; using the
representation in [24], this is essentially equivalent to the fact that the
standard multiplicative coalescent is binary. To see that this is indeed the
case, note that local minima of a continuous function are also global minima
on an interval with rational extremities; then for, three disjoint intervals
$[a,b]$, $[a^{\prime},b^{\prime}]$ and $[a^{\prime\prime},b^{\prime\prime}]$
with rational extremities, the local minima $(t,X_{t})$,
$(t^{\prime},X_{t^{\prime}})$ and $(t^{\prime\prime},X_{t^{\prime\prime}})$ on
each of these intervals have a law which absolutely continuous with respect to
the Lebesgue measure on $\mathbb{R}^{2}$, and then, are aligned with
probability zero; the union of this countable number of zero probability
events also has probability zero. Assume that
$(a,b,c,\lambda)\in\operatorname{Merge}(X)$. In this case, $b$ is a local
minimum of $X^{\lambda}$, and $X^{\lambda}$ is strictly above the (horizontal)
line connecting $(a,X_{a}^{\lambda}),(b,X_{b}^{\lambda}),(c,X_{c}^{\lambda})$
on $(a,b)\cup(b,c)$; because of the property recalled above, since
$(a,X_{a}^{\lambda})$ and $(b,X_{b}^{\lambda})$ are local minima of
$X^{\lambda}$, it is (a.s.) not the case for $(c,X_{c}^{\lambda})$, so that,
$\inf\\{X^{\lambda}_{t}:t\in(c,c+\eta)\\}<0$ for any $\eta>0$.

Given compact intervals $I\subset(0,\infty)$, $\Lambda\subset\mathbb{R}$ and a
threshold $\epsilon>0$, there are only finitely many points in
$\operatorname{Merge}(X)\cap I^{3}\times\Lambda$ with the first three
coordinates at least $\epsilon$ apart, and it suffices to consider each one
separately. Take some $(a,b,c,\lambda)\in\operatorname{Merge}(X)$, so that, in
particular $X^{\lambda}_{a}=X^{\lambda}_{b}=X^{\lambda}_{c}$. Consider, for
$\varepsilon^{\prime}>0$, $\delta>0$, the event

$\displaystyle{\mathcal{E}}((a,b,c,\lambda);\varepsilon^{\prime},\delta)=$
$\displaystyle\Big{\\{}\inf_{[0,a-\varepsilon^{\prime}]}X^{\lambda}\geq
X^{\lambda}_{a}+\delta\Big{\\}}\cap\Big{\\{}\inf_{[a+\varepsilon^{\prime},b-\varepsilon^{\prime}]}X^{\lambda}>X_{a}^{\lambda}+\delta\Big{\\}}$
$\displaystyle\cap\Big{\\{}\inf_{[b+\varepsilon^{\prime},c-\varepsilon^{\prime}]}X^{\lambda}>X^{\lambda}_{a}+\delta\Big{\\}}\cap\Big{\\{}\inf_{[c,c+\varepsilon^{\prime}]}X^{\lambda}<X^{\lambda}_{c}-\delta\Big{\\}}\,.$

Fix $\varepsilon^{\prime}>0$, $a,b,c\in(0,\infty)$ such that $|b-a|,|c-b|\geq
2\varepsilon^{\prime}$, and $\lambda\in\mathbb{R}$. Using the properties of
the local minima of the Brownian motion, for any $\epsilon>0$ there exists a
$\delta>0$ such that

${\mathbf{P}}({\mathcal{E}}((a,b,c,\lambda);\varepsilon^{\prime},\delta)~{}|~{}(a,b,c,\lambda)\in\operatorname{Merge}(X))\geq
1-\epsilon.$

Let us show that, the event
${\mathcal{E}}((a,b,c,\lambda);\varepsilon^{\prime},\delta)$, for all $n$
large enough, there must exist some vector $(a_{n},b_{n},c_{n},\lambda_{n})$
close to $(a,b,c,\lambda)$ such that
$(a_{n},b_{n},c_{n},\lambda_{n})\in\operatorname{Merge}_{n}((X^{n,\lambda})_{\lambda\in\mathbb{R}})$.
To prove this, it suffices to show that for some $\lambda^{\prime}<\lambda$
close enough to $\lambda$ there are points $(a_{n},b_{n},c_{n})\in
Z^{n,\lambda^{\prime}}$ which are close to $(a,b,c)$, while for some
$\lambda^{\prime\prime}>\lambda$ close enough to $\lambda$ there are points
$a_{n}^{\prime\prime},b_{n}^{\prime\prime}\in Z^{n,\lambda^{\prime\prime}}$
with $(a_{n}^{\prime\prime},c_{n}^{\prime\prime})$ close to $(a,c)$ but no
other point of $Z^{n,\lambda^{\prime\prime}}$ between $a_{n}^{\prime\prime}$
and $c_{n}^{\prime\prime}$. We now proceed with the details.

On the one hand, for any $\lambda^{\prime}<\lambda$ there exists
$\delta^{\prime}>0$ small enough such that
$X^{\lambda}_{b}<X^{\lambda^{\prime}}_{a}-\delta^{\prime}$ and
$X^{\lambda^{\prime}}_{c}<X^{\lambda^{\prime}}_{b}-\delta^{\prime}$. Taking
$\lambda^{\prime}$ close enough to $\lambda$ and $\delta^{\prime}>0$ even
smaller, we may also ensure that $X^{\lambda^{\prime}}>\delta^{\prime}$ on
$[a+\varepsilon^{\prime},b-\varepsilon^{\prime}]$ and
$[b+\varepsilon^{\prime},c-\varepsilon^{\prime}]$. The convergence of
$X^{n,\lambda^{\prime}}$ to $X^{\lambda^{\prime}}$ then ensures that we may
find $a_{n},b_{n},c_{n}\in Z^{n,\lambda^{\prime}}$ all within distance
$\varepsilon^{\prime}$ of $a$, $b$ or $c$.

On the other hand, for any $\lambda^{\prime}>\lambda$, we have
$X^{\lambda^{\prime}}>\delta^{\prime}$ on
$(a+\varepsilon^{\prime},c-\varepsilon^{\prime})$; we may take
$\lambda^{\prime}$ close enough to $\lambda$, and $\delta^{\prime}>0$ small
enough such that we also have
$\inf\\{X^{\lambda^{\prime}}:[c,c+\varepsilon^{\prime}]\\}<\inf\\{X^{\lambda^{\prime}}:[0,a-\varepsilon^{\prime}]\\}-\delta^{\prime}$.
The convergence of $X^{n,\lambda^{\prime}}$ to $X^{\lambda}$ then ensures that
there exists $a_{n}\in[a-\varepsilon^{\prime},a+\epsilon^{\prime}]$ and
$c_{n}\in[c-\varepsilon^{\prime},c+\varepsilon^{\prime}]$ such that
$a_{n},c_{n}\in Z^{n,\lambda^{\prime}}$ and $Z^{n,\lambda}$ does not have any
other point between $a_{n}$ and $c_{n}$.

Consider now a accumulation point $(a,b,c,\lambda)$ of
$\operatorname{Merge}_{n}((X^{n,\lambda})_{\lambda\in\mathbb{R}})$ with
$a<b<c$. Then, there exists a sequence $(a_{n},b_{n},c_{n},\lambda_{n})$
converging to $(a,b,c,\lambda)$ with $a_{n},b_{n},c_{n}\in Z^{n,\lambda_{n}}$.
It follows that

$X^{n,\lambda_{n}}_{a_{n}}=\underline{X}^{n,\lambda_{n}}_{a_{n}}=X^{n,\lambda_{n}}_{b_{n}}+n^{-1/3}=X^{n,\lambda_{n}}_{c_{n}}+2n^{-1/3}\,.$

The convergence of $X^{n,\lambda}$ to $X^{\lambda}$ then implies that
$\underline{X}^{\lambda}_{a}=X^{\lambda}_{a}=X^{\lambda}_{b}=X^{\lambda}_{c}$,
so that $a,b,c\in Z^{\lambda}$. The path properties of $X$ also imply that,
for any $\lambda^{\prime}>\lambda$, $(a,c)\cap
Z^{\lambda^{\prime}}=\varnothing$ so that
$(a,b,c,\lambda)\in\operatorname{Merge}(X)$. This completes the proof. ∎

The next step concerns the convergence of the representation of the cyclic
edges. Given $X$, let $\Xi$ be a Poisson point process with intensity
$\mathbf{1}_{{x<y}}\mathbf{1}_{{(x,y)\cap
Z^{\lambda}=\varnothing}}dxdyd\lambda$ on
$\mathbb{R}_{+}^{2}\times\mathbb{R}$.

######  Proposition 7.8 (Convergence of cyclic edges).

We have the convergence in distribution

$((X^{n,\lambda})_{\lambda\in\mathbb{R}},\Xi_{n})\xrightarrow[n\to\infty]{d}((X^{\lambda})_{\lambda\in\mathbb{R}},\Xi)\,.$

Furthermore, for any $a\in\mathbb{R}_{+}$ and $\lambda\in\mathbb{R}$, we have
jointly,

$\\#\Xi_{n}\cap\Big{(}[0,a]^{2}\times(-\infty,\lambda]\Big{)}\xrightarrow[n\to\infty]{d}\\#\Xi\cap\Big{(}[0,a]^{2}\times(-\infty,\lambda]\Big{)}\,.$

###### Proof.

Consider first the larger point process
$\Xi^{\circ}_{n}=\\{(in^{-2/3},jn^{-2/3},\lambda^{n}_{ij}):1\leq i<j\leq
n\\}$. For every set $A\subset\\{(x,y,\lambda):a\leq x<y\leq
b,\lambda_{1}\leq\lambda\leq\lambda_{2}\\}$, $|\Pi_{n}\cap A|$ is a binomial
random variable with parameters asymptotic to $n^{4/3}(b-a)^{2}/2$ and
${\mathbf{P}}(\lambda_{ij}^{n}\in[\lambda_{1},\lambda_{2}])=(\lambda_{2}-\lambda_{1})n^{-4/3}$,
and thus converges to a Poisson random variable $P_{A}$ with parameter
$(\lambda_{2}-\lambda_{1})(b-a)^{2}/2$.

Let now $A$ and $A^{\prime}$ be any two disjoint such sets; we show that the
distributional limits $P_{A}$ and $P_{A^{\prime}}$ are independent. If the
space intervals $[a,b]$ and $[a^{\prime},b^{\prime}]$ are disjoint, this is
straightforward since the random variables $(Y_{ij})$ involved in the
definitions of $\Pi_{n}$ on $A$ and $A^{\prime}$ are themselves independent.
Otherwise the space intervals do intersect, and the time intervals must then
be disjoint. Fix any $\epsilon>0$. There is some constant $K$ such that
${\mathbf{P}}(|\Pi_{n}\cap A|>K)<\epsilon$. Now, conditionaly on $\Pi_{n}\cap
A$ and $|\Pi_{n}\cap A|\leq K$, to estimate the distribution of $|\Pi_{n}\cap
A^{\prime}|$ we shall remove the pairs $(i,j)$ which correspond to a point in
$\Pi_{n}\cap A$, and correct the probability of every other to account for the
fact that they did not occur in $[\lambda_{1},\lambda_{2}]$ (for those that
indeed intersect). This removes only at most $K$ out of the
$n^{4/3}(b^{\prime}-a^{\prime})^{2}/2$ pairs, and boots the probability of
some of others by a factor $(1-n^{-4/3}(\lambda_{2}-\lambda_{1}))$. Overall,
the limit remains Poisson random variable with the same distribution. Since
$\epsilon>0$ was arbitrary, this proves that $\Xi^{\circ}_{n}$ converges to a
Poisson point process $\Xi^{\circ}$ with unit rate on
$\\{(x,z,\lambda)\in\mathbb{R}_{+}^{2}\times\mathbb{R}:x<y\\}$.

For the remainder of the proof, we consider now a probability space on which
$(X^{n,\lambda})_{\lambda\in\mathbb{R}}$ converges almost surely to
$(X^{\lambda})_{\lambda\in\mathbb{R}}$. To complete the proof, it now suffices
to show that, the set
$A^{n}=\\{(in^{-2/3},jn^{-2/3},\lambda):Z^{n,\lambda}\cap\\{i+1,\dots,j\\}=\varnothing\\}$
used to filter the points of $\Xi^{\circ}_{n}$ converges to
$A=\\{(x,y,\lambda):Z^{\lambda}\cap(x,y)=\varnothing\\}$ used to filter those
of $\Xi^{\circ}$, for the Hausdorff distance. Indeed, since $Z^{n,\lambda}$
and $Z^{\lambda}$ are both decreasing in $\lambda$ this would imply the
convergence of their Lebesgue measures. Let $(x,y,\lambda)\in A$, then
$(x,y)\cap Z^{\lambda}=\varnothing$, so that for any $\epsilon>0$ small
enough,
$\inf\\{X^{\lambda}_{s}-\underline{X}^{\lambda}_{s}:s\in(x+\epsilon,y-\epsilon)\\}>0$.
It follows that
$\inf\\{X^{n,\lambda}_{s}-\underline{X}^{n,\lambda}_{s}:s\in(x+\epsilon,x-\epsilon)\\}>0$
as well for all $n$ large enough, so that $(x+\epsilon,y-\epsilon,\lambda)\in
A^{n}$. Similarly, if $(x,y,\lambda)\not\in A$, then
$\inf\\{X_{s}^{\lambda}-\underline{X}_{s}^{\lambda}:s\in(x+\epsilon,y-\epsilon)\\}\leq
0$ for every $\epsilon>0$ small enough, and therefore
$\inf\\{X^{\lambda-\epsilon}_{s}-\underline{X}^{\lambda-\epsilon}_{s}:s\in(x+\epsilon,y-\epsilon)\\}<0$
for some $\epsilon>0$ small enough. The argument we used above implies that
$(x+\epsilon,y-\epsilon,\lambda-\epsilon)\not\in A^{n}$ for all $n$ large
enough.

For the second claim, one only needs the additional tightness of the number
$N_{n}$ of points of $\Xi_{n}$ in $[0,a]^{2}\times(-\infty,\lambda]$. Since
the discrete representation of $\Xi_{n}$ is delicate to handle, we shall
change the point of view: By the exact distribution of $N_{n}$ in Proposition
7.2, $N_{n}$ is dominated by the number of surplus edges in the connected
components that have nodes with Prim ranks at most $an^{2/3}$ at time
$p_{n}(\lambda)$. The latter is known to be tight by the results in [24]
(Corollary 20 and Section 7.2), which consider the alternative representation
for the surplus edges using a Bernoulli pointset under the graph of the
discrete reflected process $X^{n,\lambda}-\underline{X}^{n,\lambda}$ (just as
in the results of Aldous in [14]). ∎

We may also recast the results of [24] about the convergence of surpluses of
connected components in the Prim order in the present setting (see also,
[14]). Recall the discrete defined in (44). There is a continuum analog to the
surplus of a connected component, that can be defined in terms of $\Xi$. At
time $\lambda\in\mathbb{R}$, the connected components correspond to the
intervals of $\mathbb{R}_{+}\setminus Z^{\lambda}$, that are sorted in
decreasing order as $(\gamma^{\lambda}_{j})_{j\geq 1}$. We define

$\operatorname{surp}^{\lambda}_{j}=\\#\big{\\{}(x,y,\lambda^{\prime})\in\Xi:x,y\in\gamma^{\lambda}_{j},\lambda^{\prime}\leq\lambda\big{\\}}\,.$
(46)

######  Corollary 7.9.

Jointly with the convergence in Proposition 7.8, for any fixed
$\lambda\in\mathbb{R}$, and any $j\geq 1$, we have

$\operatorname{surp}_{j}^{n,\lambda}\xrightarrow[n\to\infty]{d}\operatorname{surp}_{j}^{\lambda}\,.$

We shall also want to couple the collection of uniform choices that are
associated to each merge, and used to sample the edges in discrete and the
point identifications in the limit. In the limit, merge events are
characterized by some element $(x,y,z,\lambda)$ where $0<x<y<z$ and
$\lambda\in\mathbb{R}$. Let
$\Delta:=\\{(x,y,z):0<x<y<z\\}\subseteq\mathbb{R}^{3}_{+}$. We shall
discretize the set of merge events. We decompose the collection of all
potential merge triples $\Delta$ into countably many cells (up to a Lebesgue
null set), so that each cell can contain at most one of the elements
$(x,y,z)\in\Delta$ for which there exists some $\lambda$ and
$(x,y,z,\lambda)\in\operatorname{Merge}(X)$.

For $x>0$, let $K(x)$ be the unique integer $k\in\mathbb{Z}$ such that
$x\in[1/2^{k+1},1/2^{k})$. For $k\in\mathbb{Z}$ and $(x,y,z)\in\Delta$ define

$D(x,y,z;k)=\big{(}\lfloor
x2^{k}\rfloor/2^{k},\lfloor(y-x)2^{k}\rfloor/2^{k},\lfloor(z-y)2^{k}\rfloor/2^{k}\big{)}.$

Then, the integer $M(x,y,z)=1+\max\\{K(y-x),K(z-y)\\}$ is used as the
precision at which we shall encode the triple $(x,y,z)$. For each
$(x,y,z)\in\Delta$, define

$\operatorname{Code}(x,y,z)=(M(x,y,z),D(x,y,z;M(x,y,z))\,,$

and observe that $\operatorname{Code}$ takes its values in the countable set
$\mathbb{N}\times\mathbb{Q}^{3}$. A subset $S\subset\Delta$ is called nested
if for any $(x,y,z),(x^{\prime},y^{\prime},z^{\prime})\in S$ we have either
(a) $(x,z)\cap(x^{\prime},z^{\prime})=\varnothing$, or (b)
$(x^{\prime},z^{\prime})$ is contained in either $(x,y)$ or $(y,z)$, or (c)
$(x,z)$ is contained in either $(x^{\prime},y^{\prime})$ or
$(y^{\prime},z^{\prime})$. The set $\operatorname{Merge}(X)$ is such that its
projection on the first three coordinates is nested. One easily verifies that,
the map $\operatorname{Code}$ is injective on any nested subset
$S\subset\Delta$.

######  Lemma 7.10.

Let $S\subseteq\Delta$ be nested. Then for any
$(x,y,z),(x^{\prime},y^{\prime},z^{\prime})\in S$ we have
$\operatorname{Code}(x,y,z)\neq\operatorname{Code}(x^{\prime},y^{\prime},z^{\prime})$.

###### Proof.

Suppose that
$\operatorname{Code}(x_{1},y_{1},z_{1})=\operatorname{Code}(x_{2},y_{2},z_{2})$.
Then, there is $k\in\mathbb{Z}$ and $(a,b,c)\in\mathbb{R}_{+}^{3}$ such that
for $i\in\\{1,2\\}$, we have

  1. i)

$M(x_{i},y_{i},z_{i})=k$, so that
$\min\\{|y_{i}-x_{i}|,|z_{i}-y_{i}|\\}\in[1/2^{k},1/2^{k-1})$, and

  2. ii)

$x_{i}\in[a,a+1/2^{k})$, $y_{i}-x_{i}\in[b,b+1/2^{k})$ and
$z-y\in[c,c+1/2^{k})$.

Now, since $S$ is nested, and there are two alternatives. Either
$(x_{1},z_{1})\cap(x_{2},z_{2})=\varnothing$, and without loss of generality,
$z_{1}\leq x_{2}$, and because of i) we have $|x_{1}-x_{2}|\geq 1/2^{k-1}$
which contradicts ii). Or, without of generality,
$(x_{1},z_{1})\subset[x_{2},y_{2}]$ or $(x_{1},z_{1})\subset[y_{2},z_{2}]$,
and either way, because of i), we cannot have simultaneously all the
inequalities in ii). ∎

The following lemma is straightforward:

######  Lemma 7.11.

Suppose that $(a_{n},b_{n},c_{n})_{n\geq 1}$ converges to $(a,b,c)\in\Delta$
such that neither $a$, $b-a$ nor $c-b$ are multiple of some $2^{j}$, with
$j\in\mathbb{Z}$. Then, for all $n$ large enough
$\operatorname{Code}(a_{n},b_{n},c_{n})=\operatorname{Code}(a,b,c)$.

From now on, we will see ${\mathbf{U}}$ as indexed by this
$\operatorname{Code}(\Delta)$ (which is countable) rather than by
$\mathbb{N}$.

###  7.4 A global coupling and ideal forests

We are now ready to define the probability space on which we will work. By
iterative applications of Skorohod’s representation theorem, we can find a
probability space in which we have the following almost sure convergences, as
$n\to\infty$:

  * •

$(X^{n,\lambda})_{\lambda\in\mathbb{R}}\to(X^{\lambda})_{\lambda\in\mathbb{R}}$;

  * •

$\Xi_{n}\to\Xi$;

  * •

$\operatorname{Merge}_{n}((X^{n,\lambda})_{\lambda\in\mathbb{R}})\to\operatorname{Merge}(X)$.

Our starting point is the following: for the discrete objects, we consider the
Kruskal forest $K(n,p)$ and random graphs $G(n,p)$ that are constructed from
$(X^{n,\lambda})_{\lambda\in\mathbb{R}}$, ${\mathbf{U}}$, and $\Xi_{n}$, which
also define the metric space $(M_{n},d_{n})$. The continuous objects are those
built from $X$, ${\mathbf{U}}$ and $\Xi$, in particular the metric $d$ on
$\mathbb{R}_{+}$. In the following, we shall identify the label and the Prim
ranks to ease the discussion.

These objects are crucial for us, and we will show that their macroscopic
structures are similar. Rather than trying to couple details at the scale
$\delta n^{2/3}$ in the discrete, and $\delta$ in the continuous, we shall
proceed as follows: both in the discrete and continuous setting, we can see
the metric spaces at some given time $\lambda$ (that is $K(n,p_{n}(\lambda))$
and $G(n,p_{n}(\lambda))$ in discrete, and the metric spaces induced by the
intervals of $\mathbb{R}_{+}\setminus Z^{\lambda}$ in continuous) as combining
together the metric spaces that were already present far in the past, say at
some time $\underline{\lambda}<\lambda$. We will never look any further in
time, and replace the metrics in the connected components at time
$\underline{\lambda}$ by some idealization. In general, the distribution will
be incorrect, but we will ensure that $\underline{\lambda}$ can be chosen far
enough for the distributions to be exact (or close enough) on an event of
arbitrarily large probability. Observe that, this modification at time
$\underline{\lambda}$ provides a coupling at all times
$\lambda>\underline{\lambda}$ simultaneously.

Fix any two points $s_{1},s_{2}\in(0,\infty)$ and define $s_{i}^{n}=\lfloor
s_{i}n^{2/3}\rfloor$, $i=1,2$. There is always some deterministic
$\overline{\lambda}$ large enough such that $s_{1}^{n}$ and $s_{2}^{n}$ lie in
the same connected component of $K(n,p_{n}(\overline{\lambda}))$ for $n$ large
enough with probability close to one; the path between $s_{1}^{n}$ and
$s_{2}^{n}$ we refer to is the one in this connected component (and at any
larger time). For any $\lambda\in\mathbb{R}$, let
$J^{n}_{\lambda}(s_{1},s_{2})$ denote the collection of ranks of the connected
components of $G(n,p_{n}(\lambda))$ that contain some node on the path between
$s_{1}^{n}$ and $s_{2}^{n}$; $j\in J^{n}_{\lambda}(s_{1},s_{2})$ means that
the $j$th largest connected component of $K(n,p_{n}(\lambda))$ contains some
node of the path between $s_{1}^{n}$ and $s_{2}^{n}$. Similarly, let
$J_{\lambda}(s_{1},s_{2})$ be the collection of indices of the intervals
$(\gamma^{\lambda}_{j})_{j\geq 1}$ obtained as $\mathbb{R}_{+}\setminus
Z^{\lambda}$ which contain part of $\llbracket s_{1},s_{2}\rrbracket$. By
construction, any of the connected components at time $\lambda$ are
traversersed by a single portion of the path, between two points that we will
denote by $a_{j}^{n}$, $b^{n}_{j}$ and $a_{j},b_{j}$ respectively. Then, we
have the following exact decompositions:

$\displaystyle d_{n}(s_{1}^{n},s_{2}^{n})=\sum_{j\in
J_{\lambda}^{n}(s_{1},s_{2})}d_{n}(a_{j}^{n},b_{j}^{n})+\\#J_{\lambda}^{n}(s_{1},s_{2})-1\,,$
(47)

and

$\displaystyle d(s_{1},s_{2})=\sum_{j\in
J_{\lambda}(s_{1},s_{2})}d(a_{j},b_{j})\,.$ (48)

We will simply replace the distances in the components at time $\lambda$ by
what they should be in an ideal situation; for now, we are only interested in
the definition, the verifications will come later. With this goal in mind, let
us suppose our probability space contains the following sequences of random
variables. Let $(V_{j})_{j\geq 1}$ be i.i.d. random variables uniform on
$[0,1]$. For each $m\geq 1$, let $F_{m}$ denote the distribution function of
the distance $D_{m}$ between two independent uniformly random points in a
uniformly random labelled tree on $m$ nodes. Then, for each $m\geq 1$,
$\bar{D}_{j}(m)=m^{-1/2}F_{m}^{-1}(V_{j})$. This provides a sequence of random
variables where each term $\bar{D}_{j}(m)$ is distributed like
$m^{-1/2}D_{m}$, and that converges almost surely as $m\to\infty$ to a
Rayleigh random variable $\bar{R}_{j}$ with density $xe^{-x^{2}/2}$ on
$\mathbb{R}_{+}$ (see, e.g., [11] for the convergence in distribution).

The objective is to control the matrix of pairwise distances between multiple
points, and our new approximation of the distance will depend on the entire
set of points. Let
${\mathbf{s}}=(s_{1},s_{2},\dots,s_{k})\in\mathbb{R}_{+}^{k}$. We will only
replace the distance in the connected components that only contain a single
portion of paths between these points; in all the other components, which
contain branch points of the collection of paths between the elements of
${\mathbf{s}}$, we will keep the distance unchanged. Let

$J^{n}_{\lambda}(s_{p},s_{q};{\mathbf{s}})=J^{n}_{\lambda}(s_{p},s_{q})\setminus\bigcup_{i<j,i\neq
p,j\neq q}J^{n}_{\lambda}(s_{i},s_{j})\,,$

and similarly, define the continuum analog by

$J_{\lambda}(s_{p},s_{q};{\mathbf{s}})=J_{\lambda}(s_{p},s_{q})\setminus\bigcup_{i<j,i\neq
p,j\neq q}J_{\lambda}(s_{i},s_{j})\,.$

Let $(n^{2/3}|\gamma_{j}^{n,\lambda}|)_{j\geq 1}$ denote the collection of
sizes of the connected components at time $\lambda$, just as
$(|\gamma^{\lambda}_{j}|)_{j\geq 1}$ denotes the Lebesgue measures in the
continuous setting. Define the following approximations, for $1\leq p<q\leq
k$,

$\displaystyle\tilde{d}_{n}(s_{p}^{n},s_{q}^{n})=$
$\displaystyle~{}n^{1/3}\sum_{j\in
J^{n}_{\lambda}(s_{p},s_{q};{\mathbf{s}})}|\gamma^{n,\lambda}_{j}|^{1/2}\bar{D}_{j}(n^{2/3}|\gamma^{n,\lambda}_{j}|)$
(49) $\displaystyle+\sum_{j\in J^{n}_{\lambda}(s_{p},s_{1})\setminus
J^{n}_{\lambda}(s_{p},s_{1};{\mathbf{s}})}d_{n}(a_{j}^{n},b_{j}^{n})+\\#J^{n}_{\lambda}(s_{1},s_{2})-1\,,$

and

$\tilde{d}(s_{p},s_{q})=\sum_{j\in
J_{\lambda}(s_{p},s_{q};{\mathbf{s}})}|\gamma_{j}^{\lambda}|^{1/2}\bar{R}_{j}+\sum_{j\in
J_{\lambda}(s_{p},s_{q})\setminus
J_{\lambda}(s_{p},s_{q};{\mathbf{s}})}d(a_{j},b_{j})\,.$ (50)

We first verify that these provide a suitable coupling of the pairwise
distances between the $k$ points $s_{1}^{n},s_{2}^{n},\dots,s_{k}^{n}$ and
$s_{1},s_{2},\dots,s_{k}$, respectively.

######  Proposition 7.12.

Fix some compact interval $I\in(0,\infty)$. For any $\epsilon>0$, there exists
$\lambda\in\mathbb{R}$ and an event $A$ of probability at least $1-\epsilon$,
such that, for any $k$ i.i.d. uniform points $s_{1},s_{2},\dots,s_{k}\in I$,
for all $n$ large enough, on the event $A$, we have

$(\tilde{d}_{n}(s_{p},s_{q}))_{1\leq p<q\leq k}\stackrel{{\scriptstyle
d}}{{=}}(d_{n}(s_{p}^{n},s_{q}^{n}))_{1\leq p<q\leq
k}\quad\text{and}\quad(\tilde{d}(s_{p},s_{q}))_{1\leq p<q\leq
k}\stackrel{{\scriptstyle d}}{{=}}(d(s_{p},s_{q}))_{1\leq p<q\leq k}\,.$

###### Proof.

There exists a $\lambda_{1}$ large enough that $I$ is contained in a single
connected component at time $\lambda_{1}$: with Lemma 5.2 in mind, let
$\lambda_{1}$ be the smallest $\lambda$ for which $\sup I<2\lambda+1$ and
$\exp(-c\lambda)<\epsilon/2$. This value being fixed, $I$ is contained in
$[0,2\lambda_{1}+1]$ with probability at least $1-\epsilon/2$.

Let $A=A_{\lambda}$ be the event that $\Xi$ does not contain any point in
$[0,2\lambda_{1}+1]$ with time lower than $\lambda$. From the correspondence
between the intensity of $\Xi$ and the area of
$X^{\lambda}-\underline{X}^{\lambda}$, there exists $\lambda$ small enough
that $A$ has probability at least $1-\epsilon$. The convergence of $\Xi_{n}$
to $\Xi$ implies that, on this event, for all $n$ large enough, $\Xi_{n}$ also
has no point in $I$ with times before $\lambda$ (Proposition 7.8).

We may choose $\lambda$ even smaller to ensure that,
$|\gamma^{\lambda}_{1}|<\epsilon/(4k)$, so that, the probability that some
point $s_{i}$, $1\leq i\leq k$, falls in an interval of
$\mathbb{R}_{+}\setminus Z^{\lambda}$ that is not fully contained in $I$ is at
most $\epsilon/(2|I|)$. When this occurs, conditionally on
$s_{i}\in\gamma^{\lambda}_{j}$, the position $s_{i}$ is uniformly random in
$\gamma^{\lambda}_{j}$. The same holds true for the discrete counterparts
$s_{i}^{n}$ for all $n$ large enough.

Now, on the event $A$, for all $n$ large enough, all the connected components
of the random graph $G(n,p_{n}(\lambda))$ containing nodes with label at most
$(2\lambda_{1}+1)n^{2/3}$ are all trees, which are uniformly random. These are
thus identical to the components in the Kruskal forest $K(n,p_{n}(\lambda))$.
By Proposition 7.1 and Lemma 7.4 the points $a_{j}^{n}$ and $b_{j}^{n}$ are
independent and uniformly random (their actual labels!) and independent of the
component. Since the end points $s_{i}^{n}$ are themselves uniformly random in
the connected component in which they lie, by the previous paragraph, this
proves that, on $A$, the discrete approximation $\tilde{d}(s_{p},s_{q})$,
$1\leq p<q\leq k$, has the same distribution as $d(s_{p},s_{q})$, $1\leq
p<q\leq k$.

The continuous analog follows from the calculations in Section 2.1 of [1]
saying that, conditionally on having no point under the curve, an excursion
under $\tilde{\mathbf{n}}_{\sigma}$ is distributed according to
${\mathbf{n}}_{\sigma}$, and is thus exactly a Brownian excursion; Brownian
scaling and Remark 3.23 saying that in
$\operatorname{CMT}({\textup{e}},{\mathbf{U}})$, the distance
$d_{\textup{e}}(0,V)$ between $0$ and a uniformly random point $V$ is Rayleigh
distributed, which completes the proof. ∎

###  7.5 Main proof of convergence

Finally, we are ready to prove that, in the probability space defined in the
previous section, we have convergence in probability of the pairwise distance.

######  Proposition 7.13.

Fix $I$ a compact interval of $(0,\infty)$, and let $s_{1},s_{2},\dots,s_{k}$
be i.i.d. uniform points in $I$. For any $\epsilon,\delta>0$, there exists
$\lambda\in\mathbb{R}$ such that,

$\limsup_{n\to\infty}{\mathbf{P}}\left(\sup_{1\leq p<q\leq
k}\big{|}n^{-1/3}\tilde{d}_{n}(s_{p}^{n},s_{q}^{n})-\tilde{d}(s_{p},s_{q})\big{|}>\delta\right)\leq\epsilon\,.$

###### Proof.

Fix any $\epsilon,\delta>0$. Let us first deal with the portions of paths
contained in connected components that are traversed by more than one path,
and that we did not bother coupling. Consider the event $A$ in Proposition
7.12 and the corresponding value $\lambda_{1}$ for $\lambda$ which ensures
that ${\mathbf{P}}(A^{c})\leq\epsilon/4$. By Lemma 7.6, there exists
$\lambda_{2}$ such that, for all $\lambda\leq\lambda_{2}$ and all $n$ large
enough, the probability that the maximum diameter of a connected component of
$G(n,p_{n}(\lambda))$ is larger than $n^{1/3}\delta/(3k)$ is at most
$\epsilon/4$. Furthermore, on the event $A$, each one of the $k-1$ portions of
continuum paths in the intervals $\gamma_{j}^{\lambda}$ which contain more
than one portion has a length stochastically dominated by
$|\gamma_{j}^{\lambda}|\bar{R}_{j}$ (or the diameter of the corresponding
CRT). We can choose $\lambda_{3}$ small enough such that the probability that
any of them is greater than $\delta/(3k)$ is at most $\epsilon/4$. Fix
$\lambda=\min\\{\lambda_{1},\lambda_{2},\lambda_{3}\\}$. Finally, by
Proposition 7.5, for this value of $\lambda$, there is some
$\delta^{\prime}>0$ small enough such that, with probability at least
$1-\epsilon/4$ all the connected components $C^{n,\lambda}_{j}$, $j\in
J^{n}_{\lambda}(s_{p},s_{q})$, $1\leq p<q\leq k$, contain at least
$\delta^{\prime}n^{2/3}$ nodes. The probability that either of these bad
events occur is at most $\epsilon$, and we now suppose we work on the event
$A^{\prime}$ that none occurs.

On the event $A^{\prime}$, we have from (49) and (50), for any $1\leq p<q\leq
k$,

$\displaystyle\big{|}n^{-1/3}\tilde{d}_{n}(s_{p}^{n},s_{q}^{n})-\tilde{d}(s_{p},s_{q})\big{|}$
$\displaystyle\leq\Bigg{|}\sum_{j\in
J^{n}_{\lambda}(s_{p},s_{q};{\mathbf{s}})}n^{-1/3}|\gamma^{n,\lambda}_{j}|^{1/2}\bar{D}_{j}(n^{2/3}|\gamma^{n,\lambda}_{j}|)-\sum_{j\in
J_{\lambda}(s_{p},s_{q};{\mathbf{s}})}|\gamma_{j}^{\lambda}|^{1/2}\bar{R}_{j}\Bigg{|}$
$\displaystyle\quad+n^{-1/3}\\#J^{n}_{\lambda}(s_{p},s_{q})+2\delta/3\,.$

Since the $\\#J^{n}_{\lambda}(s_{p},s_{q})$ are all tight by the proof of
Proposition 7.5, we only need to deal with the first term the right-hand side
above.

We claim that the fact that all discrete connected components
$C^{n,\lambda}_{j}$, for $j\in J^{n}_{\lambda}(s_{p},s_{q})$ for some $1\leq
p<q\leq k$ contain at least $\delta^{\prime}n^{2/3}$ nodes, the convergence of
the merge events implies that, for all $n$ large enough, we have
$J^{n}_{\lambda}(s_{p},s_{q})=J_{\lambda}(s_{p},s_{q})$ for every $1\leq
p<q\leq k$. The reason is the following: (1) for all $n$ large enough, for
every $i$, if $s_{i}\in\gamma^{\lambda}_{j}$, then $s_{i}^{n}\in
C^{n,\lambda}_{j}$, because $\\{s_{1},\dots,s_{k}\\}$ and $Z^{\lambda}$ are
almost surely disjoint. (2) The merges of large connected components do
converge because
$\operatorname{Merge}((X^{n,\lambda})_{\lambda\in\mathbb{R}})\to\operatorname{Merge}(X)$.
(3) The points random points constructed in the discrete and continuuous model
for matching merges use the same uniforms by Lemma 7.11. It follows that, for
$n$ large enough, these points themselves end up in matching pair of discrete
and continuum components. (4) The number of such merges is finite (the
$\\#J^{n}_{\lambda}(s_{p},s_{q})$, $J_{\lambda}(s_{p},s_{q})$ are tight). As a
consequence, for all $n$ large enough, we are lead to bounding

${\mathbf{P}}\left(\Bigg{|}\sum_{j\in
J_{\lambda}(s_{p},s_{q};{\mathbf{s}})}n^{-1/3}|\gamma^{n,\lambda}_{j}|^{1/2}\bar{D}_{j}(n^{2/3}|\gamma^{n,\lambda}_{j}|)-|\gamma_{j}^{\lambda}|^{1/2}\bar{R}_{j}\Bigg{|}>\delta/3\right)\,,$

but we our coupling precisely ensures that every single term of the sum
converges almost surely to zero. This completes the proof. ∎

###  7.6 Remaining proofs of convergence

Finally, we rely on the results of the previous section to complete the proofs
of the remaining results, namely that of Theorem 1.3 about the MST of a
connected graph with given surplus, and Theorem 1.4 about the dynamics for the
limit random graph and Kruskal processes.

Before going further, let us discuss the types of convergence. Proposition
7.13 implies the convergence of the distribution of the matrix of pairwise
distances between any finite number of points, and may thus be used to prove
convergence in the Gromov–Prokhorov (GP) sense (Theorem 5 of [37]): indeed,
for any $\lambda$, restriction of the $d$ to any interval
$\gamma^{\lambda}_{j}$, $j\geq 1$, is the limit of the metric of the discrete
minimum on $C^{n,\lambda}_{j}$. The reason why this suffices to also prove
convergence in the sense of Gromov–Hausdorff–Prokhorov (GHP) is that we
actually already know that the sequences are tight for GHP ([4, 2]), and that
the limit we construct has a mass measure which has full support because of
Proposition 3.22 (see [18]). In the following, we thus only discuss GP
convergence.

###### Proof of Theorem 1.4.

_i)_ Since the coupling is global, the proof of the joint convergence of the
Kruskal forest
$({\mathfrak{F}}^{n,\lambda_{1}},\dots,{\mathfrak{F}}^{n,\lambda_{k}})$ at
times $\lambda_{1}<\lambda_{2}<\dots<\lambda_{k}$ is an immediate consequence
of Proposition 7.13, and the above discussion about the GHP versus GP
convergence. The connected components at time $\lambda_{i}$ correspond to the
intervals of $\mathbb{R}_{+}\setminus Z^{\lambda_{i}}$, equipped with the
metric induced by $d$.

_ii)_ For the same reason, the proof of the joint convergence
$({\mathfrak{G}}^{n,\lambda_{1}},\dots,{\mathfrak{G}}^{n,\lambda_{k}})$ would
be complete once we have an analog of Proposition 7.13 for the random graph at
a fixed time. Proving this amounts to verifying that the joint convergence of
the minimum spanning tree and of $\Xi_{n}$ is sufficient to guarantee the
convergence of the end points of every single surplus edge.

Once we have convergence of the end points of the edges, the techniques in [2]
imply the convergence of the graph. Proving that we indeed have convergence of
the locations of the end points of edges is not immediate because the function
$d(x,y)$ is not continuous in either $x$ or $y$. However, we can find a small
$\underline{\lambda}\in\mathbb{R}$ such that the points appear between the
correct connected components at time $\underline{\lambda}$ for all $n$ large
enough (almost surely, since the points have a diffuse distribution). Since
the diameter of these components at time $\underline{\lambda}$ may be made
arbitrarily small by choice of $\underline{\lambda}$, we do have convergence
of the locations of the end points. This completes the proof of the sequence
of graphs, in the product topology for a fixed $\lambda$. The extension to a
vector of $(\lambda_{1},\dots,\lambda_{k})$ is immediate using the same
arguments as above. ∎

###### Proof of Theorem 1.3.

Consider the probability space from above, and fix some interval
$\gamma^{\lambda}_{i}$ of $\mathbb{R}_{+}\setminus Z^{\lambda}$. Recall the
discrete and continuum surplus defined in (44) and (46), respectively.
Furthermore, $\operatorname{surp}^{\lambda}_{i}$ is a Poisson random variable
with parameter the area of the process $X^{\lambda}-\underline{X}^{\lambda}$
on $\gamma^{\lambda}_{i}$. It thus follows from the calculations in Section
2.1 of [1] that,

$\displaystyle\mathbf{E}\left[f((X^{\lambda}_{t_{0}+t}-\underline{X}^{\lambda}_{t_{0}+t})_{0\leq
t\leq\sigma})~{}\Big{|}~{}\gamma_{i}^{\lambda}=(t_{0},t_{0}+\sigma),\operatorname{surp}_{j}^{\lambda}=s\right]$
$\displaystyle=\frac{\mathbf{E}[f({\textup{e}}_{\sigma})\cdot(\int_{0}^{\sigma}{\textup{e}}_{\sigma}(u)du)^{s}]}{\mathbf{E}[(\int_{0}^{\sigma}{\textup{e}}_{\sigma}(u)du)^{s}]}$

where ${\textup{e}}_{\sigma}$ is a Brownian excursion of duration $\sigma$. By
definition, the right-hand side above is nothing else than
$\mathbf{E}[f({\textup{e}}_{\sigma}^{(s)})]$. Furthermore, on the event that
$\operatorname{surp}^{\lambda}_{i}=s$, by Corollary 7.9, we have
$\operatorname{surp}^{n,\lambda}_{i}=s$ for all $n$ large enough. Therefore,
up to a trivial relabelling, $C^{n,\lambda}_{i}$ is a uniformly random
connected component with surplus $s$ and size $\gamma^{n,\lambda}_{i}$. Since
each of the values for $\operatorname{surp}^{\lambda}_{i}$ has positive
probability, Theorem 1.3 follows from Proposition 7.13, and the discussion
about the strengthening to Gromov–Hausdorff–Prokhorov convergence. ∎

Finally, we prove our main result about the entire minimum spanning tree. In
[4], it is proved that the scaling limit of the minimum spanning tree can be
constructed as the limit as $\lambda\to\infty$ of the scaling limit of the
minimum spanning tree of the largest connected component of the random graph
at $p_{n}(\lambda)$. Here, we use the limit as $\lambda\to\infty$ of the
connected component containing the vertex with Prim order $\lfloor
n^{2/3}\rfloor$. We now verify that this coincides with our definition, which
uses a connected component $H_{\lambda}$containing the point $1$ and the
measured metric space $(H_{\lambda},d,\hat{\mu}_{\lambda})$,
$\hat{\mu}_{\lambda}$ is the (image of the) probability measure which is
proportional to Lebesgue measure on $H_{\lambda}$. At this point, this should
be essentially straightforward.

###### Proof of Theorem 1.1.

Let ${\mathcal{E}}^{\star}_{\lambda}$ be the event that the largest connected
component of $\mathbb{R}_{+}\setminus Z^{\lambda}$ contains the point $1$.
Observe that, for all $\lambda\geq 2$, with $R_{\lambda}=\sup H_{\lambda}$,

$\displaystyle{\mathbf{P}}({\mathcal{E}}^{\star}_{\lambda})$
$\displaystyle\geq{\mathbf{P}}(R_{\lambda}>2\lambda-1,|\gamma^{\lambda}_{2}|\leq
2)$ $\displaystyle\geq 1-{\mathbf{P}}(R_{\lambda}\leq
2\lambda-1)-{\mathbf{P}}(|\gamma^{\lambda}_{2}|\geq 2).$

Lemma 5.2 implies that the first probability in the right-hand side above
tends to zero as $\lambda\to\infty$. The same holds for the second one, see
for instance, Proposition 5.3 of [6] which says that $|\gamma^{\lambda}_{2}|$
is $O(\lambda^{-2}\log\lambda)$ in probability. This also easily follows from
Lemma 5.8 _i)_ : indeed, for any natural number $i\geq 1$, on the event that
$R_{\lambda}>i^{3}$, we have (with the notation of Section 5.4)

$|\gamma^{\lambda}_{2}|\leq 1+\sup_{k\geq
i}\sup\\{m_{\lambda^{\prime}}:\lambda^{\prime}\in\Lambda_{k}\\}\,,$

which is at most $1+i^{-5}$ with probability at least $1-O(i^{-1/4})$. This
implies ${\mathbf{P}}(|\gamma^{\lambda}_{2}|\geq 2)=O(\lambda^{-1/12})$, and
in turn that ${\mathbf{P}}({\mathcal{E}}^{\star}_{\lambda})\to 1$ as
$\lambda\to\infty$. By Proposition 7.13, $(H_{\lambda},d,\hat{\mu}_{\lambda})$
is the Gromov–Prokhorov limit (in distribution) of the minimum spanning tree
of the connected component containing the vertex with Prim order $n^{2/3}$.
However we know by the results of [7] that the sequence of rescaled minimum
spanning trees converge for the Gromov–Hausdorff–Prokhorov topology, so that
the convergence actually holds for GHP. Together with the fact that
${\mathbf{P}}({\mathcal{E}}^{\star}_{\lambda})\to 1$ as $\lambda\to\infty$,
this proves that $(\mathscr{M},d,\mu)$ has the same distribution as
$(\mathscr{M}^{\prime},d^{\prime},\mu^{\prime})$ constructed in [7]. ∎

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