$\displaystyle\lim_{N\to\infty}\mathbb{E}F_{N}=\lim_{\varepsilon\downarrow
0}\lim_{N\to\infty}\mathbb{E}F_{N,\varepsilon}^{\mbox{\tiny
PVB}}=\lim_{\varepsilon\downarrow
0}\inf_{\sigma\in\mathcal{M}_{\varepsilon}}Q_{\varepsilon}(\sigma).$ (8.4)

Let $\sigma_{0}\in\mathcal{N}.$ Then there exists a sequence
$(\sigma_{m})_{m\geq 1}$ for $\sigma_{m}\in\mathcal{M}_{\varepsilon_{m}}$ and
$\varepsilon_{m}\downarrow 0$ such that $\sigma_{m}\to\sigma_{0}$ f.d.d.
Therefore, from Theorem 8.1, Theorem 8.5, and Lemma 8.6,

$\limsup_{N\to\infty}\mathbb{E}F_{N}\leq\lim_{m\to\infty}Q_{\varepsilon_{m}}(\sigma_{m})=Q(\sigma_{0}).$

Since this holds for any $\sigma_{0}\in\mathcal{N},$ it follows that
$\limsup_{N\to\infty}\mathbb{E}F_{N}\leq\inf_{\mathcal{N}}Q(\sigma).$
Conversely, from (8.4), for any $\eta>0,$ as long as $\varepsilon$ is small
enough,
$\liminf_{N\to\infty}\mathbb{E}F_{N}\geq\inf_{\sigma\in\mathcal{M}_{\varepsilon}}Q_{\varepsilon}(\sigma)-\eta.$
We can pick a sequence $(\sigma_{m})_{m\geq 1}$ for
$\sigma_{m}\in\mathcal{M}_{\varepsilon_{m}}$ and $\varepsilon_{m}\downarrow 0$
that again converges to some $\sigma_{0}$ f.d.d. and that for all $m\geq 1,$
$\inf_{\sigma\in\mathcal{M}_{\varepsilon_{m}}}Q_{\varepsilon_{m}}(\sigma)\geq
Q_{\varepsilon_{m}}(\sigma_{m})-\eta.$ Hence, from Theorem 8.1, Theorem 8.5,
and Lemma 8.6 again,

$\displaystyle\liminf_{N\to\infty}\mathbb{E}F_{N}\geq\lim_{m\to\infty}Q_{\varepsilon_{m}}(\sigma_{m})-2\eta=Q(\sigma_{0})-2\eta.$

Since this holds for any $\eta>0,$ the lower bound,
$\liminf_{N\to\infty}\mathbb{E}F_{N}\geq Q(\sigma),$ follows, from which,
since $\sigma_{0}\in\mathcal{N}$, we readily have that
$\lim_{N\to\infty}\mathbb{E}F_{N}=\inf_{\mathcal{N}}Q(\sigma)$ and this
establishes the first equality in (8.3).

As for the second inequality in (8.3), note that
$\inf_{\sigma\in\mathcal{N}}Q(\sigma)\geq\inf_{\sigma\in\mathcal{N}_{inv}}Q(\sigma).$
It suffices to show the reverse inequality, for which we follow the argument
of Lemma 7 in [41]. Recall that from Lemma 6 in [41] and the identity (3.1),
for any function $\sigma\in\mathcal{S}$, we have

$\displaystyle\lim_{N\to\infty}\frac{1}{N}\mathbb{E}F_{N,\varepsilon}^{\mbox{\tiny
PVB}}$ $\displaystyle\leq\ln
2+\frac{\alpha}{2}\int_{\varepsilon}^{\infty}\frac{\ln\cosh(\beta
x)}{x^{\alpha+1}}dx+\frac{1}{n}\mathbb{E}\ln\mathcal{A}_{n,\varepsilon}(\sigma)-\frac{1}{n}\mathbb{E}\ln\mathcal{B}_{n,\varepsilon,1}(\sigma),$

and from Theorem 6.3 and (8.4), passing to the limit gives

$\displaystyle\lim_{N\to\infty}\mathbb{E}F_{N}$ $\displaystyle\leq\ln
2+\frac{\alpha}{2}\int_{0}^{\infty}\frac{\ln\cosh(\beta
x)}{x^{\alpha+1}}dx+\frac{1}{n}\mathbb{E}\ln\mathcal{A}_{n}(\sigma)-\frac{1}{n}\mathbb{E}\ln\mathcal{B}_{n,1}(\sigma).$

We claim that for any $\sigma\in\mathcal{N}_{inv},$

$\displaystyle\begin{split}\mathbb{E}\ln\frac{\mathcal{A}_{n+1}(\sigma)}{\mathcal{A}_{n}(\sigma)}&=\mathbb{E}\ln\mathcal{A}_{1}(\sigma)\,\,\mbox{and}\,\,\mathbb{E}\ln\frac{\mathcal{B}_{n+1}(\sigma)}{\mathcal{B}_{n}(\sigma)}=\mathbb{E}\ln\mathcal{B}_{1}(\sigma).\end{split}$

If these are valid, then we readily have

$\inf_{\mathcal{N}_{inv}}Q(\sigma)\leq\inf_{\mathcal{N}}Q(\sigma)=\lim_{N\to\infty}\mathbb{E}F_{N}\leq\inf_{\mathcal{N}_{inv}}Q(\sigma)$

and this establishes the second equality in (8.3). To prove our claim, we will
only establish the proof for the first equation since the other one can be
treated similarly. First of all, note that from Theorem A.2, we have

$\displaystyle\mathbb{E}A_{1}(\sigma)$
$\displaystyle=\lim_{\varepsilon\downarrow
0}\mathbb{E}\ln\mathbb{E}_{u,x,\delta}\prod_{k:|\xi_{n+1,k}|\geq\varepsilon}\bigl{(}1+\tanh(\beta\xi_{n+1,k})s_{n+1,k}\delta_{n+1}\bigr{)}$

and a small modification of Theorem A.2 yields that

$\displaystyle\mathbb{E}A_{n+1}(\sigma)$
$\displaystyle=\lim_{\varepsilon\downarrow
0}\mathbb{E}\ln\mathbb{E}_{u,x,\delta}\prod_{k:|\xi_{n+1,k}|\geq\varepsilon}\bigl{(}1+\tanh(\beta\xi_{n+1,k})s_{n+1,k}\delta_{n+1}\bigr{)}$
$\displaystyle\qquad\qquad\qquad\qquad\prod_{i\leq n}\prod_{k\geq
1}\bigl{(}1+\tanh(\beta\xi_{i,k})s_{i,k}\delta_{i}\bigr{)}.$

From these, it suffices to show that for any $\varepsilon>0,$

$\displaystyle\mathbb{E}\ln\Bigl{(}\mathbb{E}_{u,x,\delta}\Bigl{[}\prod_{k:|\xi_{n+1,k}|\geq\varepsilon}\bigl{(}1+\tanh(\beta\xi_{n+1,k})s_{n+1,k}\delta_{n+1}\bigr{)}\frac{\prod_{i\leq
n}\prod_{k\geq
1}(1+\tanh(\beta\xi_{i,k})s_{i,k}\delta_{i})}{\mathbb{E}_{u,x,\delta}\prod_{i\leq
n}\prod_{k\geq 1}(1+\tanh(\beta\xi_{i,k})s_{i,k}\delta_{i})}\Bigr{]}\Bigr{)}$
$\displaystyle=\mathbb{E}\ln\Bigl{(}\mathbb{E}_{u,x,\delta}\Bigl{[}\prod_{k:|\xi_{n+1,k}|\geq\varepsilon}\bigl{(}1+\tanh(\beta\xi_{n+1,k})s_{n+1,k}\delta_{n+1}\bigr{)}\Bigr{]}\Bigr{)}.$

Denote by $\mathbb{E}^{\prime}$ the expectation conditionally on
$(\xi_{n+1,k})_{k\geq 1}.$ Observe that conditionally on $(\xi_{n+1,k})_{k\geq
1},$

$\prod_{k:|\xi_{n+1,k}|\geq\varepsilon}\bigl{(}1+\tanh(\beta\xi_{n+1,k})s_{n+1,k}\delta_{n+1}\bigr{)}$

is a finite product and it is uniformly bounded from above and is uniformly
positive. Since $x\in[c,C]\to\ln x$ for some $c,C>0$ can be uniformly
approximated by polynomials, to establish the equality above, we only need to
show that for all integer $r\geq 1$,

$\displaystyle\mathbb{E}^{\prime}\Bigl{[}\Bigl{(}\mathbb{E}_{u,x,\delta}\Bigl{[}\prod_{k:|\xi_{n+1,k}|\geq\varepsilon}(1+\tanh(\beta\xi_{n+1,k})s_{n+1,k}\delta_{n+1})\frac{\prod_{i\leq
n}\prod_{k\geq
1}(1+\tanh(\beta\xi_{i,k})s_{i,k}\delta_{i})}{\mathbb{E}_{u,x,\delta}\prod_{i\leq
n}\prod_{k\geq
1}(1+\tanh(\beta\xi_{i,k})s_{i,k}\delta_{i})}\Bigr{]}\Bigr{)}^{r}\Bigr{]}$
$\displaystyle=\mathbb{E}^{\prime}\Bigl{[}\Bigl{(}\mathbb{E}_{u,x,\delta}\Bigl{[}\prod_{k:|\xi_{n+1,k}|\geq\varepsilon}(1+\tanh(\beta\xi_{n+1,k})s_{n+1,k}\delta_{n+1})\Bigr{]}\Bigr{)}^{r}\Bigr{]}.$

Now, if we expand the first product on the left-hand side, the product on the
right-hand side, and $(\cdot)^{r}$ on the two sides, then by comparison, we
see that the quality above holds as long as we have that for any
$I_{1},\ldots,I_{r}\subseteq\\{k:|\xi_{n+1,k}|\geq\varepsilon\\},$

$\displaystyle\mathbb{E}^{\prime}\frac{\prod_{l\leq
r}U_{l,0}(\sigma)}{V_{0}(\sigma)^{r}}=\mathbb{E}^{\prime}\prod_{l\leq
r}\prod_{i\in I_{l}}s_{i}^{l},$

which follows from the assumption that $\sigma$ satisfies the invariance
principle (2.9), where

$\displaystyle U_{l,0}(\sigma)$
$\displaystyle=\mathbb{E}_{u,x,\delta}\Bigl{(}\prod_{i\leq n}\prod_{k\geq
1}(1+\tanh(\beta\xi_{i,k})s_{i,k}\delta_{i})\Bigr{)}\Bigl{(}\prod_{k\in
I_{l}}s_{i}\Bigr{)},$ $\displaystyle V_{0}(\sigma)$
$\displaystyle=\mathbb{E}_{u,x,\delta}\Bigl{(}\prod_{i\leq n}\prod_{k\geq
1}(1+\tanh(\beta\xi_{i,k})s_{i,k}\delta_{i})\Bigr{)}.$

This ends the proof of our claim.

###  8.3 Proof of Proposition 2.8

Let $0<\beta<\beta_{\alpha}$ be fixed. Recall the free energy
$F_{N,\varepsilon}^{\mbox{\tiny PVB}}(\beta)$ of the PVB model associated to
the choice of parameter $(y,\gamma)$ in (6.3) in Section 6.2. From [27], it
was already known that for any small enough $\varepsilon,$ as long as $\beta$
satisfies

$\alpha\int_{\varepsilon}^{\infty}\frac{\tanh^{2}(\beta
x)}{x^{\alpha+1}}dx=2\gamma\mathbb{E}\tanh^{2}(\beta y)<1,$

we have that

$\displaystyle\lim_{N\to\infty}\mathbb{E}\langle
R_{1,2}^{2}\rangle_{\varepsilon}^{\mbox{\tiny PVB}}=0.$ (8.5)

In particular, recalling the definition of $\beta_{\alpha}$ from (2.3), the
above inequality is satisfied as long as $\varepsilon$ is small enough.

We claim that for small enough $\varepsilon,$ any
$\sigma\in\mathcal{M}_{\varepsilon}$ satisfies
$\mathbb{E}_{x}\sigma(w,u,v,x)=0.$ a.s.. If this is valid, then we readily
have that $\mathbb{E}_{x}\sigma(w,u,v,x)=0$ a.s. for all
$\sigma\in\mathcal{N}$ and this completes our proof. To establish our claim,
let $\sigma\in\mathcal{M}_{\varepsilon}$ and let
$s_{i}^{l}=\sigma(w,u_{l},v_{i},x_{i,l})$ for all $i,l\geq 1.$ From (8.5) and
using symmetry, we readily have
$\mathbb{E}s_{1}^{1}s_{1}^{2}s_{2}^{1}s_{2}^{2}=0.$ Since we can write

$\displaystyle\mathbb{E}s_{1}^{1}s_{1}^{2}s_{2}^{1}s_{2}^{2}$
$\displaystyle=\mathbb{E}\bar{s}(w,u_{1},v_{1})\bar{s}(w,u_{2},v_{1})\bar{s}(w,u_{1},v_{2})\bar{s}(w,u_{2},v_{2})$
$\displaystyle=\mathbb{E}\bigl{[}\bigl{(}\mathbb{E}_{v}\bar{s}(w,u_{1},v)\bar{s}(w,u_{2},v)\bigr{)}^{2}\bigr{]}$

for $\bar{s}(w,u,v):=\int_{0}^{1}s(w,u,v,x)dx,$ it follows that almost surely,
$\mathbb{E}_{v}\bar{s}(w,u_{1},v)\bar{s}(w,u_{2},v)=0,$ which implies that
$\mathbb{E}[(\mathbb{E}_{u}\bar{s}(w,u,v))^{2}]=0.$ From this and Jensen’s
inequality, $\mathbb{E}[\bar{s}(w,u,v)^{2}]=0.$ Thus, $\bar{s}(w,u,v)=0$ a.s.
and this finishes the proof of our claim.

##  9 Establishing the Limiting Free Energy and Edge Alignments

###  9.1 Proof of Theorem 2.9

For any integer $k\geq 1,$ denote $[k]=\\{1,2,\ldots,k\\}.$ Let $X$ have
density

$\frac{\alpha}{2}\frac{\mathbbm{1}_{\\{|x|\geq 1\\}}}{|x|^{\alpha+1}}$

so that $X/N^{1/\alpha}$ has the same distribution as $J$. Set $Y=|X|$. For
$n\geq 1$, let $X_{1},X_{2},\ldots,X_{n}$ be i.i.d. copies of $X$ and denote
$Y_{i}=|X_{i}|$. Let $(I_{1},I_{2},\ldots,I_{n})$ be the random permutation of
$[n]$ such that

$|X_{I_{1}}|>|X_{I_{2}}|>\cdots>|X_{I_{n}}|.$

From (2.10), recall the PPP, $(\gamma_{j})_{j\geq 1}$, with intensity
$\nu(x)=\alpha x^{-(1+\alpha)}$ on $(0,\infty)$ ordered descendingly. Let
$\mathcal{A}_{\geq}$ be the following subset of $\mathbb{R}^{\infty}$,

$\mathcal{A}_{\geq}=\\{a=(a_{1},a_{2},\ldots):a_{1}\geq a_{2}\geq\cdots\geq
0\text{ such that }\lim_{i}a_{i}=0\\},$

equipped with the metric $d(a,a^{\prime})=\sup_{i\geq
1}|a_{i}-a_{i}^{\prime}|$, which ensures that $\mathcal{A}_{\geq}$ is a
complete, separable metric space. The convergence in distribution for
$\mathcal{A}_{\geq}$-valued random variables is equivalent to finite
dimensional weak convergence (see, for example, [10]).

In the following lemma, we collect some well-known results for the order
statistics of the heavy-tailed distribution, tailored to our special choice of
distribution (see [35] and [12, Lemma 2.4]). We include a proof for
completeness.

######  Lemma 9.1.

The following statements hold:

  1. (a)

For $0<\alpha<2$,

$n^{-1/\alpha}\bigl{(}Y_{I_{1}},Y_{I_{2}},\ldots,Y_{I_{n}}\bigr{)}\stackrel{{\scriptstyle
d}}{{=}}\bigl{(}\gamma_{n+1}/n)^{1/\alpha}(\gamma_{1}^{-1/\alpha},\gamma_{2}^{-1/\alpha},\ldots,\gamma_{n}^{-1/\alpha}\bigr{)}.$
(9.1)

Also,

$n^{-1/\alpha}\bigl{(}X_{1},X_{2},\ldots,X_{n}\bigr{)}\stackrel{{\scriptstyle
d}}{{=}}(\gamma_{n+1}/n)^{1/\alpha}\bigl{(}\varepsilon_{1}\gamma_{r(1)}^{-1/\alpha},\varepsilon_{2}\gamma_{r(2)}^{-1/\alpha},\ldots,\varepsilon_{n}\gamma_{r(n)}^{-1/\alpha}\bigr{)},$
(9.2)

where $\varepsilon_{1},\varepsilon_{2},\ldots,\varepsilon_{n}$ are i.i.d.
Rademacher variables and $r:[n]\to[n]$ is an independent uniform random
permutation, both independent of $(\gamma_{j})_{j\geq 1}$.

  2. (b)

For $0<\alpha<2$, as $n\to\infty$,

$n^{-1/\alpha}\bigl{(}Y_{I_{1}},Y_{I_{2}},\ldots,Y_{I_{n}},0,0,\ldots\bigr{)}\stackrel{{\scriptstyle
d}}{{\to}}\bigl{(}\gamma_{1}^{-1/\alpha},\gamma_{2}^{-1/\alpha},\ldots\bigr{)}\text{
on }\mathcal{A}_{\geq}.$

  3. (c)

Assume that $0<\alpha<1$. As $n\to\infty$,

$n^{-1/\alpha}\sum_{j=1}^{n}Y_{j}\stackrel{{\scriptstyle
d}}{{\to}}\sum_{j=1}^{\infty}\gamma_{j}^{-1/\alpha},$

where the infinite sum is finite a.s. Also, for any fixed integer $R\geq 1$,
as $n\to\infty,$

$n^{-1/\alpha}\Bigl{(}\sum_{j=1}^{R}Y_{I_{j}},\sum_{j=R+1}^{n}Y_{I_{j}}\Bigr{)}\stackrel{{\scriptstyle
d}}{{\to}}\Bigl{(}\sum_{j=1}^{R}\gamma_{j}^{-1/\alpha},\sum_{j=R+1}^{\infty}\gamma_{j}^{-1/\alpha}\Bigr{)}.$

###### Proof.

(a) It is easy to check that $Y\stackrel{{\scriptstyle d}}{{=}}U^{-1/\alpha}$,
where $U$ has uniform distribution on $[0,1]$. Therefore, by independence,

$(Y_{I_{1}},Y_{I_{2}},\ldots,Y_{I_{n}})\stackrel{{\scriptstyle
d}}{{=}}(U_{(1)}^{-1/\alpha},U_{(2)}^{-1/\alpha},\ldots,U_{(n)}^{-1/\alpha}),$

where $U_{(1)}<U_{(2)}<\cdots<U_{(n)}$ are order statistics of $n$ i.i.d.
uniform variables on $[0,1]$. (9.1) is now a consequence of the following
identity (see, for example, [23, Theorem 3.7.11]),

$\bigl{(}U_{(1)},U_{(2)},\ldots,U_{(n)}\bigr{)}\stackrel{{\scriptstyle
d}}{{=}}\bigl{(}\gamma_{1}/\gamma_{n+1},\gamma_{2}/\gamma_{n+1},\ldots,\gamma_{n}/\gamma_{n+1}\bigr{)}.$

Next, since $X$ has symmetric distribution, $\mathrm{sgn}(X)$ is a Rademacher
variable, independent of $Y=|X|$. Therefore,

$\bigl{(}X_{I_{1}},X_{I_{2}},\ldots,X_{I_{n}}\bigr{)}\stackrel{{\scriptstyle
d}}{{=}}\bigl{(}\varepsilon_{1}Y_{I_{1}},\varepsilon_{2}Y_{I_{2}},\ldots,\varepsilon_{n}Y_{I_{n}}\bigr{)},$

where $\varepsilon_{i}$’s are independent of $Y_{i}$’s. Also, the ranking
$(I_{1},I_{2},\ldots,I_{n})$ is independent of the order statistics
$(Y_{I_{1}},Y_{I_{2}},\ldots,Y_{I_{n}})$, which immediately yields (9.2) from
(9.1).

(b) By strong law of large numbers, $\gamma_{n+1}/n\to 1$ almost surely, which
yields the finite dimensional convergence of
$(\gamma_{n+1}/n)^{1/\alpha}(\gamma_{1}^{-1/\alpha},\gamma_{2}^{-1/\alpha},\ldots,\gamma_{n}^{-1/\alpha},0,0,\ldots)$
to $(\gamma_{1}^{-1/\alpha},\gamma_{2}^{-1/\alpha},\ldots)$.

(c) By (9.1),

$n^{-1/\alpha}\sum_{j=1}^{n}Y_{j}\stackrel{{\scriptstyle
d}}{{=}}(\gamma_{n+1}/n)^{1/\alpha}\sum_{j=1}^{n}\gamma_{j}^{-1/\alpha}.$

By strong law of large numbers, $\gamma_{j}/j\to 1$ almost surely as
$j\to\infty$. Thus, since $0<\alpha<1$, the series
$\sum_{j=1}^{\infty}\gamma_{j}^{-1/\alpha}$ is convergent almost surely. It
follows that

$(\gamma_{n+1}/n)^{1/\alpha}\sum_{j=1}^{n}\gamma_{j}^{-1/\alpha}\to\sum_{j=1}^{\infty}\gamma_{j}^{-1/\alpha}$

almost surely. The second part of (c) has a similar proof. ∎

Before proving the theorem, we need a simple lemma on weak convergence, whose
proof is omitted.

######  Lemma 9.2.

Let $(F_{N})_{N\geq 1},(W_{N,R})_{N,R\geq 1},(W_{N})_{N\geq 1}$ be sequences
of random variables defined on the same probability space such that

$W_{N,R}\leq F_{N}\leq W_{N}\ \ \text{ for any }N,R\geq 1.$

Suppose that for any fixed $R\geq 1$, $W_{N,R}\stackrel{{\scriptstyle
d}}{{\to}}W_{\infty,R}$ and $W_{N}\stackrel{{\scriptstyle
d}}{{\to}}W_{\infty}$ as $N\to\infty$. Also, assume that
$W_{\infty,R}\stackrel{{\scriptstyle d}}{{\to}}W_{\infty}$ as $R\to\infty$.
Then $F_{N}\stackrel{{\scriptstyle d}}{{\to}}W_{\infty}$.

###### Proof of Theorem 2.9.

Note that $Z_{N}\leq 2^{N}\exp(\beta\sum_{i<j}|J_{ij}|)$. Therefore, by Lemma
9.1(c),

$\frac{1}{N^{1/\alpha}}\ln Z_{N}\leq\frac{1}{N^{1/\alpha}}\Big{(}N\ln
2+\beta\sum_{i<j}|J_{ij}|\Big{)}\stackrel{{\scriptstyle
d}}{{\to}}\frac{\beta}{2^{1/\alpha}}\sum_{k=1}^{\infty}\gamma_{k}^{-1/\alpha}.$

For the lower bound, recall the sequence $(U(k,1),U(k,2))_{1\leq k\leq n}.$
For any fixed $R\geq 1$,

$\displaystyle Z_{N}$
$\displaystyle\geq\max_{\sigma}\exp\Bigl{(}\beta\sum_{i<j}J_{ij}\sigma_{i}\sigma_{j}\Bigr{)}$
$\displaystyle\geq\exp\Big{(}\beta\max_{\sigma}\sum_{k=1}^{R}J_{U(k,1),U(k,2)}\sigma_{U(k,1)}\sigma_{U(k,2)}-\beta\sum_{k=R+1}^{n}|J_{U(k,1),U(k,2)}|\Big{)}.$

Let $\mathcal{C}_{N,R}$ be the event that the edges $\\{(U(k,1),U(k,2)):1\leq
k\leq R\\}$ on the vertex set $[N]$ are all incident on distinct vertices.
Clearly, $\mathbb{P}(\mathcal{C}_{N,R})\to 1$ as $N\to\infty$. Given the
disorders $(J_{ij})$, on the event $\mathcal{C}_{N,R}$, we can choose $\sigma$
such that
$\mathrm{sgn}(\sigma_{I(k,1)}\sigma_{I(k,2)})=\mathrm{sgn}(J_{I(k,1),I(k,2)})$
for each $1\leq k\leq R$. Consequently, again by Lemma 9.1(c), we can write

$\displaystyle\frac{1}{N^{1/\alpha}}\ln Z_{N}$
$\displaystyle\geq\frac{\mathbbm{1}_{\mathcal{C}^{c}_{N,R}}}{N^{1/\alpha}}\ln
Z_{N}+\frac{\beta}{N^{1/\alpha}}\Big{(}\sum_{k=1}^{R}|J_{U(k,1),U(k,2)}|-\sum_{k=R+1}^{n}|J_{U(k,1),U(k,2)}|\Big{)}\mathbbm{1}_{\mathcal{C}_{N,R}}$
$\displaystyle\stackrel{{\scriptstyle
d}}{{\to}}\frac{\beta}{2^{1/\alpha}}\Big{(}\sum_{k=1}^{R}\gamma_{k}^{-1/\alpha}-\sum_{k=R+1}^{\infty}\gamma_{k}^{-1/\alpha}\Big{)}.$

The proof now follows from Lemma 9.2. ∎

###  9.2 Proof of Theorem 2.10

Recall the random edges $(U(k,1),U(k,2))$ for $1\leq k\leq n=N(N-1)/2$ defined
through (2.11). We say the edge $(i,j)$ has rank $k$, and write $r(i,j)=k$, if
$|J_{ij}|$ is the $k$-th largest among $(|J_{ij}|)_{1\leq i<j\leq N}$. In
other words, for $1\leq i<j\leq N$, $r(i,j)=k$ if and only if
$(i,j)=(U(k,1),U(k,2)).$ The rank function $r:\\{(i,j):1\leq i<j\leq
N\\}\to\\{1,\ldots,n\\}$ is a uniform bijection. We adopt the convention that
$r(i,j)=r(j,i)$ if $i>j$ and $r(i,i)=\infty$ for all $i$. The proof of Theorem
2.10 relies crucially on the following properties of the disorder matrix.

######  Lemma 9.3.

Let $A$ be a $N\times N$ matrix symmetric with $A_{ii}=0$ and
$A_{ij}=|J_{ij}|$ for $1\leq i<j\leq N$. Let $\delta>0$ be an arbitrary small
constant and let $\kappa=\min(1/2,1-\alpha)-\delta.$ Take $R=N^{\kappa}$ and
$T=N^{1/2+\delta/2}$. Then

  * (a)

With probability going to one, no row of $A$ has two distinct entries with
ranks at most $R$ and $T$ respectively. As a result, with probability going to
one, the number of entries with rank at most $R$ in each row of $A$ is either
zero or one.

  * (b)

With probability going to one, the rows of $A$ satisfy that for all $1\leq
i\leq N,$ whenever there exists some $j$ such that $r(i,j)\leq R$, the
following inequality holds

$\sum_{j:r(i,j)\leq R}|J_{ij}|-\sum_{j:r(i,j)>R}|J_{ij}|\geq
10^{-1/\alpha}(N/R)^{1/\alpha}.$

###### Proof.

(a) Let $D_{1},D_{2},\ldots,D_{N-1}$ be i.i.d. uniform variables on $[n]$. The
probability that they are all distinct is given by

$\prod_{i=0}^{N-2}\Bigl{(}1-\frac{i}{n}\Bigr{)}\geq\exp\Bigl{(}-\frac{2}{n}\sum_{i=0}^{N-2}i\Bigr{)}\geq
e^{-2},$

where we used the inequality that $1-x\geq e^{-2x}$ for $0<x\leq 1/2$. Observe
that for any row $i$, the vector $(r(i,j))_{j\neq i}$ has the same
distribution as $(D_{1},\ldots,D_{N-1})$ conditioned on them being all
distinct. Therefore, for each $i$,

$\displaystyle\mathbb{P}\bigl{(}(r(i,j))_{j\neq
i}\in\cdot\bigr{)}\leq\frac{\mathbb{P}((D_{1},\ldots,D_{N-1})\in\cdot)}{\mathbb{P}(\text{$D_{1},\ldots,D_{N-1}$
are all distinct})}.$ (9.3)

Now since the probability of the event that there exist $j_{1}\neq j_{2}$ such
that $D_{j_{1}}\leq R$ and $D_{j_{2}}\leq T$ is bounded above by
$(N-1)^{2}(R/n)(T/n)=4RTN^{-2},$ from the display above, the probability that
there exist $j_{1}\neq j_{2}$ such that $r(i,j_{1})\leq R$ and $r(i,j_{2})\leq
T$ is bounded above by $4e^{2}RTN^{-2}.$ By a union bound, the probability of
the event that there is at least one row of $A$ having two distinct entries
with ranks at most $R$ and $T$ is bounded above by $4e^{2}RTN^{-1},$ which
goes to $0$. This proves part (a).

(b) First, we claim that with probability going to one, no row of $A$ has at
least $\lceil\ln N\rceil$ entries with ranks at most $N$. To see this, from
(9.3), for a particular row, such probability is bounded above by

$e^{2}{N-1\choose\lceil\ln N\rceil}\Big{(}\frac{N}{n}\Big{)}^{\lceil\ln
N\rceil}\leq e^{2}\Big{(}\frac{e(N-1)}{\lceil\ln N\rceil}\Big{)}^{\lceil\ln
N\rceil}\Big{(}\frac{N}{n}\Big{)}^{\lceil\ln
N\rceil}=e^{2}\Big{(}\frac{2e}{\lceil\ln N\rceil}\Big{)}^{\lceil\ln N\rceil},$

where we have used the bound

${m\choose k}\leq\Bigl{(}\frac{em}{k}\Bigr{)}^{k}$

for $0<k\leq m.$ The claim now follows from taking a union bound over $N$
rows.

Next, by Lemma 9.1(a), we can write

$|J_{ij}|=(n/N)^{1/\alpha}(\gamma_{n+1}/n)^{1/\alpha}\gamma_{r(i,j)}^{-1/\alpha},\,\,1\leq
i\neq j\leq N,$

where $(\gamma_{j})_{j}$ is independent of $r$. By the strong law of large
numbers, for any sequence $j_{0}(N)\to\infty$, with probability tending to
one,

$0.9j^{-1/\alpha}\leq\gamma_{j}^{-1/\alpha}\leq 1.1j^{-1/\alpha},\,\,\forall
j\geq j_{0}(N).$

Now suppose that $i$-th row of $A$ has an entry with rank at most $R$. By
part(a) of the lemma and our claim above, with probability tending to one, the
following three statements hold: (i) the number of the entries in that row
with rank at most $R$ is exactly one, (ii) the rest of the entries in the row
has rank at least $T+1$ and (iii) except at most $\lceil\ln N\rceil$ entries,
all other entries in the row has rank at least $N+1$. On the intersection of
these events,

$\displaystyle\sum_{j:r(i,j)\leq R}|J_{ij}|-\sum_{j:r(i,j)>R}|J_{ij}|$
$\displaystyle\geq(\gamma_{n+1}/N)^{1/\alpha}\big{(}0.9R^{-1/\alpha}-\ln
N\cdot 1.1T^{-1/\alpha}-N\cdot 1.1N^{-1/\alpha}\big{)}$
$\displaystyle\geq(0.9)2^{-1/\alpha}(N-1)^{1/\alpha}\big{(}0.9R^{-1/\alpha}-\ln
N\cdot 1.1T^{-1/\alpha}-N\cdot 1.1N^{-1/\alpha}\big{)}.$ (9.4)

From the choice of $\kappa$, we have

$\frac{T^{-1/\alpha}\ln
N}{R^{-1/\alpha}}\leq\frac{(N^{1/2+\delta/2})^{-1/\alpha}\ln
N}{(N^{1/2-\delta})^{-1/\alpha}}\to
0\,\,\mbox{and}\,\,\frac{N^{1-1/\alpha}}{R^{-1/\alpha}}\leq\frac{N^{-(1-\alpha)/\alpha}}{N^{-((1-\alpha)-\delta)/\alpha}}\to
0.$

Thus, for large $N$, (9.2) is bounded below by
$10^{-1/\alpha}(N/R)^{1/\alpha}$, completing the proof. ∎

###### Proof of Theorem 2.10.

Let $\mathcal{C}_{N}$ be the event that the disorders $(J_{ij})$ satisfy the
statements (a) and (b) of Lemma 9.3 and $\mathbb{P}(\mathcal{C}_{N})\to 1$. We
assume that the event $\mathcal{C}_{N}$ holds for the rest of the proof. First
of all, define the subset $\mathcal{V}$ of $[N]$ as

$\mathcal{V}=\\{U(k,p):p\in\\{1,2\\},1\leq k\leq R\\}.$

By the statement (a) of $\mathcal{C}_{N}$, the edges $(U(k,1),U(k,2))$ for
$1\leq k\leq R$ cannot share any common vertex. Indeed, if $(U(k,1),U(k,2))$
and $(U(k^{\prime},1),U(k^{\prime},2))$ for some $k,k^{\prime}\leq R$ share a
vertex, say $a,$ then the $a$-th row of $A$ will contain two entries that are
ranked $k,k^{\prime}\leq R$, a contradiction. Hence, the elements in
$\mathcal{V}$ are distinct and the size of $\mathcal{V}$ is $2R$.

Write $\sigma\in\\{-1,1\\}^{N}$ as
$\sigma=(\sigma_{\mathcal{V}},\sigma_{\mathcal{V}^{c}})\in\\{-1,1\\}^{\mathcal{V}}\times\\{-1,1\\}^{[N]\setminus\mathcal{V}^{c}}$.
For any $B\subseteq[R]$, define

$\Sigma_{\mathcal{V}}^{B}=\Bigl{\\{}\tau\in\\{-1,1\\}^{\mathcal{V}}:\chi_{k}(\tau)=0\text{
for all }k\in B\text{ and }\chi_{k}(\tau)=1\text{ for all }k\in
B^{c}:=[R]\setminus B\Bigr{\\}}$

and set

$\Sigma_{\mathcal{V}}=\Sigma_{\mathcal{V}}^{\emptyset}=\Bigl{\\{}\tau\in\\{-1,1\\}^{\mathcal{V}}:\chi_{k}(\tau)=1\text{
for all }1\leq k\leq R\Bigr{\\}},$

where, with a slightly abuse of notation, for
$\tau\in\\{-1,1\\}^{\mathcal{V}}$ and $k\in[R],$

$\chi_{k}(\tau):=\mathbbm{1}_{\\{\mbox{sgn}(J_{U(k,1),U(k,2)}\tau_{U(k,1)}\tau_{U(k,2)})=1\\}}.$

Observe that for $\sigma\in\\{-1,1\\}^{N},$

$\chi_{k}(\sigma)=1,\ \forall\ 1\leq k\leq R\ \ \Leftrightarrow\ \
\sigma_{\mathcal{V}}\in\Sigma_{\mathcal{V}}.$ (9.5)

Note that $|\Sigma_{\mathcal{V}}^{B}|=2^{2R}$ for any $B\subseteq[R]$ . Let
$f_{B}$ be a bijection from $\Sigma_{\mathcal{V}}^{B}$ to
$\Sigma_{\mathcal{V}}$ defined as follows. For each $1\leq k\leq R,$
$f_{B}(\tau)_{U(k,2)}=\tau_{U(k,2)}$ and $f_{B}(\tau)_{U(k,1)}=-\tau_{U(k,1)}$
or $\tau_{U(k,1)}$ depending on whether $k\in B$ or $k\in B^{c}.$

Now, for any $\sigma$ such that $\sigma_{\mathcal{V}}\in\Sigma_{\mathcal{V}}$
and for any $\emptyset\neq B\subseteq[R]$, using the convention that
$J_{ij}=J_{ji}$ and $J_{ii}=0$, observe that the spin configuration,
$\bigl{(}f_{B}^{-1}(\sigma_{\mathcal{V}}),\sigma_{{\mathcal{V}}^{c}}\bigr{)},$
is obtained by keeping the spins in $\sigma$ fixed on the coordinates
$(U(k,1):k\in B^{c})$, $(U(k,2):k\in[R]),$ and $k\in\mathcal{V}^{c}$ and
flipping the sign of $\sigma$ on the coordinates $(U(k,1):k\in B).$ With this,
we readily check that

$\displaystyle-\bigl{(}H_{N}(f_{B}^{-1}(\sigma_{\mathcal{V}}),\sigma_{{\mathcal{V}}^{c}})-H_{N}(\sigma)\bigr{)}$
$\displaystyle=2\beta\sum_{k\in
B}\sum_{j\in[N]}J_{U(k,1),j}(-\sigma_{U(k,1)})\sigma_{j}$
$\displaystyle=-2\beta\sum_{k\in
B}\Bigl{(}|J_{U(k,1),U(k,2)}|+\sigma_{U(k,1)}\sum_{j:j\neq
U(k,2)}J_{U(k,1),j}\sigma_{j}\Bigr{)}$ $\displaystyle\leq-2\beta\sum_{k\in
B}\Bigl{(}|J_{U(k,1),U(k,2)}|-\sum_{j:j\neq U(k,2)}|J_{U(k,1),j}|\Bigr{)}$
$\displaystyle\leq-2\beta|B|10^{-1/\alpha}(N/R)^{1/\alpha},$

where the last inequality follows from condition (a) and (b) of
$\mathcal{C}_{N}$. Therefore, for $N$ sufficiently large, on the event
$\mathcal{C}_{N}$,

$\displaystyle\sum_{\emptyset\neq
B\subseteq[R]}\max_{\sigma:\sigma_{\mathcal{V}}\in\Sigma_{\mathcal{V}}}\frac{e^{-H_{N}(f_{B}^{-1}(\sigma_{\mathcal{V}}),\sigma_{{\mathcal{V}}^{c}})}}{e^{-H_{N}(\sigma)}}$
$\displaystyle\leq\bigl{(}1+e^{-2\beta
10^{-1/\alpha}(N/R)^{1/\alpha}}\bigr{)}^{R}-1\leq e^{Re^{-2\beta
10^{-1/\alpha}(N/R)^{1/\alpha}}}-1$ $\displaystyle\leq 2e^{-2\beta
10^{-1/\alpha}(N/R)^{1/\alpha}+\ln R}\leq 2e^{-2\beta
10^{-1/\alpha}N^{1/(2\alpha)}},$ (9.6)

where we have used the elementary inequalities $1+x\leq e^{x},x\in\mathbb{R}$
and $e^{x}\leq 1+2x,0\leq x<1$. Now from (9.5), we obtain

$\displaystyle 1-\langle\mathbbm{1}_{\\{\chi_{k}=1,\ 1\leq k\leq R\\}}\rangle$
$\displaystyle=\frac{\sum_{\sigma:\sigma_{\mathcal{V}}\not\in\Sigma_{\mathcal{V}}}e^{-H_{N}(\sigma)}}{\sum_{\sigma:\sigma_{\mathcal{V}}\in\Sigma_{\mathcal{V}}}e^{-H_{N}(\sigma)}+\sum_{\sigma:\sigma_{\mathcal{V}}\not\in\Sigma_{\mathcal{V}}}e^{-H_{N}(\sigma)}}$
$\displaystyle\leq\frac{\sum_{\emptyset\neq
B\subseteq[R]}\sum_{\sigma:\sigma_{\mathcal{V}}\in\Sigma^{B}_{\mathcal{V}}}e^{-H_{N}(\sigma_{\mathcal{V}},\sigma_{{\mathcal{V}}^{c}})}}{\sum_{\sigma:\sigma_{\mathcal{V}}\in\Sigma_{\mathcal{V}}}e^{-H_{N}(\sigma)}}$
$\displaystyle\leq\sum_{\emptyset\neq
B\subseteq[R]}\max_{\sigma:\sigma_{\mathcal{V}}\in\Sigma_{\mathcal{V}}}\frac{e^{-H_{N}(f_{B}^{-1}(\sigma_{\mathcal{V}}),\sigma_{{\mathcal{V}}^{c}})}}{e^{-H_{N}(\sigma)}}\leq
2e^{-2\beta 10^{-1/\alpha}N^{1/(2\alpha)}}.$

The theorem now follows from dominated convergence theorem. ∎

##  10 Extension for General Power-Law Distributions

Consider the Lévy Hamiltonian, (2.2), equipped with the disorder
$\tilde{J}_{ij}=\tilde{X}_{ij}/a_{N}$ for i.i.d.
$\tilde{X}_{ij}\sim\tilde{X}$, where $\tilde{X}$ and $a_{N}$ are defined in
Section 2.2.1. In this section, we will establish Theorem 2.11, i.e., the
statements of Theorems 2.1, 2.4, 2.7, 2.9, and 2.10 hold for the general Lévy
model. We need the following technical lemma for Theorems 2.1, 2.4, and 2.7,
whose proof is deferred to Appendix C.

######  Lemma 10.1.

Suppose $0<\alpha<2$. Let $f:\mathbb{R}\to\mathbb{R}$ be continuously
differentiable and let $\bar{f}(x):=(f(x)+f(-x))/2$ for $x>0$.

  * $(i)$

Suppose $f(0)=0$, $|f^{\prime}(x)|=O(|x|^{\delta})$ as $x\to 0$, and
$|f^{\prime}(x)|=O(|x|^{\varepsilon})$ as $x\to\pm\infty$ for some
$-\infty<\varepsilon<\alpha-1<\delta$. Then

$\lim_{N\to\infty}N\mathbb{E}f(\tilde{J})=\alpha\int_{0}^{\infty}\frac{\bar{f}(x)}{x^{1+\alpha}}dx.$

  * $(ii)$

Let $\eta>0.$ Suppose $|f^{\prime}(x)|=O(|x|^{\varepsilon})$ as
$x\to\pm\infty$ for some $-\infty<\varepsilon<\alpha-1$. Then

$\lim_{N\to\infty}N\mathbb{E}f(\tilde{J})\mathbbm{1}_{|\tilde{J}|\geq\eta}=\alpha\int_{\eta}^{\infty}\frac{\bar{f}(x)}{x^{1+\alpha}}dx.$

  * $(iii)$

Let $\eta>0.$ Suppose $f(0)=0$ and $|f^{\prime}(x)|=O(|x|^{\delta})$ as $x\to
0$ for some $\alpha-1<\delta<\infty$. Then

$\lim_{N\to\infty}N\mathbb{E}f(\tilde{J})\mathbbm{1}_{|\tilde{J}|\leq\eta}=\alpha\int_{0}^{\eta}\frac{\bar{f}(x)}{x^{1+\alpha}}dx.$

###  10.1 Extension of Theorems 2.1 and 2.7

Let $\tilde{F}_{N}$ be the free energies associated to the general Lévy model,
i.e.,

$\tilde{F}_{N}=\frac{1}{N}\ln\sum_{\sigma}e^{\beta\sum_{i<j}\tilde{J}_{ij}\sigma_{i}\sigma_{j}}.$

To establish the universality for Theorems 2.1 and 2.7, it suffices to show

######  Theorem 10.2 (Universality of free energy).

For $1<\alpha<2$ and $\beta>0$, we have

$\limsup_{N\to\infty}|\mathbb{E}F_{N}-\mathbb{E}\tilde{F}_{N}|=0.$

######  Remark 10.3.

[47, Proposition 1.2] established a quantitative universality for general SK-
type free energies in terms of the statistics of disorder distributions. The
following proof adapts a similar approach.

###### Proof.

Consider an arbitrary enumeration $\\{(i_{t},j_{t}):1\leq t\leq N(N-1)/2\\}$
of $\\{(i,j):1\leq i<j\leq N\\}$. For any $1\leq t\leq N(N-1)/2+1$, define

$\displaystyle\tilde{F}_{N,t}=\frac{1}{N}\ln\sum_{\sigma}\exp\Bigl{(}\beta\sum_{r<t}J_{i_{r}j_{r}}\sigma_{i_{r}}\sigma_{j_{r}}+\beta\sum_{r\geq
t}\tilde{J}_{i_{r}j_{r}}\sigma_{i_{r}}\sigma_{j_{r}}\Bigr{)}.$

Write

$\displaystyle F_{N}-\tilde{F}_{N}$
$\displaystyle=\sum_{t=1}^{N(N-1)/2}\bigl{(}\tilde{F}_{N,t+1}-\tilde{F}_{N,t}\bigr{)}$
$\displaystyle=\frac{1}{N}\sum_{t=1}^{N(N-1)/2}\bigl{(}\ln\bigl{\langle}e^{\beta
J_{i_{t}j_{t}}\sigma_{i_{t}}\sigma_{j_{t}}}\bigr{\rangle}_{t}-\ln\bigl{\langle}e^{\beta\tilde{J}_{i_{t}j_{t}}\sigma_{i_{t}}\sigma_{j_{t}}}\bigr{\rangle}_{t}\bigr{)},$

where $\langle\cdot\rangle_{t}$ is the Gibbs expectation associated to the
Hamiltonian

$\tilde{H}_{N,t}(\sigma)=\beta\sum_{r<t}J_{i_{r}j_{r}}\sigma_{i_{r}}\sigma_{j_{r}}+\beta\sum_{r>t}\tilde{J}_{i_{r}j_{r}}\sigma_{i_{r}}\sigma_{j_{r}}.$

Now, similar to (6.5), we can write

$\displaystyle\ln\bigl{\langle}e^{\beta\tilde{J}_{i_{t}j_{t}}\sigma_{i_{t}}\sigma_{j_{t}}}\bigr{\rangle}_{t}$
$\displaystyle=\ln\cosh\beta\tilde{J}_{i_{t}j_{t}}-\sum_{r=1}^{\infty}\frac{(-1)^{r}}{r}\tanh^{r}(\beta\tilde{J}_{i_{t}j_{t}})\cdot\langle\sigma_{i_{t}}\sigma_{j_{t}}\rangle_{t}^{r}.$

Let $K>0$ and $k\geq 1.$ From this and using the independence between
$\tilde{J}_{i_{t}j_{t}}$ and $\langle\cdot\rangle_{t},$

$\displaystyle\mathbb{E}\ln\bigl{\langle}e^{\beta\tilde{J}_{i_{t}j_{t}}\sigma_{i_{t}}\sigma_{j_{t}}}\bigr{\rangle}_{t}$
$\displaystyle=\mathbb{E}\bigl{[}\ln\bigl{\langle}e^{\beta\tilde{J}\sigma_{i_{t}}\sigma_{j_{t}}}\bigr{\rangle}_{t};|\tilde{J}|>K\bigr{]}+\mathbb{E}\bigl{[}\ln\cosh\beta\tilde{J};|\tilde{J}|\leq
K\bigr{]}$
$\displaystyle-\sum_{r=1}^{\infty}\frac{1}{2r}\mathbb{E}\bigl{[}\tanh^{2r}(\beta\tilde{J});|\tilde{J}|\leq
K\bigr{]}\cdot\mathbb{E}\langle\sigma_{i_{t}}\sigma_{j_{t}}\rangle_{t}^{2r},$

where $\tilde{J}$ is independent of $\langle\cdot\rangle_{t}.$ Note that
$|\ln\langle
e^{\beta\tilde{J}\sigma_{i_{t}}\sigma_{j_{t}}}\rangle_{t}|\leq\beta|\tilde{J}|$
and

$\displaystyle\sum_{r=k+1}^{\infty}\frac{1}{2r}\mathbb{E}\bigl{[}\tanh^{2r}(\beta\tilde{J});|\tilde{J}|\leq
K\bigr{]}\cdot\mathbb{E}\langle\sigma_{i_{t}}\sigma_{j_{t}}\rangle_{t}^{2r}$
$\displaystyle\leq\mathbb{E}\bigl{[}\tanh^{2}(\beta\tilde{J});|\tilde{J}|\leq
K\bigr{]}\sum_{r=k}^{\infty}\tanh^{2r}(\beta K)$
$\displaystyle=\mathbb{E}\bigl{[}\tanh^{2}(\beta\tilde{J});|\tilde{J}|\leq
K\bigr{]}\frac{\tanh^{2k}(\beta K)}{1-\tanh^{2}(\beta K)}.$

Consequently,

$\displaystyle\Bigl{|}\mathbb{E}\ln\bigl{\langle}e^{\beta\tilde{J}\sigma_{i_{t}}\sigma_{j_{t}}}\bigr{\rangle}_{t}-\mathbb{E}\bigl{[}\ln\cosh\beta\tilde{J};|\tilde{J}|\leq
K\bigr{]}+\sum_{r=1}^{k}\frac{1}{2r}\mathbb{E}\bigl{[}\tanh^{2r}(\beta\tilde{J});|\tilde{J}|\leq
K\bigr{]}\cdot\mathbb{E}\langle\sigma_{i_{t}}\sigma_{j_{t}}\rangle_{t}^{2r}\Bigr{|}$
$\displaystyle\leq\beta\mathbb{E}\bigl{[}|\tilde{J}|;|\tilde{J}|>K\bigr{]}+\mathbb{E}\bigl{[}\tanh^{2}(\beta\tilde{J});|\tilde{J}|\leq
K\bigr{]}\frac{\tanh^{2k}(\beta K)}{1-\tanh^{2}(\beta K)}.$

Using Lemma 10.1, we have

$\displaystyle\limsup_{k\to\infty}\limsup_{N\to\infty}\Bigl{|}\frac{1}{N}\sum_{t=1}^{N(N-1)/2}\mathbb{E}\ln\bigl{\langle}e^{\beta\tilde{J}_{i_{t}j_{t}}\sigma_{i_{t}}\sigma_{j_{t}}}\bigr{\rangle}_{t}-\frac{\alpha}{2}\int_{0}^{K}\frac{\ln\cosh(\beta
x)}{x^{\alpha+1}}dx$
$\displaystyle\qquad\qquad+\sum_{r=1}^{k}\frac{\alpha}{2r}\int_{0}^{K}\frac{\tanh^{2r}(\beta
x)}{x^{\alpha+1}}dx\cdot\frac{1}{N^{2}}\sum_{t=1}^{N(N-1)/2}\mathbb{E}\langle\sigma_{i_{t}}\sigma_{j_{t}}\rangle_{t}^{2r}\Bigr{|}$
$\displaystyle\leq\limsup_{k\to\infty}\Bigl{(}\frac{\alpha\beta}{2}\int_{K}^{\infty}\frac{1}{x^{\alpha}}dx+\frac{\alpha}{2}\int_{0}^{K}\frac{\tanh^{2}(\beta
x)}{x^{\alpha+1}}dx\cdot\frac{\tanh^{2k}(\beta K)}{1-\tanh^{2}(\beta
K)}\Bigr{)}$
$\displaystyle=\frac{\alpha\beta}{2}\int_{K}^{\infty}\frac{1}{x^{\alpha}}dx=\frac{\alpha\beta}{2(\alpha-1)}K^{1-\alpha}.$

Note that by substituting $\tilde{J}$ with $J$, the same computation still
carries through and yields the same inequality. As a result,

$\displaystyle\limsup_{N\to\infty}\Bigl{|}\frac{1}{N}\sum_{t=1}^{N(N-1)/2}\bigl{(}\mathbb{E}\ln\bigl{\langle}e^{\beta\tilde{J}_{i_{t}j_{t}}\sigma_{i_{t}}\sigma_{j_{t}}}\bigr{\rangle}_{t}-\mathbb{E}\ln\bigl{\langle}e^{\beta
J_{i_{t}j_{t}}\sigma_{i_{t}}\sigma_{j_{t}}}\bigr{\rangle}_{t}\bigr{)}\Bigr{|}\leq\frac{\alpha\beta}{\alpha-1}K^{1-\alpha}.$

Since this holds for any $K>0,$ sending $K\to\infty$ completes our proof. ∎

###  10.2 Extension of Theorem 2.2

#### Extension of Theorem 4.1

The modified statement should be as follows.

######  Theorem 10.4.

For $0<\alpha<2$ and $\beta>0$, we have

$\frac{1}{a_{N}^{-1}a_{{N\choose 2}}}\Bigl{(}\ln\bar{Z}_{N}-{N\choose
2}\mathbb{E}\Bigl{(}\ln\cosh(\beta\tilde{J})\mathbbm{1}_{\\{a_{N}\ln\cosh(\beta\tilde{J})\leq
a_{N\choose 2}\\}}\Bigr{)}\Bigr{)}\stackrel{{\scriptstyle d}}{{\to}}\beta
Y_{\alpha},$

as $N\to\infty$, where the characteristic function of $Y_{\alpha}$ is given by
(4.1).

###### Sketch of proof.

Let $\varepsilon>0$ and recall $n=N(N-1)/2$. Define

$\displaystyle
S_{1}^{\varepsilon}=\sum_{i<j}a_{N}\ln\cosh(\beta\tilde{J}_{ij})\mathbbm{1}_{\\{a_{N}\ln\cosh(\beta\tilde{J}_{ij})\leq\varepsilon
a_{n}\\}}\ \mathrm{and}$ $\displaystyle
S_{2}^{\varepsilon}=\sum_{i<j}a_{N}\ln\cosh(\beta\tilde{J}_{ij})\mathbbm{1}_{\\{a_{N}\ln\cosh(\beta\tilde{J}_{ij})>\varepsilon
a_{n}\\}},$

so that

$a_{N}\ln\bar{Z}_{N}=S_{1}^{\varepsilon}+S_{2}^{\varepsilon}.$

Further define auxiliary quantities

$\displaystyle
b_{N}=n\mathbb{E}\bigl{(}a_{N}\ln\cosh(\beta\tilde{J})\mathbbm{1}_{\\{a_{N}\ln\cosh(\beta\tilde{J})\leq
a_{n}\\}}\bigr{)}\ \mathrm{and}$ $\displaystyle
c_{N}=n\mathbb{E}\bigl{(}a_{N}\ln\cosh(\beta\tilde{J})\mathbbm{1}_{\\{\varepsilon
a_{n}<a_{N}\ln\cosh(\beta\tilde{J})\leq a_{n}\\}}\bigr{)}.$

Define the centered random variable

$\displaystyle\bar{S}_{1}^{\varepsilon}\colonequals
S_{1}^{\varepsilon}-(b_{N}-c_{N})$
$\displaystyle=\sum_{i<j}\bigl{(}a_{N}\ln\cosh(\beta\tilde{J})\mathbbm{1}_{\\{a_{N}\ln\cosh(\beta\tilde{J})\leq\varepsilon
a_{n}\\}}$
$\displaystyle\qquad-\mathbb{E}a_{N}\ln\cosh(\beta\tilde{J})\mathbbm{1}_{\\{a_{N}\ln\cosh(\beta\tilde{J})\leq\varepsilon
a_{n}\\}}\bigr{)}.$

Combining the above definitions yields the decomposition

$\displaystyle\frac{\ln\bar{Z}_{N}-a_{N}^{-1}b_{N}}{a_{N}^{-1}a_{n}}=\frac{\bar{S}_{1}^{\varepsilon}}{a_{n}}+\frac{S_{2}^{\varepsilon}-c_{N}}{a_{n}}.$
(10.1)

The first term in (10.1) can be controlled uniformly in $N$:

$\displaystyle\limsup_{N\to\infty}\mathbb{E}(\bar{S}_{1}^{\varepsilon}/a_{n})^{2}\leq\frac{2\beta^{\alpha}\varepsilon^{2-\alpha}}{2-\alpha}.$
(10.2)

Next, the second term in (10.1) converges to

$\displaystyle\lim_{N\to\infty}\mathbb{E}\exp(itS_{2}^{\varepsilon}/a_{n})=\exp\Bigl{(}\beta^{\alpha}\int_{\varepsilon}^{\infty}(e^{itx}-1)\alpha
x^{-1-\alpha}dx\Bigr{)}.$ (10.3)

Setting
$I_{N}(\varepsilon)\colonequals\\{(i,j):a_{N}\ln\cosh(\beta\tilde{J}_{ij})>\varepsilon
a_{n}\\}$, we can expect (10.3) since given $|I_{N}(\varepsilon)|=k$, the
quantity $S_{2}^{\varepsilon}/a_{n}$ is a sum of $k$ independent random
variables supported on the interval $[\varepsilon,\infty)$ with the tail
distribution converging to $\varepsilon^{\alpha}x^{-\alpha}$ for
$x\geq\varepsilon$, and
$|I_{N}(\varepsilon)|\sim\mathrm{Bin}(n,\mathbb{P}(a_{N}\ln\cosh(\beta\tilde{J})>\varepsilon
a_{n}))$. Lastly, a straightforward computation using the relation
$\lim_{N\to\infty}n\mathbb{P}(a_{N}\ln\cosh(\beta\tilde{J})>\varepsilon
a_{n})=\beta^{\alpha}\varepsilon^{-\alpha}$ shows

$\displaystyle\lim_{N\to\infty}\frac{c_{N}}{a_{n}}=\beta^{\alpha}\int_{\varepsilon}^{1}x\alpha
x^{-1-\alpha}dx.$ (10.4)

From (10.3) and (10.4), we have

$\displaystyle\lim_{N\to\infty}\mathbb{E}\exp\Bigl{(}it\frac{S_{2}^{\varepsilon}-c_{N}}{a_{n}}\Bigr{)}=\exp\Bigl{(}\beta^{\alpha}\int_{1}^{\infty}(e^{itx}-1)\alpha
x^{-1-\alpha}dx+\beta^{\alpha}\int_{\varepsilon}^{1}(e^{itx}-1-itx)\alpha
x^{-1-\alpha}dx\Bigr{)}.$ (10.5)

Note that the second term on the right-hand side converges as $\varepsilon\to
0$ since $e^{itx}-1-itx=O(x^{2})$ as $x\to 0$.

Combining (10.1), (10.2), and (10.5), it is straightforward to see that Lemma
4.4 yields the desired result. ∎

#### Extension of Theorem 4.2

For $\varepsilon>0$, from (C.1),

$\tilde{J}\mathbbm{1}_{\\{|\tilde{J}|>\varepsilon\\}}\stackrel{{\scriptstyle
d}}{{=}}Bg_{\varepsilon,N},$

where $B\sim\mathrm{Ber}((1+o(1))\varepsilon^{-\alpha}/N)$ and
$g_{\varepsilon,N}$ has a symmetric distribution with the tail probability
$\mathbb{P}(|g_{\varepsilon,N}|>x)=(\varepsilon/x)^{\alpha}L(a_{N}x)/L(a_{N}\varepsilon),x>\varepsilon$
and they are independent. It is easy to see that
$g_{\varepsilon,N}\stackrel{{\scriptstyle d}}{{\to}}g_{\varepsilon}$ as
$N\to\infty$, where $g_{\varepsilon}$ is as given in (4.9).

Choose $\eta>0$ small such that $(1+\eta)\beta/\beta_{\alpha}<1$. The
inequality in the statement of Lemma 4.7 should read as

$\sup_{N\geq 1}\mathbb{E}\Bigl{(}\sum_{\partial\Gamma=\emptyset,|\Gamma|\geq
m}w(\Gamma)\Bigr{)}^{2}\leq
Ce^{\sqrt{2m}}((1+\eta)\beta/\beta_{\alpha})^{\alpha m},\,\,m\geq 1.$

Similarly, the claim relevant to (4.6) is now that for sufficiently large $N$,
we have

$\sum_{\gamma:|\gamma|=m}\mathbb{E}w(\gamma)^{2}\leq\frac{1}{2m}((1+\eta)\beta/\beta_{\alpha})^{\alpha
m},$

which in the proof we use that

$\limsup_{N\to\infty}N\mathbb{E}\tanh^{2}(\beta
J)\leq((1+\eta)\beta/\beta_{\alpha})^{\alpha}$

as apparent from Lemma 10.1$(i)$.

The proof of Lemma 4.8 remains essentially unchanged.

As for the proof of main Theorem 4.2, the convergence in distribution given by
(4.10) holds since Lemmas 4.5 and 4.6 continue to be true if we change the
parameter $p$ with $p_{N}=p(1+o(1))$, and the rest of the argument remains the
same.

###  10.3 Extension of Theorem 2.4

We recall two useful facts that $a_{N}=N^{1/\alpha}\ell(N)$ for some slowly
varying function $\ell$ at $\infty$ and that slowly varying functions decay or
grow at most sub-polynomially fast at $\infty$, see [44]. Moreover, $X$ is
$L^{p}$-integrable for any $0<p<\alpha$.

#### Extension of (2.4)

We adapt the argument in Section 5.1. Recall the sequence $\gamma_{\ell}$ from
Proposition 5.1. First of all, we claim that Proposition 5.1 also holds in the
limit,

$\displaystyle\lim_{N\to\infty}\Bigl{|}\mathbb{E}\frac{1}{N}\sum_{i<j}\tilde{J}_{ij}\langle\sigma_{i}\sigma_{j}\rangle-\frac{c_{1}}{2}\bigl{(}1-\mathbb{E}\bigl{\langle}R_{\tilde{L}}^{2}\bigr{\rangle}\bigr{)}\Bigr{|}=0,$

where for $\ell\geq 1$,
$c_{\ell}:=N\mathbb{E}\tilde{J}\tanh^{\ell}(\beta\tilde{J})$ with
$\lim_{N\to\infty}c_{\ell}=\gamma_{\ell}$ and $\tilde{L}$ has the
distribution, for $k\geq 1,$

$\mathbb{P}(\tilde{L}=2k)=\frac{c_{2k-1}-c_{2k+1}}{c_{1}}.$

To show this, recall the proof of Proposition 5.1. Note that the bounds in
(5.6) and (5.7) should be replaced by $C_{0}a_{N}^{-1}\mathbb{E}|\tilde{X}|$
and that

$\mathbb{E}\bigl{[}\frac{|\tilde{J}|}{1-|\tanh(\beta\tilde{J})|};|\tilde{J}|\leq
K\bigr{]}\leq e^{2\beta K}\frac{\mathbb{E}|\tilde{X}|}{a_{N}}.$

It follows that for any $K>0$,

$\displaystyle\mathbb{E}\Bigl{|}\frac{1}{N}\sum_{i<j}\tilde{J}_{ij}\langle\sigma_{i}\sigma_{j}\rangle-\frac{1}{N}\sum_{i<j}\tilde{J}_{ij}f\bigl{(}\langle\sigma_{i}\sigma_{j}\rangle_{ij}^{\prime},\beta\tilde{J}_{ij}\bigr{)}\Bigr{|}$
$\displaystyle\leq\frac{(\mathbb{E}|\tilde{X}|)^{2}e^{2\beta
K}}{N^{2/\alpha-1}\ell(N)^{2}}+N\mathbb{E}[|\tilde{J}|;|\tilde{J}|\geq K].$

For fixed $K,$ the first vanishes as $N\to\infty$, and from Lemma 10.1$(ii)$,
the second term has the limit $\alpha(\alpha-1)^{-1}K^{-\alpha+1}.$ Therefore,
we can send $K\to\infty$ to obtain

$\displaystyle\limsup_{N\to\infty}\mathbb{E}\Bigl{|}\frac{1}{N}\sum_{i<j}\tilde{J}_{ij}\langle\sigma_{i}\sigma_{j}\rangle-\frac{1}{N}\sum_{i<j}\tilde{J}_{ij}f\bigl{(}\langle\sigma_{i}\sigma_{j}\rangle_{ij}^{\prime},\beta\tilde{J}_{ij}\bigr{)}\Bigr{|}=0.$

From (5.11) and using (5.12) for $(c_{\ell})_{\ell\geq 1}$, we obtain the
desired limit and from Lemma 10.1$(i)$, we have
$\lim_{N\to\infty}c_{\ell}=\gamma_{\ell}$, completing the proof of our claim.
To establish the proof of (2.4), we apply Griffith’s lemma as in Section 5.1
and use our claim to conclude that
$\lim_{N\to\infty}\mathbb{E}\bigl{\langle}R_{\tilde{L}}^{2}\bigr{\rangle}=0$,
which finishes our proof since for any $\ell\geq 1,$

$\displaystyle\frac{\gamma_{\ell-1}-\gamma_{\ell+1}}{\gamma_{1}}\limsup_{N\to\infty}\mathbb{E}\langle
R_{2\ell}^{2}\rangle\leq\lim_{N\to\infty}\mathbb{E}\bigl{\langle}R_{\tilde{L}}^{2}\bigr{\rangle}=0.$

#### Extension of (2.5)

We need to match the first two moments as in Proposition 5.2. With the help of
Lemma 10.1$(ii)$ and $(iii)$, for any $K>0,$ we have that

$\lim_{N\to\infty}N\mathbb{E}\tanh^{2}(\beta\tilde{J})1_{\\{|\tilde{J}|\geq
K\\}}=2C_{K}$

and that as long as $N$ is large,

$\displaystyle\mathbb{E}\tanh^{2}(\beta\tilde{J})1_{\\{|\tilde{J}|\leq K\\}}$
$\displaystyle\leq\frac{2\alpha\beta^{2}}{N(2-\alpha)}K^{2-\alpha},\,\,\mathbb{P}(|\tilde{J}|\geq
K)\leq\frac{2}{NK^{\alpha}}.$

These readily ensure that Proposition 5.2 also holds for the general Lévy
model by an identical argument. Finally, note that
$\tilde{M}:=\sum_{i<j}\mathbbm{1}_{\\{|\tilde{J}|\geq
K\\}}\sim\mbox{Binomial}(N(N-1)/2,\mathbb{P}(|\tilde{J}|\geq K))$ satisfies
$N\tilde{M}^{-1}\mathbbm{1}_{\\{\tilde{M}\geq 1\\}}\to 2K^{\alpha}$ in
probability. Following the proof in Section 5.3 yields (2.5) for the general
Lévy model.

###  10.4 Extension of Theorem 2.9

For general Lévy model, we need to modify the statement of Theorem 2.9 can as
follows.

######  Theorem 10.5.

For $0<\alpha<1$ and $\beta>0$, we have that as $N\to\infty,$

$\frac{1}{b_{N}}\ln Z_{N}\stackrel{{\scriptstyle
d}}{{\to}}\beta\sum_{j=1}^{\infty}\gamma_{j}^{-1/\alpha},$

where $b_{N}=a_{N}^{-1}a_{{N\choose 2}}$.

Our proof relies on the following generalization of Lemma 9.1. Let
$G(t)=\mathbb{P}(|\tilde{X}|>t)$ for $t>0$ and define its inverse

$G^{-1}(u)=\inf\\{y>0:G(y)\leq u\\},\ u\in(0,1).$

We can morally think of $G^{-1}$ as $x^{-1/\alpha}$ for $x>0$. Let
$(\tilde{X}_{i})_{i\geq 1}$ be i.i.d. copies of $\tilde{X}$ and let
$\tilde{Y}_{i}=|\tilde{X}_{i}|$ for $i\geq 1.$

######  Lemma 10.6.

The following statements hold:

  1. (a)

Let $0<\alpha<2$. For $n\geq 1,$

$a_{n}^{-1}\bigl{(}\tilde{Y}_{I_{1}},\tilde{Y}_{I_{2}},\ldots,\tilde{Y}_{I_{n}}\bigr{)}\stackrel{{\scriptstyle
d}}{{=}}a_{n}^{-1}\bigl{(}G^{-1}(\gamma_{1}/\gamma_{n+1}),G^{-1}(\gamma_{2}/\gamma_{n+1}),\ldots,G^{-1}(\gamma_{n}/\gamma_{n+1})\bigr{)}.$

Also, the random vector
$a_{n}^{-1}(\tilde{X}_{1},\tilde{X}_{2},\ldots,\tilde{X}_{n})$ has the same
distribution as that of

$a_{n}^{-1}\bigl{(}\varepsilon_{1}G^{-1}(\gamma_{r(1)}/\gamma_{n+1}),\varepsilon_{2}G^{-1}(\gamma_{r(2)}/\gamma_{n+1}),\ldots,\varepsilon_{n}G^{-1}(\gamma_{r(n)}/\gamma_{n+1})\bigr{)},$

where $\varepsilon_{1},\varepsilon_{2},\ldots,\varepsilon_{n}$ are i.i.d.
Rademacher variables and $r:[n]\to[n]$ is an independent uniform random
permutation, both independent of $(\gamma_{j})_{j\geq 1}$.

  2. (b)

Let $0<\alpha<2$. As $n\to\infty$,

$a_{n}^{-1}\bigl{(}\tilde{Y}_{I_{1}},\tilde{Y}_{I_{2}},\ldots,\tilde{Y}_{I_{n}},0,0,\ldots\bigr{)}\stackrel{{\scriptstyle
d}}{{\to}}\bigl{(}\gamma_{1}^{-1/\alpha},\gamma_{2}^{-1/\alpha},\ldots\bigr{)}\text{
on }\mathcal{A}_{\geq}.$

  3. (c)

Let $0<\alpha<2$. For any $\eta>0$, with probability tending to one, for all
$1\leq j\leq n/\log n,$ we have

$(1-\eta)\gamma_{j}^{-(1/\alpha+\eta)}\leq
a_{n}^{-1}G^{-1}(\gamma_{j}/\gamma_{n+1})\leq(1+\eta)\gamma_{j}^{-(1/\alpha-\eta)},$

and for all $n/\log n\leq j\leq n$,

$a_{n}^{-1}G^{-1}(\gamma_{j}/\gamma_{n+1})\leq(1+\eta)n^{-(1/\alpha-\eta)}.$

  4. (d)

Assume that $0<\alpha<1$. As $n\to\infty$,

$a_{n}^{-1}\sum_{j=1}^{n}Y_{j}\stackrel{{\scriptstyle
d}}{{\to}}\sum_{j=1}^{\infty}\gamma_{j}^{-1/\alpha},$

where the infinite sum is finite a.s.. Also, for any fixed integer $R\geq 1$,
as $n\to\infty,$

$a_{n}^{-1}\Bigl{(}\sum_{j=1}^{R}\tilde{Y}_{I_{j}},\sum_{j=R+1}^{n}\tilde{Y}_{I_{j}}\Bigr{)}\stackrel{{\scriptstyle
d}}{{\to}}\Bigl{(}\sum_{j=1}^{R}\gamma_{j}^{-1/\alpha},\sum_{j=R+1}^{\infty}\gamma_{j}^{-1/\alpha}\Bigr{)}.$

###### Proof.

The proof of Lemma 10.6 except part (c) can be traced back to [12]. Here we
only give a sketch of the proof of part (c). Define $V=1/G$ which is non-
decreasing, and let $V^{-1}(t)=\inf\\{x>0:1/G(x)\geq t\\}=G^{-1}(1/t)$ for
$t>0$. Since $G$ is regularly varying with index $-\alpha$, we have that $V$
is regularly varying with index $\alpha$. From Proposition 2.6 of [44],
$V^{-1}$ is regularly varying with index $1/\alpha$. Note that
$a_{n}=V^{-1}(n)$ for any $n\in\mathbb{N}$. Then, we have

$a_{n}^{-1}G^{-1}(\gamma_{j}/\gamma_{n+1})=\frac{V^{-1}(\gamma_{n+1}/\gamma_{j})}{V^{-1}(n)}.$
(10.6)

For $1\leq j\leq n/\log n$, we may use (C.4) for the function $V^{-1}$ and the
fact that $\lim_{n\to\infty}\gamma_{n+1}/n=1$ with probability one to conclude
the first half of the result from (10.6). For $n/\log n\leq j\leq n$, the
other half of the result follows from the monotonicity
$a_{n}^{-1}G^{-1}(\gamma_{j}/\gamma_{n+1})\leq
a_{n}^{-1}G^{-1}(\gamma_{\lfloor n/\log n\rfloor}/\gamma_{n+1}),$ and then
applying (C.4).

∎

###### Proof of Theorem 10.5.

Using Lemma 10.6(d), we have the upper bound

$\frac{1}{b_{N}}\ln Z_{N}\leq\frac{1}{a_{N}^{-1}a_{{N\choose 2}}}\Big{(}N\ln
2+\beta\sum_{i<j}|\tilde{J}_{ij}|\Big{)}\stackrel{{\scriptstyle
d}}{{\to}}\beta\sum_{k=1}^{\infty}\gamma_{k}^{-1/\alpha}.$

The lower bound can similarly be derived. ∎

###  10.5 Extension of Theorem 2.10

Due to the presence of possible atoms, we can have ties in the order
statistics of $\tilde{Y}_{i}$’s. This necessitates that we redefine the ranks
carefully. In this case, we can pick a uniform decreasing ordering $\sigma$
maintaining $\tilde{Y}_{\sigma(1)}\geq\dots\geq\tilde{Y}_{\sigma(n)}$ and
define the rank of $(i,j)$ as $r(i,j)=\sigma^{-1}(i,j)$. Note that this
definition retains the key property that the resulting rank function still
forms a uniform bijection.

The proof of Lemma 9.3(a) is irrelevant to the exact distribution of the
disorder. However, in view of Lemma 10.6, recalling $n=N(N-1)/2,$ the
inequality in Lemma 9.3(b) should be modified to

$\sum_{j:r(i,j)\leq R}|\tilde{J}_{ij}|-\sum_{j:r(i,j)>R}|\tilde{J}_{ij}|\geq
ca_{N}^{-1}a_{n}R^{-(1/\alpha+\eta)},$

where $c$ is a small absolute constant and
$0<\eta<\min\\{3\delta/(2\alpha(1-\delta/2),\delta/(\alpha(2-\alpha-\delta)),1/\alpha\\}$.
In fact, we can write

$|\tilde{J}_{ij}|=a_{N}^{-1}G^{-1}(\gamma_{r(i,j)}/\gamma_{n+1}),\ 1\leq i\neq
j\leq N,$

and (9.2) becomes

$a_{N}^{-1}a_{n}\bigl{(}0.9(1-\eta)R^{-1/\alpha-\eta}-\ln N\cdot
1.1(1+\eta)T^{-1/\alpha+\eta}-N\cdot 1.1(1+\eta)N^{-1/\alpha+\eta}\bigr{)}.$

Then, the final bound in (9.6) is

$2\exp\bigl{(}-2c\beta a_{N}^{-1}a_{n}N^{-(1/\alpha+\eta)/2}\bigr{)},$

which converges to $0$ as $N\to\infty$.

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##  Appendix A Establishing Theorem 8.1

###  A.1 Uniform Convergence of Auxiliary Functionals

In this subsection, we will state the uniform convergence of two auxiliary
functionals as a preparation for the proof of Theorem 8.1. Let $n\geq 1,$
$\varepsilon>0$, and $\lambda>0.$ From Subsection 2.1.3, recall
$s_{I},\hat{s}_{I}$ and the PPPs $(\xi_{i,k})_{k\geq 1},$
$(\xi_{i,k}^{\prime})_{k\geq 1},$ $(\xi_{k}^{\prime})_{k\geq 1}.$ For $r\geq
1$ and $I\in\mathbb{N}^{r}$, denote

$\bar{s}_{I}=\bar{\sigma}(w,u,v_{I})=\int_{0}^{1}\sigma(w,u,v_{I},x)dx.$

For any $\sigma\in\mathcal{S},$ define
$A_{n,\varepsilon}(\sigma)=\ln\mathcal{A}_{n,\varepsilon}(\sigma)$ and
$B_{n,\varepsilon,\lambda}(\sigma)=\ln\mathcal{B}_{n,\varepsilon,\lambda}(\sigma),$
where

$\displaystyle\begin{split}\mathcal{A}_{n,\varepsilon}(\sigma)&=\mathbb{E}_{u,\delta}\prod_{i\leq
n}\prod_{k:|\xi_{i,k}|\geq\varepsilon}(1+\tanh(\beta\xi_{i,k})\bar{s}_{i,k}\delta_{i}),\\\
\mathcal{B}_{n,\varepsilon,\lambda}(\sigma)&=\mathbb{E}_{u}\prod_{i\leq
n}\prod_{k:|\xi_{i,k}^{\prime}|\geq\varepsilon}(1+\tanh(\beta\xi_{i,k}^{\prime})\bar{s}_{1,i,k}\bar{s}_{2,i,k}).\end{split}$
(A.1)

To describe the limit of these objectives, we define
${A}_{n}(\sigma)=\ln\mathcal{A}_{n}(\sigma)$ and
${B}_{n,\lambda}(\sigma)=\ln\mathcal{B}_{n,\lambda}(\sigma),$ where

$\displaystyle\mathcal{A}_{n}(\sigma)$
$\displaystyle=\mathbb{E}_{u,\delta}\prod_{i\leq n}\prod_{k\geq
1}(1+\tanh(\beta\xi_{i,k})\bar{s}_{i,k}\delta_{i}),$
$\displaystyle\mathcal{B}_{n,\lambda}(\sigma)$
$\displaystyle=\mathbb{E}_{u}\prod_{i\leq n}\prod_{k\geq
1}(1+\tanh(\beta\xi_{i,k}^{\prime})\bar{s}_{1,i,k}\bar{s}_{2,i,k}).$

The following proposition justifies the well-definiteness and integrability of
the four functionals above.

######  Proposition A.1.

Let $n\geq 1$, $\varepsilon>0,$ and $\lambda>0.$ For any $\sigma$,
${A}_{n,\varepsilon}(\sigma)$, ${B}_{n,\varepsilon,\lambda}(\sigma),$
${A}_{n}(\sigma)$, and $B_{n,\lambda}(\sigma)$ are a.s. finite and integrable.

######  Theorem A.2.

Let $n\geq 1$ and $\lambda>0.$ We have that

$\displaystyle\lim_{\varepsilon\downarrow
0}\sup_{\sigma}\mathbb{E}|A_{n,\varepsilon}(\sigma)-A_{n}(\sigma)|$
$\displaystyle=0,$ (A.2) $\displaystyle\lim_{\varepsilon\downarrow
0}\sup_{\sigma}\mathbb{E}|B_{n,\varepsilon,\lambda}(\sigma)-B_{n,\lambda}(\sigma)|$
$\displaystyle=0.$

The proofs of the two results are presented in the next two subsections.

###  A.2 Proof of Proposition A.1

We only establish Proposition A.1 for $A_{n,\varepsilon}$ and $A_{n}$ since
the case for $B_{n,\varepsilon,\lambda}$ and $B_{n,\lambda}$ are nearly
identical. To simplify our notation, we shall denote
$\mathcal{A}_{n,\varepsilon}$ and $\mathcal{A}_{n}$ by
$\mathcal{A}_{\varepsilon}$ and $\mathcal{A},$ respectively. For any
measurable set $S\subseteq(0,\infty),$ define

$\mu_{S}(dx)=\frac{\alpha I(|x|\in S)}{2|x|^{\alpha+1}}dx$

and the conditional measure of $\mu$ on the set $S$ as

$\mu_{\cdot|S}(dx)=\frac{\mu_{S}(dx)}{\mu_{S}(\\{x\in\mathbb{R}:|x|\in
S\\})}.$

Note that

$\displaystyle\mu_{[\varepsilon,\infty)}=\mu_{[1,\infty)}(|x|\geq
1)\cdot\mu_{\cdot|[1,\infty)}+\mu_{[\varepsilon,1)}(\varepsilon\leq|x|<1)\cdot\mu_{\cdot|[\varepsilon,1)}=\mu_{\cdot|[1,\infty)}+(\gamma-1)\mu_{\cdot|[\varepsilon,1)}.$

Let $(\hat{g}_{i,k})$ be i.i.d. copies from $\mu_{\cdot|[1,\infty)}$ and
$(\hat{g}_{i,k,\varepsilon})$ be i.i.d. copies from
$\mu_{\cdot|(\varepsilon,1)}$ As usual, they are independent of each other and
also of other randomness. From the above identity, we can decompose
$(g_{i,k,\varepsilon})_{1\leq k\leq\pi_{i}(\gamma)}$ in distribution as
$(\hat{g}_{i,k})_{1\leq k\leq\pi_{i}(1)}\cup(\hat{g}_{i,k,\varepsilon})_{1\leq
k\leq\pi_{i}(\gamma-1)}.$ Let $\sigma\in\mathcal{S}.$ From this decomposition,
we can write

$\displaystyle\begin{split}\mathcal{A}_{\varepsilon}(\sigma)&\stackrel{{\scriptstyle
d}}{{=}}\mathbb{E}_{u,\delta}\Bigl{(}\prod_{i\leq
n}\prod_{k\leq\pi_{i}(1)}(1+\tanh(\beta\hat{g}_{i,k})\bar{\sigma}(w,u,v_{i,k})\delta_{i})\Bigr{)}\\\
&\qquad\quad\Bigl{(}\prod_{i\leq
n}\prod_{k\leq\pi_{i}(\gamma-1)}(1+\tanh(\beta\hat{g}_{i,k,\varepsilon})\bar{\sigma}(w,u,\hat{v}_{i,k})\delta_{i})\Bigr{)}.\end{split}$

From this,

$\displaystyle\begin{split}\mathcal{A}_{\varepsilon}(\sigma)&\stackrel{{\scriptstyle
d}}{{\leq}}\Bigl{(}\prod_{i\leq
n}\prod_{k\leq\pi_{i}(1)}(1+|\tanh(\beta\hat{g}_{i,k})|)\Bigr{)}W_{1,\varepsilon}(\sigma)^{2},\\\
\mathcal{A}_{\varepsilon}^{-1}(\sigma)&\stackrel{{\scriptstyle
d}}{{\leq}}\Bigl{(}\prod_{i\leq
n}\prod_{k\leq\pi_{i}(1)}(1-|\tanh(\beta\hat{g}_{i,k})|)^{-1}\Bigr{)}W_{2,\varepsilon}(\sigma)^{2},\end{split}$

where

$\displaystyle W_{1,\varepsilon}(\sigma)$
$\displaystyle=\Bigl{(}\mathbb{E}_{u,\delta}\prod_{i\leq
n}\prod_{k\leq\pi_{i}(\gamma-1)}\big{(}1+\tanh(\beta\hat{g}_{i,k,\varepsilon})\bar{\sigma}(w,u,\hat{v}_{i,k})\delta_{i}\big{)}\Bigr{)}\Bigr{)}^{1/2},$
$\displaystyle W_{2,\varepsilon}(\sigma)$
$\displaystyle=\Bigl{(}\mathbb{E}_{u,\delta}\prod_{i\leq
n}\prod_{k\leq\pi_{i}(\gamma-1)}\big{(}1+\tanh(\beta\hat{g}_{i,k,\varepsilon})\bar{\sigma}(w,u,\hat{v}_{i,k})\delta_{i}\big{)}\Bigr{)}^{-1/2}.$

Denote by $\ln_{+}x=\max(\ln x,0)$ for $x>0$. Note that $|\ln
x|\leq\ln_{+}x+\ln_{+}x^{-1},$ $\ln_{+}x\leq\sqrt{x}$ for $x>0$, and
$\ln_{+}(ab)\leq\ln_{+}(a)+\ln_{+}(b)$ for $a,b>0$. It follows that if
$x,a,b,a^{\prime},b^{\prime}\geq 0$ satisfy $x\leq ab$ and $x^{-1}\leq
a^{\prime}b^{\prime}$, then $|\ln
x|\leq\ln_{+}a+\ln_{+}a^{\prime}+\sqrt{b}+\sqrt{b^{\prime}}.$ Consequently,

$\displaystyle|A_{\varepsilon}(\sigma)|\stackrel{{\scriptstyle
d}}{{\leq}}Z_{1}+Z_{2}+W_{1,\varepsilon}(\sigma)+W_{2,\varepsilon}(\sigma),$
(A.3)

where

$\displaystyle Z_{1}$ $\displaystyle:=\sum_{i\leq
n}\sum_{k\leq\pi_{i}(1)}\ln(1+|\tanh(\beta\hat{g}_{i,k})|)\geq
0\,\,\mbox{and}\,\,Z_{2}:=\sum_{i\leq
n}\sum_{k\leq\pi_{i}(1)}-\ln(1-|\tanh(\beta\hat{g}_{i,k})|)\geq 0.$

To show that ${A}_{\varepsilon}(\sigma)\in L^{1},$ it suffices to establish
that $Z_{1},Z_{2}\in L^{1}$ and both $W_{1,\varepsilon}$ and
$W_{2,\varepsilon}$ are uniformly $L^{2}$ bounded. To this end, we bound

$\ln(1+|\tanh(u)|)\leq|u|,\quad-\ln(1-|\tanh(u)|)\leq|u|.$

Since $1<\alpha<2$, these inequalities imply that
$\mathbb{E}\ln(1+|\tanh(\beta\hat{g}_{1,1})|)<\infty$ and
$-\mathbb{E}\ln(1-|\tanh(\beta\hat{g}_{1,1})|)<\infty$, and therefore,
$\mathbb{E}Z_{1}<\infty$ and $\mathbb{E}Z_{2}<\infty.$ Next, for any
$\varepsilon>0$,

$\displaystyle\mathbb{E}W_{1,\varepsilon}(\sigma)^{2}=\mathbb{E}\prod_{i\leq
n}\prod_{k\leq\pi_{i}(\gamma-1)}\big{(}1+\tanh(\beta\hat{g}_{i,k,\varepsilon})\bar{\sigma}(w,u,\hat{v}_{i,k})\delta_{i}\big{)}=1.$
(A.4)

On the other hand, by Jensen’s inequality, we have

$\displaystyle\mathbb{E}W_{2,\varepsilon}(\sigma)^{2}$
$\displaystyle\leq\mathbb{E}\prod_{i\leq
n}\prod_{k\leq\pi_{i}(\gamma-1)}\big{(}1+\tanh(\beta\hat{g}_{i,k,\varepsilon})\bar{\sigma}(w,u,\hat{v}_{i,k})\delta_{i}\big{)}^{-1}$
$\displaystyle=\mathbb{E}_{w,u,\delta}\exp\Big{(}n(\gamma-1)\Big{(}\mathbb{E}_{g,v}\big{(}1+\tanh(\beta\hat{g}_{\varepsilon})\bar{\sigma}(w,u,v)\delta\big{)}^{-1}-1\Big{)}\Big{)}$
$\displaystyle=\mathbb{E}_{w,u,\delta}\exp\Big{(}n(\gamma-1)\Big{(}\mathbb{E}_{g,v}\Big{[}\big{(}1+\tanh(\beta\hat{g}_{\varepsilon})\bar{\sigma}(w,u,v)\delta\big{)}^{-1}-1+\tanh(\beta\hat{g}_{\varepsilon})\bar{\sigma}(w,u,v)\delta\Big{]}\Big{)}\Big{)}$
$\displaystyle=\mathbb{E}_{w,u,\delta}\exp\Big{(}n(\gamma-1)\mathbb{E}_{g,v}\frac{\tanh^{2}(\beta\hat{g}_{\varepsilon})\bar{\sigma}^{2}(w,u,v)\delta^{2}}{1+\tanh(\beta\hat{g}_{\varepsilon})\bar{\sigma}(w,u,v)\delta}\Big{)}$
$\displaystyle\leq\exp\Big{(}n\int_{-1}^{1}\frac{\tanh^{2}(\beta
x)}{1-\tanh(\beta)}\mu(dx)\Big{)}<\infty,$ (A.5)

where $\hat{g}_{\varepsilon}$ is a copy of $\hat{g}_{i,k,\varepsilon}$ and the
last inequality holds since

$\frac{\tanh^{2}(\beta\hat{g}_{\varepsilon})\bar{\sigma}^{2}(w,u,v)\delta^{2}}{1+\tanh(\beta\hat{g}_{\varepsilon})\bar{\sigma}(w,u,v)\delta}\leq\frac{\tanh^{2}(\beta\hat{g}_{\varepsilon})}{1-\tanh(\beta)}.$

These together imply that $(A_{\varepsilon}(\sigma))_{\varepsilon>0}$ is
uniformly integrable.

Next, we proceed to show that $A(\sigma)$ is a.s. finite and integrable. We
need the following lemma.

######  Lemma A.3.

For $m\geq 1$, we have
$\sup_{\sigma,\varepsilon}\mathbb{E}\mathcal{A}_{\varepsilon}(\sigma)^{m}<\infty.$

###### Proof.

From the Hölder inequality, it suffices to establish lemma only for $n=1.$ Our
assertion follows directly from

$\displaystyle\mathbb{E}\mathcal{A}_{\varepsilon}(\sigma)^{m}$
$\displaystyle=\mathbb{E}\prod_{k\leq\pi(\varepsilon^{-\alpha})}\big{(}1+\tanh(\beta
g_{1,k,\varepsilon})\bar{s}_{1,k}\delta_{1}\big{)}^{m}$
$\displaystyle=\mathbb{E}_{w,u,\delta}\exp\Big{(}\varepsilon^{-\alpha}\Big{(}\mathbb{E}_{g,v}\big{(}1+\tanh(\beta
g_{1,1,\varepsilon})\bar{s}_{1,k}\delta_{1}\big{)}^{m}-1\Big{)}\Big{)}$
$\displaystyle=\mathbb{E}_{w,u}\exp\Big{(}\varepsilon^{-\alpha}\Big{(}\mathbb{E}_{g,v}\Big{[}\sum_{1\leq\ell\leq
m/2}{m\choose 2\ell}\tanh^{2\ell}(\beta
g_{1,1,\varepsilon})\bar{s}_{1,k}^{2\ell}\Big{]}\Big{)}\Big{)}$
$\displaystyle\leq\mathbb{E}_{w,u}\exp\Big{(}\mathbb{E}_{v}\bar{s}_{1,k}^{2}2^{m}\int_{|x|>\varepsilon}\tanh^{2}(\beta
x)\mu(dx)\Big{)}$
$\displaystyle\leq\exp\Big{(}2^{m}\int_{\mathbb{R}}\tanh^{2}(\beta
x)\mu(dx)\Big{)}<\infty,$

where the third equality holds due the symmetry of $g_{1,k,\varepsilon}.$ ∎

Observe that

$\displaystyle Y_{m}(\sigma):=\prod_{i\leq n}\prod_{1\leq k\leq
m}(1+\tanh(\beta\xi_{i,k})\bar{s}_{i,k}\delta_{i}),\,\,m\geq 1$

is a positive martingale with respect to the filtration
$(\sigma(\xi_{i,k},\bar{s}_{i,k}:1\leq k\leq m,1\leq i\leq n)\bigr{)}_{m\geq
1}.$ It follows that $Y_{m}(\sigma)$ converges a.s. as $m\to\infty$. Denote
this limit by $Y(\sigma).$ Lemma A.3 further implies that this convergence
holds in $L^{m}$ for any $m\geq 1.$ This concludes that
$\mathcal{A}(\sigma)=\mathbb{E}_{u,\delta}Y(\sigma)$ is a.s. finite and
nonnegative and furthermore,

$\displaystyle\mathcal{A}(\sigma)=\mathbb{E}_{u,\delta}Y(\sigma)=\lim_{m\to\infty}\mathbb{E}_{u,\delta}Y_{m}(\sigma)=\lim_{\varepsilon\downarrow
0}\mathcal{A}_{\varepsilon}(\sigma).$ (A.6)

Hence, by using Fatou’s lemma and the fact that
$({A}_{\varepsilon}(\sigma))_{\varepsilon>0}$ is uniformly integrable, we
arrive at

$\displaystyle\mathbb{E}|A(\sigma)|$
$\displaystyle\leq\liminf_{\varepsilon\downarrow
0}\mathbb{E}|A_{\varepsilon}(\sigma)|<\infty.$ (A.7)

This bound ensures that $A(\sigma)$ is finite a.s. and is integrable,
completing our proof.

###  A.3 Proof of Theorem A.2

We will again only establish the uniform convergence only for
${A}_{n,\varepsilon}$ since the case for ${B}_{n,\varepsilon,\lambda}(\sigma)$
can be treated similarly. Recall our notation $\cal A,$ $A,$
$\mathcal{A}_{\varepsilon},$ $A_{\varepsilon},$
$W_{1,\varepsilon},W_{2,\varepsilon},Z_{1},Z_{2}$ from the previous
subsection. We need a few lemmas. First of all, we show that $\mathcal{A}$ and
$\mathcal{A}_{\varepsilon}$ are uniformly positive in $\sigma$ and
$\varepsilon$ and that $\mathcal{A}_{\varepsilon}$ converges to $\mathcal{A}$
uniformly in $\sigma.$

######  Lemma A.4.

We have

$\lim_{\eta\downarrow
0}\inf_{\sigma,\varepsilon>0}\mathbb{P}(\mathcal{A}_{\varepsilon}(\sigma)>\eta)=1\,\,\mbox{and}\,\,\lim_{\eta\downarrow
0}\inf_{\sigma}\mathbb{P}(\mathcal{A}(\sigma)>\eta)=1.$

In particular, for any $\varepsilon>0$,
$\min\\{\mathcal{A}(\sigma),\mathcal{A}_{\varepsilon}(\sigma)\\}>0$ a.s..

###### Proof.

Using (A.7), we have
$\sup_{\sigma}\mathbb{E}|A(\sigma)|\leq\sup_{\sigma,\varepsilon}\mathbb{E}|A_{\varepsilon}(\sigma)|<\infty$.
For $0<\eta<1$, we have

$\displaystyle\sup_{\sigma}\mathbb{P}(\mathcal{A}(\sigma)\leq\eta)=\sup_{\sigma}\mathbb{P}(1/\mathcal{A}(\sigma)\geq
1/\eta)\leq\frac{\sup_{\sigma}\mathbb{E}|\ln(1/\mathcal{A}(\sigma))|}{\ln(1/\eta)}$

and taking $\eta\downarrow 0$ yields the result. The other claim can be
treated similarly. ∎

######  Lemma A.5.

For any $m\geq 1$, we have $\lim_{\varepsilon\downarrow
0}\sup_{\sigma}\mathbb{E}|\mathcal{A}_{\varepsilon}(\sigma)-\mathcal{A}(\sigma)|^{m}=0.$

###### Proof.

We only consider the case that $n=1$ as the general case can be treated
similarly. Let $m\geq 1$ be fixed. Set

$\displaystyle Y(\sigma)$ $\displaystyle=\prod_{k\geq
1}(1+\tanh(\beta\xi_{1,k})\bar{s}_{1,k}\delta_{1}),\,\,Y_{\varepsilon}(\sigma)=\prod_{k:|\xi_{1,k}|>\varepsilon}(1+\tanh(\beta\xi_{1,k})\bar{s}_{1,k}\delta_{1}).$

Recall from the end of the proof of Proposition A.1 that the infinite product
in $Y(\sigma)$ converges a.s.. Thus, $Y_{\varepsilon}(\sigma)\to Y(\sigma)$
a.s. as $\varepsilon\downarrow 0.$ From Fatou’s lemma,

$\mathbb{E}|Y(\sigma)-Y_{\varepsilon}(\sigma)|^{m}\leq\liminf_{\varepsilon^{\prime}\downarrow
0}\mathbb{E}|Y_{\varepsilon^{\prime}}(\sigma)-Y_{\varepsilon}(\sigma)|^{m}.$

For $\varepsilon>0,$, we bound

$\displaystyle|Y_{\varepsilon^{\prime}}(\sigma)-Y_{\varepsilon}(\sigma)|^{m}$
$\displaystyle\leq
2^{m}\bigl{(}|Y_{\varepsilon^{\prime}}(\sigma)|^{m}+|Y_{\varepsilon}(\sigma)|^{m}\bigr{)}|Y_{\varepsilon^{\prime}}(\sigma)-Y_{\varepsilon}(\sigma)|$
$\displaystyle=2^{m}\bigl{(}|Y_{\varepsilon^{\prime}}(\sigma)|^{m}+|Y_{\varepsilon}(\sigma)|^{m}\bigr{)}|Y_{\varepsilon}(\sigma)|\Bigl{(}1-\prod_{k:\varepsilon^{\prime}<|\xi_{1,k}|\leq\varepsilon}(1+\tanh(\beta\xi_{1,k})\bar{s}_{1,k}\delta_{1})\Bigr{)}.$

From the Cauchy-Schwartz inequality and Lemma A.3, it suffices to show that

$\sup_{\sigma}\sup_{0<\varepsilon^{\prime}<\varepsilon}\mathbb{E}\Bigl{(}1-\prod_{k:\varepsilon^{\prime}<|\xi_{k}|\leq\varepsilon}(1+\tanh(\beta\xi_{1,k})\bar{s}_{1,k}\delta_{1})\Bigr{)}^{2}\xrightarrow[\varepsilon\downarrow
0]{}0.$

To see this, we compute

$\displaystyle\mathbb{E}\Bigl{(}1-\prod_{k:\varepsilon^{\prime}<|\xi_{1,k}|<\varepsilon}(1+\tanh(\beta\xi_{1,k})\bar{s}_{1,k}\delta_{1})\Bigr{)}^{2}$
$\displaystyle=1-2\mathbb{E}\prod_{k:\varepsilon^{\prime}<|\xi_{1,k}|\leq\varepsilon}(1+\tanh(\beta\xi_{1,k})\bar{s}_{1,k}\delta_{1})+\mathbb{E}\prod_{k:\varepsilon^{\prime}<|\xi_{1,k}|\leq\varepsilon}(1+\tanh(\beta\xi_{1,k})\bar{s}_{1,k}\delta_{1})^{2}.$

Note that for $g_{\varepsilon,\varepsilon^{\prime},k}$ i.i.d. sampled from
$\mu_{\cdot|(\varepsilon^{\prime},\varepsilon]}$, the random set
$\\{\xi_{1,k}:\varepsilon^{\prime}<|\xi_{1,k}|\leq\varepsilon\\}$ is in
distribution the same as $\\{g_{\varepsilon,\varepsilon^{\prime},k}:1\leq
k\leq\pi((\varepsilon^{\prime})^{-\alpha}-\varepsilon^{-\alpha})\\}.$
Therefore, using the symmetry of $g_{\varepsilon,\varepsilon^{\prime},k}$, the
first two terms together equals $-1$. As for the third term, write

$\displaystyle\mathbb{E}\prod_{k:\varepsilon^{\prime}<|\xi_{1,k}|\leq\varepsilon}(1+\tanh(\beta\xi_{1,k})\bar{s}_{1,k}\delta_{1})^{2}$
$\displaystyle=\mathbb{E}\prod_{k\leq\pi(\varepsilon^{-\alpha}-\varepsilon^{\prime-\alpha})}(1+\tanh(\beta
g_{\varepsilon,\varepsilon^{\prime},k})\bar{s}_{1,k}\delta_{1})^{2}$
$\displaystyle=\mathbb{E}_{w,u,\delta}\exp\Bigl{(}(\varepsilon^{-\alpha}-\varepsilon^{\prime-\alpha})\bigl{(}\mathbb{E}_{g,v}(1+\tanh(\beta
g_{\varepsilon,\varepsilon^{\prime},k})\bar{s}_{1,k}\delta_{1})^{2}-1\bigr{)}\Bigr{)}$
$\displaystyle=\mathbb{E}_{w,u}\exp\Bigl{(}(\varepsilon^{-\alpha}-\varepsilon^{\prime-\alpha})\mathbb{E}_{g,v}\tanh^{2}(\beta
g_{\varepsilon,\varepsilon^{\prime},k})\bar{s}_{1,k}^{2})\Bigr{)}$
$\displaystyle=\mathbb{E}_{w,u}\exp\Bigl{(}\mathbb{E}_{v}\bar{s}_{1,k}^{2}\int_{\varepsilon^{\prime}<|x|\leq\varepsilon}\tanh^{2}(\beta
x)\mu(dx)\Bigr{)}$
$\displaystyle\leq\exp\Bigl{(}\int_{|x|\leq\varepsilon}\tanh^{2}(\beta
x)\mu(dx)\Bigr{)}\xrightarrow[\varepsilon\downarrow 0]{}1,$

which completes our proof. ∎

By using the two lemmas above, we continue to show that $A_{\varepsilon}$
converges to $A$ uniformly in probability.

######  Lemma A.6.

For any $\eta>0$, $\lim_{\varepsilon\downarrow
0}\sup_{\sigma}\mathbb{P}(|A_{\varepsilon}(\sigma)-A(\sigma)|>\eta)=0.$

###### Proof.

Fix $\eta,\eta^{\prime}>0$. From Lemma A.4, we can choose
$\eta^{\prime\prime}>0$ such that
$\inf_{\sigma}\mathbb{P}(\mathcal{A}(\sigma)>\eta^{\prime\prime})>1-\eta^{\prime}$.
From Lemma A.5, we can choose $\varepsilon_{0}>0$ such that for any
$0<\varepsilon<\varepsilon_{0}$,
$\sup_{\sigma}\mathbb{E}|\mathcal{A}(\sigma)-\mathcal{A}_{\varepsilon}(\sigma)|<\eta\eta^{\prime}\eta^{\prime\prime}.$
For any $0<\varepsilon<\varepsilon_{0}$, write

$\displaystyle\mathbb{P}(|A_{\varepsilon}(\sigma)-A(\sigma)|>\ln(1+\eta))$
$\displaystyle=\mathbb{P}(\mathcal{A}_{\varepsilon}(\sigma)>\mathcal{A}(\sigma)(1+\eta))+\mathbb{P}(\mathcal{A}(\sigma)>\mathcal{A}_{\varepsilon}(\sigma)(1+\eta))$
$\displaystyle=\mathbb{P}\biggl{(}\frac{\mathcal{A}_{\varepsilon}(\sigma)-\mathcal{A}(\sigma)}{\mathcal{A}(\sigma)}>\eta\biggr{)}+\mathbb{P}\biggl{(}\frac{\mathcal{A}(\sigma)-\mathcal{A}_{\varepsilon}(\sigma)}{\mathcal{A}_{\varepsilon}(\sigma)}>\eta\biggr{)}.$

We bound the first term by

$\displaystyle\mathbb{P}\Bigl{(}\frac{\mathcal{A}_{\varepsilon}(\sigma)-\mathcal{A}(\sigma)}{\mathcal{A}(\sigma)}>\eta\Bigr{)}$
$\displaystyle\leq\mathbb{P}\biggl{(}\frac{\mathcal{A}_{\varepsilon}(\sigma)-\mathcal{A}(\sigma)}{\mathcal{A}(\sigma)}>\eta,A(\sigma)>\eta^{\prime\prime}\biggr{)}+\eta^{\prime}$
$\displaystyle\leq\mathbb{P}(\mathcal{A}_{\varepsilon}(\sigma)-\mathcal{A}(\sigma)>\eta\eta^{\prime\prime})+\eta^{\prime}$
$\displaystyle\leq\frac{\mathbb{E}|\mathcal{A}(\sigma)-\mathcal{A}_{\varepsilon}(\sigma)|}{\eta\eta^{\prime\prime}}+\eta^{\prime}\leq
2\eta^{\prime}.$

It follows that

$\sup_{\sigma}\mathbb{P}\biggl{(}\frac{\mathcal{A}_{\varepsilon}(\sigma)-\mathcal{A}(\sigma)}{\mathcal{A}(\sigma)}>\eta\biggr{)}\leq
2\eta^{\prime}.$

We can similarly bound the second term and get the desired result. ∎

The last lemma we need establishes a uniform control on the large deviation
between $A_{\varepsilon}$ and $A$.

######  Lemma A.7.

For $M>0$ and $1<\alpha<2$,

$\sup_{\sigma,\varepsilon}\mathbb{E}\bigl{[}|A_{\varepsilon}(\sigma)-A(\sigma)|;|A_{\varepsilon}(\sigma)-A(\sigma)|>M\bigr{]}$

is bounded above by

$\displaystyle
16\bigl{(}\mathbb{E}Z_{1}\mathbbm{1}_{\\{Z_{1}>M/8\\}}+Z_{2}\mathbbm{1}_{\\{Z_{2}>M/8\\}}\bigr{)}+2\mathbb{E}|A(\sigma)|\mathbbm{1}_{\\{A(\sigma)|>M/2\\}}$
$\displaystyle\qquad+\frac{128}{M}\Bigl{(}1+\exp\int_{-1}^{1}\frac{\tanh^{2}(\beta
x)}{1-\tanh(\beta)}\mu(dx)\Bigr{)}.$

###### Proof.

For any random variables $X,Y$ and $M>0,$ it is easy to see that

$\displaystyle\mathbb{E}|X+Y|\mathbbm{1}_{\\{|X+Y|>M\\}}\leq
2\bigl{(}\mathbb{E}|X|\mathbbm{1}_{\\{2|X|>M\\}}+\mathbb{E}|Y|\mathbbm{1}_{\\{2|Y|>M\\}}\bigr{)}.$

Applying the above to $A_{\varepsilon}$ and $A$ yields

$\displaystyle\mathbb{E}|A_{\varepsilon}(\sigma)-A(\sigma)|\mathbbm{1}_{\\{|A_{\varepsilon}(\sigma)-A(\sigma)|>M\\}}\leq
2\mathbb{E}|A_{\varepsilon}(\sigma)|\mathbbm{1}_{\\{|A_{\varepsilon}(\sigma)|>M/2\\}}+2\mathbb{E}|A(\sigma)|\mathbbm{1}_{\\{|A(\sigma)|>M/2\\}}.$

Recall from (A.3) that $|A_{\varepsilon}(\sigma)|\stackrel{{\scriptstyle
d}}{{\leq}}Z_{1}+Z_{2}+W_{1,\varepsilon}(\sigma)+W_{2,\varepsilon}(\sigma)$.
Then, by iteration, we have

$\displaystyle\mathbb{E}|A_{\varepsilon}(\sigma)|\mathbbm{1}_{\\{|A_{\varepsilon}(\sigma)|>M/2\\}}$
$\displaystyle\leq\mathbb{E}(Z_{1}+Z_{2}+W_{1,\varepsilon}(\sigma)+W_{2,\varepsilon}(\sigma))\mathbbm{1}_{\\{Z_{1}+Z_{2}+W_{1,\varepsilon}(\sigma)+W_{2,\varepsilon}(\sigma)>M\\}}$
$\displaystyle\leq
8(\mathbb{E}Z_{1}\mathbbm{1}_{\\{Z_{1}>M/8\\}}+\mathbb{E}Z_{2}\mathbbm{1}_{\\{Z_{2}>M/8\\}}+\mathbb{E}W_{1,\varepsilon}(\sigma)\mathbbm{1}_{\\{W_{1,\varepsilon}(\sigma)>M/8\\}}+\mathbb{E}W_{2,\varepsilon}(\sigma)\mathbbm{1}_{\\{W_{2,\varepsilon}(\sigma)>M/8\\}}).$

The proof is now complete by noticing that from (A.4) and (A.5),

$\mathbb{E}W_{1,\varepsilon}(\sigma)\mathbbm{1}_{\\{W_{1,\varepsilon}(\sigma)>M/8\\}}\leq\frac{8}{M}\mathbb{E}W_{1,\varepsilon}(\sigma)^{2}=\frac{8}{M}$

and

$\mathbb{E}W_{2,\varepsilon}(\sigma)\mathbbm{1}_{\\{W_{2,\varepsilon}(\sigma)>M/8\\}}\leq\frac{8}{M}\mathbb{E}W_{2,\varepsilon}(\sigma)^{2}\leq\frac{8}{M}\exp\Big{(}n\int_{-1}^{1}\frac{\tanh^{2}(\beta
x)}{1-\tanh(\beta)}\mu(dx)\Big{)}.$

∎

###### Proof of (A.2) in Theorem A.2.

Denote by ${L}_{\varepsilon}(\sigma)=A_{\varepsilon}(\sigma)-A(\sigma).$ Fix
$0<\eta<1$. From the statement and the proof of Proposition A.1, we see that
$A(\sigma),Z_{1},Z_{2}$ are integrable. As a result, by Lemma A.7, we can
choose $M>1$ sufficiently large so that
$\sup_{\sigma,\varepsilon}\mathbb{E}[|{L}_{\varepsilon}(\sigma)|;|{L}_{\varepsilon}(\sigma)|>M]\leq\eta.$
From Lemma A.6, we can choose $\varepsilon_{0}>0$ small such that for
$0<\varepsilon<\varepsilon_{0},$
$\sup_{\sigma}\mathbb{P}(|{L}_{\varepsilon}(\sigma)|>\eta)<\eta/M.$ Putting
these together yields that for any $0<\varepsilon<\varepsilon_{0}$,

$\displaystyle\sup_{\sigma}\bigl{|}\mathbb{E}{L}_{\varepsilon}(\sigma)\bigr{|}$
$\displaystyle\leq\sup_{\sigma}\mathbb{E}|{L}_{\varepsilon}(\sigma)|$
$\displaystyle\leq\sup_{\sigma}\mathbb{E}\bigl{[}|{L}_{\varepsilon}(\sigma)|;|{L}_{\varepsilon}(\sigma)|\geq
M\bigr{]}+\sup_{\sigma}\mathbb{E}\bigl{[}|{L}_{\varepsilon}(\sigma)|;\eta\leq|{L}_{\varepsilon}(\sigma)|<M\bigr{]}+\eta$
$\displaystyle\leq\sup_{\sigma}\mathbb{E}\bigl{[}|{L}_{\varepsilon}(\sigma)|;|{L}_{\varepsilon}(\sigma)|\geq
M\bigr{]}+M\sup_{\sigma}\mathbb{P}(\eta\leq|{L}_{\varepsilon}(\sigma)|<M)+\eta$
$\displaystyle\leq 3\eta,$

completing our proof. ∎

###  A.4 Proof of Theorem 8.1

Note that from and $U_{l,\lambda}(\sigma)$ and $V_{\lambda}(\sigma)$ are
almost surely finite and are integrable.

Proof of (8.1): First of all, from the same argument for the a.s. finiteness
and integrability for $\mathcal{A}_{n,\epsilon}$ and $\mathcal{A}_{n}$ in
Proposition A.1 (see Lemma A.3 and (A.6)), we see that $U_{l,\lambda}$ and
$V_{\lambda}$ are a.s. finite and have any finite $m$-th moments. In addition,
we can have that

$\displaystyle\sup_{\sigma}\mathbb{E}|\ln
V_{\lambda}|\leq\sup_{\sigma,\varepsilon}\mathbb{E}|\ln V_{\lambda}|<\infty.$
(A.8)

Second, from Lemma A.5, we can further see that for any $m\geq 1,$

$\displaystyle\begin{split}\lim_{\varepsilon\downarrow
0}\sup_{\sigma}\mathbb{E}|U_{l,\varepsilon,\lambda}-U_{l,\lambda}|^{m}=0,\,\,\lim_{\varepsilon\downarrow
0}\sup_{\sigma}\mathbb{E}|V_{\varepsilon,\lambda}-V_{\lambda}|^{m}=0.\end{split}$
(A.9)

Now similar to the proof of Lemma A.4, (A.8) readily implies that

$\displaystyle\lim_{\eta\downarrow
0}\bigl{(}\sup_{\sigma,\varepsilon}\mathbb{P}(V_{\varepsilon,\lambda}\leq\eta)+\mathbb{P}(V_{\lambda}\leq\eta)\bigr{)}=0.$
(A.10)

From this and noting that $|U_{l,\varepsilon,\lambda}|\leq
V_{\varepsilon,\lambda}$ implies

$\frac{\prod_{l\leq
q}|U_{l,\varepsilon,\lambda}|}{V_{\varepsilon,\lambda}^{q}}\leq
1\,\,\mbox{and}\,\,\frac{\prod_{l\leq q}|U_{l,\lambda}|}{V_{\lambda}^{q}}\leq
1,$

we have

$\displaystyle\lim_{\eta\downarrow 0}\limsup_{\varepsilon\downarrow
0}\sup_{\sigma}\mathbb{E}\biggl{|}\frac{\prod_{l\leq
q}U_{l,\varepsilon,\lambda}}{V_{\varepsilon,\lambda}^{q}}\mathbbm{1}_{\\{V_{\varepsilon,\lambda}\leq\eta\\}}-\frac{\prod_{l\leq
q}U_{l,\lambda}}{V_{\lambda}^{q}}\mathbbm{1}_{\\{V_{\lambda}\leq\eta\\}}\biggr{|}=0.$

On the other hand,

$\displaystyle\mathbb{E}\biggl{|}\frac{\prod_{l\leq
q}U_{l,\varepsilon,\lambda}}{V_{\varepsilon,\lambda}^{q}}\mathbbm{1}_{\\{V_{\varepsilon,\lambda}>\eta\\}}-\frac{\prod_{l\leq
q}U_{l,\lambda}}{V_{\lambda}^{q}}\mathbbm{1}_{\\{V_{\lambda}>\eta\\}}\biggl{|}$
$\displaystyle\leq\mathbb{E}\biggl{|}\prod_{l\leq
q}U_{l,\varepsilon,\lambda}-\prod_{l\leq
q}U_{l,\lambda}\biggr{|}\frac{\mathbbm{1}_{\\{V_{\lambda}>\eta\\}}}{V_{\lambda}^{q}}$
$\displaystyle+\mathbb{E}\biggl{(}\prod_{l\leq
q}|U_{l,\varepsilon,\lambda}|\biggr{)}\biggl{|}\frac{1}{V_{\varepsilon,\lambda}^{q}}-\frac{1}{V_{\lambda}^{q}}\biggr{|}\mathbbm{1}_{\\{V_{\lambda}>\eta\\}}$
$\displaystyle+\mathbb{E}\frac{\prod_{l\leq
q}|U_{l,\varepsilon,\lambda}|}{V_{\varepsilon,\lambda}^{q}}|\mathbbm{1}_{\\{V_{\varepsilon,\lambda}>\eta\\}}-\mathbbm{1}_{\\{V_{\lambda}>\eta\\}}|.$

Here, the first term is obviously bounded by $\eta^{-q}\mathbb{E}|\prod_{l\leq
q}U_{l,\varepsilon,\lambda}-\prod_{l\leq q}U_{l,\lambda}|$ and the second term
is bound by

$\displaystyle\mathbb{E}V_{\varepsilon,\lambda}^{q}\biggl{|}\frac{1}{V_{\varepsilon,\lambda}^{q}}-\frac{1}{V_{\lambda}^{q}}\biggr{|}\mathbbm{1}_{\\{V_{\lambda}>\eta\\}}\leq\eta^{-q}\mathbb{E}|{V_{\varepsilon,\lambda}^{q}}-{V_{\lambda}^{q}}|.$

As for the third term, it is bounded above by
$\mathbb{P}(V_{\varepsilon,\lambda}\geq\eta)+\mathbb{P}(V_{\lambda}\geq\eta).$
Our assertion then follows from (A.8), (A.9), and (A.10).

Proof of (8.2): Note that from Proposition A.1,
$P(\sigma)=\mathbb{E}A_{1}(\sigma)-\mathbb{E}B_{1,1}(\sigma)$ is finite. From
the definition of $P_{\varepsilon}$ in (7.3), since
$(g_{i,k,\varepsilon})_{1\leq k\leq\pi_{i}(\gamma)}$ and
$(g_{i,k,\varepsilon})_{1\leq k\leq\pi_{i}(\gamma/2)}$ are PPPs with mean
measures $\mu_{[\varepsilon,\infty)}$ and $\mu_{[\varepsilon,\infty)}/2$,
these imply that, in terms of random sets,

$\displaystyle\\{g_{i,k,\varepsilon}:1\leq
k\leq\pi_{i}(\gamma)\\}\stackrel{{\scriptstyle
d}}{{=}}\\{\xi_{i,k}:|\xi_{i,k}|\geq\varepsilon\,\,\mbox{for $k\geq 1$}\\},$
$\displaystyle\\{g_{i,k,\varepsilon}:1\leq
k\leq\pi_{i}(\gamma/2)\\}\stackrel{{\scriptstyle
d}}{{=}}\\{\xi_{i,k}^{\prime}:|\xi_{i,k}^{\prime}|\geq\varepsilon\,\,\mbox{for
$k\geq 1$}\\}.$

Consequently,
$P_{\varepsilon}(\sigma)=\mathbb{E}A_{1,\varepsilon}(\sigma)-\mathbb{E}B_{1,\varepsilon,1}(\sigma)$
and (8.2) follows from Theorem A.2 and (7.4).

##  Appendix B Establishing Theorem 8.5

Recall $\mathcal{A}_{n,\varepsilon}$ and $B_{n,\varepsilon,\lambda}$ from
(A.1). We need a key lemma.

######  Lemma B.1.

Suppose $\sigma_{m}\to\sigma_{0}$ f.d.d. as $m\to\infty$. For any
$\varepsilon>0$ and $\lambda>0,$ we have that as $m\to\infty,$

$\bigl{(}U_{1,\varepsilon,\lambda}(\sigma_{m}),\ldots,U_{q,\varepsilon,\lambda}(\sigma_{m}),V_{\varepsilon,\lambda}(\sigma_{m})\bigr{)}\stackrel{{\scriptstyle
d}}{{\to}}\bigl{(}U_{1,\varepsilon,\lambda}(\sigma_{0}),\ldots,U_{q,\varepsilon,\lambda}(\sigma_{0}),V_{\varepsilon,\lambda}(\sigma_{0})\bigr{)}$

and

$\displaystyle\mathcal{A}_{n,\varepsilon}(\sigma_{m})\stackrel{{\scriptstyle
d}}{{\to}}\mathcal{A}_{n,\varepsilon}(\sigma_{0}),\,\,\mathcal{B}_{n,\varepsilon,\lambda}(\sigma_{m})\stackrel{{\scriptstyle
d}}{{\to}}\mathcal{B}_{n,\varepsilon,\lambda}(\sigma_{0}).$

###### Proof.

We will only establish our first assertion since the arguments for the other
two are identical. Denote by $(\delta_{l,\ell,i})_{\ell,i\geq 1}$ for
$l,\ell,i\geq 1$ i.i.d. Rademacher random variables and
$u_{l,\ell},x_{l,\ell,i},x_{l,\ell,i,k},\hat{x}_{l,\ell,i,k}$ for
$l,\ell,i,k\geq 1$ i.i.d. uniform random variables on $[0,1]$. These are
independent of each other and everything else. Denote

$\displaystyle s_{m,l,\ell,i}$
$\displaystyle=\sigma_{m}(w,u_{l,\ell},v_{i},x_{l,\ell,i}),$ $\displaystyle
s_{m,l,\ell,i,k}$
$\displaystyle=\sigma_{m}(u,u_{l,\ell},v_{i,k},x_{l,\ell,i,k}),$
$\displaystyle\hat{s}_{m,l,\ell,i,k}$
$\displaystyle=\sigma_{m}(w,u_{l,\ell},\hat{v}_{i,k},\hat{x}_{l,\ell,i,k}).$

Note that for any $a_{1},\ldots,a_{q},b\geq 1,$ we can write by introducing
independent copies,

$\displaystyle\Bigl{(}\prod_{l\leq
q}U_{l,\varepsilon,\lambda}(\sigma_{m})^{a_{l}}\Bigr{)}V_{\varepsilon,\lambda}(\sigma_{m})^{b}$
$\displaystyle=\mathbb{E}_{u,x,\delta}L_{m},$

where

$\displaystyle L_{m}$ $\displaystyle:=\Bigl{(}\prod_{l\leq q}\prod_{\ell\leq
a_{l}}\prod_{i\in C_{l}^{1}}\delta_{l,\ell,i}\Bigr{)}\Bigl{(}\prod_{l\leq
q}\prod_{\ell\leq a_{l}+b}\prod_{i\leq
n}\prod_{k\leq\pi_{i}(2\gamma)}\big{(}1+\tanh(\beta
g_{i,k,\varepsilon})\delta_{l,\ell,i}s_{m,l,\ell,i,k}\big{)}\Bigr{)}$
$\displaystyle\qquad\Bigl{(}\prod_{l\leq q}\prod_{\ell\leq a_{l}}\prod_{i\in
C_{\ell}^{2}}s_{m,l,\ell,i}\Bigr{)}\Bigl{(}\prod_{l\leq q}\prod_{\ell\leq
a_{l}+b}\prod_{k\leq\pi(\gamma)}\bigl{(}1+\tanh(\beta\hat{g}_{k,\varepsilon}){\hat{s}}_{m,l,\ell,1,k}{\hat{s}}_{m,l,\ell,2,k}\bigr{)}\Bigr{)}.$

From the convergence of $\sigma_{m}\to\sigma_{0}$ in f.d.d., we see that
$\mathbb{E}_{u,v,x}L_{m}\to\mathbb{E}_{u,v,x}L_{0}$ as $m\to\infty.$ On the
other hand, note that

$\displaystyle|\mathbb{E}_{u,v,x}L_{m}|$ $\displaystyle\leq
2^{\sum_{i=1}^{n}\pi_{i}(2\gamma)\cdot\sum_{l=1}^{q}(a_{i}+b)+\pi(\gamma)\sum_{l=1}^{q}(a_{l}+b)},\,\,\forall
m\geq 0.$

For any $M\geq 1,$ let $I_{M}:=\mathbbm{1}_{\\{\pi_{1}(2\gamma)\leq
M,\ldots,\pi_{n}(2\gamma)\leq M,\pi(\gamma)\leq M\\}}$ and
$I_{M}^{c}:=1-I_{M}.$ It follows from the bounded convergence theorem,

$\displaystyle\mathbb{E}\Bigl{(}\prod_{l\leq
q}U_{l,\varepsilon,\lambda}(\sigma_{m})^{a_{l}}\Bigr{)}V_{\varepsilon,\lambda}(\sigma_{m})^{b}I_{M}$
$\displaystyle=\mathbb{E}\bigl{[}\mathbb{E}_{u,\delta,x}[L_{m}]I_{M}\bigr{]}=\mathbb{E}\bigl{[}\mathbb{E}_{u,v,x}[L_{m}]I_{M}\bigr{]}$
$\displaystyle\to\mathbb{E}\bigl{[}\mathbb{E}_{u,v,x}[L_{0}]I_{M}\bigr{]}=\mathbb{E}\bigl{[}\mathbb{E}_{u,\delta,x}[L_{0}]I_{M}\bigr{]}$
$\displaystyle=\mathbb{E}\Bigl{(}\prod_{l\leq
q}U_{l,\varepsilon,\lambda}(\sigma_{0})^{a_{l}}\Bigr{)}V_{\varepsilon,\lambda}(\sigma_{0})^{b}I_{M}\,\,\mbox{as
$m\to\infty$}.$

From this and noting that $U_{l,\varepsilon,\lambda}(\sigma_{m})I_{M}$ and
$V_{\varepsilon,\lambda}(\sigma_{m})I_{M}$ are $m$-uniformly bounded by a
deterministic constant, the Cramér-Wold theorem then implies that

$\bigl{(}U_{1,\varepsilon,\lambda}(\sigma_{m})I_{M},\ldots,U_{q,\varepsilon,\lambda}(\sigma_{m})I_{M},V_{\varepsilon,\lambda}(\sigma_{m})I_{M}\bigr{)}\stackrel{{\scriptstyle
d}}{{\to}}\bigl{(}U_{1,\varepsilon,\lambda}(\sigma_{0})I_{M},\ldots,U_{q,\varepsilon,\lambda}(\sigma_{0})I_{M},V_{\varepsilon,\lambda}(\sigma_{0})I_{M}\bigr{)}.$

Finally, since $\varepsilon$ is fixed, it can be checked directly that

$\sup_{m\geq
0}\mathbb{E}\bigl{|}U_{l,\varepsilon,\lambda}(\sigma_{m})\bigr{|}^{2}<\infty\,\,\mbox{and}\,\,\sup_{m\geq
0}\mathbb{E}\bigl{|}V_{\varepsilon,\lambda}(\sigma_{m})\bigr{|}^{2}<\infty,$

which implies that as $M\to\infty,$

$\displaystyle\sup_{m\geq
0}\mathbb{E}\bigl{|}U_{l,\varepsilon,\lambda}(\sigma_{m})I_{M}^{c}\bigr{|}\leq\bigl{(}\sup_{m\geq
0}\mathbb{E}\bigl{|}U_{l,\varepsilon,\lambda}(\sigma_{m})\bigr{|}^{2}\bigr{)}^{1/2}\bigl{(}\mathbb{E}I_{M}^{c}\bigr{)}^{1/2}\to
0,$ $\displaystyle\sup_{m\geq
0}\mathbb{E}V_{\varepsilon,\lambda}(\sigma_{m})I_{M}^{c}\leq\bigl{(}\sup_{m\geq
0}\mathbb{E}V_{\varepsilon,\lambda}(\sigma_{m})^{2}\bigr{)}^{1/2}\bigl{(}\mathbb{E}I_{M}^{c}\bigr{)}^{1/2}\to
0.$

From Slutsky’s theorem, our first assertion follows. ∎

###### Proof of Theorem 8.5.

It is easy to see that $V_{\varepsilon,\lambda}(\sigma_{0})>0$ a.s.. The
continuous mapping theorem and Lemma B.1 imply that

$\frac{\prod_{l\leq
q}U_{l,\varepsilon,\lambda}(\sigma_{\eta})}{V_{\varepsilon,\lambda}^{q}(\sigma_{\eta})}\xrightarrow[\varepsilon\downarrow
0]{d}\frac{\prod_{l\leq
q}U_{l,\varepsilon,\lambda}(\sigma_{0})}{V_{\varepsilon,\lambda}^{q}(\sigma_{0})}.$

Next, since these are bounded random variables, we can apply the bounded
convergence theorem to conclude the proof for our first statement. As for the
second one, note that $\mathcal{A}_{n,\varepsilon}(\sigma_{0})>0$ and
$\mathcal{B}_{n,\varepsilon,\lambda}(\sigma_{0})>0$ a.s.. It follows from the
continuous mapping theorem and Lemma B.1, $A_{n,\varepsilon}(\sigma_{\eta})$
converges to $A_{n,\varepsilon}(\sigma_{0})$ and
$B_{n,\varepsilon,\lambda}(\sigma_{0})$ converges to
$B_{n,\varepsilon,\lambda}(\sigma_{0})$ weakly as $\eta\to 0.$ To show that
their expectations also converge, note that the argument before Lemma A.3 in
Subsection A.2 readily implies that
$(A_{n,\varepsilon}(\sigma_{\eta}))_{\eta>0}$ is uniformly integrable and the
same argument also yields the uniform integrability of
$(B_{n,\varepsilon,\lambda}(\sigma_{\eta}))_{\eta>0}$. Consequently,

$\lim_{\eta\downarrow
0}\mathbb{E}{A}_{n,\varepsilon}(\sigma_{\eta})=\mathbb{E}{A}_{n,\varepsilon}(\sigma_{0})\,\,\mbox{and}\,\,\lim_{\eta\downarrow
0}\mathbb{E}{B}_{n,\varepsilon,\lambda}(\sigma_{\eta})=\mathbb{E}{B}_{n,\varepsilon,\lambda}(\sigma_{0}).$

In particular, letting $n=1$ and $\lambda=1$ implies our second assertion. ∎

##  Appendix C Proof of Lemma 10.1

Before we proceed with the proof, we first collect some useful properties. We
refer the readers to [44] as a detailed reference. First of all, by the
definition of $a_{N}$, it is easy to see that

$\displaystyle 1\geq N\mathbb{P}(|\tilde{X}|>a_{N})\to 1$ (C.1)

as $N\to\infty$. A well-known fact is that

$\displaystyle a_{N}=N^{1/\alpha}\ell(N)$ (C.2)

for some slowly varying function $\ell$ at $\infty$. Moreover, the following
uniform convergence holds for any $0<a<\infty$:

$\displaystyle\lim_{y\to\infty}\sup_{t\in[a,\infty)}\Bigl{|}\frac{\mathbb{P}(|\tilde{X}|>ty)}{\mathbb{P}(|\tilde{X}|>y)}-\frac{1}{t^{\alpha}}\Bigr{|}=0.$
(C.3)

Potter’s bound says that if $L$ is a slowly varying function, for any
$\eta>0$, there exists $y_{0}>0$ such that for all $y\geq y_{0}$ and $t\geq
1$,

$\displaystyle\frac{1-\eta}{t^{\eta}}<\frac{L(ty)}{L(y)}<\frac{1+\eta}{t^{-\eta}}.$
(C.4)

We only prove Lemma 10.1(i). Conditions on $f$ and Karamata’s theorem imply
that $\mathbb{E}|f(\tilde{J})|<\infty$, so we have
$\mathbb{E}f(\tilde{J})=\mathbb{E}f(-\tilde{J})=\mathbb{E}\bar{f}(\tilde{J})=\mathbb{E}\bar{f}(|\tilde{J}|)$
by the symmetry of $\tilde{J}$. Then, $f(0)=0$ and Fubini’s theorem gives

$\displaystyle\mathbb{E}f(\tilde{J})=\int_{0}^{\infty}(\bar{f})^{\prime}(x)\mathbb{P}(|\tilde{J}|>x)dx.$

Note that

$N\mathbb{P}(|\tilde{J}|>x)=N\mathbb{P}(|\tilde{X}|>a_{N})\frac{\mathbb{P}(|\tilde{X}|>xa_{N})}{\mathbb{P}(|\tilde{X}|>a_{N})},$

so (C.1) and (C.3) imply that for $x>0$,

$\displaystyle\lim_{N\to\infty}N\mathbb{P}(|\tilde{J}|>x)=\frac{1}{x^{\alpha}}.$
(C.5)

Inequality (C.4) implies that for any
$0<\eta<\min\\{\delta-\alpha+1,\alpha-\varepsilon-1\\}$, there exists
$x_{0}>0$ such that for all $x>0$ and $N$ large enough satisfying $a_{N}\wedge
xa_{N}>x_{0}$, we have

$\displaystyle\mathbbm{1}_{a_{N}\wedge
xa_{N}>x_{0}}\frac{\mathbb{P}(|\tilde{X}|>xa_{N})}{\mathbb{P}(|\tilde{X}|>a_{N})}\leq\frac{1+\eta}{x^{\alpha-\eta}}\mathbbm{1}_{x\geq
1}+\frac{1}{(1-\eta)x^{\alpha+\eta}}\mathbbm{1}_{x_{0}/a_{N}<x<1}.$ (C.6)

On the interval $[1,\infty)$, applying the dominated convergence theorem with
(C.6), we have

$\displaystyle\lim_{N\to\infty}\int_{1}^{\infty}(\bar{f})^{\prime}(x)N\mathbb{P}(|\tilde{J}|>x)dx=\int_{1}^{\infty}\frac{(\bar{f})^{\prime}(x)}{x^{\alpha}}dx.$
(C.7)

There exists $\delta^{\prime}>0$ small enough such that $|f^{\prime}(x)|\leq
C|x|^{\delta}$ for all $|x|<\delta^{\prime}$, where $C$ is a large enough
constant. If $N$ is large enough so that $x_{0}/a_{N}<\delta^{\prime}$,

$\displaystyle\int_{0}^{x_{0}/a_{N}}|(\bar{f})^{\prime}(x)|N\mathbb{P}(|\tilde{J}|>x)dx\leq
CN\int_{0}^{x_{0}/a_{N}}x^{\delta}dx=\frac{Cx_{0}^{1+\delta}}{1+\delta}Na_{N}^{-1-\delta}.$

(C.2) and (C.4) imply that $\lim_{N\to\infty}Na_{N}^{-1-\delta}=0$, so

$\displaystyle\lim_{N\to\infty}\int_{0}^{x_{0}/a_{N}}(\bar{f})^{\prime}(x)N\mathbb{P}(|\tilde{J}|>x)dx=0.$
(C.8)

Now, if $N$ is large enough so that $x_{0}/a_{N}<1$, then from (C.1) and
(C.6),

$\displaystyle\mathbbm{1}_{x_{0}/a_{N}<x<1}|(\bar{f})^{\prime}(x)|N\mathbb{P}(|\tilde{X}|>xa_{N})$
$\displaystyle\leq\frac{|(\bar{f})^{\prime}(x)|}{(1-\eta)x^{\alpha+\eta}}\mathbbm{1}_{x_{0}/a_{N}<x<1}$
$\displaystyle\leq\frac{Cx^{\delta}\mathbbm{1}_{0<x<\delta^{\prime}}+|(\bar{f})^{\prime}(x)|\mathbbm{1}_{\delta^{\prime}\leq
x\leq 1}}{(1-\eta)x^{\alpha+\eta}}$
$\displaystyle\leq\frac{1}{1-\eta}\bigl{(}Cx^{\delta-\alpha-\eta}+|(\bar{f})^{\prime}(x)|(\delta^{\prime})^{-\alpha-\eta}\bigr{)},$

where the last line is integrable on $(0,1)$. Thus, the dominated convergence
theorem gives

$\displaystyle\lim_{N\to\infty}\int_{x_{0}/a_{N}}^{1}(\bar{f})^{\prime}(x)N\mathbb{P}(|\tilde{J}|>x)dx=\int_{0}^{1}\frac{(\bar{f})^{\prime}(x)}{x^{\alpha}}dx.$
(C.9)

Finally, combining (C.7),(C.8), and (C.9) yields the desired result

$\displaystyle\lim_{N\to\infty}N\mathbb{E}f(\tilde{J})=\lim_{N\to\infty}\int_{0}^{\infty}(\bar{f})^{\prime}(x)N\mathbb{P}(|\tilde{J}|>x)dx=\int_{0}^{\infty}\frac{(\bar{f})^{\prime}(x)}{x^{\alpha}}dx=\alpha\int_{0}^{\infty}\frac{\bar{f}(x)}{x^{\alpha+1}}dx,$

where the assumptions on $f$ assure that the integration by parts holds in the
last equality.