$\alpha\in(\frac{2}{\gamma},Q-\frac{\gamma}{4})$.

In order to prove Proposition 4.15, we first record two simple calculus facts
which will allow us to relate the inverse gamma distribution to the FZZ
formula.

######  Lemma 4.16.

Let $x\in(-1,1)$ and $a\in\mathbb{R}\setminus\mathbb{Z}$. The following
expansion holds:

$\displaystyle\cos(a\arccos x)=\frac{a}{2}\sin(\pi
a)\sum_{n=0}^{+\infty}\frac{(-2)^{n-1}}{\pi
n!}\Gamma(\frac{n+a}{2})\Gamma(\frac{n-a}{2})x^{n}.$ (4.18)

###### Proof.

The function $T_{a}(x)=\cos(a\arccos x)$ is a generalization of Chebyshev
polynomials, which reduces to the usual Chebyshev polynomials when $a$ is a
positive integer. For $x\in(-1,1)$, Taylor expansion gives:

$\displaystyle T_{a}(x)$
$\displaystyle=a\sum_{n=0}^{+\infty}\frac{(2x)^{n}}{2(n!)}\cos(\frac{\pi}{2}(a-n))\frac{\Gamma(\frac{a+n}{2})}{\Gamma(\frac{a-n}{2}+1)}$
$\displaystyle=a\sum_{n=0}^{+\infty}\frac{(2x)^{n}}{2\pi
n!}\cos(\frac{\pi}{2}(a-n))\Gamma(\frac{a+n}{2})\Gamma(-\frac{a-n}{2})\sin(\pi\frac{a-n}{2}+\pi)$
$\displaystyle=\frac{a}{2}\sin(\pi a)\sum_{n=0}^{+\infty}\frac{(-2)^{n-1}}{\pi
n!}\Gamma(\frac{n+a}{2})\Gamma(\frac{n-a}{2})x^{n}.\qed$

######  Lemma 4.17.

Let $a,b,c,\lambda\in\mathbb{R}$ satisfying $a>0$, $b>-1$,
$c<\frac{b}{2}-\frac{1}{2}$, $0<\lambda<2a$. Then the following identity
holds:

$\displaystyle\int_{0}^{\infty}dy\;y^{b}e^{-\lambda
y}\int_{0}^{\infty}dt\;t^{c}\exp\mathopen{}\mathclose{{}\left(-y^{2}t-\frac{a^{2}}{t}}\right)=\frac{1}{2}\sum_{n=0}^{+\infty}\frac{(-\lambda)^{n}}{n!}a^{-n+2c-b+1}\Gamma(\frac{n}{2}+\frac{b}{2}-c-\frac{1}{2})\Gamma(\frac{n}{2}+\frac{b}{2}+\frac{1}{2}).$
(4.19)

###### Proof.

To compute the double integral we expand $e^{-\lambda y}$ and then perform the
change of variable $u=y^{2}t$, $v=\frac{a^{2}}{t}$. Hence
$y=\frac{\sqrt{uv}}{a}$, $t=\frac{a^{2}}{v}$,
$dydt=\frac{a}{2v\sqrt{uv}}dudv$. Therefore:

$\displaystyle\int_{0}^{\infty}dy\;y^{b}e^{-\lambda
y}\int_{0}^{\infty}dt\;t^{c}\exp\mathopen{}\mathclose{{}\left(-y^{2}t-\frac{a^{2}}{t}}\right)$
$\displaystyle=\sum_{n=0}^{+\infty}\frac{(-\lambda)^{n}}{n!}\int_{0}^{\infty}dy\;y^{b+n}\int_{0}^{\infty}dt\;t^{c}\exp\mathopen{}\mathclose{{}\left(-y^{2}t-\frac{a^{2}}{t}}\right)$
$\displaystyle=\sum_{n=0}^{+\infty}\frac{(-\lambda)^{n}}{n!}\int_{0}^{\infty}\int_{0}^{\infty}dudv\frac{a}{2v\sqrt{uv}}a^{2c}v^{-c}a^{-b-n}(uv)^{\frac{b}{2}+\frac{n}{2}}e^{-u-v}$
$\displaystyle=\frac{1}{2}\sum_{n=0}^{+\infty}\frac{(-\lambda)^{n}}{n!}a^{-n+2c-b+1}\int_{0}^{\infty}\int_{0}^{\infty}dudv\;u^{\frac{b}{2}+\frac{n}{2}-\frac{1}{2}}v^{\frac{b}{2}+\frac{n}{2}-c-\frac{3}{2}}e^{-u-v}$
$\displaystyle=\frac{1}{2}\sum_{n=0}^{+\infty}\frac{(-\lambda)^{n}}{n!}a^{-n+2c-b+1}\Gamma(\frac{n}{2}+\frac{b}{2}-c-\frac{1}{2})\Gamma(\frac{n}{2}+\frac{b}{2}+\frac{1}{2}).$

The conditions on the parameters are such to make the above integrals and
series converge. For $n=0$ in the sum, the integrals over $u$ and $v$ converge
respectively if $b>-1$ and $c<\frac{b}{2}-\frac{1}{2}$. The condition
$\lambda>0$ is obvious. The condition $\lambda<2a$ is required for the series
over $n$ to converge, as can be checked on the last line. ∎

###### Proof of Proposition 4.15.

To start off we assume that the parameters $\mu,\mu_{B}>0$ are chosen such
that

$\frac{\mu^{2}_{B}}{\mu}\sin\frac{\pi\gamma^{2}}{4}\in(0,1).$ (4.20)

This constraint will then be lifted by an analyticity argument in the variable
$\mu_{B}$. Applying integration by parts to the integration over $\ell$ in
(4.17), we get

$\displaystyle\mathopen{}\mathclose{{}\left\langle
e^{\alpha\phi(i)}}\right\rangle$ $\displaystyle=\frac{\gamma
C}{2(\alpha-Q)}\int_{0}^{\infty}\ell^{\frac{2}{\gamma}(\alpha-Q)}\mathcal{M}_{1,0}^{\mathrm{disk}}(\alpha;1)^{\\#}[(2\mu\ell
A+\mu_{B})e^{-\mu\ell^{2}A-\mu_{B}\ell}]\,d\ell$
$\displaystyle=C^{\prime}\int_{0}^{\infty}y^{\frac{2}{\gamma}(\alpha-Q)}\mathcal{M}_{1,0}^{\mathrm{disk}}(\alpha;1)^{\\#}\mathopen{}\mathclose{{}\left[(\lambda+2yA)e^{-\lambda
y-y^{2}A}}\right]dy,$ (4.21)

where $y=\sqrt{\mu}\ell$, $C^{\prime}=-\frac{\gamma
C}{2(Q-\alpha)}\mu^{\frac{Q-\alpha}{\gamma}}$ and
$\lambda=\frac{\mu_{B}}{\mu^{1/2}}$. Now we use Proposition 4.11 which
requires to assume $\alpha<Q-\frac{\gamma}{4}$ and we apply twice Lemma 4.17
with $(a,b,c,\lambda)$ equal to

$\mathopen{}\mathclose{{}\left(\frac{1}{2\sqrt{\sin\frac{\pi\gamma^{2}}{4}}},\frac{2}{\gamma}(\alpha-Q),\frac{2}{\gamma}(\alpha-Q)-1,\lambda}\right)\quad\text{and}\quad\mathopen{}\mathclose{{}\left(\frac{1}{2\sqrt{\sin\frac{\pi\gamma^{2}}{4}}},\frac{2}{\gamma}(\alpha-Q)+1,\frac{2}{\gamma}(\alpha-Q),\lambda}\right).$

The constraints on $(a,b,c,\lambda)$ required by Lemma 4.17 are satisfied
provided that $\alpha\in(\frac{2}{\gamma},Q)$ and that equation (4.20) holds.
Hence one arrives at the following power series expansions in the parameter
$\lambda$:

$\displaystyle
C^{\prime}\int_{0}^{\infty}y^{\frac{2}{\gamma}(\alpha-Q)}\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;1)^{\\#}\mathopen{}\mathclose{{}\left[\lambda
e^{-\lambda y-y^{2}A}}\right]dy$ (4.22)
$\displaystyle\quad=-\frac{C^{\prime\prime}}{2}\sum_{n=0}^{\infty}\frac{(-\lambda)^{n+1}}{n!}\mathopen{}\mathclose{{}\left(2\sqrt{\sin\frac{\pi\gamma^{2}}{4}}}\right)^{n+\frac{2}{\gamma}(Q-\alpha)+1}\Gamma(\frac{n}{2}+\frac{1}{\gamma}(\alpha-Q)+\frac{1}{2})\Gamma(\frac{n}{2}+\frac{1}{\gamma}(Q-\alpha)+\frac{1}{2}),$

and

$\displaystyle
C^{\prime}\int_{0}^{\infty}y^{\frac{2}{\gamma}(\alpha-Q)}\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;1)^{\\#}\mathopen{}\mathclose{{}\left[2yAe^{-\lambda
y-y^{2}A}}\right]dy$ (4.23)
$\displaystyle\quad=C^{\prime\prime}\sum_{n=0}^{\infty}\frac{(-\lambda)^{n}}{n!}\mathopen{}\mathclose{{}\left(2\sqrt{\sin\frac{\pi\gamma^{2}}{4}}}\right)^{n+\frac{2}{\gamma}(Q-\alpha)}\Gamma(\frac{n}{2}+\frac{1}{\gamma}(\alpha-Q)+1)\Gamma(\frac{n}{2}+\frac{1}{\gamma}(Q-\alpha)),$

where
$C^{\prime\prime}=\frac{(4\sin\frac{\pi\gamma^{2}}{4})^{\frac{2}{\gamma}(\alpha-Q)}}{\Gamma(\frac{2}{\gamma}(Q-\alpha))}C^{\prime}$.
By shifting the sum in (4.22) to start at $n=1$ and applying in (4.23) the
following identity on the Gamma function

$\Gamma(\frac{n}{2}+\frac{1}{\gamma}(\alpha-Q)+1)=\mathopen{}\mathclose{{}\left(\frac{n}{2}+\frac{1}{\gamma}(\alpha-Q)}\right)\Gamma(\frac{n}{2}+\frac{1}{\gamma}(\alpha-Q)),$

we obtain

$\displaystyle\mathopen{}\mathclose{{}\left\langle
e^{\alpha\phi(i)}}\right\rangle$
$\displaystyle=-\frac{C^{\prime\prime}}{2}\sum_{n=0}^{\infty}\frac{(-\lambda)^{n+1}}{n!}\mathopen{}\mathclose{{}\left(2\sqrt{\sin\frac{\pi\gamma^{2}}{4}}}\right)^{n+\frac{2}{\gamma}(Q-\alpha)+1}\Gamma(\frac{n}{2}+\frac{1}{\gamma}(Q-\alpha)+\frac{1}{2})\Gamma(\frac{n}{2}+\frac{1}{\gamma}(\alpha-Q)+\frac{1}{2})$
$\displaystyle+C^{\prime\prime}\sum_{n=0}^{\infty}\frac{(-\lambda)^{n}}{n!}\mathopen{}\mathclose{{}\left(2\sqrt{\sin\frac{\pi\gamma^{2}}{4}}}\right)^{n+\frac{2}{\gamma}(Q-\alpha)}\Gamma(\frac{n}{2}+\frac{1}{\gamma}(Q-\alpha))\Gamma(\frac{n}{2}+\frac{1}{\gamma}(\alpha-Q)+1)$
$\displaystyle=C^{\prime\prime}\frac{\alpha-Q}{\gamma}\sum_{n=0}^{\infty}\frac{(-\lambda)^{n}}{n!}\mathopen{}\mathclose{{}\left(2\sqrt{\sin\frac{\pi\gamma^{2}}{4}}}\right)^{n+\frac{2}{\gamma}(Q-\alpha)}\Gamma(\frac{n}{2}+\frac{1}{\gamma}(\alpha-Q))\Gamma(\frac{n}{2}+\frac{1}{\gamma}(Q-\alpha)),$

To then obtain the claimed formula $U_{\mathrm{FZZ}}(\alpha)$, one simply
needs to apply Lemma 4.16 with $a=\frac{2}{\gamma}(\alpha-Q)$ and
$x=\lambda\sqrt{\sin\frac{\pi\gamma^{2}}{4}}$. Thanks again to (4.20) and the
fact that $\alpha\in(\frac{2}{\gamma},Q)$, the conditions to apply the lemma
are satisfied. This gives

$\displaystyle\mathopen{}\mathclose{{}\left\langle
e^{\alpha\phi(i)}}\right\rangle$
$\displaystyle=2C^{\prime\prime}\frac{\alpha-Q}{\gamma}\frac{\pi\gamma}{(\alpha-Q)}\frac{1}{\sin(\frac{2\pi}{\gamma}(Q-\alpha))}\mathopen{}\mathclose{{}\left(2\sqrt{\sin\frac{\pi\gamma^{2}}{4}}}\right)^{\frac{2}{\gamma}(Q-\alpha)}\cos\mathopen{}\mathclose{{}\left(\frac{2}{\gamma}(\alpha-Q)\arccos\mathopen{}\mathclose{{}\left(\lambda\sqrt{\sin\frac{\pi\gamma^{2}}{4}}}\right)}\right)$
$\displaystyle=2C^{\prime\prime}\Gamma(\frac{2}{\gamma}(Q-\alpha))\Gamma(1-\frac{2}{\gamma}(Q-\alpha))\mathopen{}\mathclose{{}\left(2\sqrt{\sin\frac{\pi\gamma^{2}}{4}}}\right)^{\frac{2}{\gamma}(Q-\alpha)}\cos((\alpha-Q)\pi
s),$

where the definition of $s$ was given in (1.5); see also (4.24). We compute:

$\displaystyle
2C^{\prime\prime}\Gamma(\frac{2}{\gamma}(Q-\alpha))\Gamma(1-\frac{2}{\gamma}(Q-\alpha))\mathopen{}\mathclose{{}\left(2\sqrt{\sin\frac{\pi\gamma^{2}}{4}}}\right)^{\frac{2}{\gamma}(Q-\alpha)}$
$\displaystyle=\frac{4}{\gamma}(\sin\frac{\pi\gamma^{2}}{4})^{\frac{1}{\gamma}(\alpha-Q)}\mu^{\frac{Q-\alpha}{\gamma}}2^{-\frac{\alpha^{2}}{2}}\mathopen{}\mathclose{{}\left(2^{-\frac{\gamma}{2}\alpha}}\right)^{\frac{2}{\gamma}(Q-\alpha)}\mathopen{}\mathclose{{}\left(\frac{\pi}{\Gamma(1-\frac{\gamma^{2}}{4})}}\right)^{\frac{2}{\gamma}(Q-\alpha)}\Gamma(\frac{\gamma}{2}(\alpha-\frac{\gamma}{2}))\Gamma(\frac{2\alpha}{\gamma}-\frac{4}{\gamma^{2}}-1)$
$\displaystyle=\frac{4}{\gamma}2^{-\frac{\alpha^{2}}{2}}\mathopen{}\mathclose{{}\left(\frac{\pi\mu}{2^{\gamma\alpha}}\frac{\Gamma(\frac{\gamma^{2}}{4})}{\Gamma(1-\frac{\gamma^{2}}{4})}}\right)^{\frac{Q-\alpha}{\gamma}}\Gamma(\frac{\gamma\alpha}{2}-\frac{\gamma^{2}}{4})\Gamma(\frac{2\alpha}{\gamma}-\frac{4}{\gamma^{2}}-1)=U_{\mathrm{FZZ}}(\alpha)/\cos((\alpha-Q)\pi
s).$

Putting everything together we conclude that
$\mathopen{}\mathclose{{}\left\langle
e^{\alpha\phi(i)}}\right\rangle=U(\alpha)=U_{\mathrm{FZZ}}(\alpha)$ provided
that the condition (4.20) holds.

We will now extend the result by analyticity in $\mu_{B}$ to the full range of
$\mu_{B}\in(0,+\infty)$. Notice $U_{\mathrm{FZZ}}(\alpha)$ contains the factor
$\cos((\alpha-Q)\pi s)$ where $s$ and $\mu_{B}$ are related by:

$\displaystyle\mu_{B}=\mu_{B}(s)=\sqrt{\frac{\mu}{\sin\frac{\pi\gamma^{2}}{4}}}\cos\frac{\pi\gamma
s}{2}.$ (4.24)

We want to find an open domain $D\subset\mathbb{C}$ of $s$ where the map
$s\mapsto\mu_{B}(s)$ will be biholomorphic onto its image. Since
$\frac{d}{ds}_{|s=0}\cos\frac{\pi\gamma s}{2}=0$, we cannot include $0$ in
$D$. Since $\cos\frac{\pi\gamma
s}{2}=1-\frac{\pi^{2}\gamma^{2}}{8}s^{2}+o(s^{2})$ and the second order
$s^{2}$ is non-zero, we can include the sector
$S_{\epsilon,\delta}:=\\{re^{i\theta}|\theta\in(-\epsilon,\frac{\pi}{2}+\epsilon),0<r<\delta\\}$
for two small parameters $\epsilon,\delta>0$. Indeed in $S_{\epsilon,\delta}$
the angle $\theta$ is contained in an interval of length strictly less than
$\pi$. We also want $D$ to contain an open neighborhood in $\mathbb{C}$ of
$i(\frac{\delta}{2},+\infty)$ where we wish to extend the equality
$U(\alpha)=U_{\mathrm{FZZ}}(\alpha)$ and an open neighborhood of
$(\frac{\delta}{2},\frac{1}{2\gamma})$ were we have already established it.
From all these considerations we choose:

$\displaystyle
D:=S_{\epsilon,\delta}\cup\mathopen{}\mathclose{{}\left\\{x+iy|x\in(\frac{\delta}{2},\frac{1}{2\gamma}),y\in(-\frac{\delta}{4},\frac{\delta}{4})}\right\\}\cup\mathopen{}\mathclose{{}\left\\{x+iy|x\in(-\frac{\delta}{4},\frac{\delta}{4}),y\in(\frac{\delta}{2},+\infty)}\right\\}.$

For $\epsilon,\delta$ small, the map $s\mapsto\mu_{B}(s)$ is then a
biholomorphic map from $D$ onto its image, an open domain we call
$D^{\prime}$. Now the exact formula $U_{\mathrm{FZZ}}(\alpha)$ is clearly an
analytic function of $\mu_{B}$ on $D^{\prime}$. We must argue the same is true
for the probabilistic definition
$U(\alpha)=\mathopen{}\mathclose{{}\left\langle
e^{\alpha\phi(i)}}\right\rangle$. For all $\mu_{B}\in D^{\prime}$, one clearly
has $\operatorname{Re}(\mu_{B})>0$ which by Lemma 2.6 gives finiteness of
$\mathopen{}\mathclose{{}\left|\mathopen{}\mathclose{{}\left\langle
e^{\alpha\phi(i)}}\right\rangle}\right|$. Moreover,
$\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}\mathopen{}\mathclose{{}\left[\mathopen{}\mathclose{{}\left|\nu_{\phi}(\mathbbm{R})e^{-\mu\mu_{\phi}(\mathbbm{H})-\mu_{B}\nu_{\phi}(\mathbbm{R})}}\right|}\right]$
is also finite in the same region. This implies that
$\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}[-\mu_{B}\nu_{\phi}(\mathbbm{R})e^{-\mu\mu_{\phi}(\mathbbm{H})-\mu_{B}\nu_{\phi}(\mathbbm{R})}]$
is analytic in $\mu_{B}$ when $\operatorname{Re}(\mu_{B})>0$. Taking anti-
derivative in $\mu_{B}$ and use Fubini theorem to interchange the integral in
$\mu_{B}$ and $\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}$, we see that
$\mathopen{}\mathclose{{}\left\langle
e^{\alpha\phi(i)}}\right\rangle=\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}[e^{-\mu\mu_{\phi}(\mathbbm{H})-\mu_{B}\nu_{\phi}(\mathbbm{R})}-1]$
is analytic in $\mu_{B}$ when $\operatorname{Re}(\mu_{B})>0$. Since the
equality of analytic functions $U(\alpha)=U_{\mathrm{FZZ}}(\alpha)$ in the
variable $\mu_{B}$ holds on the interval given by (4.20) which is contained in
the open set $D^{\prime}$, it holds on all of $D^{\prime}$. In particular it
holds for
$\frac{\mu_{B}}{\sqrt{\mu}}\sqrt{\sin\frac{\pi\gamma^{2}}{4}}\in(1,+\infty)$.
By continuity we recover the equality at the special value
$\frac{\mu_{B}}{\sqrt{\mu}}\sqrt{\sin\frac{\pi\gamma^{2}}{4}}=1$. This
completes the proof. ∎

###  4.5 Analyticity in $\alpha$ and the proof of Theorem 1.1

In this section we prove that $U(\alpha)$ is complex analytic in $\alpha$
around $(\frac{2}{\gamma},Q)$. The proof utilizes the method given first in
[KRV20] and in [RZ22], adapted to the case where there are both area and
boundary GMC measures in the correlation function. Together with Proposition
4.15 this will conclude the proof of Theorem 1.1. We will then prove Corollary
4.19 and Theorem 1.2 that extend the range of $\alpha$ from
$(\frac{2}{\gamma},Q)$ to $(\frac{\gamma}{2},Q)$.

######  Proposition 4.18.

For any compact set $K\subset(\frac{2}{\gamma},Q)$, the function
$\alpha\rightarrow U(\alpha)$ is complex analytic on a complex neighborhood of
the set $K$.

###### Proof.

To start we use the following identity coming from Lemma 2.6

$\mathopen{}\mathclose{{}\left\langle
e^{\alpha\phi(2i)}}\right\rangle=\frac{U(\alpha)}{\mathopen{}\mathclose{{}\left|\operatorname{Im}2i}\right|^{2\Delta_{\alpha}}},$
(4.25)

to place the bulk insertion at the location $2i$, and will prove that
$\mathopen{}\mathclose{{}\left\langle e^{\alpha\phi(2i)}}\right\rangle$ is
analytic in $\alpha$. Fix a $r_{0}$ such that $e^{-r_{0}}<\frac{1}{2}$. Let
$h$ be the free boundary GFF on $\mathbb{H}$, which is normalized to have zero
average over the upper half unit circle. Thanks to the Markov property of the
field, see for instance [BP, Theorem 5.5], we can write the decomposition
$h=h_{1}+h_{2}$, where $h_{1}$ is a Dirichlet GFF on $B(2i,e^{-r_{0}})$ and
$h_{2}$ is a field which is independent of $h_{1}$ and smooth in a
neighborhood of $2i$. Notice we placed the insertion at $2i$ so
$B(2i,e^{-r_{0}})$ does not intersect the upper half unit circle. Consider now
the one-dimensional process $h_{1,r}(2i)$ obtained by taking the mean of
$h_{1}(2i)$ over the circles of radius $e^{-r}$ centered at $2i$, assuming
$r\geq r_{0}$. The main fact we will use about this field is that
$h_{1,r+t}(2i)-h_{1,r}(2i)$ is a Brownian motion independent of
$(h(z))_{z\in\mathbb{H}_{r}}$, where $\mathbb{H}_{r}:=\mathbb{H}\backslash
B(2i,e^{-r})$. Record also that $\mathbbm{E}[h_{1,r}(2i)^{2}]=r-r_{0}$ and the
notation $h_{r}:=h_{1,r}+h_{2}$. From these facts we can deduce that:

$\displaystyle\mathbb{E}\mathopen{}\mathclose{{}\left[e^{\alpha
h_{1,r+1}(2i)-\frac{\alpha^{2}}{2}\mathbb{E}[h_{1,r+1}(2i)^{2}]}F\mathopen{}\mathclose{{}\left((h(z))_{z\in\mathbb{H}_{r}}}\right)}\right]$
$\displaystyle=e^{\frac{\alpha^{2}}{2}\mathbb{E}[h_{1,r}(2i)^{2}]-\frac{\alpha^{2}}{2}\mathbb{E}[h_{1,r+1}(2i)^{2}]}\mathbb{E}\mathopen{}\mathclose{{}\left[e^{\alpha
h_{1,r+1}(2i)-\alpha
h_{1,r}(2i)}}\right]\mathbb{E}\mathopen{}\mathclose{{}\left[e^{\alpha
h_{1,r}(2i)-\frac{\alpha^{2}}{2}\mathbb{E}[h_{1,r}(2i)^{2}]}F\mathopen{}\mathclose{{}\left((h(z))_{z\in\mathbb{H}_{r}}}\right)}\right]$
$\displaystyle=\mathbb{E}\mathopen{}\mathclose{{}\left[e^{\alpha
h_{1,r}(2i)-\frac{\alpha^{2}}{2}\mathbb{E}[h_{1,r}(2i)^{2}]}F\mathopen{}\mathclose{{}\left((h(z))_{z\in\mathbb{H}_{r}}}\right)}\right]$
$\displaystyle\Rightarrow\>$
$\displaystyle\mathbb{E}\mathopen{}\mathclose{{}\left[e^{\alpha
h_{r}(2i)-\frac{\alpha^{2}}{2}\mathbb{E}[h_{r}(2i)^{2}]}F\mathopen{}\mathclose{{}\left((h(z))_{z\in\mathbb{H}_{r}}}\right)}\right]=\mathbb{E}\mathopen{}\mathclose{{}\left[e^{\alpha
h_{r+1}(2i)-\frac{\alpha^{2}}{2}\mathbb{E}[h_{r+1}(2i)^{2}]}F\mathopen{}\mathclose{{}\left((h(z))_{z\in\mathbb{H}_{r}}}\right)}\right].$

Now we can obtain $U(\alpha)$ from the following limit
$U(\alpha)=2^{-\frac{\alpha^{2}}{2}}\lim_{r\to\infty}U_{r}(\alpha)$, where we
have introduced:

$\displaystyle
U_{r}(\alpha)=\int_{\mathbb{R}}dce^{(\alpha-Q)c}\mathbb{E}\mathopen{}\mathclose{{}\left[e^{\alpha
h_{r}(2i)-\frac{\alpha^{2}}{2}\mathbb{E}[h_{r}(2i)^{2}]}\mathopen{}\mathclose{{}\left(\exp\mathopen{}\mathclose{{}\left(-e^{\gamma
c}\mu_{h}(\mathbb{H}_{r})-e^{\frac{\gamma}{2}c}\nu_{h}(\mathbb{R})}\right)-1}\right)}\right].$

When $\alpha$ is a complex number, we write $\alpha=a+ib$. We want to prove
there exists a complex neighborhood $V$ containing the set $K$ such that for
any compact set $K^{\prime}$ contained in $V$, $U_{r}(\alpha)$ converges
uniformly as $r\rightarrow+\infty$ over $K^{\prime}$. Setting
$\tilde{h}(z):=h(z)+aG_{\mathbb{H}}(z,2i)$, we have the following:

$\displaystyle|U_{r+1}(\alpha)-U_{r}(\alpha)|$
$\displaystyle=\int_{\mathbb{R}}dc\mathopen{}\mathclose{{}\left|e^{(\alpha-Q)c}\mathbb{E}\mathopen{}\mathclose{{}\left[e^{\alpha
h_{r+1}(2i)-\frac{\alpha^{2}}{2}\mathbb{E}[h_{r+1}(2i)^{2}]}\exp\mathopen{}\mathclose{{}\left(-e^{\frac{\gamma}{2}c}\nu_{h}(\mathbb{R})}\right)\mathopen{}\mathclose{{}\left(\exp\mathopen{}\mathclose{{}\left(-e^{\gamma
c}\mu_{h}(\mathbb{H}_{r+1})}\right)-\exp\mathopen{}\mathclose{{}\left(-e^{\gamma
c}\mu_{h}(\mathbb{H}_{r})}\right)}\right)}\right]}\right|$ $\displaystyle\leq
C\,e^{\frac{r+1}{2}b^{2}}\int_{\mathbb{R}}dce^{(a-Q)c}\mathbb{E}\mathopen{}\mathclose{{}\left[\exp\mathopen{}\mathclose{{}\left(-e^{\frac{\gamma}{2}c}\nu_{\tilde{h}}(\mathbb{R})}\right)\mathopen{}\mathclose{{}\left|\exp\mathopen{}\mathclose{{}\left(-e^{\gamma
c}\mu_{\tilde{h}}(\mathbb{H}_{r+1})}\right)-\exp\mathopen{}\mathclose{{}\left(-e^{\gamma
c}\mu_{\tilde{h}}(\mathbb{H}_{r})}\right)}\right|}\right]$ $\displaystyle\leq
C\,e^{\frac{r+1}{2}b^{2}}\int_{\mathbb{R}}dce^{(a-Q)c}\mathbb{E}\mathopen{}\mathclose{{}\left[\exp\mathopen{}\mathclose{{}\left(-e^{\gamma
c}\mu_{\tilde{h}}(\mathbb{H}_{r+1})}\right)-\exp\mathopen{}\mathclose{{}\left(-e^{\gamma
c}\mu_{\tilde{h}}(\mathbb{H}_{r})}\right)}\right]$
$\displaystyle=C^{\prime}e^{\frac{r+1}{2}b^{2}}\mathbbm{E}\mathopen{}\mathclose{{}\left[\mu_{\widetilde{h}}(\mathbb{H}_{r+1})^{\frac{Q-a}{\gamma}}-\mu_{\widetilde{h}}(\mathbb{H}_{r})^{\frac{Q-a}{\gamma}}}\right].$

From the second line to the third, we have applied the Girsanov theorem to the
real part of $\alpha h_{r+1}(2i)$, before moving the absolute value inside the
expression. The last equality follows from changing the integration variable
from $c$ to $y=e^{\gamma c}\mu_{\widetilde{h}}(\mathbbm{H}_{i})$ for
$i=r,r+1$. Thus,

$|U_{r+1}(\alpha)-U_{r}(\alpha)|\leq
C^{\prime}e^{\frac{r+1}{2}b^{2}}\mathbbm{E}\mathopen{}\mathclose{{}\left[\mu_{\widetilde{h}}(\mathbbm{H}_{r+1}\backslash\mathbbm{H}_{r})^{\frac{Q-a}{\gamma}}}\right]\leq
C^{\prime\prime}e^{\frac{r+1}{2}b^{2}}e^{ra(Q-a)}\mathbbm{E}\mathopen{}\mathclose{{}\left[\mu_{h}(\mathbbm{H}_{r+1}\backslash\mathbbm{H}_{r})^{\frac{Q-a}{\gamma}}}\right].$

The first inequality follows from $(a+b)^{s}\leq a^{s}+b^{s}$ for $s\in(0,1)$,
and the second since $\widetilde{h}$ and $h$ differ by roughly $ar$ on
$\mathbbm{H}_{r+1}\backslash\mathbbm{H}_{r}$.

By the multifractal scaling of GMC, see e.g. [BP, Section 3.6] or [RV14,
Section 4], we have

$\mathbbm{E}[\mu_{h}(\mathbbm{H}_{r+1}\backslash\mathbbm{H}_{r})^{\frac{Q-a}{\gamma}}]\asymp
e^{(-\gamma Qq+\frac{\gamma^{2}q^{2}}{2})r},\qquad q=\frac{Q-a}{\gamma}.$

Combining with the previous inequality, we deduce that
$|U_{r+1}(\alpha)-U_{r}(\alpha)|\lesssim
e^{r(\frac{b^{2}}{2}-\frac{1}{2}(Q-a)^{2})}$. Choosing the open set $V$ in
such a way that $\frac{b^{2}}{2}<\frac{1}{2}(Q-a)^{2}$ always holds, all the
inequalities we have done before hold true and hence we have shown that
$U_{r}(\alpha)$ converges locally uniformly. Since $U_{r}(\alpha)$ is complex
analytic in $\alpha$, this proves the analyticity result of $U(\alpha)$. ∎

###### Proof of Theorem 1.1 .

From the exact formula, $U_{\mathrm{FZZ}}(\alpha)$ is a meromorphic function
of $\alpha$ on $\mathbbm{C}$. By uniqueness of the meromorphic continuation,
Propositions 4.15 and 4.18, we have $U(\alpha)=U_{\mathrm{FZZ}}(\alpha)$ for
all $\alpha\in(\frac{2}{\gamma},Q)$. ∎

###### Proof of Theorem 1.2.

Proposition 4.11 proves the theorem for
$\alpha\in(\frac{\gamma}{2},Q-\frac{\gamma}{4})$. To complete the proof we
consider $\alpha\in[Q-\frac{\gamma}{4},Q)$. Let $L$ be sampled from the power
law
$\frac{2}{\gamma}2^{-\frac{\alpha^{2}}{2}}\overline{U}_{0}(\alpha)\ell^{\frac{2}{\gamma}(\alpha-Q)-1}\,d\ell$
where $\overline{U}_{0}$ is as in Lemma 2.7, and let $A=L^{2}X$ where $X$ is
sampled from an independent inverse gamma distribution with shape
$\frac{2}{\gamma}(Q-\alpha)$ and scale
$\frac{1}{4\sin\frac{\pi\gamma^{2}}{4}}$. Let $\Pi$ be the joint law of
$(A,L)$. Proposition 4.15 proves that $\Pi[e^{-\mu
A-\mu_{B}L}-1]=U_{\mathrm{FZZ}}(\alpha)$ for
$\alpha\in(\frac{2}{\gamma},Q-\frac{\gamma}{4})$, but the argument works
equally well for our range $\alpha\in[Q-\frac{\gamma}{4},Q)$. Thus
$\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}[e^{-\mu\mu_{\phi}(\mathbbm{H})-\mu_{B}\nu_{\phi}(\mathbbm{R})}-1]=\Pi[e^{-\mu
A-\mu_{B}L}-1]$ by Theorem 1.1, so
$\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}[\nu_{\phi}(\mathbbm{R})e^{-\mu\mu_{\phi}(\mathbbm{H})-\mu_{B}\nu_{\phi}(\mathbbm{R})}]=\Pi[Le^{-\mu
A-\mu_{B}L}]$. Therefore by standard arguments of characterization of a law by
the Laplace transform, the joint law of
$(\mu_{\phi}(\mathbbm{H}),\nu_{\phi}(\mathbbm{R}))$ under
$\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}$ is the same as that of $(A,L)$ under
$\Pi$, which concludes the proof. ∎

Since we have now established Theorem 1.2 for the whole range
$\alpha\in(\frac{\gamma}{2},Q)$, we have the following extension of the
probabilistic definition of $U(\alpha)$ and of Theorem 1.1, for which we omit
the proof.

######  Corollary 4.19.

Let $\alpha\in(\frac{\gamma}{2},Q)$. For $k\in\mathbb{N}$ and
$\alpha\in(Q-\frac{\gamma(k+1)}{2},Q-\frac{\gamma k}{2})$, we define
$U(\alpha)$ by:

$U(\alpha)=\frac{2}{\gamma}2^{-\frac{\alpha^{2}}{2}}\overline{U}_{0}(\alpha)\int_{0}^{\infty}\ell^{\frac{2}{\gamma}(\alpha-Q)-1}\mathbbm{E}\mathopen{}\mathclose{{}\left[e^{-\mu\ell^{2}A-\mu_{B}\ell}-\sum_{i=0}^{k}c_{i}(A)\ell^{i}}\right]\,d\ell.$
(4.26)

Here the expectation is with respect to the law of $A$ which is distributed
according to the inverse gamma distribution with shape
$\frac{2}{\gamma}(Q-\alpha)$ and scale
$\frac{1}{4\sin^{2}\frac{\pi\gamma^{2}}{4}}$. $\overline{U}_{0}(\alpha)$ is
given by the explicit formula (2.6). Lastly the constants $c_{i}(A)$ are
specified by the following expansion in powers of $\ell$:

$\displaystyle
e^{-\mu\ell^{2}A-\mu_{B}\ell}=\sum_{i=0}^{k}c_{i}(A)\ell^{i}+O(\ell^{i+1}).$
(4.27)

Then this definition of $U(\alpha)$ is the meromorphic extension of (1.8) on
$(Q-\frac{\gamma(k+1)}{2},Q-\frac{\gamma k}{2})$ and it obeys
$U(\alpha)=U_{\mathrm{FZZ}}(\alpha)$.

Finally, we provide a commonly used form of the FZZ formula; see [FZZ00,
(2.44)].

######  Proposition 4.20.

For $\alpha\in(\frac{\gamma}{2},Q)$ and $\ell>0$, writing $A$ for the random
area of a sample from $\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell)$, we have

$\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell)[e^{-\mu
A}]=\frac{2}{\gamma}2^{-\frac{\alpha^{2}}{2}}\overline{U}_{0}(\alpha)\ell^{-1}\frac{2}{\Gamma(\frac{2}{\gamma}(Q-\alpha))}\mathopen{}\mathclose{{}\left(\frac{1}{2}\sqrt{\frac{\mu}{\sin(\pi\gamma^{2}/4)}}}\right)^{\frac{2}{\gamma}(Q-\alpha)}K_{\frac{2}{\gamma}(Q-\alpha)}\mathopen{}\mathclose{{}\left(\ell\sqrt{\frac{\mu}{\sin(\pi\gamma^{2}/4)}}}\right).$

Here, $K_{\nu}(x)$ is the modified Bessel function of the second kind; see
e.g. [DLMF, Section 10.25].

###### Proof.

By Theorem 1.2, writing $\beta=\frac{1}{4\sin\frac{\pi\gamma^{2}}{4}}$ and
considering $\ell=1$, we have

$\displaystyle\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;1)^{\\#}[e^{-\mu A}]$
$\displaystyle=\frac{\beta^{\frac{2}{\gamma}(Q-\alpha)}}{\Gamma(\frac{2}{\gamma}(Q-\alpha))}\int_{0}^{\infty}t^{-\frac{2}{\gamma}(Q-\alpha)-1}\exp^{-\frac{\beta}{t}-\mu
t}\,dt$
$\displaystyle=\frac{2}{\Gamma(\frac{2}{\gamma}(Q-\alpha))}(\mu\beta)^{\frac{1}{\gamma}(Q-\alpha)}K_{\frac{2}{\gamma}(Q-\alpha)}(\sqrt{4\mu\beta}),$

where the last equality follows from a change of variables $s=\mu t$ and the
integral identity
$K_{\nu}(z)=\frac{1}{2}(\frac{1}{2}z)^{\nu}\int_{0}^{\infty}\exp(-s-\frac{z^{2}}{4s})\frac{ds}{s^{\nu+1}}$
[DLMF, Eq. 10.32.10] with the choice $z=\sqrt{4\mu\beta}$ and
$\nu=\frac{2}{\gamma}(Q-\alpha)$. Since for general $\ell$ we have
$\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell)^{\\#}[e^{-\mu
A}]=\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;1)^{\\#}[e^{-\mu\ell^{2}A}]$,

$\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell)^{\\#}[e^{-\mu
A}]=\frac{2}{\Gamma(\frac{2}{\gamma}(Q-\alpha))}\ell^{\frac{2}{\gamma}(Q-\alpha)}\mathopen{}\mathclose{{}\left(\frac{1}{2}\sqrt{\frac{\mu}{\sin(\pi\gamma^{2}/4)}}}\right)^{\frac{2}{\gamma}(Q-\alpha)}K_{\frac{2}{\gamma}(Q-\alpha)}\mathopen{}\mathclose{{}\left(\ell\sqrt{\frac{\mu}{\sin(\pi\gamma^{2}/4)}}}\right).$

By Lemma 2.7 we have
$|\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell)|=\frac{2}{\gamma}2^{-\frac{\alpha^{2}}{2}}\overline{U}_{0}(\alpha)\ell^{\frac{2}{\gamma}(\alpha-Q)-1}$,
so the result follows. ∎

##  5 Proof of the SLE bubble zipper with a $\gamma$-bulk insertion

In this section we prove Theorem 4.1. For technical convenience, we will
consider loops on $\mathbbm{H}$ passing through $\infty$ instead of $0$. More
precisely, let $\psi(z)=-z^{-1}$ and
$\mathrm{Bubble}_{\mathbbm{H}}(i,\infty)=\\{\eta:\psi(\eta)\in\mathrm{Bubble}_{\mathbbm{H}}(i,0)\\}$.
We will find a measure on $\mathrm{Bubble}_{\mathbbm{H}}(i,\infty)$ with the
desired property and then use $\psi$ to pull it back to get $\mathsf{m}$ in
Theorem 4.1. The proof is a limiting argument based on Lemma 3.15 and a three-
disk variant of the conformal welding result in Theorem 2.13. We set up the
framework of the proof in Section 5.1 and carry out the details in Sections
5.2 and 5.3.

###  5.1 A conformal welding of three quantum disks

Set $\kappa=\gamma^{2}$. Let $\eta_{1}$ be an
$\operatorname{SLE}_{\kappa}(\frac{\gamma^{2}}{2}-2,\frac{\gamma^{2}}{2})$
curve on $(\mathbbm{H},0,\infty)$. Let $H^{+}_{\eta_{1}}$ the be right
component of $\mathbbm{H}\setminus\eta_{1}$. Conditioning on $\eta_{1}$, let
$\eta_{2}$ be an $\operatorname{SLE}_{\kappa}(0;\frac{\gamma^{2}}{2}-2)$ on
$(\mathbbm{H}_{\eta_{1}}^{+},0,\infty)$. We denote
$\mathcal{P}(\mathbbm{H},0,\infty)$ as the law of $(\eta_{1},\eta_{2})$. For a
general simply connected domain $D$ with two boundary points $(a,b)$, we write
$\mathcal{P}(D,a,b)$ as the conformal image of
$\mathcal{P}(\mathbbm{H},0,\infty)$. As a straightforward extension of Theorem
2.13, we have the following conformal welding result.

######  Theorem 5.1.

Let $\ell,\ell^{\prime}>0$ and $(\mathbbm{H},\phi,0,\infty)$ be an embedding
of a sample from
$\mathcal{M}^{\mathrm{disk}}_{0,2}(2+\gamma^{2};\ell,\ell^{\prime})$. Let
$(\eta_{1},\eta_{2})$ be sampled from $\mathcal{P}(\mathbbm{H},0,\infty)$
independent of $\phi$. Let $\mathbbm{H}^{1}_{\eta}$, $\mathbbm{H}^{12}_{\eta}$
and $\mathbbm{H}^{2}_{\eta}$ be the left, middle, and right components of
$\mathbbm{H}\setminus(\eta_{1}\cup\eta_{2})$, respectively. The joint law of
$(\mathbbm{H}^{1}_{\eta},\phi,0,\infty)$,
$(\mathbbm{H}^{12}_{\eta},\phi,0,\infty)$, and
$(\mathbbm{H}^{2}_{\eta},\phi,0,\infty)$ viewed as marked quantum surfaces
equals

$C\iint_{0}^{\infty}\mathcal{M}_{0,2}^{\mathrm{disk}}(\frac{\gamma^{2}}{2};\ell;p)\times\mathcal{M}_{0,2}^{\mathrm{disk}}(2;p;q)\times\mathcal{M}_{0,2}^{\mathrm{disk}}(\frac{\gamma^{2}}{2};q;\ell^{\prime})\,dp\,dq,\quad\textrm{
for some }C\in(0,\infty).$ (5.1)

###### Proof.

This is [AHS20, Theorem 2.3] when $W_{1}=W_{3}=\frac{\gamma^{2}}{2}$ and
$W_{2}=2$. ∎

Fix $\delta\in(0,\frac{1}{2})$. Sample $(\phi,\mathbf{x})$ from
$\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}\times dx$ and sample
$(\eta_{1},\eta_{2})$ from $\mathcal{P}(\mathbbm{H},\mathbf{x},\infty)$, and
restrict to the event that
$\nu_{\phi}(\mathbf{x},\infty)\in(\delta,\frac{1}{2})$,
$\nu_{\phi}(\mathbbm{R})\in(1,2)$ and $i$ is in between $\eta_{1}$ and
$\eta_{2}$. Here and later, we write $\nu_{\phi}(a,b)$ to denote the
$\nu_{\phi}$-length of the interval $(a,b)$. Let $M_{\delta}$ be the law of
$(\phi,\mathbf{x},\eta_{1},\eta_{2})$ (restricted to the aforementioned
event). See Figure 3.

Figure 3:  Illustration of $(\phi,\mathbf{x},\eta_{1},\eta_{2})$ under
$M_{\delta}$. Sample $(\phi,\mathbf{x})$ from
$\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}\times dx$ and sample
$(\eta_{1},\eta_{2})$ from $\mathcal{P}(\mathbbm{H},\mathbf{x},\infty)$. To
represent geometrically the condition
$\nu_{\phi}(\mathbf{x},\infty)\in(\delta,\frac{1}{2})$, consider the points
$x_{1/2}$ and $x_{\delta}$ such that $\nu_{\phi}(x_{1/2},\infty)=1/2$ and
$\nu_{\phi}(x_{\delta},\infty)=\delta$. Then $M_{\delta}$ is obtained by
restricting to the event $\mathbf{x}\in(x_{1/2},x_{\delta})$, $i$ is in the
middle of $\eta_{1}$ and $\eta_{2}$ and $\nu_{\phi}(\mathbbm{R})\in(1,2)$.
Lemma 5.2 describes the law of the three shaded quantum surfaces.

######  Lemma 5.2.

There exists $C>0$ such that for each $\delta\in(0,\frac{1}{2})$, if
$(\phi,\mathbf{x},\eta_{1},\eta_{2})$ is sampled from $M_{\delta}$ then the
law of the three marked quantum surfaces of
$(\mathbbm{H},\phi,\eta_{1},\eta_{2},i,\mathbf{x},\infty)$ bounded by
$\eta_{1}$, $\eta_{2}$ and $\partial\mathbbm{H}$ is given by

$C\int_{\delta}^{1/2}\int_{1-b}^{2-b}\int_{0}^{\infty}\int_{0}^{\infty}\mathcal{M}_{0,2}^{\mathrm{disk}}(\frac{\gamma^{2}}{2};a,p)\times\mathcal{M}^{\mathrm{disk}}_{1,2}(2;p,q)\times\mathcal{M}_{0,2}^{\mathrm{disk}}(\frac{\gamma^{2}}{2};q,b)\,dq\,dp\,da\,db.$
(5.2)

###### Proof.

Lemma 3.15 states that if we sample $(\phi,\mathbf{x})$ from
$\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}\times dx$ then the law of the quantum
surface $(\mathbbm{H},\phi,i,\mathbf{x},\infty)$ is
$C\mathcal{M}_{1,2}^{\mathrm{disk}}(2+\gamma^{2})$ for some constant $C$. If
we further sample $(\eta_{1},\eta_{2})$ from
$\mathcal{P}(\mathbbm{H},\mathbf{x},\infty)$ and restrict to the event that
$i$ lies between $\eta_{1}$ and $\eta_{2}$, then by Definition 3.10 and
Theorem 5.1, the quantum surface
$(\mathbbm{H},\phi,\eta_{1},\eta_{2},i,\mathbf{x},\infty)$ has law

$C\iiiint_{0}^{\infty}\mathcal{M}_{0,2}^{\mathrm{disk}}(\frac{\gamma^{2}}{2};a,p)\times\mathcal{M}^{\mathrm{disk}}_{1,2}(2;p,q)\times\mathcal{M}_{0,2}^{\mathrm{disk}}(\frac{\gamma^{2}}{2};q,b)\,dq\,dp\,da\,db$

for some $C>0$. Further restricting to the event
$\nu_{\phi}(\mathbf{x},\infty)\in(\delta,\frac{1}{2})$ and
$\nu_{\phi}(\mathbbm{R})\in(1,2)$ we conclude the proof. ∎

In Section 5.2, we will show that the probability measure $M^{\\#}_{\delta}$
that is proportional to $M_{\delta}$ concentrates on the event that
$\nu_{h}(\mathbf{x},\infty)$ is of order $\delta$ and $\log\mathbf{x}$ is of
order $\log\delta^{-1}$. This effectively collapses the right boundary of
$(\mathbbm{H},\phi,\mathbf{x},\infty)$. Building on this, we will prove the
following proposition. Recall that for two probability measures $P,Q$ on the
same measure space, the total variation distance between $P$ and $Q$ is
$\sup_{A}|P(A)-Q(A)|$ where the supremum is taken over all measurable sets.

######  Proposition 5.3.

For each $\delta\in(0,\frac{1}{2})$, let $\mathsf{m}_{\delta}$ be the marginal
law of $(\mathbf{x},\eta_{1},\eta_{2})$ under $M^{\\#}_{\delta}$. Let $M$ be
$\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}$ restricted to
$\\{\nu_{\phi}(\mathbbm{R})\in(1,2)\\}$. Then the total variational distance
of $M_{\delta}^{\\#}$ and $M^{\\#}\times\mathsf{m}_{\delta}$ tends to 0 as
$\delta\to 0$.

In Section 5.3, we prove that by sending $\delta\to 0$ in the integral (5.2),
we get the conformal welding of $\mathrm{QD}_{1,1}$ and
$\mathcal{M}_{0,2}^{\mathrm{disk}}(\frac{\gamma^{2}}{2})$ as in Theorem 4.1.
Moreover, $\mathsf{m}_{\delta}$ from Proposition 5.3 has a weak limit
supported on loops rooted at $\infty$, whose pushforward under
$z\mapsto-z^{-1}$ gives the measure $\mathsf{m}$ in Theorem 4.1.

Our approximation procedure (where we conformally weld three quantum disks and
have one vanish in the $\delta\to 0$ limit) may seem more complicated than
necessary, but it is advantageous for the following reasons. Firstly, in our
setup the law of the field of $M_{\delta}$ is absolutely continuous with
respect to $M$, so the statement and proof of Proposition 5.3 can avoid
modifying the field in some way. Secondly, $M_{\delta}$ arises from a setup
where the interfaces and field are independent (i.e.
$(\phi,\mathbf{x})\sim\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}\times dx$ then
$(\eta_{1},\eta_{2})\sim\mathcal{P}(\mathbbm{H},\mathbf{x},\infty)$). Finally,
this approach allows us to avoid SLE estimates entirely.

###  5.2 An approximate bubble zipper: proof of Proposition 5.3

Suppose we are in the setting of Lemma 5.2 and Proposition 5.3. We first give
a simple description of the Radon-Nykodim derivative of
$\frac{dM_{\delta}}{dM}$. For $x\in\mathbbm{R}$, let $p(x)$ be the conditional
probability that $i$ is in between $\eta_{1}$ and $\eta_{2}$ given
$\mathbf{x}=x$. Then $p(x)$ is an even function determined by the SLE measure
$\mathcal{P}(\mathbbm{H},x,\infty)$. Define the function
$f:\mathbbm{R}\to\mathbbm{R}$ by

$f(x)=\mathopen{}\mathclose{{}\left\\{\begin{array}[]{ll}\int_{0}^{x}p(y)\,dy&\mbox{if
}x\geq 0\\\ -\int_{x}^{0}p(y)\,dy&\mbox{if }x<0.\end{array}}\right.$ (5.3)

For each $\ell\in(0,\nu_{\phi}(\mathbbm{R}))$, let

$x_{\ell}=\inf\\{x\in\mathbbm{R}:\nu_{\phi}(x,\infty)=\ell\\}.$ (5.4)

######  Lemma 5.4.

For a non-negative measurable function $F$ on $H^{-1}(\mathbbm{H})$, we have

$M_{\delta}[F(\phi)]=\int[f(x_{\delta})-f(x_{1/2})]F(\phi)\,dM.$

###### Proof.

By definition, if we sample $(\phi,\mathbf{x})$ from $M\times dx$ and then
sample $(\eta_{1},\eta_{2})$ from
$\mathcal{P}(\mathbbm{H},\mathbf{x},\infty)$, and restrict to the event that
$\mathbf{x}\in(x_{1/2},x_{\delta})$ and $i$ is in between $\eta_{1}$ and
$\eta_{2}$, then the law of $(\phi,\mathbf{x},\eta_{1},\eta_{2})$ is
$M_{\delta}$. Conditioning on $(\phi,\mathbf{x})$, the conditional probability
that $i$ lies between $\eta_{1}$ and $\eta_{2}$ is
$\int_{x_{1/2}}^{x_{\delta}}p(x)\,dx=f(x_{\delta})-f(x_{1/2})$. This gives the
result. ∎

The hardest step in proving Proposition 5.3 is to show that the
$M_{\delta}^{\\#}$-law of $\phi$ converges to that of $M^{\\#}$, i.e., to show
that a.s.
$\frac{dM_{\delta}^{\\#}}{dM^{\\#}}(\phi)=|M|(f(x_{\delta})-f(x_{1/2}))/|M_{\delta}|\to
1$ as $\delta\to 0$. In the argument below Lemma 5.5 will give that
$|M_{\delta}|$ diverges as $\log\delta^{-1}$. Lemma 5.7 will give that
$x_{\delta}=\delta^{-\frac{2}{\gamma Q}+o(1)}$ a.s. Lemmas 5.10 and 5.11 will
give the asymptotic growth of $f(x_{\delta})$. Putting them together we will
get the desired limit.

Let $A=\nu_{h}(-\infty,\mathbf{x})$ and $B=\nu_{h}(\mathbf{x},\infty)$. Let
$P$ and $Q$ be the quantum lengths of $\eta_{1}$ and $\eta_{2}$ with respect
to $\phi$.

######  Lemma 5.5.

There exists $C\in(0,\infty)$ such that $\lim_{\delta\to
0}\frac{|M_{\delta}|}{\log\delta^{-1}}=C$. Moreover, the
$M_{\delta}^{\\#}$-law of $(A,P)$ converges in total variational distance to
the probability measure on $[1,2]\times(0,\infty)$ whose density is
proportional to

$\frac{a^{4/\gamma^{2}-1}}{(a^{4/\gamma^{2}}+p^{4/\gamma^{2}})^{2}}\,dadp$

and $\lim_{\delta\to 0}M_{\delta}^{\\#}[B>\varepsilon]=\lim_{\delta\to
0}M_{\delta}^{\\#}[Q>\varepsilon]=0$ for each $\varepsilon>0$.

###### Proof.

Our proof relies on (5.2), which gives a description of the joint distribution
of $(A,P,Q,B)$ under $M_{\delta}$ in terms of
$\mathcal{M}_{0,2}^{\mathrm{disk}}(\frac{\gamma^{2}}{2})$ and
$\mathcal{M}_{1,2}^{\mathrm{disk}}(2)$, whose quantum length distributions are
given in Lemma 2.15. Indeed, write $g=\frac{4}{\gamma^{2}}>1$. By Lemma 2.15
and the definition of $\mathcal{M}^{\mathrm{disk}}_{1,2}(2;\ell,r)$, there
exist constants $C_{1},C_{2}$ such that

$|\mathcal{M}_{0,2}^{\mathrm{disk}}(\frac{\gamma^{2}}{2};\ell,r)|=C_{1}\frac{(\ell
r)^{g-1}}{(\ell^{g}+r^{g})^{2}}\quad\textrm{and}\quad|\mathcal{M}^{\mathrm{disk}}_{1,2}(2;\ell,r)|=C_{2}(\ell+r)^{-g+1}.$
(5.5)

By (5.2), the $M_{\delta}^{\\#}$-law of $(A,P,Q,B)$ is the probability measure
supported on the set
$S_{\delta}=\\{(a,p,q,b)\in(0,\infty)^{4}:b\in(\delta,1/2),a+b\in(1,2)\\}$
whose density function is proportional to

$m(a,p,q,b)=\frac{(ap)^{g-1}}{(a^{g}+p^{g})^{2}}\cdot(p+q)^{-g+1}\cdot\frac{(bq)^{g-1}}{(b^{g}+q^{g})^{2}}$
(5.6)

and

$|M_{\delta}|=C_{1}^{2}C_{2}\int_{S_{\delta}}m(a,p,q,b)\,da\,dp\,dq\,db.$
(5.7)

Using the change of variable $p=ar$ and $q=bs$, we see that

$\iint_{0}^{\infty}m(a,p,q,b)\,dp\,dq=\frac{1}{a^{g}b}\iint_{0}^{\infty}\frac{r^{g-1}}{(1+r^{g})^{2}}\cdot\frac{1}{(r+\frac{b}{a}s)^{g-1}}\cdot\frac{s^{g-1}}{(1+s^{g})^{2}}\,dr\,ds.$
(5.8)

By the monotone convergence theorem, the last integral converges to

$C_{3}=\iint_{0}^{\infty}\frac{s^{g-1}}{(1+r^{g})^{2}(1+s^{g})^{2}}\,drds<\infty\textrm{
as }b/a\to 0.$

Therefore:

$\iint_{0}^{\infty}m(a,p,q,b)\,dp\,dq=\frac{C_{3}}{a^{g}b}(1+o_{b/a}(1)).$
(5.9)

Integrating (5.9) over $b\in(\delta,1/2),a+b\in(1,2)$ and using (5.7), we get
$\lim_{\delta\to 0}\frac{|M_{\delta}|}{\log\delta^{-1}}=C$ with
$C:=\int_{1}^{2}\frac{C_{1}^{2}C_{2}C_{3}}{a^{g}}\,da$. Since
$\sup_{\delta\in(0,1/2)}M_{\delta}[B>\varepsilon]<\infty$ for each
$\varepsilon>0$, $\lim_{\delta\to 0}M_{\delta}^{\\#}[B>\varepsilon]=0$. By
(5.8), the $M_{\delta}^{\\#}$-law of $\frac{Q}{B}$ is converging to the
probability measure on $(0,\infty)$ that is proportional to
$\frac{s^{g-1}\,ds}{(1+s^{g})^{2}}$. In particular, $\lim_{\delta\to
0}M_{\delta}^{\\#}[Q>\varepsilon]=0$. Similarly, the $M_{\delta}^{\\#}$-law of
$(A,\frac{P}{A})$ converges to the probability measure on
$[1,2]\times(0,\infty)$ that is proportional to
$\frac{da}{a^{g}}\times\frac{dr}{(1+r^{g})^{2}}$. This gives the desired
limiting joint distribution of $(A,P)$. ∎

We first gather some basic facts on the quantum boundary length under the GFF
measure $P_{\mathbbm{H}}$.

######  Lemma 5.6.

Sample $h$ from $P_{\mathbbm{H}}$. For each $p>0$ we have
$P_{\mathbbm{H}}[\nu_{h}(0,1)<\frac{1}{s}]=O(s^{-p})$ as $s\to\infty$.
Moreover, there exists $C\in(0,\infty)$ such that
$P_{\mathbbm{H}}[\nu_{h}(0,y)>t]\leq Ct^{-\frac{4}{\gamma^{2}}}y$ for all
$t>0$ and $y\in(0,1)$. Finally, $\lim_{y\to
0^{+}}\frac{\mathopen{}\mathclose{{}\left|\log
y}\right|}{\mathopen{}\mathclose{{}\left|\log\nu_{h}(0,y)}\right|}=\frac{2}{\gamma
Q}$ a.s.

###### Proof.

We first show that if $G$ is a standard Gaussian independent of $h$ and
$y\in(0,1)$, then

$\nu_{h}(0,y)\stackrel{{\scriptstyle
d}}{{=}}e^{\frac{\gamma}{2}(-Q\mathopen{}\mathclose{{}\left|\log
y}\right|+\sqrt{2\mathopen{}\mathclose{{}\left|\log
y}\right|}G)}\nu_{h}(0,1).$ (5.10)

Let $c$ be the average of $h$ on $\partial B_{y}(0)\cap\mathbbm{H}$, so the
marginal law of $c$ is $N(0,2\mathopen{}\mathclose{{}\left|\log y}\right|)$.
The conditional law of $\widetilde{h}=h-c$ given $c$ is the free-boundary GFF
on $\mathbbm{H}$ normalized to have average zero on $\partial
B_{y}(0)\cap\mathbbm{H}$. Let $\psi(z)=yz$ for $z\in\mathbbm{H}$. Then
$\widetilde{h}\circ\psi$ has law $P_{\mathbbm{H}}$. Therefore

$\nu_{h}(0,y)=e^{\frac{\gamma}{2}c}\nu_{\tilde{h}}(0,y)=e^{\frac{\gamma}{2}c}\nu_{\tilde{h}\circ\psi+Q\log|\psi^{\prime}|}(0,1)\stackrel{{\scriptstyle
d}}{{=}}e^{\frac{\gamma}{2}c+\frac{\gamma}{2}Q\log y}\nu_{h}(0,1).$

This proves (5.10).

Now we address the tail bounds in the lemma. [RV14, Theorem 2.12] gives the
bound $P_{\mathbbm{H}}[\nu_{h}(0,1)<\frac{1}{s}]=O(s^{-p})$ as $s\to\infty$.
[Won20, Theorem 1.1] gives $P_{\mathbbm{H}}[\nu_{h}(0,1)>s]\leq
Cs^{-\frac{4}{\gamma^{2}}}$ for all $s>0$, so using (5.10) and
$s=te^{\frac{\gamma}{2}(Q\mathopen{}\mathclose{{}\left|\log
y}\right|-\sqrt{2\mathopen{}\mathclose{{}\left|\log y}\right|}G)}$ gives

$P_{\mathbbm{H}}[\nu_{h}(0,y)>t]\leq
C\mathbbm{E}[(te^{\frac{\gamma}{2}(Q\mathopen{}\mathclose{{}\left|\log
y}\right|-\sqrt{2\mathopen{}\mathclose{{}\left|\log
y}\right|}G)})^{-\frac{4}{\gamma^{2}}}]=Ct^{-\frac{4}{\gamma^{2}}}y.$

For the last assertion, let $y_{n}=e^{-(\log n)^{4}}$ and
$\varepsilon_{n}=\frac{1}{\log n}$ so that $y_{n}$ and $\varepsilon_{n}\to 0$
as $n\to\infty$. Let $E_{n}=\\{\nu_{h}(0,y_{n})\in(y_{n}^{\frac{\gamma
Q}{2}+2\varepsilon_{n}},y_{n}^{\frac{\gamma Q}{2}-2\varepsilon_{n}})\\}$. By
the tail bounds, we have

$P_{\mathbbm{H}}[E_{n}^{c}]\leq\mathbbm{P}[e^{\frac{\gamma}{2}\sqrt{2\mathopen{}\mathclose{{}\left|\log
y_{n}}\right|}G}\not\in(y_{n}^{\varepsilon_{n}},y_{n}^{-\varepsilon_{n}})]+P_{\mathbbm{H}}[\nu_{h}(0,1)\not\in(y_{n}^{\varepsilon_{n}},y_{n}^{-\varepsilon_{n}})]\leq
2e^{-O(\varepsilon_{n}^{2}\mathopen{}\mathclose{{}\left|\log
y_{n}}\right|)}+O(y_{n}^{\frac{4}{\gamma^{2}}\varepsilon_{n}}).$

Here, the first term is bounded from above by the standard Gaussian tail
bound. We conclude that $\sum_{n}P_{\mathbbm{H}}[E_{n}^{c}]<\infty$, so the
Borel-Cantelli lemma yields that a.s. $E_{n}$ occurs for all sufficiently
large $n$. Since $\lim_{n\to\infty}\frac{\mathopen{}\mathclose{{}\left|\log
y_{n+1}}\right|}{\mathopen{}\mathclose{{}\left|\log y_{n}}\right|}=1$, this
gives that $\lim_{y\to\infty}\frac{\mathopen{}\mathclose{{}\left|\log
y}\right|}{\mathopen{}\mathclose{{}\left|\log\nu_{h}(0,y)}\right|}=\frac{2}{\gamma
Q}$ a.s. ∎

We now give an asymptotic estimate for $\log x_{\delta}$.

######  Lemma 5.7.

$M$-almost everywhere, $\lim_{\delta\to 0}\frac{\log
x_{\delta}}{\log\delta^{-1}}=\frac{2}{\gamma Q}$.

###### Proof.

Recall from Definition 2.3 that $\phi$ sampled from
$\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}$ is given by
$\phi=h-2Q\log|z|_{+}+\gamma G_{\mathbbm{H}}(\cdot,i)+\mathbf{c}$ where
$(h,\mathbf{c})$ is sampled from
$2^{-\gamma^{2}/2}P_{\mathbbm{H}}\times[e^{(\gamma-Q)c}dc]$. Thus
$\frac{\nu_{\phi}(0,y)}{e^{\frac{\gamma}{2}\mathbf{c}}\nu_{h}(0,y)}\in(e^{-\frac{\gamma^{2}}{2}G_{\mathbbm{H}}(i,1)},1)$
for all $y\in(0,1)$, and so $\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}$-almost
everywhere $\lim_{y\to 0^{+}}\frac{\log\nu_{\phi}(0,y)}{\log\nu_{h}(0,y)}=1$.
Therefore by the last assertion of Lemma 5.6, we have

$\lim_{y\to 0^{+}}\frac{\mathopen{}\mathclose{{}\left|\log
y}\right|}{\mathopen{}\mathclose{{}\left|\log\nu_{\phi}(0,y)}\right|}=\frac{2}{\gamma
Q}\quad\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}\text{-almost everywhere}.$
(5.11)

By Lemma 2.4 applied to the inversion map $z\mapsto-z^{-1}$, this yields that
$\lim_{y\to\infty}\frac{\log y}{-\log\nu_{\phi}(y,\infty)}=\frac{2}{\gamma
Q}$. Setting $y=x_{\delta}$ so that $\nu_{\phi}(y,\infty)=\delta$, we get
$\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}$-almost everywhere $\lim_{\delta\to
0}\frac{\log x_{\delta}}{\log\delta^{-1}}=\frac{2}{\gamma Q}$. Hence the same
limit holds for $M$ by restricting to $\nu_{\phi}(\mathbbm{R})\in(1,2)$. ∎

The following tail bound is quite loose, but suffices for our purposes.

######  Lemma 5.8.

For $\zeta\in(0,1)$, there exists $C_{\zeta}\in(0,\infty)$ such that

$M[\\{x_{t}\geq y\\}]<C_{\zeta}(t^{4/\gamma^{2}}y)^{-(1-\zeta)}\quad\text{for
all }t>0\text{ and }y>1.$ (5.12)

###### Proof.

By the definition of $x_{t}$, (5.12) is equivalent to
$M[\\{\nu_{\phi}(y,\infty)>t\\}]<C_{\zeta}(t^{4/\gamma^{2}}y)^{-(1-\zeta)}$
for all $t>0$ and $y>1$. Using Lemma 2.4 with the inversion map
$z\mapsto-z^{-1}$, this is equivalent to

$M[\\{\nu_{\phi}(0,y)>t\\}]<C_{\zeta}(t^{-4/\gamma^{2}}y)^{(1-\zeta)}\quad\text{for
all }t>0\text{ and }y\in(0,1).$ (5.13)

Recall that $M$ is $\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}$ restricted to
$\\{\nu_{\phi}(\mathbbm{R})\in(1,2)\\}$. By Definition 2.3, $\phi$ sampled
from $\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}$ is
$\phi(z)=h(z)-2Q\log|z|_{+}+\gamma G_{\mathbbm{H}}(z,i)+\mathbf{c}$ where
$(h,\mathbf{c})$ is sampled from
$2^{-\gamma^{2}/2}P_{\mathbbm{H}}\times[e^{(\gamma-Q)c}dc]$. Since
$-2Q\log|z|_{+}+\gamma G_{\mathbbm{H}}(z,i)\in(-\log 2,0)$ for $z\in[-1,1]$,
we have
$2^{-\gamma^{2}/2}e^{\frac{\gamma}{2}\mathbf{c}}\nu_{h}(I)<\nu_{\phi}(I)<e^{\frac{\gamma}{2}\mathbf{c}}\nu_{h}(I)$
for any $I\subset[-1,1]$. Since $M$-a.s. we have $\nu_{\phi}(-1,0)<2$, for any
$r\in\mathbbm{R}$ we have

$\displaystyle 2^{\gamma^{2}/2}M[\\{\nu_{\phi}(0,y)>t\\}]$
$\displaystyle\leq\int_{\mathbbm{R}}P_{\mathbbm{H}}\mathopen{}\mathclose{{}\left[e^{\frac{\gamma}{2}c}\nu_{h}(0,y)>t\text{
and
}2^{-\gamma^{2}/2}e^{\frac{\gamma}{2}c}\nu_{h}(-1,0)<2}\right]e^{(\gamma-Q)c}\,dc$
$\displaystyle\leq\int_{-\infty}^{r}P_{\mathbbm{H}}[e^{\frac{\gamma}{2}c}\nu_{h}(0,y)>t]e^{(\gamma-Q)c}\,dc+\int_{r}^{\infty}P_{\mathbbm{H}}[e^{\frac{\gamma}{2}c}\nu_{h}(-1,0)<2^{\gamma^{2}/2+1}]e^{(\gamma-Q)c}\,dc.$

The first term can be bounded using the second assertion in Lemma 5.6

$\int_{-\infty}^{r}P_{\mathbbm{H}}[e^{\frac{\gamma}{2}c}\nu_{h}(0,y)>t]e^{(\gamma-Q)c}\,dc\leq
C\int_{-\infty}^{r}t^{-\frac{4}{\gamma^{2}}}ye^{\frac{2}{\gamma}c}\cdot
e^{(\gamma-Q)c}\,dc=\frac{2}{\gamma}t^{-\frac{4}{\gamma^{2}}}ye^{\frac{\gamma}{2}r},$

where $C\in(0,\infty)$ is a constant that can change from line to line. For
the second term, by Lemma 5.6 for any $p>0$ we have:

$\int_{r}^{\infty}P_{\mathbbm{H}}[e^{\frac{\gamma}{2}c}\nu_{h}(-1,0)<2^{\gamma^{2}/2+1}]e^{(\gamma-Q)c}\,dc\leq
C\int_{r}^{\infty}2^{(\gamma^{2}/2+1)p}e^{-\frac{\gamma}{2}pc}e^{(\gamma-Q)c}\,dc\leq
Ce^{(\gamma-Q-\frac{\gamma}{2}p)r}.$

Combining the two bounds gives:

$M[\\{\nu_{\phi}(0,y)>t\\}]\leq
Ct^{-\frac{4}{\gamma^{2}}}ye^{\frac{\gamma}{2}r}+Ce^{(\gamma-Q-\frac{\gamma}{2}p)r}.$

Choosing $r=\frac{\frac{4}{\gamma^{2}}\log
t+\mathopen{}\mathclose{{}\left|\log
y}\right|}{\frac{2}{\gamma}+\frac{\gamma}{2}p}$ gives (5.13) with
$\zeta=\frac{\frac{\gamma}{2}}{\frac{2}{\gamma}+\frac{\gamma}{2}p}$. Varying
$p\in(0,\infty)$, we get all $\zeta\in(0,1)$. ∎

The next lemma allows us to control $f(x_{1/2})$.

######  Lemma 5.9.

We have $M[|f(x_{1/2})|]<\infty$.

###### Proof.

We first show there is a constant $C>0$ such that

$f(y)<C\log y\quad\text{ for all }y>2.$ (5.14)

Write $p=M[\\{x_{1/2}<0\\}]$. By Lemma 5.7, for $\delta$ small enough,
$M[\\{x_{\delta}<\delta^{-\frac{1}{\gamma Q}}\\}]<\frac{p}{2}$. Let
$F_{\delta}=\\{x_{\frac{1}{2}}<0\text{ and
}x_{\delta}>\delta^{-\frac{1}{\gamma Q}}\\}$ so that
$M[F_{\delta}]\geq\frac{p}{2}$. By Lemma 5.4 we have
$|M_{\delta}|=M[f(x_{\delta})-f(x_{1/2})]\geq
M[1_{F_{\delta}}(f(x_{\delta})-f(x_{1/2}))]$. Since $f$ is increasing and
$f(0)=0$, we have

$M[1_{F_{\delta}}(f(x_{\delta})-f(x_{1/2}))]\geq f(\delta^{-\frac{1}{\gamma
Q}})M[F_{\delta}]\geq\frac{p}{2}f(\delta^{-\frac{1}{\gamma Q}}).$

By Lemma 5.5, $|M_{\delta}|=(1+o_{\delta}(1))\mathfrak{C}\log\delta^{-1}$.
Therefore there exists $C\in(0,\infty)$ such that $f(\delta^{-\frac{1}{\gamma
Q}})\leq C\log\delta^{-1}$ for all $\delta\in(0,1/2)$. Setting
$y=\delta^{-\frac{1}{\gamma Q}}$ and possibly choosing $C$ larger we obtain
(5.14).

Now we bound $M[|f(x_{1/2})|]$. By (5.3) $f$ is an increasing odd function, so

$M[1_{x_{1/2}>1}f(x_{1/2})]=\sum_{n=1}^{\infty}M[1_{2^{n}\geq
x_{1/2}>2^{n-1}}f(x_{1/2})]\leq\sum_{n=1}^{\infty}f(2^{n})M[\\{x_{1/2}>2^{n-1}\\}].$
(5.15)

By (5.14) and the tail bound of $x_{1/2}$ in Lemma 5.8, for a possibly larger
$C$ we have

$M[1_{x_{1/2}>1}f(x_{1/2})]\leq\sum_{n=1}^{\infty}C\log
2^{n}M[\\{x_{1/2}>2^{n-1}\\}]\leq C^{2}\sum_{n=1}^{\infty}n2^{-n/2}<\infty.$

Since $0<f(t)<f(2)$ for $t\in(0,2)$, we have $M[1_{1\geq x_{1/2}\geq
0}f(x_{1/2})]\leq f(2)|M|<\infty$ hence $M[1_{x_{1/2}\geq
0}|f(x_{1/2})|]=M[1_{x_{1/2}\geq 0}f(x_{1/2})]<\infty$. As $M$-almost
everywhere $\nu_{\phi}(\mathbbm{R})\geq 1$, we conclude

$M[x_{1/2}<-y]=M[\nu_{\phi}(-y,\infty)\leq\frac{1}{2}]\leq
M[\nu_{\phi}(-\infty,-y)\geq\frac{1}{2}]=M[x_{1/2}>y]\textrm{ for }y>0.$

Since $f$ is an odd function, we have that $M[1_{x_{1/2}<0}|f(x_{1/2})|]\leq
M[1_{x_{1/2}\geq 0}|f(x_{1/2})|]<\infty$. ∎

Now, we give the asymptotic upper bound on $f$.

######  Lemma 5.10.

For any $s>0$ we have $\limsup_{\delta\to 0}|M|f(\delta^{-\frac{2}{\gamma
Q}+s})/|M_{\delta}|\leq 1$.

###### Proof.

By Lemma 5.4, we have

$\displaystyle|M_{\delta}|$ $\displaystyle=M[f(x_{\delta})-f(x_{1/2})]\geq
M[1_{\\{x_{\delta}>\delta^{-\frac{2}{\gamma
Q}+s}\\}}(f(x_{\delta})-f(x_{1/2}))]$ $\displaystyle\geq
f(\delta^{-\frac{2}{\gamma
Q}+s})(|M|-M[\\{x_{\delta}\leq\delta^{-\frac{2}{\gamma
Q}+s}\\}])-M[|f(x_{1/2})|].$

Since $\lim_{\delta\to 0}M[\\{x_{\delta}\leq\delta^{-\frac{2}{\gamma
Q}+s}\\}]=0$ by Lemma 5.7, $|M_{\delta}|\to\infty$ by Lemma 5.5 and
$M[|f(x_{1/2})|]<\infty$ by Lemma 5.9 we are done. ∎

We next obtain the matching asymptotic lower bound on $f$.

######  Lemma 5.11.

For any $s>0$ we have $\liminf_{\delta\to 0}|M|f(\delta^{-\frac{2}{\gamma
Q}-s})/|M_{\delta}|\geq 1$.

###### Proof.

Let $E_{\delta}=\\{x_{\delta}>\delta^{-\frac{2}{\gamma Q}-s}\\}$. By the
monotonicity of $f$ we may assume $\frac{2}{\gamma Q}+s<\frac{8}{\gamma^{2}}$,
and by (5.14) and monotonicity we have

$\displaystyle|M_{\delta}|$ $\displaystyle=M[f(x_{\delta})-f(x_{1/2})]\leq
M[1_{E_{\delta}}f(x_{\delta})]+M[1_{E_{\delta}^{c}}f(x_{\delta})]+M[|f(x_{1/2})|]$
$\displaystyle\leq
CM[1_{E_{\delta}}\log(x_{\delta})]+f(\delta^{-\frac{2}{\gamma
Q}-s})(|M|-M[E_{\delta}])+M[|f(x_{1/2})|].$

We claim that $M[1_{E_{\delta}}\log x_{\delta}]=o(\log\delta^{-1})$. To see
this, we write

$M[1_{E_{\delta}}\log x_{\delta}]=\int_{0}^{\infty}M[E_{\delta}\cap\\{\log
x_{\delta}>w\\}]\,dw\leq\frac{8}{\gamma^{2}}\log\delta^{-1}M[E_{\delta}]+\int_{\frac{8}{\gamma^{2}}\log\delta^{-1}}^{\infty}M[\\{\log
x_{\delta}>w\\}]\,dw.$

By Lemma 5.7 the first term is $o(\log\delta^{-1})$, and for the second term,
let $\zeta\in(0,1)$ and use Lemma 5.8:

$\int_{\frac{8}{\gamma^{2}}\log\delta^{-1}}^{\infty}M[\\{\log
x_{\delta}>w\\}]\,dw\leq\int_{\frac{8}{\gamma^{2}}\log\delta^{-1}}^{\infty}C_{\zeta}(\delta^{\frac{4}{\gamma^{2}}}e^{w})^{-1+\zeta}\,dw=(1-\zeta)^{-1}C_{\zeta}\delta^{\frac{(1-\zeta)4}{\gamma^{2}}}.$

Thus $M[1_{E_{\delta}}\log x_{\delta}]=o(\log\delta^{-1})$. Moreover,
$\lim_{\delta\to 0}M[E_{\delta}]=0$ by Lemma 5.7,
$|M_{\delta}|\gtrsim\log\delta^{-1}$ by Lemma 5.5 and $M[|f(x_{1/2})|]<\infty$
by Lemma 5.9. This gives the desired result. ∎

Now we are ready to conclude the proof of Proposition 5.3. We start by the
marginal law of $\phi$.

######  Lemma 5.12.

The $M^{\\#}_{\delta}$-law of $\phi$ converges to $M^{\\#}$ in total
variational distance as $\delta\to 0$.

###### Proof.

Let $R_{\delta}(\phi)=\frac{dM_{\delta}^{\\#}}{dM^{\\#}}(\phi)$ be the Radon
derivative of the marginal of $\phi$ under $M^{\\#}_{\delta}$ with respect to
$M$. By Lemma 5.4, we have

$R_{\delta}(\phi)=\frac{|M|}{|M_{\delta}|}(f(x_{\delta})-f(x_{1/2}))\textrm{
almost surely in }M^{\\#}.$

By Lemmas 5.7, 5.10 and 5.11, we see that $\lim_{\delta\to
0}\frac{|M|}{|M_{\delta}|}f(x_{\delta})=1$ a.s. in $M^{\\#}$. Since
$\lim_{\delta\to 0}|M_{\delta}|=\infty$ and $-\infty<f(x_{1/2})<\infty$ a.s.
in $M^{\\#}$, we see that $\lim_{\delta\to 0}R_{\delta}=1$ a.s. in $M^{\\#}$.
Since $M^{\\#}[R_{\delta}(\phi)]=1$ for each $\delta$, we have
$\lim_{\delta\to 0}M^{\\#}[|R_{\delta}(\phi)-1|]=0$. Therefore, with the
supremum taken over measurable sets $A$,
$\sup_{A}|M_{\delta}^{\\#}[A]-M^{\\#}[A]|=\sup_{A}\mathopen{}\mathclose{{}\left|M^{\\#}[1_{A}(R_{\delta}(\phi)-1)]}\right|\leq
M^{\\#}[|R_{\delta}(\phi)-1|]\to 0$, which concludes the proof. ∎

We now deal with the joint law of $(\phi,\mathbf{x})$.

######  Lemma 5.13.

Let $\hat{m}_{\delta}=1_{\\{0<x<\delta^{-\frac{2}{\gamma Q}}\\}}p(x)\,dx$. The
$M_{\delta}^{\\#}$-law of $(\phi,\mathbf{x})$ is $o_{\delta}(1)$-close in
total variational distance to $M^{\\#}\times\hat{m}^{\\#}_{\delta}$.

###### Proof.

The conditional law of $\mathbf{x}$ given $\phi$ is
$1_{x_{1/2}<x<{x_{\delta}}}(f(x_{\delta})-f(x_{1/2}))^{-1}p(x)dx$. Let
$d_{\delta}(\phi)$ be the total variational distance between this distribution
and $\hat{m}^{\\#}_{\delta}$. Let
$E_{\delta}=\\{|\frac{x_{\delta}}{\log\delta^{-1}}-\frac{2}{\gamma Q}|<1\\}$.
By Lemmas 5.7 and 5.12, $\lim_{\delta\to 0}M_{\delta}[E_{\delta}^{c}]=0$.
Moreover, by Lemmas 5.5, 5.10 and 5.11, $\lim_{y\to\infty}(\log y)^{-1}f(y)$
is a positive constant, and hence $\lim_{\delta\to
0}M_{\delta}[1_{E_{\delta}}d_{\delta}(\phi)]=0$. We conclude that
$\lim_{\delta\to 0}M_{\delta}[d_{\delta}(\phi)]=0$ as needed. ∎

###### Proof of Proposition 5.3.

Since the conditional law of $(\eta_{1},\eta_{2})$ under $M_{\delta}^{\\#}$
given $(\phi,\mathbf{x})$ only depends on $\mathbf{x}$, Lemma 5.13 implies
Proposition 5.3. ∎

######  Remark 5.14.

The above argument implies that there exists a constant
$\mathfrak{C}\in(0,\infty)$ such that

$\lim_{\delta\to
0}\frac{|M_{\delta}|}{\log\delta^{-1}}=\mathfrak{C}\quad\textrm{and}\quad\lim_{y\to\infty}\frac{f(y)}{\log
y}=\frac{\gamma Q\mathfrak{C}}{2|M|}.$ (5.16)

Moreover, the constant $\mathfrak{C}$ can be made explicit by keeping track of
the various constants in Lemma 5.5 (see [AHS21, Proposition 5.2] for these
constants). We find it interesting that though the function $f$ is defined in
terms of SLE, its asymptotics are derived using properties of the Liouville
field $\phi$.

###  5.3 Passing to the limit: proof of Theorem 4.1

The arguments in this section are technical but standard and can be skipped on
a first reading. Sample a pair of quantum surfaces
$(\mathcal{D}_{1},\mathcal{D}_{2})$ from
$\int_{1}^{2}\int_{0}^{\infty}\mathcal{M}_{0,2}^{\mathrm{disk}}(\frac{\gamma^{2}}{2};a,p)\times\mathrm{QD}_{1,1}(p)\,dp\,da$.
Let $\mathcal{D}_{1}\oplus\mathcal{D}_{2}$ be the curve-decorated quantum
surface obtained by conformally welding the right boundary of
$\mathcal{D}_{1}$ and the total boundary of $\mathcal{D}_{2}$. The surface
$\mathcal{D}_{1}\oplus\mathcal{D}_{2}$ naturally carries a marked interior
point and a marked boundary point. See Figure 4. Let
$(\mathbbm{D},\phi_{\mathbbm{D}},\eta_{\mathbbm{D}},0,i)$ be the unique
embedding of $\mathcal{D}_{1}\oplus\mathcal{D}_{2}$ in $(\mathbbm{D},0,i)$. We
denote the law of $(\phi_{\mathbbm{D}},\eta_{\mathbbm{D}})$ by
$M_{\mathbbm{D}}$. Let $f:\mathbbm{H}\to\mathbbm{D}$ be the conformal map with
$f(i)=0$ and $f(\infty)=i$. We will prove Theorem 4.1 by showing the
following.

######  Proposition 5.15.

Under the probability measure $M_{\mathbbm{D}}^{\\#}$ proportional to
$M_{\mathbbm{D}}$, $\phi_{\mathbbm{D}}$ and $\eta_{\mathbbm{D}}$ are
independent, and moreover, the law of $\phi_{\mathbbm{D}}\circ
f+Q\log|f^{\prime}|$ is proportional to
$\int_{1}^{2}\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}(\ell)d\ell$.

Recall $M_{\delta}$ from Lemma 5.2 and Proposition 5.3. Let
$(\phi,\mathbf{x},\eta_{1},\eta_{2})$ be sampled from $M_{\delta}^{\\#}$. Let
$\mathcal{D}_{1,\delta}$, $\mathcal{D}_{2,\delta}$, $\mathcal{D}_{3,\delta}$
be the left, right, and middle marked quantum surface in the sample space of
$M_{\delta}$, so that $(\mathbbm{H},\phi,\eta_{1},\eta_{2},i,\mathbf{x})$ is
the embedding of the conformally welded surface
$\mathcal{D}_{1,\delta}\oplus\mathcal{D}_{2,\delta}\oplus\mathcal{D}_{3,\delta}$.
Let $\phi^{\delta}=\phi\circ f^{-1}+Q\log|(f^{-1})^{\prime}|$ and
$\eta^{\delta}=f\circ\eta_{1}$, so that $(\mathbbm{D},\phi^{\delta},0,i)$ is
the embedding of
$\mathcal{D}_{1,\delta}\oplus\mathcal{D}_{2,\delta}\oplus\mathcal{D}_{3,\delta}$
on $(\mathbbm{D},0,i)$ by forgetting the interfaces and $\eta^{\delta}$ is the
interface between $\mathcal{D}_{1,\delta}$ and $\mathcal{D}_{2,\delta}$. See
Figure 4. The following lemma and Proposition 5.3 immediately give Proposition
5.15.

######  Lemma 5.16.

There exists a coupling between $M_{\delta}^{\\#}$ and $M_{\mathbbm{D}}^{\\#}$
such that the following holds. We can find random simply-connected domains
$U_{\delta},\widetilde{U}_{\delta}\subset\mathbbm{D}$ and a conformal map
$g_{\delta}:\widetilde{U}_{\delta}\to U_{\delta}$ such that with probability
$1-o_{\delta}(1)$,

  * •

$\phi_{\mathbbm{D}}(z)=\phi^{\delta}\circ
g_{\delta}(z)+Q\log|g^{\prime}_{\delta}(z)|\textrm{ for
}z\in\widetilde{U}_{\delta}$,

  * •

$\mathrm{diam}(\mathbbm{D}\backslash U_{\delta})=o_{\delta}(1)$,
$\mathrm{diam}(\mathbbm{D}\backslash\widetilde{U}_{\delta})=o_{\delta}(1)$,

  * •

$\sup_{z\in K}|g_{\delta}(z)-z|=o_{\delta}(1)\textrm{ for any compact
}K\subset\mathbbm{D}.$

To prove Lemma 5.16, we use the following basic coupling result on quantum
disks. Let $\varepsilon>0$. For a simply-connected quantum surface
$\mathcal{D}$ decorated with one bulk and one boundary point, let
$\mathcal{D}^{\varepsilon}$ be the quantum surface obtained by embedding
$\mathcal{D}$ as $(\mathbbm{H},\phi,i,-1)$ and setting
$\mathcal{D}^{\varepsilon}:=(\mathbbm{H}_{\varepsilon},\phi,i,-1,-1-2\varepsilon)$
where $\mathbbm{H}_{\varepsilon}=\mathbbm{H}\backslash
B_{\varepsilon}(-1-\varepsilon)$ with
$B_{\varepsilon}(-1-\varepsilon)=\\{z\in\mathbbm{C}:|z+1+\varepsilon|\leq\varepsilon\\}$.

######  Lemma 5.17.

For $\varepsilon>1$ and $\ell>0$, when $\mathcal{D}$ and
$\widetilde{\mathcal{D}}$ are sampled from $\mathrm{QD}_{1,1}(\ell)^{\\#}$ and
$\mathrm{QD}_{1,1}(\widetilde{\ell})^{\\#}$ respectively, the law of
$\widetilde{\mathcal{D}}^{\varepsilon}$ converges to that of
$\mathcal{D}^{\varepsilon}$ in total variation distance as
$\widetilde{\ell}\to\ell$.

###### Proof.

This can be proved directly via the explicit Liouville field description
$\mathrm{QD}_{1,1}(\ell)$ in Proposition 3.9. However, the corresponding
statement for $\mathrm{QD}_{0,2}(\ell)^{\\#}$ and
$\mathrm{QD}_{0,2}(\widetilde{\ell})^{\\#}$ is already proved in [AHS20,
Proposition 2.23]. (In fact, Proposition 2.23 there proved the general result
for $\mathcal{M}_{0,2}^{\mathrm{disk}}(W)$.) By Definition 2.10, we can
transfer the result for $\mathrm{QD}_{0,2}(\ell)^{\\#}$ into the desired
result for $\mathrm{QD}_{1,1}(\ell)^{\\#}$. ∎

Figure 4:  Left: $(\phi_{\mathbbm{D}},\eta_{\mathbbm{D}})$ is obtained by
embedding $\mathcal{D}_{1}\oplus\mathcal{D}_{2}$ into $(\mathbbm{D},0,i)$, and
$(\phi^{\delta},\eta^{\delta})$ is obtained by embedding
$\mathcal{D}_{1,\delta}\oplus\mathcal{D}_{2,\delta}\oplus\mathcal{D}_{3,\delta}$
into $(\mathbbm{D},0,i)$. Right: We can couple $M_{\mathbbm{D}}^{\\#}$ and
$M_{\delta}^{\\#}$ so the pink and blue quantum surfaces agree with high
probability. The domains $\widetilde{U}_{\delta}$ and $U_{\delta}$ are the
unions of the pink and blue regions.

###### Proof of Lemma 5.16.

Recall the marked quantum surfaces $\mathcal{D}_{1},\mathcal{D}_{2}$ in the
definition of $(\phi_{\mathbbm{D}},\eta_{\mathbbm{D}})$ and
$M_{\mathbbm{D}}^{\\#}$. Let $\widetilde{A}$ and $\widetilde{P}$ be the left
and right boundary lengths of $\mathcal{D}_{1}$, respectively. The law of
$(\widetilde{A},\widetilde{P})$ is the probability measure on
$[1,2]\times(0,\infty)$ proportional to

$|\mathcal{M}_{0,2}^{\mathrm{disk}}(\frac{\gamma^{2}}{2};a,p)||\mathrm{QD}_{1,1}(p)|\propto
a^{\frac{4}{\gamma^{2}}-1}/(a^{\frac{4}{\gamma^{2}}}+p^{\frac{4}{\gamma^{2}}})^{2},$
(5.17)

where we have used (5.5) for the expression of
$|\mathcal{M}_{0,2}^{\mathrm{disk}}(\frac{\gamma^{2}}{2};a,p)|$. Conditioning
on $(\widetilde{A},\widetilde{P})$, the conditional law of
$(\mathcal{D}_{1},\mathcal{D}_{2})$ is
$\mathcal{M}_{0,2}^{\mathrm{disk}}(\frac{\gamma^{2}}{2};\widetilde{A},\widetilde{P})^{\\#}\times\mathrm{QD}_{1,1}(\widetilde{P})^{\\#}$.

Similarly, recall that the law of $\mathcal{D}_{1,\delta}$,
$\mathcal{D}_{2,\delta}$, $\mathcal{D}_{3,\delta}$ is given in (5.2). Let
$A_{\delta}$ and $P_{\delta}$ be the left and right boundary lengths of
$\mathcal{D}_{1,\delta}$, respectively. Let $Q_{\delta}$ and $B_{\delta}$ be
the left and right boundary lengths of $\mathcal{D}_{3,\delta}$, respectively.
Then, by Lemma 5.5, the $M_{\delta}^{\\#}$-law of $(A_{\delta},P_{\delta})$
weakly converge to the the $M_{\mathbbm{D}}^{\\#}$-law of
$(\widetilde{A},\widetilde{P})$, and $\lim_{\delta\to
0}M_{\delta}^{\\#}[Q_{\delta}>\varepsilon]=\lim_{\delta\to
0}M_{\delta}^{\\#}[B_{\delta}>\varepsilon]=0$ for all $\varepsilon>0$.

By Lemma 5.5 we can couple $M_{\delta}^{\\#}$ and $\widetilde{M}^{\\#}$ such
that $(\widetilde{A},\widetilde{P})=(A_{\delta},P_{\delta})$ with probability
$1-o_{\delta}(1)$. By Proposition 2.11, a sample from
$\mathcal{M}_{1,2}^{\mathrm{disk}}(2;p,q)^{\\#}$ can be obtained by sampling a
quantum surface from $\mathrm{QD}_{1,1}(p+q)^{\\#}$, then marking a second
boundary point at quantum length $p$ counterclockwise from the marked boundary
point. Since $(\widetilde{A},\widetilde{P})=(A_{\delta},P_{\delta})$ with
probability $1-o_{\delta}(1)$ and $Q_{\delta}\to 0$ in probability as
$\delta\to 0$, by Lemma 5.17, we can extend our coupling such that
$\lim_{\delta\to
0}\mathbbm{P}[(\mathcal{D}_{1},\mathcal{D}_{2}^{\varepsilon})=(\mathcal{D}_{1,\delta},\mathcal{D}_{2,\delta}^{\varepsilon})]=1$
for any fixed $\varepsilon>0$; consequently the same holds when $\varepsilon$
is e.g. a piecewise constant function of $\delta$ which decays to 0
sufficiently slowly as $\delta\to 0$.

Let $U_{\delta}$ (resp. $\widetilde{U}_{\delta}$) be the domain parametrizing
the conformal welding of $\mathcal{D}_{1}$ and $\mathcal{D}_{2}^{\varepsilon}$
(resp. $\widetilde{\mathcal{D}}_{1}$ and
$\widetilde{\mathcal{D}}_{2}^{\varepsilon}$) in the aforementioned embedding
of $\mathcal{D}_{1}\oplus\mathcal{D}_{2}$ (resp.
$\mathcal{D}_{1,\delta}\oplus\mathcal{D}_{2,\delta}\oplus\mathcal{D}_{3,\delta}$)
in $(\mathbbm{D},0,i)$. More precisely, $U_{\delta}$ is the interior of the
union of the closures of the domains corresponding to $\mathcal{D}_{1}$ and
$\mathcal{D}_{2}^{\varepsilon}$ in our embedding, and the analogous statement
holds for $\widetilde{U}_{\delta}$. By definition, the marked quantum surfaces
$(U_{\delta},\phi^{\delta},0,i)$ and
$(\widetilde{U}_{\delta},\phi_{\mathbbm{D}},0,i^{-})$ agree with probability
$1-o_{\delta}(1)$, where $i^{-}$ be the boundary point immediately to the left
of $i$. On this event, by the agreement of quantum surfaces, there exists a
unique conformal map $g_{\delta}:\widetilde{U}_{\delta}\rightarrow U_{\delta}$
such that $\phi_{\mathbbm{D}}=\phi^{\delta}\circ
g_{\delta}+Q\log|g^{\prime}_{\delta}|$, $g_{\delta}(0)=0$ and
$g_{\delta}(i^{-})=i$.

Note that the simply connected open sets $\widetilde{U}_{\delta}$ are
determined by $(\phi_{\mathbbm{D}},\eta_{\mathbbm{D}})$ and

$M^{\\#}_{\mathbbm{D}}\textrm{-almost surely
}(\mathbbm{D}\cup\partial\mathbbm{D})\setminus\widetilde{U}_{\delta}\textrm{
is decreasing as }\delta\to 0\textrm{ and their intersection equals }\\{i\\}.$

Therefore $M^{\\#}_{\mathbbm{D}}$-almost surely $\lim_{\delta\to
0}\operatorname{diam}(\mathbbm{D}\setminus\widetilde{U}_{\delta})=0$ . Hence
in our coupling of $M_{\delta}^{\\#}$ and $M_{\mathbbm{D}}^{\\#}$,
$\mathrm{diam}(\mathbbm{D}\backslash\widetilde{U}_{\delta})=o_{\delta}(1)$
with probability $1-o_{\delta}(1)$. As a basic deterministic fact in complex
analysis, $\operatorname{diam}(\mathbbm{D}\setminus\widetilde{U}_{\delta})=0$
if and only if the harmonic measure of
$\mathbbm{D}\backslash\widetilde{U}_{\delta}$ viewed from $0$ in
$\widetilde{U}_{\delta}$ tend to $0$ as $\delta\rightarrow 0$. Therefore, in
our coupling the harmonic measure of
$\mathbbm{D}\backslash\widetilde{U}_{\delta}$ viewed from $0$ in
$\widetilde{U}_{\delta}$ is $o_{\delta}(1)$ with probability
$1-o_{\delta}(1)$. By conformal invariance, the same holds for the harmonic
measure of $\mathbbm{D}\backslash U_{\delta}$ viewed from $0$ in $U_{\delta}$.
So $\operatorname{diam}(\mathbbm{D}\backslash U_{\delta})=o_{\delta}(1)$ with
probability $1-o_{\delta}(1)$. This gives the second condition for the
coupling.

Finally, since $g_{\delta}(0)=0,g_{\delta}(i_{-})=i$, and
$\operatorname{diam}(\mathbbm{D}\backslash
U_{\delta}),\operatorname{diam}(\mathbbm{D}\backslash\widetilde{U}_{\delta})\to
0$ with probability as $1-o_{\delta}(1)$, standard conformal distortion
estimates yield the third condition. ∎

###### Proof of Proposition 5.15.

Sending $\delta\to 0$ in Lemma 5.17 and Proposition 5.3, we see that the law
of $\phi_{\mathbbm{D}}\circ f+Q\log|f^{\prime}|$ under $M_{\mathbbm{D}}^{\\#}$
is the same as that of $\phi$ under $M^{\\#}$, which is the probability
measure proportional to
$\int_{1}^{2}\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}(\ell)\,d\ell$. Moreover,
the independence of $\phi_{\mathbbm{D}}$ and $\eta_{\mathbbm{D}}$ follows from
the asymptotic independence of $\phi$ and $\eta_{1}$ under $M_{\delta}^{\\#}$
established in Proposition 5.3. ∎

###### Proof of Theorem 4.1.

Let $\psi(z)=-z^{-1}$ and
$\mathrm{Bubble}_{\mathbbm{H}}(i,\infty)=\\{\eta:\psi(\eta)\in\mathrm{Bubble}_{\mathbbm{H}}(i,0)\\}$.
By Proposition 5.15, the law of $(\phi_{\mathbbm{D}}\circ
f+Q\log|f^{\prime}|,f^{-1}\circ\eta_{\mathbbm{D}})$ can be written as the
probability measure proportional to
$\int_{1}^{2}\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}(\ell)\,d\ell\times\mathsf{m}^{\infty}$
where $\mathsf{m}^{\infty}$ is the probability measure on
$\mathrm{Bubble}_{\mathbbm{H}}(i,\infty)$ describing the marginal law of
$f^{-1}\circ\eta_{\mathbbm{D}}$. Therefore for some $C\in(0,\infty)$ we have

$\int_{1}^{2}\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}(\ell)\,d\ell\times\mathsf{m}^{\infty}=C\int_{1}^{2}\int_{0}^{\infty}\mathrm{QD}_{1,1}(r)\times\mathcal{M}_{0,2}^{\mathrm{disk}}(\frac{\gamma^{2}}{2};\ell,r)\,dr\,d\ell,$
(5.18)

in the sense that when a field and curve are sampled from the left hand side
of (5.18), the law of the two quantum surfaces cut out by the curve is the
right hand side of (5.18).

A scaling argument shows that (5.18) holds when the integration interval
$[1,2]$ is replaced by $[e^{\frac{\gamma}{2}c},2e^{\frac{\gamma}{2}c}]$ for
any $c\in\mathbbm{R}$. Indeed, from Lemma 2.2 and the change of coordinates
$\mathbf{c}^{\prime}=\mathbf{c}+c$ it is immediate that when $\phi$ is sampled
from $\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}(\ell)$ then the law of $\phi+c$
is
$e^{(Q-\frac{\gamma}{2})c}\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}(e^{\frac{\gamma}{2}c}\ell)$.
Similarly, if $(\mathbbm{H},\phi,i,0)$ has law $\mathrm{QD}_{1,1}(r)$ then
$(\mathbbm{H},\phi+c,i,0)$ has law
$e^{(Q-\gamma)c}\mathrm{QD}_{1,1}(e^{\frac{\gamma}{2}c}r)$, and if
$(\mathbbm{H},\phi,0,1)$ has law
$\mathcal{M}_{0,2}^{\mathrm{disk}}(\frac{\gamma^{2}}{2};\ell,r)$ then
$(\mathbbm{H},\phi+c,0,1)$ has law $e^{\gamma
c}\mathcal{M}_{0,2}^{\mathrm{disk}}(\frac{\gamma^{2}}{2};e^{\frac{\gamma}{2}c}\ell,e^{\frac{\gamma}{2}c}r)$.
Therefore, adding $c$ to the fields in (5.18) and changing variables
$(\ell^{\prime},r^{\prime})=(e^{\frac{\gamma}{2}c}\ell,e^{\frac{\gamma}{2}c}r)$
gives (5.18) with $[1,2]$ replaced with
$[e^{\frac{\gamma}{2}c},2e^{\frac{\gamma}{2}c}]$. Summing over the intervals
$[2^{n},2^{n+1}]$ for integer $n$ yields (5.18) with $[1,2]$ replaced by
$(0,\infty)$.

Finally, since $\psi:z\mapsto-z^{-1}$ satisfies $|\psi^{\prime}(i)|=1$, by
Lemma 2.4 $\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}$ is mapped to itself by the
coordinate change of $\psi$. Recall the definition of
$\mathcal{M}_{0,2}^{\mathrm{disk}}(\frac{\gamma^{2}}{2};\cdot,r)$ from (2.10).
Reparametrizing (5.18) via $\psi$ yields Theorem 4.1. ∎

##  Appendix A Brownian motion in cones

For $\phi\in(0,2\pi)$ let $\mathcal{C}_{\phi}=\\{z\in\mathbbm{C}:\arg
z\in(0,\phi)\\}$ be the cone of angle $\phi$. Following [LW04, Section 3] we
will define various measures corresponding to Brownian motion in
$\mathcal{C}_{\phi}$ conditioned on certain probability zero events via
limiting procedures. These constructions are in the same spirit as [LW04,
Section 3], but we also consider Brownian path measures in a cone with an
endpoint at its vertex. For more details on the validity of these
constructions see [LW04, Section 3].

Let $\mathcal{K}$ be the collection of continuous planar curves $Z$ defined on
a finite time interval $[0,T_{Z}]$, where $T_{Z}$ is the _duration_ of the
curve. Endow $\mathcal{K}$ with the metric
$d_{\mathcal{K}}(Z_{1},Z_{2})=\inf_{\theta}\\{\sup_{0<t<T_{Z_{1}}}|t-\theta(t)|+|Z_{1}(t)-Z_{2}(\theta(t))|\\}$
with the infimum taken over increasing homeomorphisms
$\theta:[0,T_{Z_{1}}]\to[0,T_{Z_{2}}]$. Our Brownian path measures will be
non-probability measures on $\mathcal{K}$ equipped with the Borel
$\sigma$-algebra associated to $d_{\mathcal{K}}$.

Let $\mathsf{m}(z,w;t)$ denote the measure on $\mathcal{K}$ such that
$|\mathsf{m}(z,w;t)|=\frac{1}{2\pi t}e^{-|z-w|^{2}/2t}$ and
$\mathsf{m}(z,w;t)^{\\#}$ is the Brownian bridge from $z$ to $w$ with duration
$t$. Let $\mathsf{m}_{\mathcal{C}_{\phi}}(z,w;t)$ be the restriction of
$\mathsf{m}(z,w;t)$ to paths staying in $\mathcal{C}_{\phi}$, and let
$\mathsf{m}_{\mathcal{C}_{\phi}}(z,w)=\int_{0}^{\infty}\mathsf{m}_{\mathcal{C}_{\phi}}(z,w;t)\,dt$.

######  Lemma A.1.

For each $w\in\mathcal{C}_{\phi}$ and $t>0$, pick any $\psi\in(0,\phi)$ and
define the measure $\mathsf{m}_{\mathcal{C}_{\phi}}(w,0;t):=\lim_{s\to
0}\frac{s^{-\frac{\pi}{\phi}}}{\sin(\frac{\pi\psi}{\phi})}\mathsf{m}_{\mathcal{C}_{\phi}}(w,se^{i\psi};t)$.
This limit exists and does not depend on the choice of $\psi$. Moreover there
is a constant $C=C(\phi)$ such that for all $w,t$,

$|\mathsf{m}_{\mathcal{C}_{\phi}}(w,0;t)|=Ct^{-1-\frac{\pi}{\phi}}|w|^{\frac{\pi}{\phi}}e^{-\frac{|w|^{2}}{2t}}\sin(\frac{\pi}{\phi}\arg
w).$ (A.1)

###### Proof.

This result follows from [Shi85, Theorem 2] except that theorem is about the
Brownian motion in $\mathcal{C}_{\phi}$ which starts at 0 and ends at $w$. Our
statement follows from the time reversal symmetry of Brownian path. ∎

Define
$\mathsf{m}_{\mathcal{C}_{\phi}}(w,0):=\int_{0}^{\infty}\mathsf{m}_{\mathcal{C}_{\phi}}(w,0;t)\,dt$
and, for $u>0$, $\mathsf{m}_{\mathcal{C}_{\phi}}(u,0):=\lim_{\varepsilon\to
0}\varepsilon^{-1}\mathsf{m}_{\mathcal{C}_{\phi}}(u+i\varepsilon,0)$.

######  Corollary A.2.

Suppose $\phi\in(0,2\pi),\theta\in(0,\phi)$ and $r>0$. There is a constant
$c_{\phi,\theta}\in(0,\infty)$ such that
$|\mathsf{m}_{\mathcal{C}_{\phi}}(re^{i\theta},0)|=c_{\phi,\theta}\cdot
r^{-\frac{\pi}{\phi}}$. Moreover, the law of the duration of a sample from
$\mathsf{m}_{\mathcal{C}_{\phi}}(re^{i\theta},0)^{\\#}$ is the inverse gamma
distribution with shape $\frac{\pi}{\phi}$ and scale $\frac{r^{2}}{2}$ (recall
(2.2)). Similarly, for $u>0$,
$|\mathsf{m}_{\mathcal{C}_{\phi}}(u,0)|=c_{\phi}u^{-\frac{\pi}{\phi}-1}$.
Moreover, the law of the duration of a sample from
$\mathsf{m}_{\mathcal{C}_{\phi}}(u,0)^{\\#}$ is the inverse gamma distribution
with shape $\frac{\pi}{\phi}$ and scale $\frac{u^{2}}{2}$.

###### Proof.

These claims are immediate from Lemma A.1. ∎

Suppose $\theta\in(0,\phi)$, we describe a path decomposition for
$\mathsf{m}_{\mathcal{C}_{\phi}}(u,0)^{\\#}$ where we split the path at the
first time it hits the ray $e^{i\theta}\mathbbm{R}_{+}$. For
$z\in\mathcal{C}_{\theta}$, let $\mathsf{m}_{\mathcal{C}_{\theta}}(z)$ denote
the probability measure corresponding to Brownian motion started at $z$ and
killed when it exits $\mathcal{C}_{\theta}$. For $y>0$, let
$E_{y,\varepsilon}$ be the event that Brownian motion exits
$\mathcal{C}_{\theta}$ on the boundary interval
$(ye^{i\theta},(y+\varepsilon)e^{i\phi})$, and let
$\mathsf{m}_{\mathcal{C}_{\theta}}(z,ye^{i\theta})=\lim_{\varepsilon\to
0}\varepsilon^{-1}\mathsf{m}_{\mathcal{C}_{\theta}}(z)|_{E_{y,\varepsilon}}$.
For $x>0$ define
$\mathsf{m}_{\mathcal{C}_{\theta}}(x,ye^{i\theta})=\lim_{\varepsilon\to
0}\varepsilon^{-1}\mathsf{m}_{\mathcal{C}_{\theta}}(x+\varepsilon
i,ye^{i\theta})$.

######  Lemma A.3.

For $0<\theta<\phi<2\pi$ and $u>0$, we have

$\mathsf{m}_{\mathcal{C}_{\phi}}(u,0)=\int_{0}^{\infty}\mathsf{m}_{\mathcal{C}_{\theta}}(u,re^{i\theta})\times\mathsf{m}_{\mathcal{C}_{\phi}}(re^{i\theta},0)\,dr$

in the sense that when a sample from the right hand side is concatenated to
obtain a path from $u$ to $0$, the law of this concatenated path is the left
hand side.

###### Proof.

Let $\psi=\frac{\phi+\theta}{2}$. For $\delta,\varepsilon>0$, by the strong
Markov property of Brownian motion we have

$\mathsf{m}_{\mathcal{C}_{\phi}}(u+\varepsilon i,\delta
e^{i\psi})=\int_{0}^{\infty}\mathsf{m}_{\mathcal{C}_{\theta}}(u+\varepsilon
i,re^{i\theta})\times\mathsf{m}_{\mathcal{C}_{\phi}}(re^{i\theta},\delta
e^{i\psi})\,dr.$

Multiplying both sides by
$\frac{\delta^{-\frac{\pi}{\phi}}}{\sin(\frac{\pi\psi}{\phi})}\varepsilon^{-1}$
and sending $\delta,\varepsilon\to 0$ yields the assertion. ∎

######  Lemma A.4.

For $0<\theta<2\pi$, there exists a constant $C$ such that

$|\mathsf{m}_{\mathcal{C}_{\theta}}(u,re^{i\theta})|=C\frac{(ur)^{\frac{\pi}{\theta}-1}}{(u^{\frac{\pi}{\theta}}+r^{\frac{\pi}{\theta}})^{2}}\quad\text{for
}u,r>0.$

###### Proof.

This is equivalent to [AHS20, Lemma C.2] after the shear transform $\Lambda$
in Proposition 2.14. ∎

######  Lemma A.5.

Suppose $0<\theta<\phi<2\pi$. Let $L$ be sampled as in (4.15). Then for
$\varepsilon\in(0,\frac{\pi}{\phi})$ we have
$\mathbbm{E}[(\frac{L}{u})^{\varepsilon}]<1$.

###### Proof.

By (4.14), Corollary A.2 and Lemma A.4 the law of $\frac{L}{u}$ is
proportional to

$1_{x>0}|\mathsf{m}_{\mathcal{C}_{\theta}}(u,uxe^{i\theta})||\mathsf{m}_{\mathcal{C}_{\phi}}(uxe^{i\theta},0)|dx\;\propto\;1_{x>0}x^{\pi/\theta-1}(1+x^{\pi/\theta})^{-2}\cdot
x^{-\frac{\pi}{\phi}}\,dx.$

Set $f(x):=\frac{x}{\sin x}$. When $p\in(-1,\frac{2\pi}{\theta}-1)$, we have

$\int_{0}^{\infty}\frac{x^{p}}{(1+x^{\pi/\theta})^{2}}\,dx=\frac{\theta}{\pi}\frac{\pi-\theta(p+1)}{\sin(\pi-\theta(p+1))}=\frac{\theta}{\pi}f(\pi-\theta(p+1)).$

Therefore

$\mathbbm{E}\mathopen{}\mathclose{{}\left[\mathopen{}\mathclose{{}\left(\frac{L}{u}}\right)^{\varepsilon}}\right]=\int_{0}^{\infty}\frac{x^{\frac{\pi}{\theta}-1-\frac{\pi}{\phi}+\varepsilon}}{(1+x^{\pi/\theta})^{2}}\,dx\bigg{/}\int_{0}^{\infty}\frac{x^{\frac{\pi}{\theta}-1-\frac{\pi}{\phi}}}{(1+x^{\pi/\theta})^{2}}\,dx=\frac{f(\frac{\pi\theta}{\phi}-\theta\varepsilon)}{f(\frac{\pi\theta}{\phi})}.$

Since $f(x)$ is increasing on $(0,\pi)$ and
$\frac{\pi\theta}{\phi}\in(0,\pi)$, we obtain the lemma. ∎

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