special case of Theorem 5.2 where $\alpha_{1}=\alpha_{2}=\alpha_{3}=\gamma$.

######  Lemma 5.3.

Theorem 5.2 holds for $\alpha_{1}=\alpha_{2}=\alpha_{3}=\gamma$.

###### Proof.

By the relation between $\mathrm{LF}_{\mathbbm{C}}^{(\gamma,z_{i})_{i}}$ and
$\mathrm{QS}_{3}$ from Theorem 2.9 and the relation between
$\mathrm{QD}_{1,0}(\ell)$ and $\mathcal{M}_{1}^{\mathrm{disk}}(\gamma;\ell)$
from Theorem 2.13, we see that Lemma 5.3 follows from Theorem 3.7. ∎

We first explain how to go from $\alpha_{1}=\alpha_{2}=\alpha_{3}=\gamma$ to
the following case.

######  Proposition 5.4.

Theorem 5.2 holds for $\alpha_{1}=\alpha\in(Q-\frac{\gamma}{4},Q)$ and
$\alpha_{2}=\alpha_{3}=\gamma$.

We will prove Proposition 5.4 from Lemma 5.3 by a reweighting procedure. By
the definition of $\mathcal{M}^{\mathrm{disk}}_{1}(\alpha;\ell)$, if we sample
a field $X$ from $\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}(\ell)$, then the law
of $(\mathbbm{H},X,i)/{\sim_{\gamma}}$ is
$\mathcal{M}^{\mathrm{disk}}_{1}(\alpha;\ell)$. We now recall a fact from
[ARS21] about the reweighting of
$\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}(\ell)$ by “$e^{(\alpha-\gamma)X}$”.

######  Lemma 5.5 ([ARS21, Lemma 4.6]).

For any $\ell>0,\varepsilon\in(0,1)$ and for any nonnegative measurable
function $f$ of $X$ that depends only on $X|_{\mathbbm{H}\setminus
B_{\varepsilon}(i)}$, we have

$\int f(X|_{\mathbbm{H}\setminus
B_{\varepsilon}(i)})\times\varepsilon^{\frac{1}{2}(\alpha^{2}-\gamma^{2})}e^{(\alpha-\gamma)X_{\varepsilon}(i)}\,d{\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}(\ell)}=\int
f(X|_{\mathbbm{H}\setminus
B_{\varepsilon}(i)})\,d\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}(\ell),$

where $X$ is a sample in $H^{-1}(\mathbbm{H})$ and $X_{\varepsilon}(i)$ means
the average of $X$ on the boundary of the ball
$B_{\varepsilon}(i)=\\{z:|z-i|<\varepsilon\\}$.

The key to the proof of Proposition 5.4 is the following reweighting result on
$\mathrm{LF}_{\mathbbm{C}}^{(\gamma,z_{i})_{i}}$.

######  Lemma 5.6.

Let $\eta_{1}$ be a simple loop separating $z_{1}$ from $z_{2}$ and $z_{3}$.
Let $D_{\eta_{1}}$ be the connected component of
$\mathbbm{C}\setminus\eta_{1}$ containing $z_{1}=0$. Let $p$ be a point on
$\eta_{1}$ and let $\psi:\mathbbm{H}\to D_{\eta_{1}}$ be the conformal map
with $\psi(i)=z_{1}$ and $\psi(0)=p$. For $\varepsilon\in(0,\frac{1}{4})$, let
$\mathbbm{C}_{\eta_{1},p,\varepsilon}=\mathbbm{C}\setminus\psi(B_{\varepsilon}(i))$.
For $\phi$ sampled from
$\mathrm{LF}_{\mathbbm{C}}^{(\gamma,z_{1}),(\gamma,z_{2}),(\gamma,z_{3})}$,
let $X=\phi\circ\psi+Q\log|\psi^{\prime}|$ so that
$(\mathbbm{H},X,i,0)/{\sim_{\gamma}}=(D_{\eta_{1}},\phi,z_{1},p)/{\sim_{\gamma}}$.
Then for a fixed $\alpha\in(Q-\frac{\gamma}{4},Q)$ and for any nonnegative
measurable function $f$ of $\phi$ that depends only on
$\phi|_{\mathbbm{C}_{\eta_{1},p,\varepsilon}}$ we have

$\displaystyle\int
f(\phi)\times\varepsilon^{\frac{1}{2}(\alpha^{2}-\gamma^{2})}e^{(\alpha-\gamma)X_{\varepsilon}(i)}\,d\mathrm{LF}_{\mathbbm{C}}^{(\gamma,z_{1}),(\gamma,z_{2}),(\gamma,z_{3})}=\int
f(\phi)\mathopen{}\mathclose{{}\left(\frac{\mathrm{CR}(\eta_{1},z_{1})}{2}}\right)^{-\frac{\alpha^{2}}{2}+Q\alpha-2}d\mathrm{LF}_{\mathbbm{C}}^{(\alpha,z_{1}),(\gamma,z_{2}),(\gamma,z_{3})}.$

###### Proof.

Let $\theta_{\varepsilon}$ be the uniform probability measure on $\partial
B_{\varepsilon}(i)$ and
$\widehat{\theta}_{\varepsilon}:=\psi_{*}\theta_{\varepsilon}$. Recall
notations in Section 2.2.1, where $\mathbbm{P}_{\mathbbm{C}}$ is the
probability measure for the GFF on $\mathbbm{C}$. Let $\mathbbm{E}$ be the
expectation for $\mathbbm{P}_{\mathbbm{C}}$, then
$G_{\mathbbm{C}}(z,w)=\mathbbm{E}[h(x)h(y)]=-\log|z-w|+\log|z|_{+}+\log|w|_{+}$.
For a sample $h$ from $\mathbbm{P}_{\mathbbm{C}}$, we set
$\widetilde{h}:=h-2Q\log|\cdot|_{+}+\sum_{j=1}^{3}\gamma
G_{\mathbbm{C}}(\cdot,z_{j})$. Lemma 5.6 will follow from three identities.

$e^{(\alpha-\gamma)X_{\varepsilon}(i)}=\mathopen{}\mathclose{{}\left(\frac{\mathrm{CR}(\eta_{1},0)}{2}}\right)^{(\alpha-\gamma)Q}e^{(\alpha-\gamma)(\phi,\widehat{\theta}_{\varepsilon})}.$
(5.4) $\int f(\phi)\times
e^{(\alpha-\gamma)(\phi,\widehat{\theta}_{\varepsilon})}\,d\mathrm{LF}_{\mathbbm{C}}^{(\gamma,z_{1}),(\gamma,z_{2}),(\gamma,z_{3})}=\mathopen{}\mathclose{{}\left(\frac{\varepsilon\mathrm{CR}(\eta_{1},0)}{2}}\right)^{-(\alpha-\gamma)\gamma}\int\mathbbm{E}[e^{(\alpha-\gamma)(h,\widehat{\theta}_{\varepsilon})}f(\widetilde{h}+c)]e^{(\alpha+2\gamma-2Q)c}\,dc.$
(5.5)
$\int\mathbbm{E}[e^{(\alpha-\gamma)(h,\widehat{\theta}_{\varepsilon})}f(\widetilde{h}+c)]e^{(\alpha+2\gamma-2Q)c}\,dc=\mathopen{}\mathclose{{}\left(\frac{\varepsilon\mathrm{CR}(\eta_{1},0)}{2}}\right)^{-\frac{1}{2}(\alpha-\gamma)^{2}}\int
f(\phi)\mathrm{LF}_{\mathbbm{C}}^{(\alpha,z_{1}),(\gamma,z_{2}),(\gamma,z_{3})}.$
(5.6)

To prove (5.4), note that
$X_{\varepsilon}(i)=(X,\theta_{\varepsilon})=(\phi\circ\psi+Q\log|\psi^{\prime}|,\theta_{\varepsilon})$.
Since $\psi^{\prime}$ is holomorphic and $\log|\psi^{\prime}|$ is harmonic, we
have $(\log|\psi^{\prime}|,\theta_{\varepsilon})=\log|\psi^{\prime}(i)|$ hence
$(X,\theta_{\varepsilon})=(\phi,\widehat{\theta}_{\varepsilon})+Q\log|\psi^{\prime}(i)|$.
Since $|\psi^{\prime}(i)|=\frac{1}{2}\mathrm{CR}(\eta_{1},0)$, we get (5.4).
To prove (5.5), let $\eta^{\varepsilon}:=\psi(\partial B_{\varepsilon}(i))$.
Since $\widehat{\theta}_{\varepsilon}$ is the harmonic measure on
$\eta^{\varepsilon}$ viewed from $0$, we have
$(\log|\cdot|,\widehat{\theta}_{\varepsilon})=\log\mathrm{CR}(\eta^{\varepsilon},0)=\log\frac{\varepsilon\mathrm{CR}(\eta_{1},0)}{2}$.
Since $\varepsilon<\frac{1}{4}$, the curve $\eta^{\varepsilon}$ is contained
in the unit disk hence $(\log|\cdot|_{+},\widehat{\theta}_{\varepsilon})=0$.
Let
$G_{\mathbbm{C}}^{\varepsilon}(0,z):=(G_{\mathbbm{C}}(\cdot,z),\widehat{\theta}_{\varepsilon})$.
Then $G_{\mathbbm{C}}^{\varepsilon}(0,z)=G_{\mathbbm{C}}(0,z)$ for
$z\in\mathbbm{C}_{\eta_{1},p,\varepsilon}$ and
$G_{\mathbbm{C}}^{\varepsilon}(0,0)=-\log\frac{\varepsilon\mathrm{CR}(\eta_{1},0)}{2}$.
Since $(z_{1},z_{2},z_{3})=(0,1,e^{i\pi/3})$, we have
$(-2Q\log|\cdot|_{+}+\sum_{j}\gamma
G_{\mathbbm{C}}(\cdot,z_{j}),\widehat{\theta}_{\varepsilon})=-\gamma\log\frac{\varepsilon\mathrm{CR}(\eta_{1},0)}{2}$.
Therefore
$(\widetilde{h}+c,\widehat{\theta}_{\varepsilon})=(h,\widehat{\theta}_{\varepsilon})-\gamma\log\frac{\varepsilon\mathrm{CR}(\eta_{1},0)}{2}+c$.
Recall from Definition 2.4 that
$\mathrm{LF}_{\mathbbm{C}}^{(\gamma,z_{1}),(\gamma,z_{2}),(\gamma,z_{3})}$ is
the law of $\widetilde{h}+c$ under
$e^{(3\gamma-2Q)c}dc\mathbbm{P}_{\mathbbm{C}}(dh)$. This gives (5.5).

To prove (5.6), note that
$\mathbbm{E}[h(z)(h,\widehat{\theta}_{\varepsilon})]=(G_{\mathbbm{C}}(\cdot,z),\widehat{\theta}_{\varepsilon})=G_{\mathbbm{C}}^{\varepsilon}(0,z)$,
which equals $G_{\mathbbm{C}}(z,0)$ for
$z\in\mathbbm{C}_{\eta_{1},p,\varepsilon}$. By Girsanov’s theorem and the fact
that $f(\phi)$ depends only on $\phi|_{\mathbbm{C}_{\eta,p,\varepsilon}}$, we
have

$\displaystyle\int\mathbbm{E}[e^{(\alpha-\gamma)(h,\widehat{\theta}_{\varepsilon})}f(\widetilde{h}+c)]e^{(\alpha+2\gamma-2Q)c}\,dc$
$\displaystyle=\mathbbm{E}[e^{(\alpha-\gamma)(h,\widehat{\theta}_{\varepsilon})}]\int\mathbbm{E}[f(\widetilde{h}+(\alpha-\gamma)G^{\varepsilon}_{\mathbbm{C}}(\cdot,0)+c)]e^{(\alpha+2\gamma-2Q)c}\,dc$
$\displaystyle=\mathbbm{E}[e^{(\alpha-\gamma)(h,\widehat{\theta}_{\varepsilon})}]\int\mathbbm{E}[f(\widetilde{h}+(\alpha-\gamma)G_{\mathbbm{C}}(\cdot,0)+c)]e^{(\alpha+2\gamma-2Q)c}\,dc.$

Since
$\operatorname{Var}((h,\widehat{\theta}_{\varepsilon}))=(G_{\mathbbm{C}}^{\varepsilon}(\cdot,0),\widehat{\theta}_{\varepsilon})=-\log\frac{\varepsilon\mathrm{CR}(\eta_{1},0)}{2}$,
we have
$\mathbbm{E}[e^{(\alpha-\gamma)(h,\widehat{\theta}_{\varepsilon})}]=\mathopen{}\mathclose{{}\left(\frac{\varepsilon\mathrm{CR}(\eta_{1},0)}{2}}\right)^{-\frac{1}{2}(\alpha-\gamma)^{2}}$.
Since the law of $\widetilde{h}+c$ under
$e^{(\alpha+2\gamma-2Q)c}dc\mathbbm{P}_{\mathbbm{C}}(dh)$ is
$\mathrm{LF}_{\mathbbm{C}}^{(\alpha,z_{1}),(\gamma,z_{2}),(\gamma,z_{3})}$, we
get (5.6).

Combining (5.4), (5.5) and (5.6), and collecting the prefactors via

$\mathopen{}\mathclose{{}\left(\frac{\mathrm{CR}(\eta_{1},0)}{2}}\right)^{(\alpha-\gamma)Q}\mathopen{}\mathclose{{}\left(\frac{\varepsilon\mathrm{CR}(\eta_{1},0)}{2}}\right)^{-(\alpha-\gamma)\gamma}\mathopen{}\mathclose{{}\left(\frac{\varepsilon\mathrm{CR}(\eta_{1},0)}{2}}\right)^{-\frac{1}{2}(\alpha-\gamma)^{2}}=\varepsilon^{-\frac{1}{2}(\alpha^{2}-\gamma^{2})}\mathopen{}\mathclose{{}\left(\frac{\mathrm{CR}(\eta_{1},0)}{2}}\right)^{-\frac{\alpha^{2}}{2}+Q\alpha-2},$

we conclude the proof of Lemma 5.6. ∎

We also need the following fact when dealing with uniform weldings that
involve $\mathcal{M}^{\mathrm{disk}}_{1}(\alpha;\ell)$.

######  Lemma 5.7.

For $\alpha>\frac{\gamma}{2}$ and $\ell>0$, let $(D,h,z)$ be an embedding of a
sample from $\mathcal{M}^{\mathrm{disk}}_{1}(\alpha;\ell)$. Given $(D,h,z)$,
let $p$ be a point sampled from the harmonic measure on $\partial D$ viewed
from $z$, then the law of $(D,h,z,p)/{\sim_{\gamma}}$ equals that of
$(\mathbbm{H},X,i,0)/{\sim_{\gamma}}$ where $X$ is sampled from
$\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}(\ell)$.

###### Proof.

We assume that $(D,h,z)=(\mathbbm{H},h,i)$ where $h$ is a sample from
$\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}(\ell)$. Let
$\psi_{p}:\mathbbm{H}\to\mathbbm{H}$ be the conformal map with $\psi_{p}(i)=i$
and $\psi_{p}(p)=0$ and set
$X=h\circ\psi_{p}^{-1}+Q\log|(\psi_{p}^{-1})^{\prime}|$. Then by the
coordinate change for Liouville fields on $\mathbbm{H}$ (see e.g. [ARS21,
Lemma 2.4]), the law of $X$ is also
$\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}(\ell)$. Since
$(D,h,z,p)/{\sim_{\gamma}}=(\mathbbm{H},X,i,0)/{\sim_{\gamma}}$ we are done. ∎

We are now ready to prove Proposition 5.4. For notational simplicity for
$\ell>0$ we let $\mathfrak{M}_{\ell}$ be the measure on decorated quantum
surfaces corresponding to

$\frac{\gamma^{2}}{4\pi^{4}(Q-\gamma)^{4}}\iint_{0}^{\infty}\ell_{2}\ell_{3}\mathrm{Weld}(\mathcal{M}_{1}^{\mathrm{disk}}(\gamma;\ell_{2}),\mathcal{M}_{1}^{\mathrm{disk}}(\gamma;\ell_{3}),\mathrm{QP}(\ell,\ell_{2},\ell_{3}))\,d\ell_{2}\,d\ell_{3},$

so that the relevant integral for Proposition 5.4 is
$\int_{0}^{\infty}\ell\mathrm{Weld}(\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell),\mathfrak{M}_{\ell})\,d\ell$.
We sample a decorated quantum surface from

$\int_{0}^{\infty}\ell\mathrm{Weld}(\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell),\mathfrak{M}_{\ell})\,d\ell$
(5.7)

and let $(\mathbbm{C},\phi,\eta_{1},\eta_{2},\eta_{3},z_{1},z_{2},z_{3})$ be
its embedding; since we specify the locations of the three marked points, the
tuple $(\phi,\eta_{1},\eta_{2},\eta_{3})$ is uniquely specified by the
decorated quantum surface. Let $D_{\eta_{1}}$ and $D^{c}_{\eta_{1}}$ be the
connected components of $\mathbbm{C}\backslash\eta_{1}$ such that
$D_{\eta_{1}}$ contains $z_{1}$. Let $p\in\eta_{1}$ be a point sampled from
the harmonic measure of $\partial D_{\eta_{1}}$ viewed from $z_{1}$ and set
$\mathcal{D}_{1}=(D_{\eta_{1}},\phi,z_{1},p)/{\sim_{\gamma}}$. Let
$\psi:\mathbbm{H}\to D_{\eta_{1}}$ be the conformal map with $\psi(i)=z_{1}$
and $\psi(0)=p$. Let $X=\phi\circ\psi+Q\log|\psi^{\prime}|$ so that
$\mathcal{D}_{1}=(\mathbbm{H},X,i,0)/{\sim_{\gamma}}$. Let
$\mathcal{D}^{c}_{1}$ be the decorated quantum surface
$(D^{c}_{\eta_{1}},\phi,\eta_{2},\eta_{3},z_{2},z_{3},p)/{\sim_{\gamma}}$. See
Figure 4 for an illustration. The next lemma describes the law of
$(X,\mathcal{D}^{c}_{1})$.

######  Lemma 5.8.

Given a decorated quantum surface $\mathcal{S}$ sampled from
$\mathfrak{M}_{\ell}$, we write $\mathcal{S}^{\bullet}$ as the decorated
quantum surface obtained by further sampling a point on the boundary of
$\mathcal{S}$ according to the probability measure proportional to its quantum
boundary length measure. Let $\mathfrak{M}_{\ell}^{\bullet}$ be the law of
$\mathcal{S}^{\bullet}$. Then the law of $(X,\mathcal{D}^{c}_{1})$ defined
right above is
$\int_{0}^{\infty}\ell\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}(\ell)\times\mathfrak{M}_{\ell}^{\bullet}d\ell$.

###### Proof.

By the definition of uniform welding, conditioning on $\mathcal{D}_{1}$, the
conditional law of $\mathcal{D}^{c}_{1}$ is the probability measure
proportional to $\mathfrak{M}_{\ell}^{\bullet}$ with $\ell$ being the boundary
length of $\mathcal{D}_{1}$. Since the marked boundary point of
$\mathcal{D}_{1}$ is sampled from the harmonic measure, Lemma 5.7 now yields
Lemma 5.8. ∎

Figure 4:  Illustration for Lemma 5.8 and the proof of Proposition 5.4. The
region $D_{\eta_{1}}$ surrounded by $\eta_{1}$ is colored grey. Both
$(D_{\eta_{1}},\phi,z_{1},p)$ and $(\mathbbm{H},X,0,i)$ are embeddings of
$\mathcal{D}_{1}$, which are related by $\psi$. The domain
$\mathbbm{C}_{\eta_{1},p,\varepsilon}$ equals
$\mathbbm{C}\setminus\psi(B_{\varepsilon}(i))$, the region outside the red
curve on the left.

###### Proof of Proposition 5.4.

In the setting of Lemma 5.8 with $\alpha=\gamma$, by Lemma 5.3 the law of
$(\phi,\eta_{1},\eta_{2},\eta_{3},p)$ is
$\mathrm{Harm_{z_{1},\eta_{1}}}(dp)\,\mathrm{LF}_{\mathbbm{C}}^{(\gamma,z_{1}),(\gamma,z_{2}),(\gamma,z_{3})}(d\phi)\,\mathsf{m}_{3}(d\eta_{1},d\eta_{2},d\eta_{3})$,
where $\mathrm{Harm}_{z_{1},\eta_{1}}$ means the harmonic measure on
$\eta_{1}$ viewed from $z_{1}$. Therefore Lemma 5.8 with $\alpha=\gamma$ can
be written as

$\mathrm{LF}_{\mathbbm{C}}^{(\gamma,z_{1}),(\gamma,z_{2}),(\gamma,z_{3})}(d\phi)\,\mathrm{Harm_{z_{1},\eta_{1}}}(dp)\,\mathsf{m}_{3}(d\eta_{1},d\eta_{2},d\eta_{3})=\int_{0}^{\infty}\ell\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}(\ell)\times\mathfrak{M}_{\ell}^{\bullet}d\ell,$
(5.8)

in the sense that $(\phi,\eta_{1},\eta_{2},\eta_{3},p)$ and
$(X,\mathcal{D}^{c}_{1})$ determine each other, and the two sides of (5.8)
give the laws of $(\phi,\eta_{1},\eta_{2},\eta_{3},p)$ and
$(X,\mathcal{D}^{c}_{1})$, respectively.

Recall the notations in Lemma 5.6. For $\varepsilon\in(0,\frac{1}{4})$, for
any nonnegative measurable function $f$ of
$\phi|_{\mathbbm{C}_{\eta_{1},p,\varepsilon}}$, and any nonnegative measure
function $g$ of $(\eta_{1},\eta_{2},\eta_{3})$, we get from (5.8) that

$\displaystyle\int
f(\phi|_{\mathbbm{C}_{\eta_{1},p,\varepsilon}})g(\eta_{1},\eta_{2},\eta_{3})\varepsilon^{\frac{1}{2}(\alpha^{2}-\gamma^{2})}e^{(\alpha-\gamma)X_{\varepsilon}(i)}\mathrm{LF}_{\mathbbm{C}}^{(\gamma,z_{1}),(\gamma,z_{2}),(\gamma,z_{3})}(d\phi)\,\mathrm{Harm_{z_{1},\eta_{1}}}(dp)\,\mathsf{m}_{3}(d\eta_{1},d\eta_{2},d\eta_{3})$
$\displaystyle=\int_{0}^{\infty}\mathopen{}\mathclose{{}\left(\int
f(\phi|_{\mathbbm{C}_{\eta_{1},p,\varepsilon}})g(\eta_{1},\eta_{2},\eta_{3})\varepsilon^{\frac{1}{2}(\alpha^{2}-\gamma^{2})}e^{(\alpha-\gamma)X_{\varepsilon}(i)}\ell\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}(\ell)\times\mathfrak{M}_{\ell}^{\bullet}}\right)d\ell.$
(5.9)

Recall that
$\mathsf{m}_{3}^{\alpha,\gamma,\gamma}=\mathopen{}\mathclose{{}\left(\frac{1}{2}\mathrm{CR}(\eta_{1},z_{1})}\right)^{-\frac{\alpha^{2}}{2}+Q\alpha-2}\mathsf{m}_{3}$.
By Lemma 5.6, the left side of (5.9) equals

$\displaystyle\int
fg\mathopen{}\mathclose{{}\left(\frac{1}{2}\mathrm{CR}(\eta_{1},z_{1})}\right)^{-\frac{\alpha^{2}}{2}+Q\alpha-2}\mathrm{LF}_{\mathbbm{C}}^{(\alpha,z_{1}),(\gamma,z_{2}),(\gamma,z_{3})}(d\phi)\,\mathrm{Harm_{z_{1},\eta_{1}}}(dp)\,\mathsf{m}_{3}(d\eta_{1},d\eta_{2},d\eta_{3})$
$\displaystyle=$ $\displaystyle\int
fg\mathrm{LF}_{\mathbbm{C}}^{(\alpha,z_{1}),(\gamma,z_{2}),(\gamma,z_{3})}(d\phi)\mathrm{Harm_{z_{1},\eta_{1}}}(dp)\,\mathsf{m}_{3}^{\alpha,\gamma,\gamma}\,(d\eta_{1},d\eta_{2},d\eta_{3}).$
(5.10)

Here we write $f=f(\phi|_{\mathbbm{C}_{\eta_{1},p,\varepsilon}})$ and
$g=g(\eta_{1},\eta_{2},\eta_{3})$ to ease the notation.

By Lemma 5.5, the right side of (5.9) equals

$\int_{0}^{\infty}\mathopen{}\mathclose{{}\left(\int
fg\ell\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}(\ell)\times\mathfrak{M}_{\ell}^{\bullet}}\right)d\ell.$
(5.11)

Since $\varepsilon,f,g$ are arbitrary, comparing (5.10) and (5.11) we get in
the same sense as in (5.8) that

$\mathrm{LF}_{\mathbbm{C}}^{(\alpha,z_{1}),(\gamma,z_{2}),(\gamma,z_{3})}(d\phi)\mathrm{Harm_{z_{1},\eta_{1}}}(dp)\,\mathsf{m}_{3}^{\alpha,\gamma,\gamma}(d\eta_{1},d\eta_{2},d\eta_{3})=\int_{0}^{\infty}\ell\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}(\ell)\times\mathfrak{M}_{\ell}^{\bullet}d\ell.$

Forgetting about the point $p$ we get Proposition 5.4. ∎

###### Proof of Theorem 5.2.

The case $(\alpha_{1},\alpha_{2},\alpha_{3})=(\alpha,\gamma,\gamma)$ was
proved in Proposition 5.4 by reweighting from the $(\gamma,\gamma,\gamma)$
case. The exact same argument in Lemma 5.6 gives the following extension:

$\displaystyle\int
f(\phi|_{\mathbbm{C}_{\eta_{1},p,\varepsilon}})\times\varepsilon^{\frac{1}{2}(\alpha^{2}-\alpha_{1}^{2})}e^{(\alpha-\alpha_{1})X_{\varepsilon}(i)}\,d\mathrm{LF}_{\mathbbm{C}}^{(\alpha_{1},z_{1}),(\alpha_{2},z_{2}),(\alpha_{3},z_{3})}$
$\displaystyle=$ $\displaystyle\int
f(\phi|_{\mathbbm{C}_{\eta_{1},p,\varepsilon}})\mathopen{}\mathclose{{}\left(\frac{1}{2}\mathrm{CR}(\eta_{1},z_{1})}\right)^{-\frac{\alpha_{1}^{2}}{2}+Q\alpha_{1}-2}\mathrm{LF}_{\mathbbm{C}}^{(\alpha,z_{1}),(\alpha_{2},z_{2}),(\alpha_{3},z_{3})}.$
(5.12)

By symmetry (5.12) holds if the roles of $z_{1},z_{2},z_{3}$ are permutated.
This allows us to obtain the general $(\alpha_{1},\alpha_{2},\alpha_{3})$ case
from the $(\gamma,\gamma,\gamma)$ case by reweighing around the three points
sequentially. ∎

###  5.2 Matching the quantum area: proof of Theorem 1.2

Theorem 1.2 is an immediate consequence of Lemma 5.1 and the following
proposition.

######  Proposition 5.9.

Fix $(z_{1},z_{2},z_{3})=(0,1,e^{i\pi/3})$. There is a constant
$C=C(\gamma)\in(0,\infty)$ such that for
$\alpha_{1},\alpha_{2},\alpha_{3}\in(Q-\frac{\gamma}{4},Q)$, and
$\mathsf{m}_{3}^{\alpha_{1},\alpha_{2},\alpha_{3}}$ as defined in (5.1) we
have

$\mathopen{}\mathclose{{}\left(\mathrm{LF}_{\mathbbm{C}}^{(\alpha_{i},z_{i})_{3}}\times\mathsf{m}_{3}^{\alpha_{1},\alpha_{2},\alpha_{3}}}\right)[e^{-\mu_{\phi}(\mathbbm{C})}]=C\prod_{i=1}^{3}\frac{2^{\alpha_{i}^{2}/2-Q\alpha_{i}}\Gamma(\frac{\gamma\alpha_{i}}{2}-\frac{\gamma^{2}}{4})}{\Gamma(\frac{2}{\gamma}(Q-\alpha_{i}))\cos(\frac{2\pi}{\gamma}(Q-\alpha_{i}))}\mathopen{}\mathclose{{}\left(\frac{\pi\Gamma(\frac{\gamma^{2}}{4})}{\Gamma(1-\frac{\gamma^{2}}{4})}}\right)^{-\frac{\alpha_{i}}{\gamma}}.$
(5.13)

###### Proof.

Recall $\mathrm{QP}_{3}(\ell_{1},\ell_{2},\ell_{3})[e^{-A}]$ from Theorem 4.2.
Recall $\mathcal{M}_{1}^{\mathrm{disk}}(\alpha_{i};\ell_{i})[e^{-A_{i}}]$ from
Theorem 2.16, where $A_{i}$ is the quantum area of a sample from
$\mathcal{M}_{1}^{\mathrm{disk}}(\alpha_{i};\ell_{i})$. By Theorem 5.2, for
some $\gamma$-dependent constants $C_{1},C_{2},C_{3}$, we have that
$\mathopen{}\mathclose{{}\left(\mathrm{LF}_{\mathbbm{C}}^{(\alpha_{i},z_{i})_{3}}\times\mathsf{m}_{3}^{\alpha_{1},\alpha_{2},\alpha_{3}}}\right)[e^{-\mu_{\phi}(\mathbbm{C})}]$
equals

$\displaystyle C_{1}$
$\displaystyle\iiint_{0}^{\infty}\ell_{1}\ell_{2}\ell_{3}\mathrm{QP}_{3}(\ell_{1},\ell_{2},\ell_{3})[e^{-A}]\prod_{i=1}^{3}\mathcal{M}_{1}^{\mathrm{disk}}(\alpha_{i};\ell_{i})[e^{-A_{i}}]\,d\ell_{1}\,d\ell_{2}\,d\ell_{3}$
$\displaystyle=C_{2}\prod_{i=1}^{3}\frac{\overline{U}(\alpha_{i})(4\sin(\frac{\pi\gamma^{2}}{4}))^{\alpha_{i}/\gamma}}{2^{\alpha_{i}^{2}/2}\Gamma(\frac{2}{\gamma}(Q-\alpha_{i}))}\int_{0}^{\infty}\frac{1}{\sqrt{\ell_{i}}}e^{-\ell_{i}\sqrt{\frac{1}{\sin(\pi\gamma^{2}/4)}}}K_{\frac{2}{\gamma}(Q-\alpha_{i})}\mathopen{}\mathclose{{}\left(\ell_{i}\sqrt{\frac{1}{\sin(\pi\gamma^{2}/4)}}}\right)\,d\ell_{i}$
$\displaystyle=C_{3}\prod_{i=1}^{3}\frac{\overline{U}(\alpha_{i})(4\sin(\frac{\pi\gamma^{2}}{4}))^{\alpha_{i}/\gamma}}{2^{\alpha_{i}^{2}/2}\Gamma(\frac{2}{\gamma}(Q-\alpha_{i}))\cos(\frac{2\pi}{\gamma}(Q-\alpha_{i}))},$

where $\overline{U}$ is as in (2.4) and (2.7) is used to evaluate the integral
on the second line. Expanding $\overline{U}(\alpha_{i})$ from (2.4) and
substituting $\sin(\frac{\pi\gamma^{2}}{4})$ by
$\frac{\pi}{\Gamma(\frac{\gamma^{2}}{4})\Gamma(1-\frac{\gamma^{2}}{4})}$, we
get (5.13). ∎

###### Proof of Theorem 1.2.

We divide the proof into four cases depending on the parameter range.

Case I: $\kappa<4$ and
$\lambda_{i}\in(\frac{3\kappa}{32}-1+\frac{2}{\kappa},\frac{\kappa}{8}-1+\frac{2}{\kappa})$
for all $i=1,2,3$. In this case we can find
$\alpha_{i}\in(Q-\frac{\gamma}{4},Q)$ satisfying
$\lambda_{i}=-\frac{\alpha_{i}^{2}}{2}+Q\alpha_{i}-2$ for each $i$. Comparing
Lemma 5.1 and Proposition 5.9 yields that for some $\gamma$-dependent constant
$C(\gamma)$ we have

${C^{\operatorname{CLE}}_{\kappa}(\lambda_{1},\lambda_{2},\lambda_{3})}=\frac{C(\gamma)}{C_{\gamma}^{\mathrm{DOZZ}}(\alpha_{1},\alpha_{2},\alpha_{3})}\prod_{i=1}^{3}\frac{\Gamma(\frac{\gamma\alpha_{i}}{2}-\frac{\gamma^{2}}{4})}{\Gamma(\frac{2}{\gamma}(Q-\alpha_{i}))\cos(\frac{2\pi}{\gamma}(Q-\alpha_{i}))}\mathopen{}\mathclose{{}\left(\frac{\pi\Gamma(\frac{\gamma^{2}}{4})}{\Gamma(1-\frac{\gamma^{2}}{4})}}\right)^{-\frac{\alpha_{i}}{\gamma}}.$

Now we can evaluate $C(\gamma)$ by setting
$\alpha_{1}=\alpha_{2}=\alpha_{3}=\gamma$, in which case $\lambda_{i}=0$ and
$C_{\kappa}^{\operatorname{CLE}}(0,0,0)=1$. This gives (1.3) for
${C^{\operatorname{CLE}}_{\kappa}(\lambda_{1},\lambda_{2},\lambda_{3})}$ and
(1.4) for the factor $N_{\gamma}(\alpha)$.

For the next two cases set $(z_{1},z_{2},z_{3})=(0,1,e^{i\pi/3})$ so that
$C_{\kappa}^{\operatorname{CLE}}(\lambda_{1},\lambda_{2},\lambda_{3})=\mathbbm{E}[\prod_{i=1}^{3}(\mathrm{CR}(\eta_{i},z_{i}))^{\lambda_{i}}]$.
Since the distance between $z_{i}$ and $\eta_{i}$ is less than $1$, Koebe 1/4
theorem yields $\frac{1}{4}\mathrm{CR}(\eta_{i},z_{i})\leq 1$. Thus
$\mathbbm{E}[\prod_{i=1}^{3}(\frac{1}{4}\mathrm{CR}(\eta_{i},z_{i}))^{\lambda_{i}}]\leq\mathbbm{E}[\prod_{i=1}^{3}(\frac{1}{4}\mathrm{CR}(\eta_{i},z_{i}))^{\widetilde{\lambda}_{i}}]$
for any real $\lambda_{i},\widetilde{\lambda}_{i}$ such that
$\lambda_{i}\geq\widetilde{\lambda}_{i}$. Namely

$4^{-\lambda_{1}-\lambda_{2}-\lambda_{3}}C_{\kappa}^{\operatorname{CLE}}(\lambda_{1},\lambda_{2},\lambda_{3})\leq
4^{-\widetilde{\lambda}_{1}-\widetilde{\lambda}_{2}-\widetilde{\lambda}_{3}}C_{\kappa}^{\operatorname{CLE}}(\widetilde{\lambda}_{1},\widetilde{\lambda}_{2},\widetilde{\lambda}_{3}).$
(5.14)

Case II: $\kappa<4$ and all
$\lambda_{i}>\frac{3\kappa}{32}-1+\frac{2}{\kappa}$. By (5.14) on
$\\{(\lambda_{1},\lambda_{2},\lambda_{3})\in\mathbbm{C}^{3}:\operatorname{Re}\lambda_{i}>\frac{3\kappa}{32}-1+\frac{2}{\kappa}\\}$,
the function
${C^{\operatorname{CLE}}_{\kappa}(\lambda_{1},\lambda_{2},\lambda_{3})}$ is
finite, hence analytic. On the other hand, the right hand side of (1.3) is an
explicit meromorphic function in $(\alpha_{1},\alpha_{2},\alpha_{3})$. Now
Case II follows from Case I.

Case III: $\kappa<4$ and
$\lambda_{i}\leq\frac{3\kappa}{32}-1+\frac{2}{\kappa}$ for some $i$. We will
show that
${C^{\operatorname{CLE}}_{\kappa}(\lambda_{1},\lambda_{2},\lambda_{3})}=\infty$
in this case. By the monotonicity (5.14) and symmetry, it suffices to prove
$C_{\kappa}^{\operatorname{CLE}}(\lambda_{1},\lambda_{2},\lambda_{3})=\infty$
for $\lambda_{1}\leq\frac{3\kappa}{32}-1+\frac{2}{\kappa}$ and
$\lambda_{2},\lambda_{3}>\frac{3\kappa}{32}-1+\frac{2}{\kappa}$. For
$\lambda>\frac{3\kappa}{32}-1+\frac{2}{\kappa}$ we have

$4^{-\lambda_{1}-\lambda_{2}-\lambda_{3}}C_{\kappa}^{\operatorname{CLE}}(\lambda_{1},\lambda_{2},\lambda_{3})>4^{-\lambda-\lambda_{2}-\lambda_{3}}C_{\kappa}^{\operatorname{CLE}}(\lambda,\lambda_{2},\lambda_{3}),$

Suppose $(\alpha,\alpha_{2},\alpha_{3})\in(Q-\frac{\gamma}{4},Q)^{3}$
correspond to $(\lambda,\lambda_{2},\lambda_{3})$. Then
$\lambda\downarrow\frac{3\kappa}{32}-1+\frac{2}{\kappa}$ means
$\alpha\downarrow Q-\frac{\gamma}{4}$. Recall the explicit formula for
$C_{\kappa}^{\operatorname{CLE}}(\lambda,\lambda_{2},\lambda_{3})$ in (1.3)
proven in Case II. Since $\lim_{\alpha\downarrow
Q-\frac{\gamma}{4}}N_{\gamma}(\alpha)=\infty$ and
$C_{\gamma}^{\mathrm{DOZZ}}(\alpha,\alpha_{2},\alpha_{3})\in(0,\infty)$ as
$(\alpha,\alpha_{2},\alpha_{3})$ satisfies the Seiberg bounds (2.2), we must
have
$C_{\kappa}^{\operatorname{CLE}}(\lambda,\lambda_{2},\lambda_{3})\rightarrow\infty$
as $\lambda\downarrow\frac{3\kappa}{32}-1+\frac{2}{\kappa}$. Therefore
$C_{\kappa}^{\operatorname{CLE}}(\lambda_{1},\lambda_{2},\lambda_{3})=\infty$
as desired.

Case IV: $\kappa=4$. This follows from Lemma A.5 via the continuity as
$\kappa\uparrow 4$. ∎

##  6 Electrical thickness of the SLE loop via conformal welding

In this section we prove Theorem 1.3. We will work on the horizontal cylinder
$\mathcal{C}=\mathbbm{R}\times[0,2\pi]/{\sim}$ with $x\sim x+2\pi i$ for
$x\in\mathbbm{R}$. Recall the shape measure $\mathcal{L}_{\kappa}$ of
$\operatorname{SLE}^{\mathrm{loop}}_{\kappa}$. Let
$\mathcal{L}_{\kappa}(\mathcal{C})$ be the pullback of $\mathcal{L}_{\kappa}$
under the map $z\mapsto e^{-z}$, which is a probability measure on loops
$\eta$ in $\mathcal{C}$ separating $\pm\infty$ and satisfying
$\max_{z\in\eta}\operatorname{Re}z=0$. For a loop $\eta$ sampled from
$\mathcal{L}_{\kappa}(\mathcal{C})$, we write $\vartheta(\eta)$ for the
electrical thickness of $\exp(\eta)$, namely
$-\log\mathrm{CR}(\exp(\eta),0)-\log\mathrm{CR}(\exp(-\eta),0)$. Then Theorem
1.3 is equivalent to

$\mathbbm{E}[e^{\lambda\vartheta(\eta)}]=\mathopen{}\mathclose{{}\left\\{\begin{array}[]{ll}\frac{\sin(\pi(1-\kappa/4))}{\pi(1-\kappa/4)}\frac{\pi\sqrt{(1-\kappa/4)^{2}+\lambda\kappa/2}}{\sin(\pi\sqrt{(1-\kappa/4)^{2}+\lambda\kappa/2})}&\mbox{if
}\lambda<1-\frac{\kappa}{8}.\\\ \infty&\mbox{if }\lambda\geq
1-\frac{\kappa}{8}\end{array}}\right.$ (6.1)

Consider $\alpha<Q$ and set $\lambda=\frac{\alpha^{2}}{2}-Q\alpha+2$. Let
$\mathcal{L}_{\kappa}^{\alpha}$ be defined by the following reweighting of
$\mathcal{L}_{\kappa}(\mathcal{C})$:

$\frac{d\mathcal{L}_{\kappa}^{\alpha}}{d\mathcal{L}_{\kappa}(\mathcal{C})}(\eta)=(\frac{1}{4}\mathrm{CR}(\exp(\eta),0)\mathrm{CR}(\exp(-\eta),0))^{-\frac{\alpha^{2}}{2}+Q\alpha-2}=2^{2\lambda}e^{\lambda\vartheta(\eta)}.$
(6.2)

Then proving Theorem 1.3 amounts to computing the total mass
$|\mathcal{L}_{\kappa}^{\alpha}|$. Similarly as in the proof of Theorem 1.2 in
Section 5, we need a conformal welding result, Proposition 6.2, which involves
$\mathcal{L}_{\kappa}^{\alpha}$. It is a generalization of Proposition 2.20
where marked points have $\alpha$-singularities. To state it we first recall
the $\alpha$-generalization of the two-pointed quantum sphere; see e.g.
[AHS20, Section 2].

######  Definition 6.1.

For $\gamma\in(0,2)$, let $(B_{s})_{s\geq 0}$ be a standard Brownian motion
conditioned on $B_{s}-(Q-\alpha)s<0$ for all $s>0$, and
$(\widetilde{B}_{s})_{s\geq 0}$ an independent copy of $(B_{s})_{s\geq 0}$.
Let

$Y_{t}=\mathopen{}\mathclose{{}\left\\{\begin{array}[]{ll}B_{t}-(Q-\alpha)t&\mbox{if
}t\geq 0\\\ \widetilde{B}_{-t}+(Q-\alpha)t&\mbox{if }t<0\end{array}}\right.$

Let $h^{1}(z)=Y_{\operatorname{Re}z}$ for each $z\in\mathcal{C}$. Let
$h^{2}_{\mathcal{C}}$ be independent of $h^{1}$ and have the law of the
lateral component of the GFF on $\mathcal{C}$. Let
$\hat{h}=h^{1}+h^{2}_{\mathcal{C}}$. Let $\mathbf{c}\in\mathbbm{R}$ be sampled
from $\frac{\gamma}{2}e^{2(\alpha-Q)c}dc$ independent of $\hat{h}$. Let
$\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)$ be the infinite measure describing
the law of the decorated quantum surface
$(\mathcal{C},\hat{h}+\mathbf{c},-\infty,+\infty)/{\sim_{\gamma}}$.

Consider $\kappa\in(0,4)$ and $\gamma=\sqrt{\kappa}$. Sample a pair
$(\eta,\mathbf{t})$ from $\mathcal{L}_{\kappa}^{\alpha}\times dt$ where $dt$
is the Lebesgue measure on $\mathbbm{R}$. Let
$\operatorname{SLE}_{\kappa}^{\mathrm{sep},\alpha}$ be the law of the
translated loop $\eta+\mathbf{t}$. Let $\mathbb{F}$ be the law of
$\hat{h}+\mathbf{c}$ as in Definition 6.1, so that the law of
$(\mathcal{C},\hat{h}+\mathbf{c},-\infty,+\infty)/{\sim_{\gamma}}$ is
$\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)$. Now sample $(h,\eta)$ from
$\mathbb{F}\times\operatorname{SLE}_{\kappa}^{\mathrm{sep},\alpha}$ and write
$\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)\otimes\operatorname{SLE}_{\kappa}^{\mathrm{sep,\alpha}}$
as the law of $(\mathcal{C},h,\eta,-\infty,+\infty)/{\sim_{\gamma}}$. We are
now ready to state the conformal welding result needed for the proof of
Theorem 1.3.

######  Proposition 6.2.

For $\alpha\in(\frac{\gamma}{2},Q)$ and some $\gamma$-dependent constant
$C\in(0,\infty)$ we have

$C(Q-\alpha)^{2}\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)\otimes\operatorname{SLE}_{\kappa}^{\mathrm{sep},\alpha}=\int_{0}^{\infty}\ell\cdot\mathrm{Weld}(\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell),\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell))\,d\ell,$
(6.3)

where
$\mathrm{Weld}(\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell),\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell))$
is defined as $\mathrm{Weld}(\mathrm{QD}_{1,0}(\ell),\mathrm{QD}_{1,0}(\ell))$
in Proposition 2.20.

We postpone the proof of Proposition 6.2 to Section 6.1 and proceed to the
proof of Theorem 1.3. Similarly as in the proof of Theorem 1.2, we would like
to evaluate the average of $e^{-A}$ on both sides of (6.3) and compare the
expressions to obtain $|\mathcal{L}^{\alpha}_{\kappa}|$. However, since
$\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)\otimes\operatorname{SLE}_{\kappa}^{\mathrm{sep},\alpha}[e^{-A}]=\infty$,
we have to consider an alternative observable. Note that
$\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)\otimes\operatorname{SLE}_{\kappa}^{\mathrm{sep},\alpha}$
is a measure on quantum surfaces decorated by two (ordered) marked points and
a loop separating them. The loop separates the quantum surface into two
connected components. For $0<\varepsilon<\delta$, let $E_{\delta,\varepsilon}$
be the event that the connected component containing the first marked point
has quantum area at least 1 and the loop has quantum length in
$(\varepsilon,\delta)$. We use the size of $E_{\delta,\varepsilon}$ as our
obervable, whose asymptotic is easy to obtain using Proposition 6.2.

######  Lemma 6.3.

Let $\alpha\in(\frac{\gamma}{2},Q)$. Recall $\overline{U}(\alpha)$ from
Proposition 2.10. With $C$ from Proposition 6.2 and an error term
$o_{\delta,\varepsilon}(1)$ satisfying $\lim_{\delta\to
0}\limsup_{\varepsilon\to 0}|o_{\delta,\varepsilon}(1)|=0$, we have

$\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)\otimes\operatorname{SLE}_{\kappa}^{\mathrm{sep},\alpha}[E_{\delta,\varepsilon}]=\frac{1}{C(Q-\alpha)^{2}}\times\frac{(1+o_{\delta,\varepsilon}(1))\log\varepsilon^{-1}}{\frac{2}{\gamma}(Q-\alpha)\Gamma(\frac{2}{\gamma}(Q-\alpha))}\mathopen{}\mathclose{{}\left(\frac{2}{\gamma}2^{-\frac{\alpha^{2}}{2}}\overline{U}(\alpha)}\right)^{2}\mathopen{}\mathclose{{}\left(4\sin\frac{\pi\gamma^{2}}{4}}\right)^{-\frac{2}{\gamma}(Q-\alpha)}.$

###### Proof.

By Proposition 6.2,
$C(Q-\alpha)^{2}\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)\otimes\operatorname{SLE}_{\kappa}^{\mathrm{sep},\alpha}[E_{\delta,\varepsilon}]$
equals the mass of $E_{\delta,\varepsilon}$ under the measure
$\int_{0}^{\infty}\ell\cdot\mathrm{Weld}(\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell),\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell))\,d\ell$.
This mass can be expressed as

$\int_{\varepsilon}^{\delta}\ell\cdot\mathopen{}\mathclose{{}\left|\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell)}\right|\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell)[A>1]\,d\ell=\int_{\varepsilon}^{\delta}\ell\mathopen{}\mathclose{{}\left|\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell)}\right|^{2}\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell)^{\\#}[A>1]\;d\ell,$
(6.4)

where $A$ is the quantum area of a sample from
$\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell)$. From the scaling relation
between quantum area and length, and the explicit density of $A$ under
$\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;1)^{\\#}$ given by the FZZ formula
Theorem 2.16, we see that

$\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell)^{\\#}[A>1]=\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;1)^{\\#}[A>\ell^{-2}]=\frac{1+o_{\ell}(1)}{\frac{2}{\gamma}(Q-\alpha)\Gamma(\frac{2}{\gamma}(Q-\alpha))}\mathopen{}\mathclose{{}\left(\frac{\ell^{2}}{4\sin\frac{\pi\gamma^{2}}{4}}}\right)^{\frac{2}{\gamma}(Q-\alpha)}\textrm{
as }\ell\to 0.$

By Proposition 2.10,
$|\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell)|=\frac{2}{\gamma}2^{-\frac{\alpha^{2}}{2}}\overline{U}(\alpha)\ell^{\frac{2}{\gamma}(\alpha-Q)-1}$.
Therefore the right side of (6.4) equals

$\frac{1}{\frac{2}{\gamma}(Q-\alpha)\Gamma(\frac{2}{\gamma}(Q-\alpha))}\mathopen{}\mathclose{{}\left(\frac{2}{\gamma}2^{-\frac{\alpha^{2}}{2}}\overline{U}(\alpha)}\right)^{2}\mathopen{}\mathclose{{}\left(4\sin\frac{\pi\gamma^{2}}{4}}\right)^{-\frac{2}{\gamma}(Q-\alpha)}\int_{\varepsilon}^{\delta}\ell^{-1}(1+o_{\ell}(1))\,d\ell.$

Writing $\int_{\varepsilon}^{\delta}\ell^{-1}(1+o_{\ell}(1))\,d\ell$ as
$(1+o_{\delta,\varepsilon}(1))\log\varepsilon^{-1}$, we have $\lim_{\delta\to
0}\limsup_{\varepsilon\to 0}|o_{\delta,\varepsilon}(1)|=0$ as desired. ∎

Given Lemma 6.3, the following proposition is the key to our proof of Theorem
1.3. It gives the size of $E_{\delta,\varepsilon}$ in terms of
$|\mathcal{L}_{\kappa}^{\alpha}|$ and the reflection coefficient for LCFT.

######  Proposition 6.4.

Let $\alpha\in(\frac{\gamma}{2},Q)$. With an error term
$o_{\delta,\varepsilon}(1)$ satisfying $\lim_{\delta\to
0}\limsup_{\varepsilon\to 0}|o_{\delta,\varepsilon}(1)|=0$, we have

$(\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)\otimes\operatorname{SLE}_{\kappa}^{\mathrm{sep},\alpha})[E_{\delta,\varepsilon}]=(1+o_{\delta,\varepsilon}(1))\frac{\overline{R}(\alpha)}{2(Q-\alpha)^{2}}|\mathcal{L}_{\kappa}^{\alpha}|\log\varepsilon^{-1},$

where $\overline{R}(\alpha)$ is the unit-volume reflection coefficient for
LCFT on the sphere [KRV17, RV19]:

$\overline{R}(\alpha)=-\mathopen{}\mathclose{{}\left(\frac{\pi\Gamma(\frac{\gamma^{2}}{4})}{\Gamma(1-\frac{\gamma^{2}}{4})}}\right)^{\frac{2}{\gamma}(Q-\alpha)}\frac{1}{\frac{2}{\gamma}(Q-\alpha)}\frac{\Gamma(-\frac{\gamma}{2}(Q-\alpha))}{\Gamma(\frac{\gamma}{2}(Q-\alpha))\Gamma(\frac{2}{\gamma}(Q-\alpha))}.$
(6.5)

We postpone the proof of Proposition 6.4 to Section 6.2 and proceed to prove
Theorem 1.3.

######  Proposition 6.5.

There exists a $\gamma$-dependent constant $C\in(0,\infty)$ such that

$2^{-\alpha^{2}+2Q\alpha}|\mathcal{L}_{\kappa}^{\alpha}|=C\frac{\frac{\gamma}{2}(Q-\alpha)}{\sin(\frac{\gamma\pi}{2}(Q-\alpha))}\quad\textrm{for
all }\alpha\in(\frac{\gamma}{2},Q).$ (6.6)

###### Proof.

By Lemma 6.3 and Proposition 6.4, we get

$\overline{R}(\alpha)|\mathcal{L}_{\kappa}^{\alpha}|=\frac{C^{\prime}}{(Q-\alpha)\Gamma(\frac{2}{\gamma}(Q-\alpha))}\mathopen{}\mathclose{{}\left(2^{-\frac{\alpha^{2}}{2}}\overline{U}(\alpha)}\right)^{2}\mathopen{}\mathclose{{}\left(4\sin\frac{\pi\gamma^{2}}{4}}\right)^{-\frac{2}{\gamma}(Q-\alpha)}$

for some $\gamma$-dependent constant $C^{\prime}$. Plugging in
$\overline{R}(\alpha)$ and $\overline{U}(\alpha)$ from (6.5) and (2.4), and
simplifying via the identities $\Gamma(z+1)=z\Gamma(z)$ and
$\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin(\pi z)}$, we arrive at (6.6). ∎

###### Proof of Theorem 1.3.

We break the proof of (6.1) hence Theorem 1.3, into three cases.

Case I: $\kappa<4$ and
$\lambda\in(1-\frac{\kappa}{8}-\frac{2}{\kappa},1-\frac{\kappa}{8})$. Set
$\alpha=Q-\sqrt{Q^{2}-4+2\lambda}$. Then $\alpha\in(\frac{\gamma}{2},Q)$ in
this case. Since $\lambda=\frac{\alpha^{2}}{2}-Q\alpha+2$, by (6.2) and
Proposition 6.5 we have

$\mathbbm{E}[e^{\lambda\vartheta(\eta)}]=2^{-2\lambda}|\mathcal{L}_{\kappa}^{\alpha}|=C\frac{\frac{\gamma}{2}(Q-\alpha)}{\sin(\frac{\gamma\pi}{2}(Q-\alpha))}=C\frac{\sqrt{(1-\frac{\kappa}{4})^{2}+\frac{\lambda\kappa}{2}}}{\sin(\pi\sqrt{(1-\frac{\kappa}{4})^{2}+\frac{\lambda\kappa}{2}})}$

for some $\gamma$-dependent constant $C$. Since $\kappa\in(0,4)$, we have
$0\in(1-\frac{\kappa}{8}-\frac{2}{\kappa},1-\frac{\kappa}{8})$. Thus we can
obtain the value of $C$ by considering
$1=\mathbbm{E}[e^{0}]=\frac{C(1-\kappa/4)}{\sin(\pi(1-\kappa/4))}$. This
yields (6.1) in this case.

Case II: $\kappa<4$ and $\lambda\in\mathbbm{R}$. Since $\vartheta(\eta)\geq
0$, the function $\lambda\mapsto\mathbbm{E}[e^{\lambda\vartheta(\eta)}]$ is
increasing. Thus for $\lambda<0$ we have
$\mathbbm{E}[e^{\lambda\vartheta(\eta)}]\leq\mathbbm{E}[e^{0\cdot\vartheta(\eta)}]=1$.
Since $1-\frac{\kappa}{8}-\frac{2}{\kappa}<0$ we can use analytic continuation
to extend (6.1) from
$\lambda\in(1-\frac{\kappa}{8}-\frac{2}{\kappa},1-\frac{\kappa}{8})$ to
$\lambda\in(-\infty,1-\frac{\kappa}{8})$. On the other hand, for any
$\lambda\geq 1-\frac{\kappa}{8}$ we have
$\mathbbm{E}[e^{\lambda\vartheta(\eta)}]\geq\lim_{\lambda^{\prime}\uparrow
1-\frac{\kappa}{8}}\mathbbm{E}[e^{\lambda^{\prime}\vartheta(\eta)}]$, which
equals $\infty$ using the explicit formula of
$\mathbbm{E}[e^{\lambda^{\prime}\vartheta(\eta)}]$.

Case III: $\kappa=4$. By Lemma A.4,
$\vartheta(\eta_{\kappa})\to\vartheta(\eta_{4})$ in law as $\kappa\uparrow 4$,
where $\eta_{\kappa}$ is a sample from $\mathcal{L}_{\kappa}$. Fix
$\lambda<\frac{1}{2}$ and $\lambda^{\prime}\in(\lambda,\frac{1}{2})$. Since
$\mathbbm{E}[e^{\lambda\vartheta(\eta_{\kappa})}]\leq\mathbbm{E}[e^{\lambda^{\prime}\vartheta(\eta_{\kappa})}]$
and $\mathbbm{E}[e^{\lambda^{\prime}\vartheta(\eta_{\kappa})}]$ is uniformly
bounded in $\kappa\in[3,4)$, the family
$\\{e^{\lambda\vartheta(\eta_{\kappa})}:\kappa\in[3,4)\\}$ is uniformly
integrable. Therefore $\lim_{\kappa\uparrow
4}\mathbbm{E}[e^{\lambda\vartheta(\eta_{\kappa})}]=\mathbbm{E}[e^{\lambda\vartheta(\eta_{4})}]$
as desired. For $\lambda\geq\frac{1}{2}$, the same argument as in Case II
gives $\mathbbm{E}[e^{\lambda\vartheta(\eta)}]=\infty$. ∎

It remains to prove Propositions 6.2 and 6.4. The common starting point of the
proofs is the relation between $\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)$ and
the Liouville field on $\mathcal{C}$ that we now recall from [AHS21].

######  Definition 6.6.

Let $P_{\mathcal{C}}$ be the law of the GFF on the cylinder $\mathcal{C}$
defined above Definition 2.7. Let $\alpha\in\mathbbm{R}$. Sample
$(h,\mathbf{c})$ from $P_{\mathcal{C}}\times[e^{(2\alpha-2Q)c}\,dc]$ and let
$\phi(z)=h(z)-(Q-\alpha)\mathopen{}\mathclose{{}\left|\operatorname{Re}z}\right|+\mathbf{c}$.
We write $\mathrm{LF}_{\mathcal{C}}^{(\alpha,\pm\infty)}$ as the law of
$\phi$.

We have the following description of
$\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)\otimes\operatorname{SLE}_{\kappa}^{\mathrm{sep,\alpha}}$
in terms of $\mathrm{LF}_{\mathcal{C}}^{(\alpha,\pm\infty)}$ and
$\mathcal{L}_{\kappa}^{\alpha}$.

######  Proposition 6.7.

For $(\phi,\eta)$ sampled from
$\mathrm{LF}_{\mathcal{C}}^{(\alpha,\pm\infty)}\times\mathcal{L}_{\kappa}^{\alpha}$,
the law of $(\mathcal{C},\phi,\eta,-\infty,+\infty)/{\sim_{\gamma}}$ is

$C(Q-\alpha)^{2}\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)\otimes\operatorname{SLE}_{\kappa}^{\mathrm{sep},\alpha}\quad\text{for
some $\gamma$-dependent constant }C\in(0,\infty).$ (6.7)

###### Proof.

This is an immediate consequence of the definition of
$\operatorname{SLE}_{\kappa}^{\mathrm{sep},\alpha}$ and [AHS21, Theorem 2.11],
which says the following: let $h$ be the field $\hat{h}+\mathbf{c}$ in
Definition 6.1, let $T\in\mathbbm{R}$ be sampled from Lebesgue measure
independently of $h$, and set $\phi:=h(\cdot+T)$. Then $\phi$ has law
$\frac{\gamma}{4(Q-\alpha)^{2}}\mathrm{LF}_{\mathcal{C}}^{(\alpha,\pm\infty)}$.
∎

We now proceed to prove Propositions 6.2 and 6.4 in the next two subsections.

###  6.1 Conformal welding with two generic insertions: proof of Proposition
6.2

Our proof of Proposition 6.2 closely follows that of Theorem 5.2. We start
from the case $\alpha=\gamma$.

######  Lemma 6.8.

If $(\phi,\eta)$ is sampled from
$\mathrm{LF}_{\mathcal{C}}^{(\gamma,\pm\infty)}\times\mathcal{L}_{\kappa}(\mathcal{C})$,
then the law of $(\mathcal{C},\phi,\eta,-\infty,+\infty)/{\sim_{\gamma}}$ is
$C\int_{0}^{\infty}\ell\cdot\mathrm{Weld}(\mathcal{M}_{1}^{\mathrm{disk}}(\gamma;\ell),\mathcal{M}_{1}^{\mathrm{disk}}(\gamma;\ell))\,d\ell$
some $\gamma$-dependent constant $C\in(0,\infty)$.

###### Proof.

Recall $\operatorname{SLE}_{\kappa}^{\mathrm{sep}}$ from Proposition 2.20,
which is $\operatorname{SLE}_{\kappa}^{\mathrm{loop}}$ restricted to loops
separating $0$ and $\infty$. By the definition of $\mathcal{L}_{\kappa}$, for
some $\gamma$-dependent constant $C^{\prime}\in(0,\infty)$, the pull back of
$\operatorname{SLE}_{\kappa}^{\mathrm{sep}}$ via the map $z\mapsto e^{-z}$ is
$C^{\prime}\operatorname{SLE}_{\kappa}^{\mathrm{sep},\gamma}$. Therefore,
$\mathrm{QS}_{2}\otimes\operatorname{SLE}_{\kappa}^{\mathrm{sep}}=C^{\prime}\mathcal{M}_{2}^{\mathrm{sph}}(\gamma)\otimes\operatorname{SLE}_{\kappa}^{\mathrm{sep,\gamma}}$.
Now Propositions 6.7 and 2.20 yield Lemma 6.8. ∎

Suppose $\eta$ is a simple curve in $\mathcal{C}$ separating $+\infty$ and
$-\infty$ with two marked points $p^{-},p^{+}\in\eta$. Let $D^{\pm}_{\eta}$ be
the connected component of $\mathcal{C}\backslash\eta$ containing $\pm\infty$,
and let $\psi^{\pm}_{\eta}:\mathbbm{H}\to D^{\pm}_{\eta}$ be the conformal
maps sending $(i,0)$ to $(\pm\infty,p^{\pm})$. We need the following variant
of Lemma 5.6.

######  Lemma 6.9.

Fix $\alpha\in(\frac{\gamma}{2},Q)$ and $\varepsilon\in(0,\frac{1}{4})$. Let
$\eta$ be a simple curve in $\mathcal{C}$ separating $\pm\infty$ with two
marked points $p^{-},p^{+}\in\eta$. Let
$\mathcal{C}_{\eta,p^{\pm},\varepsilon}=\mathcal{C}\setminus(\psi^{-}_{\eta}(B_{\varepsilon}(i))\cup\psi^{+}_{\eta}(B_{\varepsilon}(i)))$.
For $\phi$ sampled from $\mathrm{LF}_{\mathcal{C}}^{(\gamma,\pm\infty)}$, let
$X^{+}=\phi\circ\psi^{+}_{\eta}+Q\log|(\psi^{+}_{\eta})^{\prime}|$ and
$X^{-}=\phi\circ\psi^{-}_{\eta}+Q\log|(\psi^{-}_{\eta})^{\prime}|$. Then for
any nonnegative measurable function $f$ of $\phi$ that depends only on
$\phi|_{\mathcal{C}_{\eta,p^{\pm},\varepsilon}}$, we have

$\displaystyle\int
f(\phi)\times\varepsilon^{\alpha^{2}-\gamma^{2}}e^{(\alpha-\gamma)(X^{-}_{\varepsilon}(i)+X^{+}_{\varepsilon}(i))}\,d\mathrm{LF}_{\mathcal{C}}^{(\gamma,\pm\infty)}$
$\displaystyle=\int
f(\phi)\mathopen{}\mathclose{{}\left(\frac{1}{4}\mathrm{CR}(\exp(\eta),0)\mathrm{CR}(\exp(-\eta),0)}\right)^{-\frac{\alpha^{2}}{2}+Q\alpha-2}\,d\mathrm{LF}_{\mathcal{C}}^{(\alpha,\pm\infty)}.$

###### Proof.

Let $g:\mathbbm{C}\to\mathbbm{C}$ be given by $g(z)=\frac{z}{z-1}$ and let
$G:\mathcal{C}\to\mathbbm{C}$ be given by $G=g\circ\exp$. By [AHS21, Lemma
2.13], if $\phi$ is sampled from
$\mathrm{LF}_{\mathcal{C}}^{(\gamma,\pm\infty)}$ then $\hat{\phi}:=\phi\circ
G^{-1}+Q\log|(G^{-1})^{\prime}|$ has law
$\mathrm{LF}_{\mathbbm{C}}^{(\gamma,0),(\gamma,-1)}$, and the same is true
when $\gamma$ is replaced by $\alpha$. Let
$(\hat{\eta},\hat{p}^{-},\hat{p}^{+})=(G(\eta),G(p^{-}),G(p^{+}))$. Since
$g^{\prime}(0)=-1$ and $\frac{d}{dz}(g(\frac{1}{z}))|_{z=0}=1$, we see that
$\mathrm{CR}(\exp(\eta),0)=\mathrm{CR}(\hat{\eta},0)$ and
$\mathrm{CR}(\exp(-\eta),0)=\mathrm{CR}(\hat{\eta},-1)$. Let
$\mathbbm{C}_{\hat{\eta},\hat{p}^{\pm},\varepsilon}:=\mathbbm{C}\backslash(G(\psi^{-}(B_{\varepsilon}(i)))\cup
G(\psi^{+}(B_{\varepsilon}(i))))$. Then Lemma 6.9 is equivalent to the
following: for any nonnegative measurable function $\hat{f}$ of $\hat{\phi}$
that depends only on
$\hat{\phi}|_{\mathbbm{C}_{\hat{\eta},\hat{p}^{\pm},\varepsilon}}$, we have

$\displaystyle\int\hat{f}(\hat{\phi})\times\varepsilon^{\alpha^{2}-\gamma^{2}}e^{(\alpha-\gamma)(X^{-}_{\varepsilon}(i)+X^{+}_{\varepsilon}(i))}\,d\mathrm{LF}_{\mathbbm{C}}^{(\gamma,0),(\gamma,-1)}$
$\displaystyle=\int\hat{f}(\hat{\phi})\mathopen{}\mathclose{{}\left(\frac{1}{4}\mathrm{CR}(\hat{\eta},0)\mathrm{CR}(\hat{\eta},-1)}\right)^{-\frac{\alpha^{2}}{2}+Q\alpha-2}d\mathrm{LF}_{\mathbbm{C}}^{(\alpha,0),(\alpha,-1)}.$
(6.8)

Indeed, by the exact same argument, Lemma 5.6 holds with
$\mathrm{LF}_{\mathbbm{C}}^{(\alpha,u_{1}),(\gamma,u_{2}),(\gamma,u_{3})}$
replaced by $\mathrm{LF}_{\mathbbm{C}}^{(\gamma,0),(\gamma,-1)}$, namely

$\displaystyle\int\hat{f}(\hat{\phi})\times\varepsilon^{\frac{1}{2}(\alpha^{2}-\gamma^{2})}e^{(\alpha-\gamma)X^{-}_{\varepsilon}(i)}\,d\mathrm{LF}_{\mathbbm{C}}^{(\gamma,0),(\gamma,-1)}=\int\hat{f}(\hat{\phi})\mathopen{}\mathclose{{}\left(\frac{1}{2}\mathrm{CR}(\hat{\eta},0)}\right)^{-\frac{\alpha^{2}}{2}+Q\alpha-2}d\mathrm{LF}_{\mathbbm{C}}^{(\alpha,0),(\gamma,-1)}.$

Applying the argument of Lemma 5.6 again to change the insertion at $-1$, we
get (6.8). ∎

Similarly as in the proof of Proposition 5.4, for a curve $\eta$ in
$\mathcal{C}$ separating $\pm\infty$, we let $\mathrm{Harm}_{-\infty,\eta}$
(resp. $\mathrm{Harm}_{+\infty,\eta}$) be the harmonic measure on $\eta$
viewed from $-\infty$ (resp., $+\infty$).

######  Lemma 6.10.

There is a $\gamma$-dependent constant $C\in(0,\infty)$ such that the
following holds. Suppose $\alpha\in(\frac{\gamma}{2},Q)$. Sample
$(\phi,\eta,p^{-},p^{+})$ from the measure

$C\cdot\mathrm{LF}_{\mathcal{C}}^{(\alpha,\pm\infty)}(d\phi)\,\mathcal{L}_{\kappa}^{\alpha}(d\eta)\,\mathrm{Harm}_{-\infty,\eta}(dp^{-})\,\mathrm{Harm}_{+\infty,\eta}(dp^{+}).$

Let $X^{+}=\phi\circ\psi^{+}_{\eta}+Q\log|(\psi^{+}_{\eta})^{\prime}|$ and
$X^{-}=\phi\circ\psi^{-}_{\eta}+Q\log|(\psi^{-}_{\eta})^{\prime}|$. Let $\tau$
be the quantum length of the clockwise arc from $p^{-}$ to $p^{+}$ in
$D^{+}_{\eta}$. Then the law of $(X^{-},X^{+},\tau)$ is

$\int_{0}^{\infty}\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}(\ell)\times\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}(\ell)\times[1_{\tau\in(0,\ell)}d\tau]\,d\ell.$

###### Proof.

We first prove the case $\alpha=\gamma$, namely there is a $\gamma$-dependent
constant $C$ such that

$C\cdot\mathrm{LF}_{\mathcal{C}}^{(\gamma,\pm\infty)}(d\phi)\,\mathcal{L}^{\gamma}_{\kappa}(d\eta)\,\mathrm{Harm}_{-\infty,\eta}(dp^{-})\,\mathrm{Harm}_{+\infty,\eta}(dp^{+})=\int_{0}^{\infty}\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}(\ell)\times\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}(\ell)\times[1_{\tau\in(0,\ell)}d\tau]\,d\ell,$
(6.9)

where we identify the left hand side as the measure on
$H^{-1}(\mathbbm{H})\times H^{-1}(\mathbbm{H})\times[0,\infty)$ describing the
law of $(X^{-},X^{+},\tau)$. Indeed, (6.9) is an immediate consequence of
Lemma 6.8 and 5.7.

Now, let $\varepsilon\in(0,\frac{1}{4})$ and
$\mathbbm{H}_{\varepsilon}:=\mathbbm{H}\backslash B_{\varepsilon}(i)$. Let $g$
be a nonnegative measurable function of
$(X^{-}|_{\mathbbm{H}_{\varepsilon}},X^{+}|_{\mathbbm{H}_{\varepsilon}},\tau)$.
Similarly as in the proof of Proposition 5.4, reweighting (6.9) gives

$\displaystyle C\int
g(X^{-}|_{\mathbbm{H}_{\varepsilon}},X^{+}|_{\mathbbm{H}_{\varepsilon}},\tau)\varepsilon^{\alpha^{2}-\gamma^{2}}e^{(\alpha-\gamma)(X^{-}_{\varepsilon}(i)+X^{+}_{\varepsilon}(i))}\mathrm{LF}_{\mathcal{C}}^{(\gamma,\pm\infty)}(d\phi)\mathcal{L}^{\gamma}_{\kappa}(d\eta)\mathrm{Harm}_{-\infty,\eta}(dp^{+})\mathrm{Harm}_{-\infty,\eta}(dp^{+})$
$\displaystyle=\int_{0}^{\infty}\mathopen{}\mathclose{{}\left(\int
g(X^{-}|_{\mathbbm{H}_{\varepsilon}},X^{+}|_{\mathbbm{H}_{\varepsilon}},\tau)\varepsilon^{\alpha^{2}-\gamma^{2}}e^{(\alpha-\gamma)(X^{-}_{\varepsilon}(i)+X^{+}_{\varepsilon}(i))}\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}(\ell)\times\mathrm{LF}_{\mathbbm{H}}^{(\gamma,i)}(\ell)\times[1_{\tau\in(0,\ell)}d\tau]}\right)\,d\ell.$
(6.10)

By Lemma 6.9 and the reweighting definition of $\mathcal{L}_{\kappa}^{\alpha}$
in (6.1), the left hand side of (6.10) equals

$C\int
g(X^{-}|_{\mathbbm{H}_{\varepsilon}},X^{+}|_{\mathbbm{H}_{\varepsilon}},\tau)\mathrm{LF}_{\mathcal{C}}^{(\alpha,\pm\infty)}(d\phi)\mathcal{L}_{\kappa}^{\alpha}(d\eta)\mathrm{Harm}_{-\infty,\eta}(dp^{-})\mathrm{Harm}_{+\infty,\eta}(dp^{+}).$

By Lemma 5.5, the right hand side equals

$\int_{0}^{\infty}\mathopen{}\mathclose{{}\left(\int
g(X^{-}|_{\mathbbm{H}_{\varepsilon}},X^{+}|_{\mathbbm{H}_{\varepsilon}},\tau)\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}(\ell)\times\mathrm{LF}_{\mathbbm{H}}^{(\alpha,i)}(\ell)\times[1_{\tau\in(0,\ell)}d\tau]}\right)\,d\ell.$

Since the above two expressions agree for every $\varepsilon$ and $g$, we
obtain the result. ∎

###### Proof of Proposition 6.2.

By Proposition 6.7, the law of
$(\mathcal{C},\phi,\eta,\pm\infty)/{\sim_{\gamma}}$ from Lemma 6.10 equals
$C(Q-\alpha)^{2}\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)\otimes\operatorname{SLE}_{\kappa}^{\mathrm{sep},\alpha}$
. On the other hand, Lemma 6.10 implies that the joint law of
$((\mathbbm{H},X^{-},i)/{\sim_{\gamma}},(\mathbbm{H},X^{+},i)/{\sim_{\gamma}},\tau)$
is
$\int_{0}^{\infty}\ell\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell)\times\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell)\times[1_{\tau\in(0,\ell)}\ell^{-1}d\tau]\,d\ell$.
The uniform distribution $[1_{\tau\in(0,\ell)}\ell^{-1}d\tau]$ means that the
two surfaces are uniformly welded. Therefore the law of
$(\mathcal{C},\phi,\eta,\pm\infty)/{\sim_{\gamma}}$ is
$\int_{0}^{\infty}\ell\mathrm{Weld}(\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell),\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\ell))\,d\ell$
as desired. ∎

###  6.2 The appearance of $\overline{R}(\alpha)$ and
$|\mathcal{L}_{\kappa}^{\alpha}|$: proof of Proposition 6.4

We will prove Proposition 6.4 via a particular embedding of
$\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)$. Let $h$ be the field
$\hat{h}+\mathbf{c}$ in Definition 6.1 so that the law of
$(\mathcal{C},h,-\infty,+\infty)/{\sim_{\gamma}}$ is
$\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)$. Now we restrict to the event
$\\{\mu_{h}(\mathcal{C})>1\\}$ and set $\phi:=h(\cdot-a)$, where
$a\in\mathbbm{R}$ is such that $\mu_{h}((-\infty,a)\times[0,2\pi])=1$. Namely,
we shift $h$ horizontally such that
$\mu_{\phi}(\\{z\in\mathcal{C}:\operatorname{Re}z\leq 0\\})=1$. Let $M$ be the
law of $\phi$ under this restriction. Then we can represent
$\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)\otimes\operatorname{SLE}_{\kappa}^{\mathrm{sep},\alpha}[E_{\delta,\varepsilon}]$
in Proposition 6.4 as follows.

######  Lemma 6.11.

Given a simple closed curve $\eta$ on $\mathcal{C}$ separating $\pm\infty$,
let $D_{\eta}^{+}$ (resp. $D_{\eta}^{-}$) be the connected component of
$\mathcal{C}\backslash\eta$ containing $+\infty$ (resp. $-\infty$). For
$t\in\mathbbm{R}$, let $\eta+t$ be the curve on $\mathcal{C}$ obtained by
shifting $\eta$ by $t$. Now sample $(\phi,\mathbf{t},\eta^{0})$ from $M\times
dt\times\mathcal{L}_{\kappa}^{\alpha}$ and set $\eta=\eta^{0}+\mathbf{t}$. Let

$E^{\prime}_{\delta,\varepsilon}:=\\{(\phi,\eta):\varepsilon<\ell_{\phi}(\eta)<\delta\\}\text{
and }\mu_{\phi}(D_{\eta}^{-})>1\\}$ (6.11)

where $\ell_{\phi}(\eta)$ is the quantum length of $\eta$. Then
$\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)\otimes\operatorname{SLE}_{\kappa}^{\mathrm{sep},\alpha}[E_{\delta,\varepsilon}]=(M\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[E^{\prime}_{\delta,\varepsilon}].$

###### Proof.

Since the measure $\operatorname{SLE}_{\kappa}^{\mathrm{sep},\alpha}$ is
invariant under translations along the cylinder, the law of
$(\mathcal{C},\phi,\eta,+\infty,-\infty)/{\sim_{\gamma}}$ is the restriction
of
$\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)\otimes\operatorname{SLE}_{\kappa}^{\mathrm{sep},\alpha}$
to the event that the total quantum area is larger than $1$. Now Lemma 6.11
follows from the definition of $E_{\delta,\varepsilon}$. ∎

The next lemma explains how the reflection coefficient $\overline{R}(\alpha)$
shows up in Proposition 6.4.

######  Lemma 6.12.

The law of the quantum area of a sample from
$\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)$ is
$1_{a>0}\frac{1}{2}\overline{R}(\alpha)a^{\frac{2}{\gamma}(\alpha-Q)-1}\,da.$
As a corollary,
$|M|=\int_{1}^{\infty}\frac{1}{2}\overline{R}(\alpha)a^{\frac{2}{\gamma}(\alpha-Q)-1}\,da=\frac{\gamma\overline{R}(\alpha)}{4(Q-\alpha)}$.

###### Proof.

For $0<a<a^{\prime}$, with $\widehat{h}$ as in Definition 6.1, we have

$\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)[\mu_{\widehat{h}+c}(\mathcal{C})\in(a,a^{\prime})]=\mathbbm{E}\mathopen{}\mathclose{{}\left[\int_{-\infty}^{\infty}\mathbf{1}_{e^{\gamma
c}\mu_{\widehat{h}}(\mathcal{C})\in(a,a^{\prime})}\frac{\gamma}{2}e^{2(\alpha-Q)c}\,dc}\right]=\mathbbm{E}\mathopen{}\mathclose{{}\left[\int_{a}^{a^{\prime}}\frac{\gamma}{2}\mathopen{}\mathclose{{}\left(\frac{y}{\mu_{\widehat{h}}(\mathcal{C})}}\right)^{\frac{2}{\gamma}(\alpha-Q)}\frac{1}{\gamma
y}\,dy}\right]$

where we have used the change of variables $y=e^{\gamma
c}\mu_{\widehat{h}}(\mathcal{C})$. By [KRV17, Theorem 3.5] and [RV19,
(1.10)–(1.12)], for $\alpha\in(\frac{\gamma}{2},Q)$ we have
$\mathbbm{E}[\mu_{\widehat{h}}(\mathcal{C})^{\frac{2}{\gamma}(Q-\alpha)}]=\overline{R}(\alpha)$.
Interchanging the expectation and integral gives the result. ∎

By Lemmas 6.11 and 6.12, Proposition 6.4 is reduced to the following
proposition.

######  Proposition 6.13.

Let $\alpha\in(\frac{\gamma}{2},Q)$. With an error term
$o_{\delta,\varepsilon}(1)$ satisfying $\lim_{\delta\to
0}\limsup_{\varepsilon\to 0}|o_{\delta,\varepsilon}(1)|=0$, we have

$(M\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[E^{\prime}_{\delta,\varepsilon}]=(1+o_{\delta,\varepsilon}(1))|M||\mathcal{L}_{\kappa}^{\alpha}|\frac{2\log\varepsilon^{-1}}{\gamma(Q-\alpha)}.$
(6.12)

###### Proof of Proposition 6.4 given Proposition 6.13.

By Lemma 6.11 and 6.12, we have

$(\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)\otimes\operatorname{SLE}_{\kappa}^{\mathrm{sep},\alpha})[E_{\delta,\varepsilon}]=(1+o_{\delta,\varepsilon}(1))|M||\mathcal{L}_{\kappa}^{\alpha}|\frac{2\log\varepsilon^{-1}}{\gamma(Q-\alpha)}=(1+o_{\delta,\varepsilon}(1))\frac{\overline{R}(\alpha)}{2(Q-\alpha)^{2}}|\mathcal{L}_{\kappa}^{\alpha}|\log\varepsilon^{-1}.\qed$

Figure 5:  Illustration of the proof of Proposition 6.13. On the left
$(\phi,\mathbf{t},\eta^{0})$ is sampled from $M\times
dt\times\mathcal{L}^{\alpha}_{\kappa}$ and $\eta=\eta^{0}+\mathbf{t}$. The
curve $\eta^{0}$ satisfies $\sup_{z\in\eta_{0}}\operatorname{Re}z=0$ and the
field $\phi$ satisfies $\mu_{\phi}(\\{z\in\mathcal{C}:\operatorname{Re}z\leq
0\\})=1$. The domain $D^{-}_{\eta}$ is colored grey. The event
$E^{\prime}_{\delta,\varepsilon}$ is
$\\{\ell_{\phi}(\eta)\in(\varepsilon,\delta)\textrm{ and
}\mu_{\phi}(D_{\eta}^{-})>1\\}$. If $E^{\prime}_{\delta,\varepsilon}$ occurs,
we have $\ell_{\phi}(\eta)\approx e^{\frac{\gamma}{2}X_{\bf t}}$. Therefore
$\mathbf{t}$ is in an interval close to the red one on the right figure, which
is approxiately
$(\frac{2}{\gamma(Q-\alpha)}\log\delta^{-1},\frac{2}{\gamma(Q-\alpha)}\log\varepsilon^{-1})$.
The proofs of Lemmas 6.17 and 6.19 are done by quantifying this statement in
two directions.

The high level idea for proving Proposition 6.13 is the following. Suppose
$(\phi,\mathbf{t},\eta^{0})$ is sampled from $M\times
dt\times\mathcal{L}_{\kappa}^{\alpha}$. For $s\geq 0$, let $X_{s}$ be the
average of $\phi$ on $[s,s+2\pi i]/{\sim}$. For most realizations of
$(\phi,\eta^{0})$, the occurrence of $E^{\prime}_{\delta,\varepsilon}$ is
equivalent to the event that $\mathbf{t}$ lies in some interval of length
$(1+o_{\delta,\varepsilon}(1))\frac{2\log\varepsilon^{-1}}{\gamma(Q-\alpha)}$
determined by $X$. (See Figure 5.) Hence the mass of
$E^{\prime}_{\delta,\varepsilon}$ is this length times
$|M||\mathcal{L}_{\kappa}^{\alpha}|$. In the rest of this section, we first
prove a few properties for $X$ in Section 6.2.1 and then prove Proposition
6.13 in Section 6.2.2.

####  6.2.1 The field average process

We need the following description of the law of $\phi|_{\mathcal{C}_{+}}$ and
$(X_{s})_{s\geq 0}$, where
$\mathcal{C}_{+}:=\\{z\in\mathcal{C}\>:\>\operatorname{Re}z>0\\}$ is the right
half cylinder.

######  Lemma 6.14.

Let $\phi$ be a sample from $M$. Conditioned on $\phi|_{\mathcal{C}_{-}}$, the
conditional law of $\phi|_{\mathcal{C}_{+}}$ is the law of
$\phi_{0}+\hat{\mathfrak{h}}-(Q-\alpha)\operatorname{Re}(\cdot)$, where
$\phi_{0}$ is a zero boundary GFF on $\mathcal{C}_{+}$, and
$\hat{\mathfrak{h}}$ is a harmonic function determined by
$\phi|_{\mathcal{C}_{-}}$ whose average on the circle $[s,s+2\pi i]/{\sim}$
does not depend on $s$. Moreover, let $X_{s}$ be the average of $\phi$ on
$[s,s+2\pi i]/{\sim}$. Conditioned on $X_{0}$, the conditional law of
$(X_{s}-X_{0})_{s\geq 0}$ is the law of $(B_{s}-(Q-\alpha)s)_{s\geq 0}$ where
$B_{s}$ is a Brownian motion.

###### Proof.

The first statement is the sphere analog of [AG21, Lemmas 2.10 and 2.11] based
on the domain Markov property of Gaussian free field. The proof is identical
so we omit it. The second statement on $(X_{s})_{s\geq 0}$ follows from the
first statement. ∎

Proposition 6.13 essentially follows from the fact that the field average
process $(X_{s})_{s\geq 0}$ looks like a line of slope $-(Q-\alpha)$. We now
introduce two random times to quantify this. For $y>0$, let

$\sigma_{y}=\inf\\{s>0\>:\>X_{s}<\frac{2}{\gamma}\log
y\\}\quad\textrm{and}\quad\tau_{y}=\sup\\{s>0\>:\>X_{s}>\frac{2}{\gamma}\log
y\\}.$ (6.13)

######  Lemma 6.15.

$M$-a.e. the field $\phi$ satisfies $\lim_{y\to 0}\frac{\sigma_{y}}{\log
y^{-1}}=\lim_{y\to 0}\frac{\tau_{y}}{\log y^{-1}}=\frac{2}{\gamma(Q-\alpha)}$.

###### Proof.

Given Lemma 6.14, this is a straightforward fact about drifted Brownian
motion. ∎

The proof of the upper bound in Proposition 6.13 will rely on the following
lemma.

######  Lemma 6.16.

Sample $(\phi,\mathbf{t},\eta^{0})$ from $M\times
dt\times\mathcal{L}_{\kappa}^{\alpha}$ and set $\eta=\eta^{0}+\mathbf{t}$. Let
$\mathfrak{l}=\ell_{\phi}(\eta)$, which is the boundary length of the quantum
surface $(D_{\eta}^{-},\phi)/{\sim_{\gamma}}$. Fix $\zeta\in(0,1)$. Then there
exists a constant $C>0$ and a function $\mathrm{err}(\ell)$ such that
$\lim_{\ell\downarrow 0}\mathrm{err}(\ell)=0$, and conditioned on
$(D_{\eta}^{-},\phi)/{\sim_{\gamma}}$, the conditional probability of
$\\{\mathbf{t}\in(\sigma_{{\mathfrak{l}}^{1-\zeta}}-C,\tau_{{\mathfrak{l}}^{1+\zeta}})\\}$
is at least $1-\mathrm{err}(\mathfrak{l})$.

###### Proof.

We first introduce $C$ and $\mathrm{err},$ and then show that they satisfy
Lemma 6.16. Recall that $\eta^{0}$ is a loop in $\mathcal{C}$ separating
$\pm\infty$ and with $\sup_{z\in\eta^{0}}\operatorname{Re}z=0$. Let
$D_{\eta^{0}}^{+}$ be the connected component of
$\mathcal{C}\backslash\eta^{0}$ containing $+\infty$. Let
$f:\mathcal{C}_{+}\to D_{\eta^{0}}^{+}$ be the unique conformal map such that
$f(+\infty)=+\infty$ and $\lim_{z\to+\infty}(f(z)-z)\in\mathbbm{R}$. By
standard conformal distortion estimates (e.g. [MS19, Lemma 2.4]), for
$C=e^{10}$ we have

$|f(z)-z|<\frac{C}{3}\quad\textrm{and}\quad|f^{\prime\prime}(z)|<\frac{1}{2}<|f^{\prime}(z)|<2\quad\textrm{for
}\operatorname{Re}z>\frac{C}{3}.$ (6.14)

Let $g$ be a fixed smooth function on $\mathcal{C}$ supported on
$\\{\operatorname{Re}z\in(C-1,C)\\}$ such that $\int g(z)\,dz=1$ and $g$ is
invariant under rotations against the axis of $\mathcal{C}$. Let $S$ be the
collection of smooth functions $\xi$ that are supported on
$\\{\operatorname{Re}z\in[\frac{2}{3}C-1,\frac{4}{3}C]\\}\subset\mathcal{C}_{+}$
and satisfy $\|\xi\|_{\infty}\leq 4\|g\|_{\infty}$ and
$\|\nabla\xi\|_{\infty}\leq 8(\|g\|_{\infty}+\|\nabla g\|_{\infty})$. Let

$\mathrm{err}(\ell):=\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;1)^{\\#}\Big{[}\sup_{\xi\in
S}|(h,\xi)|+Q\log 2\geq-\frac{2\zeta}{\gamma}\log\ell\Big{]}$ (6.15)

where $(\mathcal{C}_{+},h,+\infty)$ is an embedding of a sample from
$\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;1)^{\\#}$, the probablity measure
proportional to $\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;1)$. Since the space
$S$ is invariant under rotations against the axis of $\mathcal{C}$, the
probability in (6.15) does not depend on the choice of the embedding.
Moreover, since $h$ is a.s. in the Sobolev space of index $-1$, we see that
$\sup_{\xi\in S}|(h,\xi)|<\infty$ a.s. hence $\lim_{\ell\to
0}\mathrm{err}(\ell)=0$.

We now show that $C$ and $\mathrm{err}$ satisfy Lemma 6.16. Set
$\phi^{0}(\cdot)=\phi(\cdot+\mathbf{t})$ and $\hat{\phi}=\phi^{0}\circ
f+Q\log|f^{\prime}|-\frac{2}{\gamma}\log\mathfrak{l}$. Then
$(D_{\eta}^{+},\phi,+\infty)/{\sim_{\gamma}}=(D_{\eta_{0}}^{+},\phi^{0},+\infty)/{\sim_{\gamma}}$
hence
$(\mathcal{C}_{+},\hat{\phi},+\infty)/{\sim_{\gamma}}=(D_{\eta}^{+},\phi-\frac{2}{\gamma}\log\mathfrak{l},+\infty)/{\sim_{\gamma}}$.
Moreover,

$(\phi^{0},g)-\frac{2}{\gamma}\log\mathfrak{l}=(\hat{\phi}\circ
f^{-1}+Q\log|(f^{-1})^{\prime}|,g)=(\hat{\phi},|f^{\prime}|^{2}g\circ
f)+(Q\log|(f^{-1})^{\prime}|,g).$ (6.16)

By (6.14) and the definition of $S$, we have $|f^{\prime}|^{2}g\circ f\in S$.
Then by (6.14) and (6.16), we have

$\big{|}(\phi^{0},g)-\frac{2}{\gamma}\log\mathfrak{l}\big{|}\leq|(\hat{\phi},|f^{\prime}|^{2}g\circ
f)|+|(Q\log|(f^{-1})^{\prime}|,g)|\leq\sup_{\xi\in S}|(\hat{\phi},\xi)|+Q\log
2.$

Recall from the proof of Lemma 6.11 that the law of
$(\mathcal{C},\phi,\eta,+\infty,-\infty)/{\sim_{\gamma}}$ is the restriction
of
$\mathcal{M}_{2}^{\mathrm{sph}}(\alpha)\otimes\operatorname{SLE}_{\kappa}^{\mathrm{sep},\alpha}$
to the event that the total quantum area is larger than $1$. By Proposition
6.2, conditioning on $(D_{\eta}^{-},\phi)/{\sim_{\gamma}}$, the conditional
law of $(D_{\eta}^{+},\phi,+\infty)/{\sim_{\gamma}}$ is
$\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;\mathfrak{l})^{\\#}$, hence the
conditional law of $(\mathcal{C}_{+},\hat{\phi},+\infty)/{\sim_{\gamma}}$ is
$\mathcal{M}_{1}^{\mathrm{disk}}(\alpha;1)^{\\#}$. Therefore, by (6.15) and
(6.16), conditioning on $(D_{\eta}^{-},\phi)/{\sim_{\gamma}}$, the conditional
probability of the event
$\mathopen{}\mathclose{{}\left|(\phi^{0},g)-\frac{2}{\gamma}\log\mathfrak{l}}\right|<-\frac{2\zeta}{\gamma}\log\mathfrak{l}$
is at least $1-\mathrm{err}(\mathfrak{l})$. Since $g$ is supported on
$\\{\operatorname{Re}z\in(C-1,C)\\}$, if
$\mathopen{}\mathclose{{}\left|(\phi^{0},g)-\frac{2}{\gamma}\log\mathfrak{l}}\right|<-\frac{2\zeta}{\gamma}\log\mathfrak{l}$,
then there exists $s\in[C-1,C]$ such that the average of $\phi^{0}$ on
$[s,s+2\pi i]$ lies in
$((1+\zeta)\frac{2}{\gamma}\log\mathfrak{l},(1-\zeta)\frac{2}{\gamma}\log\mathfrak{l})$.
This gives
$X_{\mathbf{t}+s}\in((1+\zeta)\frac{2}{\gamma}\log\mathfrak{l},(1-\zeta)\frac{2}{\gamma}\log\mathfrak{l})$
hence
$\mathbf{t}+s\in(\sigma_{{\mathfrak{l}}^{1-\zeta}},\tau_{{\mathfrak{l}}^{1+\zeta}})$
for some $s\in[C-1,C]$. Therefore
$\mathbf{t}\in(\sigma_{{\mathfrak{l}}^{1-\zeta}}-C,\tau_{{\mathfrak{l}}^{1+\zeta}})$,
which gives Lemma 6.16 with our choice of $C$ and $\mathrm{err}$. ∎

####  6.2.2 Proof of Proposition 6.13

We refer to Figure 5 and the paragraph above Section 6.2.1 for an illustration
of our proof ideas. We will write $o_{\delta,\varepsilon}(1)$ as an error term
satisfying $\lim_{\delta\to 0}\limsup_{\varepsilon\to
0}|o_{\delta,\varepsilon}(1)|=0$ which can change from place to place. We
first prove the lower bound for $(M\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[E^{\prime}_{\delta,\varepsilon}]$.

######  Lemma 6.17.

We have $(M\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[E^{\prime}_{\delta,\varepsilon}]\geq(1+o_{\delta,\varepsilon}(1))|M||\mathcal{L}_{\kappa}^{\alpha}|\frac{2\log\varepsilon^{-1}}{\gamma(Q-\alpha)}.$

We will prove Lemma 6.17 by a coupling of the probability measure
$M^{\\#}=M/|M|$ and cylindrical GFF measure $P_{\mathcal{C}}$. Recall that for
a sample $h$ from $P_{\mathcal{C}}$, it can be written as $h=h_{1}+h_{2}$,
where $h_{1}$ is constant on vertical circles and $h_{2}$ is the lateral
component that has mean zero on all such circles.

######  Lemma 6.18.

There exists a coupling of $h$ sampled from $P_{\mathcal{C}}$ and $\phi$
sampled from $M^{\\#}$ such that $h_{2}$ is independent of $(X_{s})_{s\geq
0}$, and moreover, $\sup_{\operatorname{Re}z>1}|{\mathfrak{h}}(z)|<\infty$
where $\mathfrak{h}(z)=\phi(z)-h_{2}(z)-X_{\operatorname{Re}z}$.

###### Proof.

By [MS17, Proposition 2.8], on $\mathcal{C}_{+}$ the field $h$ can be written
as $h_{0}+\tilde{\mathfrak{h}}$ where $h_{0}$ is a zero boundary GFF on
$\mathcal{C}_{+}$ and $\tilde{\mathfrak{h}}$ is an independent harmonic
function on $\mathcal{C}_{+}$ such that
$\sup_{\operatorname{Re}z>1}|\tilde{\mathfrak{h}}(z)|<\infty$. Similarly, by
Lemma 6.14, $\phi(z)+(Q-\alpha)\operatorname{Re}z$ on $\mathcal{C}_{+}$ can be
written as $\phi_{0}+\hat{\mathfrak{h}}$ with the same properties. Coupling
$h$ and $\phi$ such that $h_{0}=\phi_{0}$ and $\tilde{\mathfrak{h}}$ is
independent of $\phi$, we are done. ∎

###### Proof of Lemma 6.17.

Let $\mathbbm{P}$ be the probability measure corresponding to the law of
$(\phi,h)$ as coupled in Lemma 6.18. Let $(\phi,h,\mathbf{t},\eta^{0})$ be a
sample from $\mathbbm{P}\times dt\times\mathcal{L}_{\kappa}^{\alpha}$. Then
the law of $(\phi,\mathbf{t},\eta^{0})$ is $M^{\\#}\times
dt\times\mathcal{L}_{\kappa}^{\alpha}$, where $M^{\\#}=M/|M|$. Let $h_{2}$ and
$\mathfrak{h}$ be defined as in Lemma 6.18 so that
$\phi=h_{2}+X_{\operatorname{Re}\cdot}+\mathfrak{h}$. Fix $\zeta\in(0,0.1)$
which will be sent to zero later. Let $I_{\delta,\varepsilon}$ be the interval
$((1+3\zeta)\frac{2}{\gamma(Q-\alpha)}\log\delta^{-1}+\delta^{-1},(1-3\zeta)\frac{2}{\gamma(Q-\alpha)}\log\varepsilon^{-1})$.
Let

$G_{\delta,\varepsilon}:=\\{I_{\delta,\varepsilon}\subset(\tau_{\delta^{1+2\zeta}}+\delta^{-1},\sigma_{\varepsilon^{1-2\zeta}})\quad\textrm{and}\quad\sup_{\operatorname{Re}z>1}|{\mathfrak{h}}(z)|<\zeta\cdot\frac{2}{\gamma}\log\delta^{-1}\\},$

where the random times $\sigma_{y},\tau_{y}$ are as in (6.13). By Lemma 6.15
we have
$\mathbbm{P}[I_{\delta,\varepsilon}\subset(\tau_{\delta^{1+2\zeta}}+\delta^{-1},\sigma_{\varepsilon^{1-2\zeta}})]=1-o_{\delta,\varepsilon}(1)$.
Since $\sup_{\operatorname{Re}z>1}|{\mathfrak{h}}(z)|<\infty$ almost surely,
we have $\mathbbm{P}[G_{\delta,\varepsilon}]=1-o_{\delta,\varepsilon}(1)$.

Fix $n>0$. Let $A_{n}=\\{\inf_{z\in\eta^{0}}\operatorname{Re}z>-n\\}$. Then
$\mathcal{L}^{\alpha}_{\kappa}[A_{n}]<\infty$. Let $\eta=\eta^{0}+\mathbf{t}$
and $\ell_{h_{2}}(\eta)$ be the quantum length of $\eta$ with respect to
$h_{2}$. Define the event

$E^{\prime}_{\delta,\varepsilon}(n):=\\{A_{n}\textrm{
occurs},\quad\mathbf{t}\in
I_{\delta,\varepsilon},\quad\textrm{and}\quad\ell_{h_{2}}(\eta)\in(\varepsilon^{\zeta},\delta^{-\zeta})\\}.$

Suppose $n<\delta^{-1}<\frac{2\zeta}{\gamma(Q-\alpha)}\log\varepsilon^{-1}$
and the event $E^{\prime}_{\delta,\varepsilon}(n)\cap G_{\delta,\varepsilon}$
occurs. Since $\mathbf{t}\in I_{\delta,\varepsilon}$ and
$G_{\delta,\varepsilon}\cap A_{n}$ occurs, we have
$\operatorname{Re}z\in(\tau_{\delta^{1+2\zeta}},\sigma_{\varepsilon^{1-2\zeta}})$
for each $z\in\eta$, hence
$\mathcal{C}_{-}:=\\{z\in\mathcal{C}:\operatorname{Re}z\leq 0\\}\subset
D^{-}_{\eta}$. By the definition of $\phi$ we have
$\mu_{\phi}(D_{\eta}^{-})\geq\mu_{\phi}(\mathcal{C}_{-})\geq 1$. Moreover,
from the bound on $\sup_{\operatorname{Re}z>1}|{\mathfrak{h}}(z)|$ we have
$\frac{\gamma}{2}(X_{\operatorname{Re}z}+\mathfrak{h}(z))\in((1-\zeta)\log\varepsilon,(1+\zeta)\log\delta)$
for each $z\in\eta$. Now
$\ell_{h_{2}}(\eta)\in(\varepsilon^{\zeta},\delta^{-\zeta})$ yields
$\ell_{\phi}(\eta)\in(\varepsilon,\delta)$. This gives
$E^{\prime}_{\delta,\varepsilon}(n)\cap G_{\delta,\varepsilon}\subset
E^{\prime}_{\delta,\varepsilon}=\\{\mu_{\phi}(D_{\eta}^{-})\geq
1,\ell_{\phi}(\eta)\in(\varepsilon,\delta)\\}$. Therefore

$(\mathbbm{P}\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[E^{\prime}_{\delta,\varepsilon}]\geq(\mathbbm{P}\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[E^{\prime}_{\delta,\varepsilon}(n)]-(\mathbbm{P}\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[E^{\prime}_{\delta,\varepsilon}(n)\setminus
G_{\delta,\varepsilon}].$ (6.17)

We claim that

$(\mathbbm{P}\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[E^{\prime}_{\delta,\varepsilon}(n)]=(1-o_{\delta,\varepsilon}(1))|I_{\delta,\varepsilon}|\mathcal{L}^{\alpha}_{\kappa}[A_{n}]\textrm{
and }(\mathbbm{P}\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[E^{\prime}_{\delta,\varepsilon}(n)\setminus
G_{\delta,\varepsilon}]=o_{\delta,\varepsilon}(1)|I_{\delta,\varepsilon}|.$
(6.18)

Since the law of $h_{2}$ is translation invariant, namely
$h_{2}\stackrel{{\scriptstyle d}}{{=}}h_{2}(\cdot-t)$ for each
$t\in\mathbbm{R}$, we have

$\displaystyle(\mathbbm{P}\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[E^{\prime}_{\delta,\varepsilon}(n)]=|I_{\delta,\varepsilon}|(\mathbbm{P}\times\mathcal{L}^{\alpha}_{\kappa})[\ell_{h_{2}}(\eta^{0})\in(\varepsilon^{\zeta},\delta^{-\zeta}),A_{n}]$
$\displaystyle=|I_{\delta,\varepsilon}|(\mathbbm{P}\times\mathcal{L}^{\alpha}_{\kappa})[A_{n}]-|I_{\delta,\varepsilon}|(\mathbbm{P}\times\mathcal{L}^{\alpha}_{\kappa})[\ell_{h_{2}}(\eta^{0})\notin(\varepsilon^{\zeta},\delta^{-\zeta}),A_{n}]=(1-o_{\delta,\varepsilon}(1))|I_{\delta,\varepsilon}|\mathcal{L}^{\alpha}_{\kappa}[A_{n}].$

In the last line we used
$(\mathbbm{P}\times\mathcal{L}^{\alpha}_{\kappa})[A_{n}]=\mathcal{L}^{\alpha}_{\kappa}[A_{n}]<\infty$
and
$(\mathbbm{P}\times\mathcal{L}^{\alpha}_{\kappa})[\ell_{h_{2}}(\eta^{0})\notin(\varepsilon^{\zeta},\delta^{-\zeta})|A_{n}]=o_{\delta,\varepsilon}(1)$,
the latter of which holds because $\ell_{h_{2}}(\eta^{0})<\infty$ a.e. in
$\mathbbm{P}\times\mathcal{L}^{\alpha}_{\kappa}$. This gives the first
identity in (6.18).

The second identity in (6.18) is proved by as follows:

$\displaystyle(\mathbbm{P}\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[E^{\prime}_{\delta,\varepsilon}(n)\setminus
G_{\delta,\varepsilon}]\leq(\mathbbm{P}\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[A_{n}\setminus
G_{\delta,\varepsilon}\textrm{ and }\mathbf{t}\in I_{\delta,\varepsilon}]$
$\displaystyle=\mathbbm{P}[G_{\delta,\varepsilon}\textrm{ does not
occur}]\times|I_{\delta,\varepsilon}|\times\mathcal{L}^{\alpha}_{\kappa}[A_{n}]=o_{\delta,\varepsilon}(1)|I_{\delta,\varepsilon}|\mathcal{L}^{\alpha}_{\kappa}[A_{n}].$

By (6.17) and (6.18), for a fixed $n$, we have $(\mathbbm{P}\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[E^{\prime}_{\delta,\varepsilon}]\geq(1-o_{\delta,\varepsilon}(1))|I_{\delta,\varepsilon}||\mathcal{L}^{\alpha}_{\kappa}[A_{n}]$
hence

$\frac{1}{|M|}(M\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[E^{\prime}_{\delta,\varepsilon}]\geq(1-o_{\delta,\varepsilon}(1))|I_{\delta,\varepsilon}|\mathcal{L}^{\alpha}_{\kappa}[A_{n}]\geq(1-o_{\delta,\varepsilon}(1))\mathcal{L}^{\alpha}_{\kappa}[A_{n}](1-4\zeta)\frac{2\log\varepsilon^{-1}}{\gamma(Q-\alpha)}.$

Sending $\varepsilon\to 0$, $\delta\to 0$, and $\zeta\to 0$ in order, we see
that

$\lim_{\delta\to 0}\liminf_{\varepsilon\to 0}\frac{(M\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[E^{\prime}_{\delta,\varepsilon}]\times\gamma(Q-\alpha)}{2\log\varepsilon^{-1}}\geq|M|\mathcal{L}^{\alpha}_{\kappa}[A_{n}].$

Since
$\mathcal{L}^{\alpha}_{\kappa}[A_{n}]\rightarrow|\mathcal{L}^{\alpha}_{\kappa}|$,
sending $n\to\infty$ we conclude the proof. ∎

It remains to prove the upper bound for $(M\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[E^{\prime}_{\delta,\varepsilon}]$.

######  Lemma 6.19.

We have $(M\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[E^{\prime}_{\delta,\varepsilon}]\leq(1+o_{\delta,\varepsilon}(1))|M||\mathcal{L}_{\kappa}^{\alpha}|\frac{2\log\varepsilon^{-1}}{\gamma(Q-\alpha)}.$

###### Proof.

Recall $\sigma$ and $\tau$ defined in (6.13). For $C>0$, consider the event

$F_{x,y,C}=\\{(\phi,\mathbf{t},\eta^{0})\>:\>\mathbf{t}\in(\sigma_{x}-C,\tau_{y})\\}.$
(6.19)

We will prove Lemma 6.19 by first bounding $(M\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[F_{x,y,C}]$ and then comparing
$E^{\prime}_{\delta,\varepsilon}$ to
$F_{\delta^{1-\zeta},\varepsilon^{1+\zeta},C}$. More precisely, we will prove
two estimates. First, for each $x>0$, we have

$(M\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[F_{x,y,C}]\leq(1+o_{y}(1))|M||\mathcal{L}_{\kappa}^{\alpha}|(\frac{2\log
y^{-1}}{\gamma(Q-\alpha)}+C)\quad\textrm{as }y\to 0,$ (6.20)

where $o_{y}(1)\to 0$ uniformly in $C$. Moreover, for a fixed $\zeta\in(0,1)$
there exists some $C>0$ such that

$(M\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[F_{\delta^{1-\zeta},\varepsilon^{1+\zeta},C}\mid
E^{\prime}_{\delta,\varepsilon}]=1-o_{\delta,\varepsilon}(1).$ (6.21)

Given (6.20) and (6.21), we see that for any fixed $\zeta>0$ there exists some
$C>0$ such that

$(M\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[E^{\prime}_{\delta,\varepsilon}]\leq(1+o_{\delta,\varepsilon}(1))(M\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[F_{\delta^{1-\zeta},\varepsilon^{1+\zeta},C}]\leq(1+o_{\delta,\varepsilon}(1))|M||\mathcal{L}_{\kappa}^{\alpha}|\frac{(1+\zeta)2\log\varepsilon^{-1}}{\gamma(Q-\alpha)},$

Sending $\zeta\to 0$ we will get Lemma 6.19.

We first prove (6.20). Note that $(M\times
dt\times\mathcal{L}_{\kappa}^{\alpha})[F_{x,y,C}]=M^{\\#}[\tau_{y}-\sigma_{x}+C]|M||\mathcal{L}^{\alpha}_{\kappa}|$
from the definition of $F_{x,y,C}$. Hence (6.20) is equivalent to
$M^{\\#}[\tau_{y}-\sigma_{x}]\leq(1+o_{y}(1))\frac{2\log
y^{-1}}{\gamma(Q-\alpha)}$. By Lemma 6.14, the process $X_{s}$ evolves as
Brownian motion with drift $-(Q-\alpha)$. Let $T=\inf\\{s\geq
0:B^{x}_{s}-(Q-\alpha)s=\frac{2}{\gamma}\log y\\}$ where $(B^{x}_{s})_{s\geq
0}$ is a Brownian motion starting from $\frac{2}{\gamma}\log x$. Then
$M^{\\#}[\tau_{y}-\sigma_{x}]\leq\mathbbm{E}[T]$. Since
$\mathbbm{E}[T]=(1+o_{y}(1))\frac{2\log y^{-1}}{\gamma(Q-\alpha)}$, we get
(6.20).

Since the event $E^{\prime}_{\delta,\varepsilon}$ is determined by
$(D_{\eta}^{-},\phi)/{\sim_{\gamma}}$, and
$\ell_{\phi}(\eta)\in(\varepsilon,\delta)$ on
$E^{\prime}_{\delta,\varepsilon}$, Lemma 6.16 yields (6.21) with the constant
$C$ from Lemma 6.16. This concludes the proof. ∎

##  Appendix A Continuity of $\operatorname{CLE}_{\kappa}$ as $\kappa\uparrow
4$

In this appendix we supply the continuity in $\kappa$ needed to extend
Theorems 1.2 and 1.3 from $\kappa<4$ to $\kappa=4$. We start with a
monotonicity statement for $\operatorname{CLE}_{\kappa}$ proved in [SW12].

######  Lemma A.1 ([SW12]).

There exists a coupling of $\operatorname{CLE}_{\kappa}$ on the unit disk
$\mathbbm{D}$ for $\kappa\in(\frac{8}{3},4]$ such that a.s. each outermost
loop of $\operatorname{CLE}_{\kappa_{1}}$ is surrounded by an outermost loop
of $\operatorname{CLE}_{\kappa_{2}}$ if $\kappa_{1}<\kappa_{2}\leq 4$.

###### Proof.

By [SW12, Theorems 1.5, 1.6], the law of the boundaries of outermost loop
clusters in a Brownian loop soup with intensity
$c_{\kappa}=(3\kappa-8)(6-\kappa)/2\kappa$ is given by the outermost loops of
$\operatorname{CLE}_{\kappa}$. Now the monotonicity of $c_{\kappa}$ in
$\kappa\in(\frac{8}{3},4]$ yields the desired monotonicity in Lemma A.1. ∎

We recall the formula from [SSW09] for the conformal radii of CLE.

######  Theorem A.2 ([SSW09]).

For $\kappa\in(8/3,8)$, let $\eta_{\kappa}$ be the outermost loop surrounding
$0$ of a $\operatorname{CLE}_{\kappa}$ on $\mathbbm{D}$. Let
$\mathrm{CR}(\eta_{\kappa},0)$ be the conformal radius of the region
surrounded by $\eta_{\kappa}$ viewed from 0. Then

$\mathbbm{E}[\mathrm{CR}(\eta_{\kappa},0)^{\lambda}]=\frac{-\cos(\frac{4\pi}{\kappa})}{\cos\mathopen{}\mathclose{{}\left(\pi\sqrt{(1-\frac{4}{\kappa})^{2}-\frac{8\lambda}{\kappa}}}\right)}\quad\textrm{for
}\lambda>\frac{3\kappa}{32}+\frac{2}{\kappa}-1.$

Recall that the Hausdorff distance between two closed sets $E_{1},E_{2}$ on a
metric space $(X,d)$ is given by $\max\\{\sup_{x\in
E_{1}}\\{d(x,E_{2})\\},\sup_{x\in E_{2}}\\{d(x,E_{1})\\}\\}$. Lemma A.1 and
Theorem A.2 yield the following continuity result.

######  Lemma A.3.

Suppose we are in the coupling in Lemma A.1. For each $z\in\mathbbm{D}$, let
$\eta_{\kappa}(z)$ be the outermost loop around $z$ of the
$\operatorname{CLE}_{\kappa}$. For any fixed $z$, viewed as closed sets,
$\eta_{\kappa}(z)$ converges almost surely to $\eta_{4}(z)$ in the Hausdorff
metric as $\kappa\uparrow 4$.

###### Proof.

By the conformal invariance of $\operatorname{CLE}_{\kappa}$ we assume $z=0$
because the same argument will work for a general $z$. In this case we simply
write $\eta_{\kappa}(0)$ as $\eta_{\kappa}$. Since $\eta_{\kappa_{1}}$ is
surrounded by $\eta_{\kappa_{2}}$ if $\kappa_{1}<\kappa_{2}\leq 4$. we see
that $\mathrm{CR}(\eta_{\kappa},0)$ is increasing in $\kappa$. By the explicit
formula in Theorem A.2, we have $\lim_{\kappa\uparrow
4}\mathbbm{E}[\mathrm{CR}(\eta_{\kappa},0)]=\mathbbm{E}[\mathrm{CR}(\eta_{\kappa},0)]$.
Thus we must have $\lim_{\kappa\uparrow
4}\mathrm{CR}(\eta_{\kappa},0)=\mathrm{CR}(\eta_{4},0)$ a.s.

Let $D_{\kappa}$ be the region surrounded by $\eta_{\kappa}$ and
$D_{4^{-}}=\cup_{\kappa<4}D_{\kappa}$. The conformal radius of $D_{4^{-}}$
must be between $\lim_{\kappa\uparrow 4}\mathrm{CR}(\eta_{\kappa},0)$ and
$\mathrm{CR}(\eta_{4},0)$, hence equals $\mathrm{CR}(\eta_{4},0)$ a.s. This
means that $D_{\kappa}\uparrow D_{4}$ a.s. hence $\eta_{\kappa}\to\eta_{4}$
a.s. in the Hausdorff metric in the coupling in Lemma A.1. ∎

We now prove the continuity of the shape measure $\mathcal{L}_{\kappa}$ of the
SLE loop in Theorem 1.3.

######  Lemma A.4.

We have $\lim_{\kappa\uparrow 4}\mathcal{L}_{\kappa}=\mathcal{L}_{4}$ weakly
with respect to the Hausdorff metric.

###### Proof.

We only explain that under the natural parameterization, chordal
$\operatorname{SLE}_{\kappa}$ converges to $\operatorname{SLE}_{4}$ as
$\kappa\uparrow 4$. Once this is done we get that the two-sided whole plane
$\operatorname{SLE}_{\kappa}$ curve
$\operatorname{SLE}_{\kappa}^{p\rightleftharpoons q}$ converges to
$\operatorname{SLE}_{4}^{p\rightleftharpoons q}$ under natural parametrization
as well, since the two-sided curve is characterized by its re-sampling
property: conditioned on one of the curve segments between $p$ and $q$, the
conditional law of the other is a chordal $\operatorname{SLE}_{\kappa}$ in the
complementary domain. From this and the definition of
$\operatorname{SLE}_{\kappa}^{\mathrm{loop}}$ we reviewed in Section 2.4, we
get the convergence of $\mathcal{L}_{\kappa}$.

We now show that the law of chordal $\operatorname{SLE}_{\kappa}$ on
$\mathbbm{H}$ from $0$ to $\infty$ under natural parametrization is continuous
as $\kappa\uparrow 4$. We first recall that this family of measures is tight
in the local uniform topology of parametrized curves thanks to their Hölder
regularity established by Zhan [Zha19]. On the other hand the natural
parametrization of $\operatorname{SLE}_{\kappa}$ is characterized by a
conformal invariance and domain Markov property considered by Lawler and
Sheffield [LS11]. Therefore all subsequential limits of the chordal
$\operatorname{SLE}_{\kappa}$ measure agree with $\operatorname{SLE}_{4}$. ∎

######  Lemma A.5.

Suppose Theorem 1.2 holds for $\kappa\in(8/3,4)$. Then it holds for $\kappa=4$
as well.

###### Proof.

The right hand side of (1.3) is continuous as $\kappa\uparrow 4$. Therefore
$\\{\mathrm{CR}(z_{i},\eta_{i})\\}_{1\leq i\leq 3}$ for $\kappa<4$ converges
in law as $\kappa\uparrow 4$ and the moment generating function of the limit
is given by
${C^{\operatorname{CLE}}_{\kappa}(\lambda_{1},\lambda_{2},\lambda_{3})}$ with
$\kappa=4$. It remains to show that the limit is given by
$\\{\mathrm{CR}(z_{i},\eta_{i})\\}_{1\leq i\leq 3}$ with $\kappa=4$.

Fix a small $\varepsilon>0$. Let $S$ be the set of simple loops in
$\widehat{\mathbbm{C}}$ separating $z_{1}$ from $\\{z_{2},z_{3}\\}$. Let
$S_{\varepsilon}=\\{\eta\in S:\mathrm{CR}(z_{1},\eta)>\varepsilon\\}$. For
$\eta\in S$, let $D_{\eta}$ be the connected component of
$\widehat{\mathbbm{C}}\backslash\eta$ containing $z_{2},z_{3}$. For
$\kappa\in(8/3,4]$, the law of $(\eta_{1},\Gamma)$ restricted to the event
$\eta_{1}\in S_{\varepsilon}$ is the same as that of $(\eta,\Gamma)$ under
$1_{\eta=\eta_{1}}\mathrm{Count}_{S_{\varepsilon}\cap\Gamma}(d\eta)\operatorname{CLE}_{\kappa}(d\Gamma)$,
where $\mathrm{Count}_{S_{\varepsilon}\cap\Gamma}(d\eta)$ is the counting
measure on $S_{\varepsilon}\cap\Gamma$. Given a sample of $(\eta,\Gamma)$ of
$\mathrm{Count}_{S_{\varepsilon}\cap\Gamma}(d\eta)\operatorname{CLE}_{\kappa}(d\Gamma)$,
let $\Gamma_{+}$ be the subset of $\Gamma$ in $D_{\eta}$ and $F_{\Gamma_{+}}$
be the event that no loop in $\Gamma_{+}$ surrounds both $z_{2}$ and $z_{3}$.
Then the event $\eta=\eta_{1}$ is the same as $F_{\Gamma_{+}}$. By Proposition
3.8, for $\kappa\in(8/3,4]$, the law of $\eta$ under
$1_{\eta=\eta_{1}}\mathrm{Count}_{S_{\varepsilon}\cap\Gamma}(d\eta)\operatorname{CLE}_{\kappa}(d\Gamma)$
is $\operatorname{CLE}_{\kappa}^{D_{\eta}}[F_{\Gamma_{+}}]1_{\eta\in
S_{\varepsilon}}\cdot\operatorname{SLE}_{\kappa}^{\mathrm{loop}}(d\eta)$,
where $\operatorname{CLE}_{\kappa}^{D_{\eta}}$ is the law of a
$\operatorname{CLE}_{\kappa}$ in $D_{\eta}$.

By Lemmas A.4 and A.3, as $\kappa\uparrow 4$, the measure
$\operatorname{CLE}_{\kappa}^{D_{\eta}}[F_{\Gamma_{+}}]1_{\eta\in
S_{\varepsilon}}\cdot\operatorname{SLE}_{\kappa}^{\mathrm{loop}}(d\eta)$ on
loops converges weakly with respect to the Hausdorff metric to
$\operatorname{CLE}_{4}^{D_{\eta}}[F_{\Gamma_{+}}]1_{\eta\in
S_{\varepsilon}}\cdot\operatorname{SLE}_{4}^{\mathrm{loop}}(d\eta)$. Therefore
the law of $\eta_{1}$ conditioned on the event $\eta_{1}\in S_{\varepsilon}$
as $\kappa\uparrow 4$ converges weakly to the same law with $\kappa=4$. Since
$\lim_{\varepsilon\to 0}\mathbbm{P}[\Gamma\cap
S_{\varepsilon}\neq\emptyset]=\lim_{\varepsilon\to
0}\mathbbm{P}[\mathrm{CR}(z_{1},\eta_{1})>\varepsilon]=1$ uniformly for
$\kappa\in(\kappa_{0},4]$ for a fixed $\kappa_{0}\in(8/3,4)$, we can remove
the restriction $\eta_{1}\in S_{\varepsilon}$ and conclude that as
$\kappa\uparrow 4$ the law of $\eta_{1}$ converges weakly to the same law when
$\kappa=4$.

The conditional law of $(\eta_{2},\eta_{3})$ given $\eta_{1}$ is the law of
the two outermost loops surrounding $z_{2},z_{3}$ under the CLE measure
$\operatorname{CLE}_{\kappa}^{D_{\eta_{1}}}$ conditioned on these two loops
being distinct. By Lemma A.3, the joint law of $(\eta_{1},\eta_{2},\eta_{3})$
has the desired continuity as $\kappa\uparrow 4$. ∎

##  Appendix B Identities for the modified Bessel function of the second kind

###### Proof of Equation (2.7).

We will prove the $c=1$ case; the general case then follows from a change of
variables. Recall the integral definition of $K_{\nu}(x)$ in (2.6). Using
$1+\cosh t=2(\cosh\frac{t}{2})^{2}$ and the fact that $\cosh$ is even, we can
express $\int_{0}^{\infty}\frac{1}{\sqrt{x}}e^{-x}K_{\nu}(x)\,dx$ as

$\iint_{0}^{\infty}\frac{1}{\sqrt{x}}e^{-x(1+\cosh t)}\cosh(\nu
t)\,dx\,dt=\sqrt{\pi}\int_{0}^{\infty}\frac{\cosh(\nu t)}{\sqrt{1+\cosh
t}}\,dt=\frac{1}{2}\sqrt{\pi/2}\int_{-\infty}^{\infty}\frac{\cosh(\nu
t)}{\cosh(t/2)}\,dt.$

Since $\int\frac{e^{\nu t}}{\cosh(t/2)}\,dt=\int\frac{e^{-\nu
t}}{\cosh(t/2)}\,dt$, we have $\int\frac{\cosh(\nu
t)}{\cosh(t/2)}\,dt=\int\frac{e^{\nu t}}{\cosh(t/2)}\,dt$. For a random
variable $X$ following the _hyperbolic secant distribution_
$\frac{dx}{\pi\cosh x}$, it is known that
$\mathbbm{E}[e^{tX}]=\frac{1}{\cos(\pi t/2)}$ for $|t|<1$; see e.g. [Fis14,
Section 1.3]. Therefore $\int_{-\infty}^{\infty}\frac{e^{\nu
t}}{\cosh(t/2)}\,dt=\frac{2\pi}{\cos(\pi\nu)}.$ This concludes the proof. ∎

###### Proof of Equation (2.8).

For $r>0$, using the change of coordinates $s=\sqrt{b}$, we get

$\displaystyle\int_{0}^{\infty}\frac{1}{\sqrt{b}(1+b)(1+b+br)}\,db=\int_{0}^{\infty}\frac{1}{s(1+s^{2})(1+s^{2}(1+r))}2s\,ds=\int_{-\infty}^{\infty}\frac{1}{(1+s^{2})(1+s^{2}(1+r))}\,ds$
$\displaystyle=\frac{1}{r}\int_{-\infty}^{\infty}\frac{1+r}{1+s^{2}(1+r)}-\frac{1}{1+s^{2}}\,ds=\pi\frac{\sqrt{1+r}-1}{r}=\frac{\pi}{\sqrt{1+r}+1}.$

Therefore, using (2.6) and the identity $1+\cosh t=2(\cosh\frac{t}{2})^{2}$,
the integral we must evaluate is

$\displaystyle\iiint_{0}^{\infty}\frac{1}{\sqrt{b}(b+1)}e^{-x(b+1+b\cosh
t)}\cosh(\nu
t)\,dt\,db\,dx=\iint_{0}^{\infty}\frac{1}{\sqrt{b}(1+b)(1+b+b\cosh
t)}\cosh(\nu t)\,db\,dt$ $\displaystyle=\pi\int_{0}^{\infty}\frac{\cosh(\nu
t)}{\sqrt{1+\cosh
t}+1}\,dt=\frac{\pi}{2}\int_{-\infty}^{\infty}\frac{\cosh(\nu
t)}{\sqrt{2}\cosh\frac{t}{2}+1}\,dt=\frac{\pi}{2}\int_{-\infty}^{\infty}\frac{e^{\nu
t}}{\sqrt{2}\cosh\frac{t}{2}+1}\,dt.$

For a random variable $X$ following the _Perk’s distribution_ whose density is
$\frac{\sin\lambda}{2\lambda}\frac{dx}{\cosh x+\cos(\lambda)}$ with
$\lambda\in[0,\pi)$, it is known that
$\mathbbm{E}[e^{itX}]=\frac{\pi\sinh(\lambda t)}{\lambda\sinh(\pi t)}$ for
$t\in\mathbbm{R}$; see e.g. [Fis14, Section 2.1]. Taking
$\lambda=\frac{\pi}{4}$, we have $\int_{-\infty}^{\infty}\frac{e^{\nu
t}}{\sqrt{2}\cosh\frac{t}{2}+1}\,dt=\frac{\pi}{\cos(\pi\nu)\cos(\frac{1}{2}\pi\nu)}.$
This concludes the proof. ∎

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