which shows that $z\notin \partial {\cluster{}{t}}$. Hence, we have shown that a.s., it holds for each $t>0$ that $\mu_h$-a.e. $z\in \partial {\cluster{}{t}}$ does not belong to $\partial{\cluster{}{t}}$, which means that $\mu_h(\partial{\cluster{}{t}}) = 0$.

§ UNIQUENESS OF HARMONIC BALLS

In this section we show that there is only one family of harmonic balls satisfying the conditions given by Theorem <ref>.

§.§ Uniqueness of subharmonic balls

The uniqueness of subharmonic balls (defined in (<ref>)) for the Lebesgue measure is well known, see, , <cit.>,
and the proof extends verbatim to $\gamma$-LQG subharmonic balls.
The idea is that every subharmonic ball generates a supersolution to the obstacle problem.

Let $A$ be a domain strictly contained in $B_R$ and let

\begin{equation} \label{eq:obstacle-comp}
f(x) = \mu_h(A) G_{B_R}(x,0) - \int_{A} G_{B_R}(x,y)  \mu_h(d y).
\end{equation}

* If $A$ is a subharmonic ball, then $f \geq 0$ on $B_R$.

* If $A$ is a harmonic ball, then $f = 0$ on $B_R \backslash A$.

For all $x \in B_R$, the function $g(z) := G_{B_R}(z, x)$
is superharmonic in $A$, so $-g$ is subharmonic in $A$. Then, since $A$ is a subharmonic ball,

\begin{equation}
\int_{A} g(z) \mu_h(dz) \leq \mu_h(A) g(0),
\end{equation}

and so

\begin{equation}
\int_{A} G_{B_R}(z, x) \mu_h(dz) \leq \mu_h(A) G_{B_R}(x,0) ,\quad \forall x \in B_R .
\end{equation}

In this case, the function $g(z) = G_{B_R}(z, x)$ is harmonic in $A$ for all $x \in B_R \backslash A$.
Hence, since $A$ is a harmonic ball,

\begin{equation}
\int_{A} G_{B_R}(z, x) \mu_h(dz) = \mu_h(A) G_{B_R}(x,0) ,\quad \forall x \in B_R \backslash A .
\end{equation}

The prior lemma implies uniqueness (up to sets of $\mu_h$-measure zero) of subharmonic balls.

The following holds on an event of probability 1. Let $A \Subset B_R$ be a subharmonic ball with $\mu_h(A)=t$ for some $t, R > 0$. Then, ${\regcluster{}{t}} \subset A$ and $\mu_h(A \backslash {\regcluster{}{t}}) = 0$.

Let $f$ be given by (<ref>) and note that by Lemma <ref> (and since every subharmonic ball is a harmonic ball),
$f  \geq 0$ in $B_R$ and $f = 0$ on $B_R \backslash A$.  Since $f \geq 0$ and $\Delta f \leq -\mu_h(A) \delta_0 + \mu_h$ on $B_R$, we have that $(f(\cdot)- t G_{B_R}(0,\cdot)) \in \mclS{B_R}{t}$ and hence $f \geq \odometer{B_R}{t}$. This implies, together with $f = 0$ on $B_R \backslash A$, that  $\cluster{B_R}{t} \subset A$.
Since $t = \mu_h(A) = \mu_h({\regcluster{}{t}})$, this completes the proof.

We do not use Proposition <ref> in this paper but decided to include it explicitly as it may be used in future work.

§.§ Comparing harmonic and subharmonic balls

We show that any harmonic ball for $\mu_h$ (in the sense of (<ref>)) must have a boundary contained in the closure of
some ${\regcluster{}{t}}$. Similar arguments have appeared in <cit.>,
<cit.>, and <cit.>.

Almost surely, every harmonic ball  $A \Subset B_R$ with $t = \mu_h(A) > 0$  satisfies $\partial A \subset \overline{{\regcluster{}{t}}}$.

Let $A$ be a harmonic ball with $t = \mu_h(A) > 0$ and $A \Subset B_R$. Take $R$ possibly larger so that
${\regcluster{}{t}} \Subset B_R$ and hence by Theorem <ref>,  ${\regcluster{}{t}}$ is a subharmonic ball in $B_R$.
We show that $\partial A \subset \overline{{\regcluster{}{t}}}$ by considering the auxiliary function

\begin{equation} \label{eq:auxiallary-function}
u(x) := \int_{A} G_{B_R}(x,y)  \mu_h(d y) - \int_{{\regcluster{}{t}}} G_{B_R}(x,y)  \mu_h(d y)
\end{equation}

which satisfies $\Delta u = (1_{{\regcluster{}{t}}} - 1_{A}) \mu_h$ on $B_R$.

Step 1: $u \geq 0$ in $B_R$.

By Lemma <ref>, 	as $A$ is a harmonic ball,

\begin{equation} \label{eq:harmonic-ball-equality}
t G_{B_R}(x,0) = \int_{A} G_{B_R}(x,y)  \mu_h(d y) \quad \mbox{on $B_R \backslash A$}.
\end{equation}

and as ${\regcluster{}{t}}$ is a subharmonic ball,

\begin{equation} \label{subharmonic-ball-inequality}
t G_{B_R}(x,0) \geq \int_{{\regcluster{}{t}}} G_{B_R}(x,y)  \mu_h(d y) \quad \mbox{on $B_R$}.
\end{equation}

Combining (<ref>) and (<ref>) shows

\begin{equation} \label{eq:intermediate-harmonic-ball-inequality}
u \geq 0 \quad \mbox{on $B_R \backslash A$}.
\end{equation}

As $u$ is superharmonic on $A$, (<ref>) together with the minimum
principle shows $u \geq 0$ on $B_R$.

Step 2: $u = 0$ on $B_R \cap {{\regcluster{}{t}}}^c \cap A^c$.

As ${\regcluster{}{t}}$ is also a harmonic ball, by Lemma <ref>,

\begin{equation} \label{eq:subharmonic-ball-equality}
t G_{B_R}(x,0) = \int_{{\regcluster{}{t}}} G_{B_R}(x,y)  \mu_h(d y) \quad \mbox{on $B_R \backslash {\regcluster{}{t}}$}.
\end{equation}

Step 2 follows by combining (<ref>) and (<ref>)
with the definition of $u$.

Step 3: Conclude.

We use  the fact $\Delta u = (1_{{\regcluster{}{t}}} - 1_{A}) \mu_h$ on $B_R$.
Suppose for sake of contradiction there is $x_0 \in \partial A \backslash \overline{{\regcluster{}{t}}}$.
Then, $u$ is superharmonic in a neighborhood of $x_0$ as $x_0 \in (\overline{{\regcluster{}{t}}})^c$.
By Step 2, $u(x_0) = 0$, which, together with Step 1 and the strong minimum principle, shows that $u$ is identically 0 in a neighborhood of $x_0$.
This in turn implies that $u$ is harmonic in a neighborhood of $x_0$.

However, as $x_0 \in \overline{{\regcluster{}{t}}}^c$, we have that $\Delta u = -1_{A} \mu_h$ in a neighborhood of $x_0$. As $\mu_h$ assigns positive mass to every open set, $A$ is open, and $x_0 \in \partial A$, $\Delta u$ is strictly negative on an open subset of every neighborhood of $x_0$, which supplies the desired contradiction.

§.§ Strong uniqueness

We first show that two regular open sets
which coincide $\mu_h$-a.e. are in fact equal.
This fact is the reason for our assumption that $\mathrm{int}(\overline{\Lambda}_t) = \Lambda_t$ in Theorem <ref>.

Almost surely, the following is true. Let $X$ and $Y$ be two open subsets of $\C$ such that  $\mathrm{int}(\overline{X}) = X$ and $\mathrm{int}(\overline{Y}) = Y$.
If $\mu_h(X \backslash Y) = \mu_h(Y \backslash X) = 0$, then $X = Y$.

We use the fact that a.s. $\mu_h$ assigns positive mass to every open set to show that $X \backslash Y = \emptyset$. A symmetric argument shows $Y \backslash X = \emptyset$.

Suppose for the sake of contradiction there is $x_0 \in X \cap Y^c$. As $X$ is open, $B_r(x_0) \subset X$ for all $r < r_0$, where $r_0$ is some small radius.
As $\mu_h(X \cap Y^c) = 0$, $B_r(x_0) \not \subset Y^c$ for all $r < r_0$. Thus,
$x_0$ is a limit point of $Y$ and so $x_0 \in \partial Y$ by definition.
Since $\mathrm{int}(\overline{Y}) = Y$, $x_0$ cannot be in the interior of $\overline Y$ so $x_0$ must be an accumulation point of $\overline Y^c$.
Since $\overline Y^c$ is open, this implies that every neighborhood of $x_0$ contains an open subset of $\overline Y^c$. Hence $X\cap \overline Y^c \subset X\cap Y^c$ contains a non-empty open set.
This implies $B_r(x_0) \cap Y^c$ contains a non-empty open set, and hence that $\mu_h(X \cap Y^c) > 0$, a contradiction.

We now use Lemma <ref> to show uniqueness of the family of harmonic balls satisfying the conditions of Theorem <ref>.
Our proof is inspired by the proof of Theorem 10.13 in <cit.>.

Let $t_0 > 0$ be given and suppose $A_{t_0} \Subset B_R$ for some $R > 0$.
By Lemma <ref> it suffices to show that $\mu_h(A_{t_0} \backslash \regcluster{}{t_0})  = \mu_h( \regcluster{}{t_0} \setminus A_{t_0}) = 0$. As we have assumed $\mu_h(A_{t_0}) = t_0$, we only need to show that $\mu_h(A_{t_0} \backslash \regcluster{}{t_0}) = 0$.

Suppose for the sake of contradiction that $\mu_h(A_{t_0} \backslash \regcluster{}{t_0}) > 0$.
Since $\mu_h(\partial \regcluster{}{t_0}) = 0$, this implies $\mu_h(A_{t_0} \backslash \overline{\regcluster{}{t_0}})  >0$.
In particular, as $A_{t_0} \backslash \overline{\regcluster{}{t_0}}$ is open, there exists a non-empty ball $B \subset A_{t_0} \backslash \overline{\regcluster{}{t_0}}$
which lies at positive distance from the origin. We will show that $B$ cannot exist by monotonicity

\begin{equation} \label{eq:cluster-monotonicity}
A_a \subseteq  A_b  \quad \text{and} \quad \regcluster{}{a} \subseteq \regcluster{}{b} , \quad \forall a \leq b
\end{equation}

and Lemma <ref>,

\begin{equation} \label{eq:boundary-contained}
\partial A_{t} \subset \overline{{\regcluster{}{t}}}, \quad \forall t > 0.
\end{equation}

As the family $\{A_t\}_{t > 0}$ continuously decreases to the origin as $t \to 0$,
there exists some $0 < s_0 < t_0$ for which

\begin{equation} \label{eq:ball-contained-in-initial-complement}
B \subset A_{s_0}^c.
\end{equation}

Now let, $s_0 < s_1 < \cdots < s_m = t_0$ be a sequence of points satisfying,

\begin{equation}
s_{n} < s_{n+1} \leq s_n + \mu_h(B)/2  \quad \forall n \in \{0,\dots,m-1\}.
\end{equation}

We show by induction on $n$ that for each $0 \leq n \leq m$,

\begin{equation} \label{eq:b-contained-in-next-cluster}
B \subset A_{s_n}^c,
\end{equation}

which contradicts $B \subset A_{t_0}$. The base case $n = 0$ is established by (<ref>).

Now assume (<ref>) holds for $n \in \{0,\dots,m-1\}$.
We show that for

\begin{equation} \label{eq:time-interval}
s_{n+1} \in \left[ s_n ,  \min(s_n + \mu_h(B)/2, t_0) \right]
\end{equation}

we must have

\begin{equation} \label{eq:b-contained-in-next-cluster-induct}
B \subset A_{s_{n+1}}^c.
\end{equation}

First note that the interval (<ref>) is non-empty as $B \subset A_{t_0}$, $B \subset A_{s_n}^c$,
and $\mu_h(A_t) = t$ for all $t > 0$.
Moreover, as $B \subset  (\overline{\regcluster{}{t_0}})^c$ and $t_0 \geq s_{n+1}$, by monotonicity (<ref>), $B \subset  (\overline{\regcluster{}{s_{n+1}}})^c$.
This together with (<ref>) implies  $B \cap \partial A_{s_{n+1}} = \emptyset$; equivalently

\begin{equation} \label{eq:two-cases}
B \subset A_{s_{n+1}} \quad \mbox{or} \quad B \subset \overline{A_{s_{n+1}}}^c.
\end{equation}

The former case in (<ref>) is impossible as

\begin{align*}
\mu_h(A_{s_{n+1}} \cap B) &\leq \mu_h(A_{s_{n} + \mu_h(B)/2} \cap B)  \quad \mbox{($s_{n+1} \leq s_{n} + \mu_h(B)/2$)} \\
&= \mu_h(A_{s_{n}} \cap B) + \mu_h((A_{s_{n} + \mu_h(B)/2} \backslash A_{s_{n}} ) \cap B) \quad \mbox{(additivity of measure)} \\
&\leq \mu_h(A_{s_{n}} \cap B) + \mu_h(B)/2  \quad \mbox{(monotonicity and $\mu_h(A_t) = t, \forall t$)} \\
&= \mu_h(B)/2 \quad \mbox{(by \eqref{eq:b-contained-in-next-cluster})}
\end{align*}

which shows (<ref>).

§ BOUNDARY CURVES OF HARMONIC BALLS

In this section we show that harmonic balls have boundaries which are simple loops.
We first consider clusters ${\cluster{}{t}} \Subset B_1$ and then rescale to achieve the result
for all $\{{\regcluster{}{t}}\}_{t > 0}$.

Our main result is essentially a consequence of the following lemma, which limits
how many times clusters cross annuli — see Figure <ref>.

A situation ruled out by Lemma <ref>. The annulus $\A_{\rho/3, 2 \rho}$ is in light blue and the connected components of ${\cluster{}{t}} \cap \A_{\rho/3, 2 \rho}$
which cross the annulus are in gray.

Almost surely, for each small enough $\epsilon > 0$
it holds for each $z \in B_1 \backslash B_{20 \sqrt{\epsilon}}$ that there is a $\rho \in [\epsilon, \epsilon^{1/2}]$
(depending on $z$) such that the cluster ${\cluster{}{t}}$ does not cross $\A_{\rho/3, 2\rho}(z)$ more than  $K := \lceil 1/\alpha \rceil$ times for all ${\cluster{}{t}} \Subset B_1$
where $\alpha = \alpha(\gamma) \in (0,1)$ is as in Proposition <ref>.
That is, there are at most $K$
connected components of ${\cluster{}{t}} \cap B_{2 \rho}(z)$
whose closures intersect both $\partial B_{\rho/3}(z)$ and $\partial B_{2\rho}(z)$.

Let $t>0$ such that ${\cluster{}{t}} \Subset B_1$. By Proposition <ref> and the Borel-Cantelli lemma, a.s. there exists a random $M_0 \in \N$ and a deterministic $\alpha \in (0,1)$ such that for all $\epsilon \in \{2^{-n}\}_{n \geq M_0}$ we have that for each $x_0 \in (B_{1 +\epsilon} \backslash B_{10 \sqrt{\epsilon}}) \cap \frac{\epsilon}{100} \Z^2$
there is a $\rho \in [\epsilon, \epsilon^{1/2}]$ for which $\overline E_\rho(x_0)$ occurs.

Fix $\epsilon \in \{2^{-n}\}_{n \geq M_0}$ and let $z \in B_1 \backslash B_{20 \sqrt{\epsilon}}$. Let $x_0$ be a point of $(B_{1 + \epsilon} \backslash B_{10 \sqrt{\epsilon}}) \cap \frac{\epsilon}{100} \Z^2$
with $|x_0 - z| < \epsilon/50$ and let $\rho$ be a radius in $[\epsilon, \epsilon^{1/2}]$ such that $\overline E_\rho(x_0)$ occurs, as in the statement of Proposition <ref>.
Note that since $\rho > \epsilon$ and $|x_0-z| < \epsilon/50$,

\begin{equation} \label{eq:annulus-grid-contains}
B_{\rho}(x_0) \subset B_{2 \rho}(z).
\end{equation}

As ${\cluster{}{t}}$ is open and connected (Lemma <ref>), and contains the origin (Proposition <ref>), (<ref>) shows that any non-empty connected component of ${\cluster{}{t}} \cap B_{2 \rho}(z)$
with closure intersecting $\partial B_{\rho/3}(z)$ and $\partial B_{2 \rho}(z)$
must intersect $B_{\rho}(x_0)$ and have closure intersecting $\partial B_{\rho}(x_0)$ and $\partial B_{\rho/2}(x_0)$.

Also note that as $\overline E_\rho(x_0)$ occurs, Proposition <ref> implies that

\begin{equation} \label{eq:harnack-contrapositive}
\begin{aligned}
\overline{\tilde \Lambda_t} \cap \partial B_{\rho/2}(x_0) \neq \emptyset  &\implies \mu_h(\tilde \Lambda_t \cap \A_{\rho/2, \rho}(x_0)) \geq \alpha \mu_h(\A_{\rho/2, \rho}(x_0)) \\
&\quad \forall  \tilde \Lambda_t, \mbox{ connected component of ${\cluster{}{t}} \cap B_{\rho}(x_0)$}.
\end{aligned}
\end{equation}

By (<ref>), every connected component of ${\cluster{}{t}} \cap B_{2 \rho}(z)$
with closure intersecting $\partial B_{\rho/3}(z)$
intersected with $B_{\rho}(x_0)$ decomposes into a union of connected components of
${\cluster{}{t}} \cap B_{\rho}(x_0)$ each of which has closure intersecting $\partial B_{\rho}(x_0)$ and $\partial B_{\rho/2}(x_0)$.
Thus, the relations (<ref>) and (<ref>) together imply that

\begin{equation} \label{eq:harnack-type-contrapositive-consequence}
\begin{aligned}
\mu_h(\hat \Lambda_t \cap \A_{\rho/2, \rho}(x_0)) &\geq \alpha \mu_h(\A_{\rho/2, \rho}(x_0)) \\
&\qquad \forall \hat \Lambda_t, \mbox{ connected component of ${\cluster{}{t}} \cap B_{2 \rho}(z)$} \\
&\quad \mbox{such that $\overline{\hat \Lambda_t} \cap \partial B_{2 \rho}(z) \neq \emptyset$ and $\overline{\hat \Lambda_t} \cap \partial B_{\rho/3}(z) \neq \emptyset$}.
\end{aligned}
\end{equation}

By (<ref>), this is a lower bound on the mass of each connected component of ${\cluster{}{t}} \cap B_{2 \rho}(z)$.
In particular, (<ref>) implies that the number of such connected components with closures which intersect
$\partial B_{\rho/3}(z)$ and $\partial B_{2 \rho}(z)$ is at most $K := \lceil 1/\alpha \rceil$.
This implies ${\cluster{}{t}}$ can cross $\A_{\rho/3, 2 \rho}(z)$ (as in the statement of Lemma <ref>) at most $K$ times.

We recall the definition of a loop.

A set $\Gamma \subset \C$ is a loop if $\Gamma = \{\varphi(\zeta): \zeta \in \mathbb T\}$
for a continuous function $\varphi$ from the unit circle $\mathbb T$ to $\C$. The set $\Gamma$ is an arc if $\Gamma = \{\varphi(\zeta): \alpha \leq \zeta \leq \beta\}$.
$\Gamma$ is a simple loop [arc] if $\Gamma$ is a loop [arc] and $\varphi$ is also injective.

We recall the definition of a locally connected set.

A set $X \subset \C$ is locally connected if every
neighborhood of each $x \in X$ with respect to $X$ contains a connected neighborhood of $x$.

We also recall the definition of cut points.

Let $A$ be a compact, connected, and locally connected set. A point $a \in A$ is a cut point
if $A \backslash \{a\}$ is no longer connected.

A visual aid to the proof of Lemma <ref>. The annulus $A$ is transparent, $U$ is the gray simply connected domain,
$C$ is the connected violet domain, the components of $(C' \backslash C) \cap A$ are in yellow, and $P$ is the path with two boundary points $a$ and $b$; both $a$ and $b$ are in $\Gamma$, the boundary of $U$.

We will use Lemma <ref> together with some basic properties of the cluster to show
that the boundaries of the complementary connected components of the cluster are simple loops. Before doing so,
we prove a topological lemma. Recall from the statement of Lemma <ref> that a set $X$ crosses an annulus $\A_{s_1, s_2}(z)$ if
$\overline{X} \cap \partial B_{s_1}(z) \neq \emptyset$ and $\overline{X} \cap \partial B_{s_2}(z) \neq \emptyset$.

Let $A \Subset B_1$ be an annulus, $U$ a simply connected component of ${\overline{{{\cluster{}{t}}}}}^c$,
and $\Gamma = \partial U \Subset B_1$. Every connected component of $U^c \cap A$ which crosses
$A$ contains a connected component of ${\cluster{}{t}} \cap A$ which crosses $A$.

See Figure <ref>.

Let $C$ be a connected component of $U^c \cap A$ which crosses $A$. The `filling' of $C$, $C'$, is the union of $C$ and the regions which are disconnected
in $A$ from the inner boundary of the annulus, the outer boundary of the annulus, or both by $C$. Note that $C'$ is a topological rectangle
with two boundary segments which are part of the inner and outer boundaries of the annulus respectively, and two boundary segments contained in $\Gamma$.

By, , the Poincaré-Miranda theorem <cit.>, there is either a simple arc in ${\cluster{}{t}} \cap C'$ between the inner and outer boundaries
of the annulus, or there is a simple arc in ${\cluster{}{t}}^c \cap C'$ between the two boundary segments contained in $\Gamma$.
In the former case, ${\cluster{}{t}}$ crosses $A$ and thus there is a connected component of ${\cluster{}{t}} \cap A$ in $C$ which crosses $A$. Hence it suffices to rule out the latter case, which we now do.

Suppose for sake of contradiction there is a simple open arc $P$ in ${\cluster{}{t}}^c \cap C'$ between the two boundary segments contained in $\Gamma$. That is, $P$ is the interior of a simple arc contained in ${\cluster{}{t}}^c \cap C'$ and $\overline{P} = P \cup \{a\} \cup \{b\}$
with $\{a, b\} \subset \Gamma$.

As $\mathrm{int}(U^c)$ is simply connected, by <cit.>,
$\mathrm{int}(U^c) \backslash P$ has exactly two components $G_0$ and $G_1$ and these satisfy

\begin{equation}
\mathrm{int}(U^c) \cap \partial G_0 = \mathrm{int}(U^c) \cap \partial G_1 = P.
\end{equation}

As $\partial G_0$ and $\partial G_1$ contain points in $\Gamma$, the boundary of a simply connected component of
${\cluster{}{t}}^c$, by definition of component, both $G_0$ and $G_1$ must contain points in ${\cluster{}{t}}$.  However, this contradicts the fact ${\cluster{}{t}}$ is connected, Lemma <ref>.

\[
\partial G_0 \cup \partial G_1 \subset \partial U \cup P \subset {\cluster{}{t}}^c
\]

which implies, as $G_0 \cap  {\cluster{}{t}}$ and $G_1 \cap {\cluster{}{t}}$ are open and disjoint,
that they must lie in different connected components of ${\cluster{}{t}}$.

Almost surely, for all $t > 0$ such that ${\cluster{}{t}} \Subset B_1$, each of the connected components of $B_1 \backslash \overline{{{\cluster{}{t}}}}$
has a boundary that is a simple loop.

Suppose ${\cluster{}{t}} \Subset B_1$ and let $U$ be a component of ${\overline{{{\cluster{}{t}}}}}^c$.
Note that $U$ and ${\cluster{}{t}}$ are disjoint and $\partial U \subseteq \partial {\cluster{}{t}} \Subset B_1$.

Let $\Gamma = \partial U \Subset B_1$.
By Caratheodory's theorem  <cit.> to show $\Gamma$ is a Jordan loop
it suffices to show that $\Gamma$ is locally connected and has no cut points.

Step 1: Locally connected.

This follows from Lemma <ref> and the definition of locally connected:

Suppose for the sake of contradiction that $\Gamma$ is not locally connected.
By <cit.>, this implies $U^c$ is not locally connected.
By definition, this implies the existence
of a point $z \in U^c$ and $s > 0$ so that for every sub-neighborhood $V \subset B_{s}(z)$ containing
$z$, the set $V \cap U^c$ is not connected.

As $U^c$ is connected, the closure of every component of $U^c \cap B_s(z)$ has non-empty intersection with $\partial B_s(z)$.
Since $U^c$ is closed, every such component not containing $z$ must lie at positive distance from $z$.
Hence, for each $\epsilon \in (0,s)$ the number of such components intersecting $B_\epsilon(z)$ must be infinite, as
otherwise we could take $V$ to be $B_{\epsilon}(z)$ minus the other components which do not contain $z$ which intersect $B_{\epsilon}(z)$.

This implies that for all $\rho > 0$ sufficiently small there is a ball $B_{2\rho}(z)$ with infinitely many distinct components of $U^c \cap B_{2 \rho}(x_0)$ with closures intersecting $\partial B_{2\rho}(z)$ and $\partial B_{\rho/3}(z)$.
Every connected component of $U^c \cap B_{2 \rho}(z)$ intersected with $\A_{\rho/3, 2 \rho}(z)$ decomposes into a union of connected components
of $U^c \cap \A_{\rho/3, 2 \rho}(z)$. Hence, by Lemma <ref> applied to each connected
component of $U^c \cap \A_{\rho/3, 2 \rho}(z)$, every connected component of $U^c \cap B_{2 \rho}(z)$ which crosses
$\A_{\rho/3, 2 \rho}(z)$ contains a connected component of ${\cluster{}{t}} \cap \A_{\rho/3, 2 \rho}(z)$ which crosses $\A_{\rho/3, 2 \rho}(z)$.

The previous paragraph implies there is a ball $B_{2\rho}(z)$ with infinitely many distinct components of ${\cluster{}{t}} \cap B_{2 \rho}(x_0)$ with closures intersecting $\partial B_{2\rho}(z)$ and $\partial B_{\rho/3}(z)$, contradicting Lemma <ref>.

Step 2: No cut points.

Let $\psi : B_1 \to U$ be a conformal map, which exists since $U$ is connected with connected complement, so is simply connected. Since $\Gamma$ is locally connected, <cit.> implies that $\psi$ extends to a continuous map $\overline B_1 \to \overline U = U \cup \Gamma$.

Now, assume by way of contradiction that $\Gamma$ has a cut point $a \in \Gamma$. By <cit.>, $\# \psi^{-1}(a) \geq 2$ (in principle $\#\psi^{-1}(a) $ could be infinite, even uncountable).
Furthermore, if $\mathcal I$ is the set of connected components of $\partial B_1 \setminus \phi^{-1}(a)$, then the set of connected components of $\Gamma \setminus \{a\}$ is $\{\psi(I) : I\in \mathcal I\}$.

Fix some $I\in\mathcal I$ and let $J$ be equal to $\partial B_1 \setminus I$ minus its endpoints.
Then $I$ and $J$ are disjoint open arcs of $\partial B_1$ and their common endpoints are distinct points of $\psi^{-1}(a)$.
Furthermore, the preceding paragraph implies that $\psi(\bar I) \cap \psi(J) = \emptyset$ and  $\psi(I) \cap \psi(\bar J) = \emptyset$

Since $\Gamma$ disconnects $0$ from $y$, the homotopy class of the loop $\psi|_{\partial B_1}$ in $(\C \cup \infty) \setminus \{0,y\}$ is non-trivial.
Since $\psi$ maps the endpoints of $I$ and $J$ to $a$, each of $\psi|_I$ and $\psi|_J$ is a loop in $\C$, and $\psi|_{\partial B_1}$ is the concatenation of these two loops.
The concatenation of two homotopically trivial loops is also homotopically trivial.
Therefore, one of $\psi|_I$ or $\psi|_J$ is not homotopic to a point in $(\C \cup \infty)\setminus \{0,y\}$.
This implies that one of $\psi(\bar I)$ or $\psi(\bar J)$ disconnects $0$ from $y$.

Assume without loss of generality that $\psi(\bar I)$ disconnects $0$ from $y$.
Since ${\cluster{}{t}} \ni 0$ is connected (Lemma <ref>) and $U \ni y$ is connected by definition, ${\cluster{}{t}}$ and $U$ are contained in different connected components of $\C\setminus \psi(\bar I)$.
But, every point of $\psi(J) \subset \Gamma$ is an accumulation point of both ${\cluster{}{t}}$ and $U$, so $\psi(J) \subset \psi(\bar I)$.
Since $J$ is non-empty by construction, this contradicts the fact that $\psi(\bar I) \cap \psi(J) = \emptyset$.
We conclude that $\Gamma$ has no cut points.

The desired claim follows immediately from a scaling argument.

Almost surely, for all $t > 0$, each of the connected components of $\C \backslash \overline{{\regcluster{}{t}}}$
has a boundary that is a simple loop.

Combine Lemma <ref> together with Lemma <ref>,  a union bound,
and the relation between ${\regcluster{}{t}}$ and $\{\cluster{B_r}{t}\}_{r > 0}$ given in Theorem <ref>.

§ NOVELTY OF HARMONIC BALLS

In this section we show that typical harmonic balls are too rough to have Lipschitz boundaries
yet differ in a quantitative way from LQG-metric balls.

§.§ Not Lipschitz

In this section we show that a `typical' harmonic ball is not a Lipschitz domain.
Roughly, a Lipschitz domain is a domain whose boundary can be locally represented
by a graph of a Lipschitz function. Note that as every convex function is locally Lipschitz, see, for example,
<cit.>, every convex domain, ,  a ball or polygon, is a Lipschitz domain.

Let $A$ be a non-empty connected open set. $A$ is a  Lipschitz domain
if for every point $x_0 \in \partial A$, there exists $r > 0$ and a Lipschitz function
$f: \R \to \R$ such that — upon relabeling and reorienting the coordinate axes if necessary —
we have

\[
U \cap B_r(x_0) = \{ z \in B_r(x_0) : \Im(z) > f(\Re(z)) \}.
\]

In this section we prove the following.

Almost surely, ${\regcluster{}{t}}$ is not a Lipschitz domain for Lebesgue-a.e. $t$.

To that end, we show that `typical' points on the boundary of the cluster
do not satisfy the `cone condition'. We define these terms.

A cone $\mathcal Q \subset \C$ is a non-empty open set strictly contained in $\C$ which can be written as $\mathcal Q = \{c v + d w: c, d > 0\}$ for extremal directions $v,w \in \C \setminus \{0\}$. We also define complements of cones to be cones.

Note that any cone $\mathcal Q$ is scale invariant, that is, for any $\theta > 0$, $\theta \mathcal Q = \mathcal Q$.
For a cone $\mathcal Q$ and $z \in \C$, we write $\mathcal Q(z) := \mathcal Q + z$.

A domain $A \subset \C$ satisfies the interior (respectively, exterior) cone condition at $z \in \partial A$
if there is a radius $R > 0$ and a cone $\mathcal Q$ such that $\mathcal Q(z) \cap B_R(z) \subset A$ (respectively, $\mathcal Q(z) \cap B_R(z) \subset A^c$).
$A$ satisfies the cone condition at $z \in \partial A$ if it satisfies both the interior and exterior cone conditions at $z$.

A domain which satisfies the cone condition of Definition <ref>.
The domain is in gray and the two cones are displayed in red.

The main input in the proof of Proposition <ref> is the following lemma.

Almost surely, for $\mu_h$-a.e. $z \in \C$, if $t > 0$ such that $z \in \partial {\regcluster{}{t}}$, then
${\regcluster{}{t}}$ does not satisfy the cone condition at $z$.

To show that that ${\regcluster{}{t}}$ does not satisfy the cone condition, we study an event
concerning the oscillation of the Liouville measure across sectors of annuli.
Before doing so, we motivate the event we consider by reformulating some of the harmonic comparison lemmas
from Section <ref> into the following.

There exists a universal constant $C > 0$ such that the following holds for each $r > 0$ and $z \in \C$ such that
$\overline B_{10 r}(z) \cap \{0\} = \emptyset$.
$\A_{4 r , 5r }(z) \cap {{\regcluster{}{t}}}^c \neq \emptyset$, we have

\[
\mu_h(\A_{r,2 r}(z) \cap {\regcluster{}{t}}) \leq C \times \SGrho{3r, 6r}{z}.
\]

where $\SGrho{3r, 6r}{z}$ is as in (<ref>).

We use Lemma <ref> and Lemma <ref> which were both
stated for $\cluster{B_1}{t}$, $\odometer{B_1}{t}$ but whose proofs and statements extend verbatim to $\cluster{B_R}{t}$, $\odometer{B_R}{t}$ for any $R > 0$.
In the following chain of inequalities $C$ refers to a universal constant
which may change from line to line,

\begin{align*}
&\mu_h({\regcluster{}{t}} \cap \A_{r, 2r}(z)) \\
&= \mu_h(\cluster{B_R}{t} \cap \A_{r, 2 r}(z)) \qquad \mbox{(as for some $R > 0$, $\mu_h({\regcluster{}{t}} \triangle \cluster{B_R}{t})=0$ by Theorem \ref{theorem:full-theorem-minus-uniqueness})} \\
&\leq C \times \sup_{\A_{0.5 r, 4 r}(z)} \odometer{B_R}{t} \qquad \mbox{(by Lemma \ref{lemma:odometer-annuli} with $(s_1,s_2,s_3,s_4)=(0.5, 1, 2,4)$)} \\
&\leq C \times \sup_{\partial B_{4 r}(z)} \odometer{B_R}{t} \qquad \mbox{($\odometer{B_R}{t}$ is subharmonic in $B_{10 r}(z)$)} \\
&\leq C \times \SGrho{3r, 6r}{z} \qquad \mbox{(by Lemma \ref{lemma:odometer-bound-lqg-mass} with $(s_1,s_2,s_3,s_4)=(3,4,5,6)$)}.
\end{align*}

For a cone $\mathcal Q$, $M > 0$,  $z \in \C$, and $r > 0$, consider the event

\begin{equation} \label{eq:mu-h-wild}
\tilde G_{r}(z) = \tilde G_{r}(z; M, \mathcal Q) := \left\{\mu_{h}(\A_{r, 2r}(z) \cap \mathcal Q(z)) \geq  M \times \SGrho{3r, 6r}{z} \right\}
\end{equation}

and observe that $\tilde G_{r}(z) \in \sigma(h |_{\A_{r, 6r}(z)})$.
We abbreviate

\begin{equation}
\tilde G_n(z) = \tilde G_{2^{-n}}(z), \quad \forall n\in\N
\end{equation}

and prove an analogue of Lemma <ref> for rare events.

Let $\mathcal Q$ be a cone and let $M > 0$. There exists $\delta = \delta(M) \in (0,1)$ such that a.s. for $\mu_h$-a.e. $z\in  \C$, it holds for each large enough $N\in \N$ (depending on $z$) that

\begin{equation} \label{eq:mu_h-bad}
\#\left\{ n \in [N+1,2N] \cap \Z  : \mbox{$\tilde G_n(z; M, \mathcal Q)$ occurs} \right\}  \geq \delta N.
\end{equation}

In order to prove this lemma, we first prove the following.

Let $\mathcal Q$ be a cone and let $M > 0$. There exists a constant $p_M \in (0,1)$ so  that

\begin{equation} \label{eq:desired-claim}
P[ \tilde G_n(z; M, \mathcal Q)] \geq p_M, \quad \mbox{$\forall z$ s.t. $\dist(\A_{2^{-n}, 6 \times 2^{-n}}(z), \{0\}) > 2^{-n}/100$,  $\forall n \geq N(z)$.}
\end{equation}

Recall from (<ref>) that $h = h^\C - \boldsymbol{\alpha}_0 \log |\cdot|$.
We first show (<ref>) in the case $\boldsymbol{\alpha}_0 = 0$,
and then use Weyl scaling to get the general case.

In the following, we write $SG_{3r ,6r}^h(z)$ instead of $\SGrho{3r,6r}{z}$ to indicate the dependence on the underlying field $h$.
Note that $SG^h_{3r ,6r }(z)$ depends only on $h$ restricted to $\A_{3r, 6r}(z)$.

Step 1: $\boldsymbol{\alpha}_0 = 0$.

By definition, if $\boldsymbol{\alpha}_0 = 0$ then $h$ is a whole-plane GFF.
We have chosen the event $\tilde G_n$ so that it is a.s. determined by $h $ viewed modulo additive constant. Hence, by the scale and translation invariance of the law of the whole-plane GFF (<ref>) it suffices to bound $\P[\tilde G_1(0)]$ from below. This lower bound is achieved via the `adding a bump function' technique.

Since the random variables involved are finite and positive, there are $\gamma$-dependent constants $C_1$ and $C_2$ so that

\begin{equation} \label{eq:l-and-r-bounds}
\P\left[\mu_h(\A_{1,2} \cap \mathcal Q) \geq C_1 \mbox{ and }  SG^h_{3,6} \leq C_2\right] \geq 1/2.
\end{equation}

Let $\phi$ be a smooth, non-negative bump function
which is identically equal to $\gamma^{-1} \log \frac{M C_2}{C_1}$ on $\A_{1,2}$ and identically equal to 0 on $\A_{0.5,2.5}^c$. On the event in (<ref>)

\begin{align*}
\mu_{h+\phi}(\A_{1,2} \cap \mathcal Q) &= M \times \mu_h(\A_{1,2} \cap \mathcal Q) \times  \frac{C_2}{C_1} \qquad \mbox{(by Weyl scaling)} \\
&\geq M \times C_2 \qquad \mbox{($\mu_h(\A_{1,2} \cap \mathcal Q) \geq C_1$ by the event)} \\
&\geq M \times SG^h_{3,6}  \qquad \mbox{($ C_2 \geq SG^h_{3,6}$ by the event \eqref{eq:l-and-r-bounds})} \\
&= M \times SG^{h+\phi}_{3,6} \qquad \mbox{($\phi \equiv 0$ on $\A_{3, 6}$)}.
\end{align*}

Since the laws of $h$ and $h+\phi$ are mutually absolutely continuous viewed modulo additive constant <cit.> this implies

\[
\P[\tilde G_1(0)] := p_M > 0,
\]

completing the proof of Step 1.

Step 2: $\boldsymbol{\alpha}_0 \in (-\infty, Q)$.

Write $r := 2^{-n}$.
In the general case, we fix $z$ and take $n \geq N(z)$ sufficiently large so that

\[
\dist(\A_{r, 6 r}(z), \{0\}) > r/100.
\]

By Weyl scaling, Fact <ref>,

\[
\frac{\mu_h(\A_{r,2r}(z) \cap \mathcal Q)}{SG^h_{3r,6r}(z)} \geq C \frac{\mu_{h^\C}(\A_{r,2r}(z) \cap \mathcal Q)}{SG^{h^\C}_{3r,6r}(z)}, \quad \mbox{for a universal constant $C > 0$}.
\]

By Step 1, with probability $p_{M/C}$,

\[
\frac{\mu_{h^\C}(\A_{r,2r} \cap \mathcal Q)}{SG^{h^\C}_{3r,6r}} \geq M/C,
\]

completing the proof.

This leads to a proof of Lemma <ref>.

The proof is nearly identical to that of Lemma <ref>, the only difference
being the event under consideration only has positive probability meaning we get $\delta N$ instead of $(1-\zeta) N$ `good' scales.

Specifically, we may carry out the proof of Lemma <ref> and substitute Lemma <ref>
as the bound in (<ref>).
This shows that there exists $\delta = \delta(M) \in (0,1)$ such that it holds for each large enough $N\in \N$

\begin{equation} \label{eq:log-singularity-event}
\#\left\{ n \in [N+1,2N]\cap \Z   : \mbox{$\tilde G_n(0; M, \mathcal Q)$ occurs} \right\}  \geq \delta N.
\end{equation}

Then the argument in the proof of Lemma <ref> together with
(<ref>) implies a.s. the lemma statement holds for $\mu_h$-a.e. $z\in B_1$ and then by scaling
for $\mu_h$-a.e. $z\in \C$.

We use Lemma <ref> to show that `$\mu_h$-typical points' on the boundaries of clusters do not satisfy the cone condition.

Take $M = 2 C$ where $C$ is the universal constant from Lemma <ref>.
Condition on $h$ and sample $z$ from $\mu_h$.

By Lemma <ref>, a.s. for every cone $\mathcal Q$ with rational extremal directions $v,w$ there exist arbitrarily large $n \in \N$  (depending on $z$ and $\mathcal Q$) for which the event $\tilde G_n(z; M, \mathcal{Q})$ occurs.
Since every cone not equal to all of $\C$ is contained within a cone with rational extremal directions, we get that a.s. for every cone $\mathcal Q$ there exist arbitrarily large $n \in \N$  (depending on $z$ and $\mathcal Q$) for which the event $\tilde G_n(z; M, \mathcal{Q})$ occurs.

Suppose $t>0$ is such that $z \in \partial {\regcluster{}{t}}$.
and let $\mathcal Q(z)$ be a cone with apex at $z$. Let $n$ be such that $\tilde G_n(z; M, \mathcal{Q})$ holds
and $B_{10 \times 2^{-n} }(z) \cap \{0\} = \emptyset$. Write $r = 2^{-n}$. By the contrapositive of Lemma <ref>,

\begin{equation} \label{eq:large-mass-implies-full}
\mu_h(A_{r, 2r}(z) \cap {\regcluster{}{t}}) > C \times \SGrho{3r,6r}{z} \implies \A_{4r, 5r}(z) \cap {{\regcluster{}{t}}}^c = \emptyset.
\end{equation}

Since the event $\tilde G_n(z; M, \mathcal{Q})$ occurs with $M = 2C$,

\begin{equation} \label{eq:mu_h-is-bad}
\mu_h(A_{3r, 5r}(z) \cap \mathcal{Q}(z)) \geq M \times \SGrho{3r,6r}{z} >  C \times \SGrho{3r,6r}{z}.
\end{equation}

We, however, also have the following implications of the cone condition:

\begin{equation} \label{eq:interior-cone-implies-full}
B_R(z) \cap \mathcal{Q}(z) \subset {\regcluster{}{t}}    \implies \mu_h(A_{r, 2r}(z) \cap {\regcluster{}{t}}) \geq 		\mu_h(A_{r, 2r}(z) \cap \mathcal{Q}(z)), \quad \forall R > 2 r
\end{equation}

and for any cone $\mathcal{Q}'$,

\begin{equation} \label{eq:exterior-cone-implies-empty}
B_R(z) \cap \mathcal{Q}'(z) \subset {{\regcluster{}{t}}}^c    \implies \A_{4 r, 5 r}(z) \cap {\regcluster{}{t}}^c \neq \emptyset, \quad \forall R > 5 r.
\end{equation}

For any $R > 0$, we can take $n$ sufficiently large so that $R > 5\times 2^{-n}=  5 r$. Thus, the  interior cone condition (<ref>) together with (<ref>) implies by
(<ref>) that  $\A_{4r, 5r}(z) \cap {{\regcluster{}{t}}}^c = \emptyset$ which is incompatible
with the exterior cone condition (<ref>).

We use this to show that for Lebesgue a.e. $t$, ${\regcluster{}{t}}$ is not a Lipschitz domain.

The set ${\regcluster{}{t}}$ is parameterized so that $\mu_h({\regcluster{}{t}})  = t$, so if $A$ is a Lebesgue measurable subset of $[0, \infty)$, then

\begin{equation}\label{eq:mass-times}
\mu_h(\{z : z \in \partial {\regcluster{}{t}} \mbox{ for some $t \in A$}\}) = \mbox{Leb}(A) ,
\end{equation}

where $\mbox{Leb}$ denotes one-dimensional Lebesgue measure. This follows from the standard machine. Indeed, (<ref>) holds for intervals,  $A = [a,b]$,

\[
\mu_h(\{z : z \in \partial {\regcluster{}{t}} \mbox{ for some $t \in [a,b]$}\}) = \mu_h( \overline{\regcluster{}{b}} \backslash
\regcluster{}{a}) =
\]

by Theorem <ref>.
By approximation, this implies (<ref>) holds for all Lebesgue measurable subsets of $[0,\infty)$.

By (<ref>) applied to the set

\[
A = \{t > 0 : \mbox{ ${\regcluster{}{t}}$ satisfies the cone condition at each $z\in \partial {\regcluster{}{t}}$}\}
\]

together with Lemma <ref>, we get that a.s. the Lebesgue measure of the set of $t>0$ for which ${\regcluster{}{t}}$ satisfies the cone condition at each boundary point is zero. One easily gets from Definition <ref> that every Lipschitz domain satisfies the cone condition at each of its boundary points. Hence, a.s. ${\regcluster{}{t}}$ is not a Lipschitz domain for a.e. $t>0$.

§.§ Small diameter and large LQG mass

Recall from Section <ref> that $D_h$ denotes the $\gamma$-LQG metric associated with $h$.
We will eventually show that ${\regcluster{}{t}}$ is not an LQG metric ball for a.e. $t>0$ by showing that LQG metric balls do not satisfy the Harnack-type condition of Section <ref>. For this purpose we will need to force an LQG metric ball to contain certain sets of large $\mu_h$-mass. In order to do this, we will need the following proposition, which we prove in this subsection.

Let $U \subset V \subset W\subset  \C$ be bounded, connected open sets such that $\overline V\subset W$ and $W$ does not intersect the unit circle $\partial  B_1(0)$.
For each $C>\epsilon > 0$, it holds with positive probability (depending on $U,V,W,C,\epsilon$) that

\begin{equation} \label{eq:mass-diam-gen}
\sup_{u,v\in V} D_h\left( u , v ; W \right) \leq \epsilon \quad \text{and} \quad \mu_h(U) \geq C ,
\end{equation}

where here we use the notation for the internal metric from (<ref>).

It is not obvious how to apply the `adding a bump function' technique used in Section <ref> to prove Proposition <ref> since if the bump function $\phi$ is positive, then adding $\phi$ tends to increase both $D_h$ and $\mu_h$, and the reverse is true if $\phi$ is negative. So, some work is needed to simultaneously make the $D_h$-diameter small and the $\mu_h$-mass large.

We first prove a version of Proposition <ref> for dyadic squares.

Illustration of the proof of Lemma <ref>. The squares $\mathsf S_1 \subset \mathsf S_2$ are shown in pink and gray, respectively (only part of $\mathsf S_2$ is shown). The smaller dyadic squares $S\in\mathcal  S_n^3$ are shown in yellow and the corresponding larger squares $\hat S$ are shown in light blue. The set $G$ is equal to $\mathsf S_2$ minus the squares $\hat S$ for $S\in \mathcal  S_N^i$, for an appropriate deterministic choice of $N \geq n_0$ and $i\in \{1,2,3,4\}$.

Fix two closed dyadic squares $\mathsf S_1 \subset \mathsf S_2$ such that $\mathsf S_2$ does not intersect the Euclidean unit circle $\partial  B_1(0)$.
For each $C >\epsilon > 0$, it holds with positive probability (depending on $\mathsf S_1, \mathsf S_2, C,\epsilon$) that

\begin{equation} \label{eq:mass-diam}
\sup_{u,v\in \mathsf S_2 } D_h\left( u , v ; \mathsf S_2 \right) \leq \epsilon \quad \text{and} \quad \mu_h(\mathsf S_1) \geq C .
\end{equation}

See Figure <ref> for an illustration.
Let $n_0 \in \mathbb Z$ be chosen so that the side length of $\mathsf S_1$ is $2^{-n_0}$.
For $n \geq n_0$, let $\mathcal  S_n$ be the set of closed $2^{-n  } \times 2^{-n }$ squares $S$ which are contained in $\mathsf S_1$.
Since $\mathsf S_1$ is dyadic, $\mathsf S_1$ is the union of the squares in $\mathsf S_1$.
Furthermore, $\mathcal  S_n$ is the disjoint union of the following four sets of squares:

\begin{align*}
\mathcal  S_n^1 &:= \left\{[(k-1) 2^{-n} ,k 2^{-n}] \times [(m-1) 2^{-n} , m 2^{-n}] \in \mathcal  S_n : \text{$k$ is even, $m$ is even} \right\} \\
\mathcal  S_n^2 &:= \left\{[(k-1) 2^{-n} ,k 2^{-n}] \times [(m-1) 2^{-n} , m 2^{-n}] \in \mathcal  S_n : \text{$k$ is even, $m$ is odd} \right\} \\
\mathcal  S_n^3 &:= \left\{[(k-1) 2^{-n} ,k 2^{-n}] \times [(m-1) 2^{-n} , m 2^{-n}] \in \mathcal  S_n : \text{$k$ is odd, $m$ is even} \right\} \\
\mathcal  S_n^4 &:= \left\{[(k-1) 2^{-n} ,k 2^{-n}] \times [(m-1) 2^{-n} , m 2^{-n}] \in \mathcal  S_n : \text{$k$ is odd, $m$ is odd} \right\} .
\end{align*}

As $\mu_h(\mathsf S_1)$ is a strictly positive random variable, there exists a constant $C_1 > 0$ so that $\P[\mu_h(\mathsf S_1) \geq C_1] \geq 1/2$.
Let $\phi : [0,1] \to \mathbb C$ be a smooth bump function which is identically equal to 1 on $\mathsf S_1$ and which is identically equal to zero outside
a neighborhood of $\overline{\mathsf S_1}$ and let $\tilde h := h + \frac{\phi}{\gamma} \log(4 C/C_1)$. By the Weyl scaling property of $\mu_h$, Fact <ref>, on the positive-probability event $\{\mu_h(\mathsf S_1) \geq C_1\}$, we have
$\{\mu_{\tilde h}(\mathsf S_1) \geq 4 C\}$. Therefore, by absolute continuity, there is $p= p(\mathsf S_1, C) > 0$ such that with probability at least $p$,

\begin{equation} \label{eqn-square-mass-lower}
\mu_h(\mathsf S_1) \geq 4C .
\end{equation}

For $S\in\mathcal  S_n$, define the larger square

\begin{equation}
\hat S := \left( \text{square of side length $2^{-n}  + 2^{-n-2}$ with same center as $S$} \right).
\end{equation}

By, , <cit.> and a union bound over all $S\in\mathcal  S_n$, the supremum over all $S\in \mathcal  S_n$ of the $D_h(\cdot,\cdot ; \hat S \cap \mathsf S_2 )$ diameter of $\hat S \cap \mathsf S_2$ tends to zero in probability as $n\to\infty$. Consequently, we can find a deterministic $N \geq n_0$ such that with probability at least $p/2$, (<ref>) holds and also

\begin{equation} \label{eqn-small-square-diam}
\sup_{S\in\mathcal  S_N} \sup_{u,v\in  \hat S \cap \mathsf S_2  } D_h(u,v ; \hat S \cap \mathsf S_2  ) \leq \frac{\epsilon}{2} .
\end{equation}

Since $\mathsf S_1$ is the union of $\bigcup_{S\in \mathcal  S_n^i} S$ for $i=1,2,3,4$, on the event that (<ref>) holds there exists $i\in \{1,2,3,4\}$ such that $\mu_h\left(  \bigcup_{S\in \mathcal  S_n^i} S \right) \geq C$. Since (<ref>) and (<ref>) hold simultaneously with probability at least $p/2$, we can find a deterministic choice of $i\in \{1,2,3,4\}$ such that with probability at least $p/8$,

\begin{equation} \label{eqn-small-square-mass-diam}
\mu_h\left(  \bigcup_{S\in \mathcal  S_N^i} S \right) \geq C \quad \text{and} \quad \sup_{S\in\mathcal  S_N^i} \sup_{u,v\in  \hat S \cap \mathsf S_2 } D_h(u,v ; \hat S \cap \mathsf S_2 ) \leq \frac{\epsilon}{2}  .
\end{equation}

Any two squares in $S_N^i$ lie at Euclidean distance at least $2^{-n}$ from each other, so the set

\begin{equation}
G := \mathsf S_2 \setminus \bigcup_{S\in\mathcal  S_N^i} \hat S
\end{equation}

is connected. Since $\overline G$ is a finite union of closed Euclidean squares, it follows from, for example, Lemma 3.9 in <cit.> that a.s. the $D_h(\cdot,\cdot;\overline G)$ diameter of $\overline G$ is finite. Hence, we can find a deterministic $A > 0$ such that with probability at least $p/16$, (<ref>) holds and also

\begin{equation} \label{eqn-exterior-diam}
\sup_{u,v\in \overline G} D_h(u,v;\overline G) \leq A .
\end{equation}

Let $\phi : \mathbb C\to [0,1]$ be a smooth compactly supported bump function which is identically equal to 1 on $\overline G$ and which is identically equal to zero on $\partial  B_1(0) \cup \bigcup_{S \in \mathcal  S_N^i} S$. Let

\begin{equation}
\tilde h := h - \frac{\phi}{\xi} \log(2 A /  \epsilon).
\end{equation}

By the Weyl scaling properties of $\mu_h$ and $D_h$, if (<ref>) and (<ref>) hold (which happens with probability at least $p/16$), then

\begin{equation} \label{eq:use-weyl}
\mu_{\tilde h}\left(  \bigcup_{S\in \mathcal  S_N^i} S \right) \geq C ,\quad
\sup_{S\in\mathcal  S_N^i} \sup_{u,v\in  \hat S \cap \mathsf S_2  } D_{\tilde h}(u,v ; \hat S \cap \mathsf S_2   ) \leq \frac{\epsilon}{2} , \quad \text{and}\quad
\sup_{u,v\in \overline G} D_{\tilde h}(u,v;\overline G) \leq \frac{\epsilon}{2} .
\end{equation}

The second and third conditions in (<ref>) together with the triangle inequality imply that

\begin{equation} \label{eqn-shifted-field-weyl}
\sup_{u,v\in \mathsf S_2 } D_{\tilde h}\left( u , v ;\mathsf S_2 \right) \leq \epsilon .
\end{equation}

Furthermore, the first condition in (<ref>) implies that $\mu_{\tilde h}(\mathsf S_1) \geq C$.

Since $\phi\equiv 0$ on $\partial  B_1(0)$, the average of $\tilde h$ over $\partial  B_1(0)$ is zero. By standard absolute continuity results for the GFF (see, e.g., <cit.>), the laws of $h$ and $\tilde h$ are mutually absolutely continuous. The previous paragraph tells us that with probability at least $p/16$, (<ref>) holds with $\tilde h$ in place of $h$. Therefore, (<ref>) holds with positive probability for $h$.

Since $U$ is open, we can find deterministic closed dyadic squares $\mathsf S_1 \subset \mathsf S_2 \subset U$ with the property that $\mathsf S_1$ is contained in the interior of $\mathsf S_2$. By Lemma <ref>, it holds with positive probability that

\begin{equation} \label{eqn-use-mass-diam}
\sup_{u,v\in \mathsf S_2 } D_h\left( u , v ; \mathsf S_2 \right) \leq \frac{\epsilon}{2}  \quad \text{and} \quad \mu_h(\mathsf S_1) \geq C .
\end{equation}

Our hypotheses on $V,W$ and $\mathsf S_1,\mathsf S_2$ imply that the closure of $V\setminus \mathsf S_2$ is contained in the interior of $W\setminus \mathsf S_1$. Hence, we can find an intermediate open set $O$ such that

\begin{equation}
\overline{V \setminus \mathsf S_2} \subset  O \quad \text{and} \quad \overline O \subset  W\setminus \mathsf S_1 .
\end{equation}

Since $D_h$ induces the Euclidean topology and $V\setminus \mathsf S_2$ is connected, there exists a deterministic $A > 0$ such that with positive probability, (<ref>) holds and also

\begin{equation} \label{eqn-exterior-diam-gen}
\sup_{u,v\in V\setminus \mathsf S_2} D_h(u,v; O) \leq A .
\end{equation}

We now use a `subtracting a bump function' argument similar to the one at the end of the proof of Lemma <ref>.
Let $\phi : [0,1] \to \mathbb C$ be a smooth bump function which is identically equal to 1 on $O$ and which is identically equal to zero outside of $W\setminus \mathsf S_1$.

\begin{equation}
\tilde h := h - \frac{\phi}{\xi} \log(2 A / \epsilon) .
\end{equation}

By the Weyl scaling properties of $\mu_h$ and $D_h$, on the positive-probability event that (<ref>) and (<ref>) hold,

\begin{equation} \label{eq:use-weyl-gen}
\sup_{u,v\in \mathsf S_2 } D_{\tilde h}\left( u , v ; \mathsf S_2 \right) \leq \frac{\epsilon}{2} ,
\quad  \mu_{\tilde h}(\mathsf S_1) \geq C  ,   \quad \text{and} \quad
\sup_{u,v\in V\setminus \mathsf S_2} D_{\tilde h}(u,v; O) \leq \frac{\epsilon}{2} .
\end{equation}

By the triangle inequality, (<ref>) implies (<ref>) with $\tilde h$ in place of $h$. Since the laws of $h$ and $\tilde h$ are mutually absolutely continuous <cit.>, we conclude the proof.

§.§ Not LQG metric balls

In this subsection we prove the following.

Almost surely, ${\regcluster{}{t}}$ is not an LQG-metric ball for Lebesgue-a.e. $t > 0$.

We follow a strategy similar to that of Section <ref>
although the arguments are slightly more complicated. To that end,
let $U \Subset T$ and $Q' \Subset Q$ be connected open sets as shown in Figure <ref>. We require that $U$ is a ball contained in $\A_{1,2}$, that $Q'$ and $Q$ are contained in $\A_{4,5} \setminus \overline T$, and that $T$ is a keyhole-shaped region contained in $\A_{1,5}$ with $\partial B_{5} \subset \partial T$, as shown in the figure.
For a set $A$, $r>0$, and $z\in\C$, we write

\[
A_r(z) := r A +  z.
\]

Sets $U,V,W,T,Q, Q'$ used in the proof of Lemma <ref> and to define the event (<ref>).
The origin is the black dot; the set $T$ is the green key-hole shaped set;
$Q' \Subset Q$ are gray ellipses; and $U \Subset V \Subset W$ are gray concentric balls.
The picture is not drawn to scale and only part of the set $T$ is shown.
The sets $V$ and $W$ are only used in the proof of Lemma <ref>.

For $z\in\C$ and $M>0$, let $\hat G_r(z) = \hat G_r(z; M) $ denote the event that the following holds:

\begin{equation}\label{eq:pinch-point-event}
\begin{aligned}
D_h(Q'_r(z), \partial Q_r(z)) &> \sup_{x \in \partial B_{5 r}(z)} D_h(z, x) \\
D_h(z, \partial B_{5 r}(z)) &>  \sup_{x, y \in T_r(z)} D_h(x, y) \\
\mu_h(U_r(z)) &> M \times \SGrho{3r,6r}{z} .
\end{aligned}
\end{equation}

\[
\hat G_n(z) := \hat G_{2^{-n}}(z) .
\]

Showing that this event occurs with positive probability is somewhat technical and hence the proof of the following will be postponed to the end of this subsection.

In the case $\boldsymbol{\alpha}_0 = 0$, as in (<ref>), $\P[\tilde G_1(0)] > 0$.

The preceding lemma implies the following.

Let $M > 0$. There exists $\delta \in (0,1)$ such that a.s. for $\mu_h$-a.e. $z\in  \C$, it holds for each large enough $N\in \N$ (depending on $z$) that

\begin{equation} \label{eq:mu_h-pinch-bad}
\#\left\{ n \in [N+1,2N]\cap \Z  : \mbox{$\hat G_n(z; M)$ occurs} \right\}  \geq \delta N.
\end{equation}

Lemma <ref> implies, by an argument similar to the proof of Lemma <ref>,
existence of a $\delta = \delta(M) \in (0,1)$ such that (<ref>) holds for $z = 0$ for each large enough $N\in \N$.
The exact same argument outlined in the proof of Lemma <ref> then leads to the lemma statement.

We now observe a deterministic consequence of the event $\hat G_r(z)$ on the shape of an LQG metric ball whose boundary contains $z$. For the statement, we recall that $\mathcal B_u(0;D_h)$ denotes the open LQG metric ball of radius $u$ centered at 0.

The following holds a.s. for each $M > 0$, each $z \in \C$, and each $r > 0$ such that $B_{10r}(z)$ does not contain the origin.
If $u > 0$ is such that $\hat G_r(z)$ occurs and $z \in \partial \mathcal{B}_u(0; D_h)$, then
$Q'_r(z) \subset (\mathcal{B}_u(0; D_h))^c$ and $T_r(z) \subset \mathcal{B}_u(0; D_h)$.

The reader is encouraged to refer to Figure <ref> as a visual aid during the proof.
For notational convenience, write $\mathcal{B}_u := \mathcal{B}_u(0; D_h)$.
Assume that $z\in \partial \mathcal B_u$ and $\hat G_r(z)$ occurs, where $r$ is sufficiently small so that $B_{10r}(z)$ does not contain the origin.
Since $z \in \partial \mathcal{B}_u$, we have $D_h(0, z) = u$.

Step 1: $Q'_r(z) \subset (\mathcal{B}_u)^c$

Recall the definition of LQG geodesics between compact sets from just after (<ref>).
As $B_{10 r}(z)$ does not contain the origin, any geodesic from $0$ to $\overline{Q'_r(z)}$ can be decomposed into
geodesics from $0 \to q_1 \in \partial B_{5 r}(z)$, from $q_1$ to $ q_2 \in \partial Q_r(z)$, and from $q_2$ to $q_3 \in \partial Q'_r(z)$.
From this and the definition (<ref>) of $\hat G_r(z)$, we obtain

\begin{align*}
D_h(0, Q'_r(z))
&= D_h(0, q_1) + D_h(q_1, q_2) + D_h(q_2, q_3)  \\
&> D_h(0, q_1) + D_h(q_2, q_3) \qquad \mbox{($D_h(q_1, q_2) > 0$)} \\
&> D_h(0, q_1) + D_h(q_1, z) \qquad \mbox{($D_h(Q'_r(z), \partial Q_r(z)) > \sup_{x \in \partial B_{5 r}(z)} D_h(z, x)$)} \\
& \geq D_h(0, z) \qquad \mbox{(triangle inequality)}.
\end{align*}

Since $D_h(0,z) = u$, this implies $Q'_r(z) \subset (\mathcal{B}_u)^c$.

Step 2: $T_r(z) \subset \mathcal{B}_u$

The proof is similar to Step 1. Let $y \in T_r(z)$. Also let $z_1$ be a point of $\partial B_{5r}(z)$ which is hit by a geodesic from 0 to $z$ (such a point exists since $0\notin B_{10 r}(z)$). Then

\begin{equation} \label{eq:geodesic3-decomp}
D_h(0, z) = D_h(0, z_1) + D_h(z_1, z).
\end{equation}

We now use the definition (<ref>) of $\hat G_r(z)$ to get

\begin{align*}
D_h(0, z) &= D_h(0, z_1) + D_h(z_1, z) \quad \mbox{\eqref{eq:geodesic3-decomp}} \\
&\geq D_h(0, z_1) + D_h(z, \partial B_{5 r}(z)) \qquad \mbox{($z_1 \in \partial B_{5 r}(z)$)} \\
&> D_h(0, z_1) + D_h(z_1, y) \\
&( D_h(z, \partial B_{5 r}(z)) >  \sup_{x, y \in T_r(z)} D_h(x, y) \mbox{ and }\partial T_r(z) \supset \partial B_{5 r}(z) )  \\
& \geq D_h(0, y) \qquad \mbox{(triangle ineq.)}.
\end{align*}

Since $D_h(0,z) = u$, this implies $T_r(z) \subset \mathcal{B}_u$.

Geodesic decomposition used in proof of Lemma <ref>. Geodesics are drawn as dashed lines. Sets are not drawn to scale
and not all of $T_r(z)$ is shown. Note that various geodesics in the figure merge into each other. This property is called confluence of geodesics <cit.> and is not needed for our proofs.

Lemmas <ref> and <ref> immediately lead to a proof of Proposition <ref>
via a similar argument as the proof of Lemma <ref>.

By the exact same argument as in the beginning of the proof of Proposition <ref>, it suffices to show that a.s. for $\mu_h$-a.e. $z$, no cluster ${\regcluster{}{t}}$
coincides with a LQG-metric ball containing $z$ on its boundary.

Throughout the proof we take $M = C$, where $C > 0$ is the universal constant from Lemma <ref>. Almost surely, the conclusion of Lemma <ref> (with this choice of $M$) holds for $\mu_h$-a.e. $z$. So, it suffices to consider a $z$ such that the conclusion of Lemma <ref> holds and show that a.s. no cluster ${\regcluster{}{t}}$ coincides with an LQG metric ball which has $z$ on its boundary.

By Lemma <ref>, a.s. there exist arbitrarily large $n \in \N$  (depending on $z$) for which the event $\hat G_n(z; C)$ occurs.
Consider such an $n$ which is large enough so that $0\notin B_{10 \times 2^{-n}}(z)$ and write $r := 2^{-n}$.

Since $\hat G_n(z;C)$ occurs, Lemma <ref>
implies that $Q_r'(z) \subset \mathcal{B}^c$ and $U_r (z) \subset T_r(z) \subset \mathcal{B}$. In particular, by the third inequality of the event (<ref>),

\begin{equation} \label{eq:not-metric-balls-mass}
\mu_h( \A_{r, 2r}(z) \cap \mathcal{B}) \geq \mu_h(U_r(z)) > C \times \SGrho{3r,6r}{z}.
\end{equation}

On the other hand, the contrapositive of Lemma <ref> shows that for each $t>0$,

\begin{equation} \label{eq:large-mass-implies-full-2}
\mu_h(\A_{r, 2r}(z) \cap {\regcluster{}{t}}) > C \times \SGrho{3r,6r}{z} \implies \A_{4r, 5r}(z) \cap {{\regcluster{}{t}}}^c = \emptyset .
\end{equation}

Since $Q_r'(z) \Subset \A_{4r, 5r}(z)$, (<ref>) implies that it cannot be the case that $\mu_h(\A_{r, 2r}(z) \cap {\regcluster{}{t}}) >  C \times \SGrho{3r,6r}{z}$ and $Q_r'(z) \subset {{\regcluster{}{t}}}^c$. Since $Q_r'(z) \subset \mathcal B^c$ and by (<ref>), we get that a.s. $\mathcal B$ is not equal to ${\regcluster{}{t}}$ for any $t>0$.

It remains to prove Lemma <ref>, which we do in several steps.
In the remainder of the subsection, let $T,U,V,W, Q', Q$ be as in Figure <ref>.
In particular, $U \Subset V \Subset W \Subset T$ and $U,V,W$ are concentric Euclidean balls contained in $\A_{1,2}$.

Fix $\delta_0 \in (0, 1/100)$ sufficiently small so that:

\begin{align} \label{eq:metric-delta0-choice}
B_{10 \delta_0}(T) \Subset \A_{1, 6} ,\quad
B_{10 \delta_0}(U) \Subset V ,\quad
B_{10 \delta_0}(V) \Subset W ,\notag\\
B_{10 \delta_0}(W) \Subset \A_{1,2} ,\quad
B_{10 \delta_0}(Q') \Subset Q ,\quad
B_{10 \delta_0}(Q) \Subset \A_{4,5} \backslash B_{10 \delta_0}(T) .
\end{align}

Recall the notation for the internal metric from (<ref>).
To be succinct, we write, for a set $A$ and $\delta > 0$,

\begin{equation}
\diam^\delta_h(A) := \sup_{x, y \in A} D_h(x,y; B_{\delta}(A)).
\end{equation}

To prove Lemma <ref>, we will show that several auxiliary events occur with positive probability.
We let $\tilde E_1$ be the event that

\begin{equation} \label{eq:first-event}
\begin{aligned}
D_h(Q', \partial Q) &> \sup_{x \in \partial B_{5}} D_h(0, x; (B_{2 \delta_0}(Q \cup W))^c) \\
D_h(0, \partial B_{\delta_0}) &> \diam^{\delta_0}_h(T \backslash B_{3 \delta_0}(W)) .
\end{aligned}
\end{equation}

We let $\tilde E_2 := \tilde E_2(M)$ be the event that

\begin{equation} \label{eq:second-event}
\begin{aligned}
D_h(\partial B_{\delta_0}, \partial B_{2 \delta_0}) &> \diam^{\delta_0}_h(V)  \\
\mu_h(U) &> M \times \SGrho{3,6}{0} .
\end{aligned}
\end{equation}

We let $\tilde E_3$ be the event that

\begin{equation} \label{eq:third-event}
D_h(\partial B_{2 \delta_0}, \partial B_{3 \delta_0}) > \diam_h^{\delta_0}(B_{3\delta_0}(W) \backslash V).
\end{equation}

We will successively prove the following lemmas.

We have $\P[\tilde E_1] > 0$.

For each $M  > 0$, $\P[\tilde E_1 \cap \tilde E_2(M)] > 0$.

For each $M  > 0$, $\P[\tilde E_1 \cap \tilde E_2(M) \cap \tilde E_3] > 0$.

Before we prove these lemmas, we show that Lemma <ref> implies Lemma <ref>.

Let $M > 0$ be given.
By Lemma <ref>,
it then suffices to show that

\begin{equation} \label{eq:events-inclusion}
\{\tilde E_1 \cap \tilde E_2(M) \cap \tilde E_3\} \subset \hat G_1(0; M).
\end{equation}

Henceforth assume that $\tilde E_1\cap \tilde E_2(M) \cap \tilde E_3$ occurs.
The first inequality in the definition (<ref>) of $\tilde E_1$ implies

\[
D_h(Q', \partial Q) > \sup_{x \in \partial B_{5}} D_h(0, x; (B_{2 \delta_0}(Q \cup W))^c) \geq \sup_{x \in \partial B_{5}} D_h(0, x)
\]

which is the first inequality in the definition (<ref>) of the event $\hat G_1(0; M)$.
We have

\begin{equation} \label{eq:ball-inclusion}
T = (T \backslash B_{3 \delta_0}(W)) \cup V \cup  (B_{3 \delta_0}(W) \backslash V).
\end{equation}

Thus, the second inequality in the definition (<ref>) of  $\tilde E_1$, the first inequality in the definition (<ref>) of $\tilde E_2$, and the definition (<ref>) of $\tilde E_3$ imply

\begin{align*}
D_h(0, \partial B_{5}) &> D_h(0, \partial B_{\delta_0}) + D_h(\partial B_{\delta_0}, \partial B_{2 \delta_0})
+ D_h(\partial B_{2 \delta_0}, \partial B_{3 \delta_0}) \quad \text{(triangle ineq.)}
\\
&\geq \diam^{\delta_0}_h(T \backslash B_{3 \delta_0}(W))
+  \diam^{\delta_0}_h(V)
+  \diam^{\delta_0}_h(B_{3 \delta_0}(W))
\\
&\geq \sup_{x, y \in T} D_h(x,y) \quad \mbox{(\eqref{eq:ball-inclusion} and triangle ineq.)}
\end{align*}

which is the second inequality in the definition of $\hat G_1(0; M)$. The second inequality in the definition (<ref>) of $\tilde E_2$ is the third and final inequality of $\hat G_1(0; M)$.

We will now show that $\P[\tilde E_1] > 0$ by adding an appropriate bump function to $h$.

Since the random variables involved are strictly positive and finite,
there exists positive finite constants $C_1, C_2, C_3, C_4$
so that the event

\begin{equation} \label{eq:stuff-finite}
\begin{aligned}
D_h(Q', \partial Q) &\geq C_1 \\
\sup_{x \in \partial B_{5}} D_h(0, x; (B_{2 \delta_0}(Q \cup W))^c) &\leq C_2 \\
D_h(0, \partial B_{\delta_0/2}) &\geq C_3  \\
\diam^{\delta_0}_h(T \backslash B_{3 \delta_0}(W)) &\leq C_4
\end{aligned}
\end{equation}

satisfies	$\P[\eqref{eq:stuff-finite}] > 0$. Henceforth assume that the event in (<ref>) occurs.

Take smooth compactly supported bump functions $\phi_1, \phi_2 : \mathbb C\to [0,1]$
so that

\[
\phi_1 \equiv \begin{cases}
1 &\mbox{on $Q$} \\
0 &\mbox{on $(B_{\delta_0}(Q))^c$}
\end{cases}
\]

\[
\phi_2 \equiv \begin{cases}
1 &\mbox{on  $B_{\delta_0/2}$} \\
0 &\mbox{on  $B_{\delta_0}^c$}.
\end{cases}
\]

With $\xi = \xi(\gamma)$ as in (<ref>), let

\begin{equation}
\tilde h := h + \frac{\phi_1}{\xi} \log\left( (\frac{2 C_2}{C_1} \vee 1) \times (\frac{2 C_4}{C_3} \vee 1)\right) + \frac{\phi_2}{\xi} \log\left(\frac{2 C_4}{C_3} \vee 1\right).
\end{equation}

Suppose (<ref>) holds.

\begin{align} \label{eq:first-event-pf1}
D_{\tilde h}(0, B_{\delta_0}) &\geq D_{\tilde h}(0, B_{\delta_0/2}) \qquad \mbox{(positivity of length)} \notag \\
&\geq \frac{2 C_4}{C_3} D_{h}(0, B_{\delta_0/2}) \qquad \mbox{(Weyl scaling, Fact~\ref{fact:lqg-metric})}  \notag\\
&> \diam^{\delta_0}_h(T \backslash B_{3 \delta_0}(W)) \qquad \mbox{(the event \eqref{eq:stuff-finite})}  \notag \\
&= \diam^{\delta_0}_{\tilde h}(T \backslash B_{3 \delta_0}(W)) \qquad \mbox{($\phi_1 + \phi_2 \equiv 0$ on $B_{\delta_0}(T)$)}.
\end{align}

Also note that by Weyl scaling and since the bump function $\phi_2$ has support contained in $(B_{2 \delta_0}(Q \cup W))^c$,

\begin{equation} \label{eq:weyl-scaling-leak}
\begin{aligned}
&\sup_{x \in \partial B_{5}} D_{\tilde h}(0, x; (B_{2 \delta_0}(Q \cup W))^c)  \\
&\qquad\qquad\leq (\frac{2 C_4}{C_3} \vee 1) \times  \sup_{x \in \partial B_{5}} D_{h}(0, x; (B_{2 \delta_0}(Q \cup W))^c).
\end{aligned}
\end{equation}

\begin{align}  \label{eq:first-event-pf2}
D_{\tilde h}(Q', \partial Q) &\geq \frac{2 C_2}{C_1} \times (\frac{2 C_4}{C_3} \vee 1) \times D_{h}(Q', \partial Q) \qquad \mbox{(Weyl scaling)}  \notag \\
&> (\frac{2 C_4}{C_3} \vee 1) \times \sup_{x \in \partial B_{5}} D_{h}(0, x; (B_{2 \delta_0}(Q \cup W))^c)  \qquad \mbox{(the event \eqref{eq:stuff-finite})}  \notag \\
&\geq \sup_{x \in \partial B_{5}} D_{\tilde h}(0, x; (B_{2 \delta_0}(Q \cup W))^c) \qquad \mbox{(by \eqref{eq:weyl-scaling-leak})}.
\end{align}

By (<ref>) and (<ref>) and since $\P[\eqref{eq:stuff-finite}] > 0$, the event $\tilde E_1$ occurs with positive probability with $\tilde h$ in place of $h$.
By <cit.>, the laws of $h$ and $\tilde h$ are mutually absolutely continuous, viewed modulo additive constant. By Weyl scaling the occurrence of $\tilde E_1$ is unaffected by adding a constant to $h$.
Thus, the fact that $\tilde E_1$ occurs with positive probability with $\tilde h$ instead of $h$ implies that $\P[\tilde E_1] > 0$.

We next show that $\P[\tilde E_1] > 0 \implies \P[\tilde E_1 \cap \tilde E_2(M)] > 0$ using the domain Markov property
with a set selected to be disjoint from the domain of dependence of $\tilde E_1$.

By the domain Markov property of the GFF, <cit.>, we can decompose

\begin{equation} \label{eq:use-gff-markov}
h = h_0 + \varphi
\end{equation}

where $h_0$ is a zero-boundary GFF on $W$, $\varphi$ is harmonic on $W$,
and $h_0$ and  $\varphi$ are independent.

Since the random variables involved are strictly positive and finite and $\P[\tilde E_1] > 0$,
there exists positive finite constants $C_1, C_2$ so that the event

\begin{equation} \label{eq:stuff-finite-2}
\begin{aligned}
D_h(\partial B_{\delta_0}, \partial B_{2 \delta_0}) &\geq C_1 \\
M \times \SGrho{3,6}{0} &\leq C_2
\end{aligned}
\end{equation}

satisfies	$\P[\tilde E_1 \cap \eqref{eq:stuff-finite-2}] > 0$.

We will also need to consider the event

\begin{equation} \label{eq:large-mass-low-diam}
\begin{aligned}
\diam_{h_0}^{\delta_0}(V) &\leq e^{-\xi \sup_{W} \varphi} C_1 \\
\mu_{h_0}(U) &\geq e^{-\gamma \inf_{U} \varphi} C_2 .
\end{aligned}
\end{equation}

By the locality properties of $\mu_h$ and $D_h$ (Facts <ref> and <ref>), the events (<ref>) and $\tilde E_1$ are both measurable with respect to the restriction of $h$ to

\[
Q \cup B_{2 \delta_0}(Q \cup W)^c \cup B_{2 \delta_0} \cup B_{2 \delta_0}(W)^c  \cup \A_{3,6} ,
\]

which is a compact subset of $W^c$.
Furthermore, the event (<ref>) is measurable with respect to the restriction of $h$ to $B_{\delta_0}(V)$ and the function $\phi$ (which is measurable with respect to $h|_{W^c}$).

By standard absolute continuity results
for the GFF (see, , <cit.>) together with (<ref>),
the conditional law of $h|_{B_{\delta_0}(V)}$ given $h|_{W^c}$
is mutually absolutely continuous with respect to its marginal law.
From this and Lemma <ref>, we obtain

\[
\P[\eqref{eq:large-mass-low-diam} \,|\,  \eqref{eq:stuff-finite-2} \cap \tilde E_1] > 0 .
\]

Since $\P[\tilde E_1 \cap \eqref{eq:stuff-finite-2}] > 0$, we thus have

\[
\P[\eqref{eq:large-mass-low-diam} \cap \eqref{eq:stuff-finite-2} \cap \tilde E_1] > 0.
\]

We will now conclude the proof by showing that $\eqref{eq:large-mass-low-diam} \cap \eqref{eq:stuff-finite-2} \subset \tilde E_2$.
Assume that $\eqref{eq:large-mass-low-diam} \cap \eqref{eq:stuff-finite-2}$ occurs. Then

\begin{align*}
\diam^{\delta_0}_h(V) &\leq   e^{\xi \sup_{B_{\delta_0}(V)} \varphi}  \diam^{\delta_0}_{h_0}(V) \quad \mbox{(Weyl scaling)} \\
&\leq D_h(\partial B_{\delta_0}, \partial B_{2 \delta_0}) \quad \mbox{(the events \eqref{eq:large-mass-low-diam} and \eqref{eq:stuff-finite-2})}
\end{align*}

\begin{align*}
\mu_h(U)&\geq  e^{\gamma \inf_{U} \varphi}  \mu_{h_0}(U) \quad \mbox{(Weyl scaling)} \\
&\geq M \times \SGrho{3,6}{0} \quad \mbox{(the events \eqref{eq:large-mass-low-diam} and \eqref{eq:stuff-finite-2})}
\end{align*}

which is exactly the event $\tilde E_2$.

We finally show that $\P[\tilde E_1 \cap \tilde E_2(M) ] > 0 \implies \P[\tilde E_1 \cap \tilde E_2(M) \cap \tilde E_3] > 0$.
The proof involves adding a bump function to make $\tilde E_3$ to occur and then checking that the events
$\tilde E_1 \cap \tilde E_2(M)$ still occur after adding the bump function.

Let $\phi : \mathbb C\to [0,1]$ be a smooth compactly supported bump function such that

\[
\phi \equiv \begin{cases}
1 &\mbox{on $B_{3 \delta_0}(W) \backslash B_{\delta_0}(U)$  } \\
0 &\mbox{on $(U \cup B_{4 \delta_0}(W))^c$}.
\end{cases}
\]

Since we know that $\P[\tilde E_1\cap \tilde E_2(M)] > 0$ (Lemma <ref>) and the quantities involved are a.s. finite and positive, we can find finite positive constants $C_1, C_2$ (depending on $M$) such that the event

\begin{equation} \label{eq:stuff-finite-3}
\begin{aligned}
D_h(\partial B_{2 \delta_0}, \partial B_{3 \delta_0}) &\geq C_1 \\
\diam^{\delta_0}_h(B_{3 \delta_0}(W) \backslash V) &\leq C_2
\end{aligned}
\end{equation}

satisfies	$\P[\tilde E_1 \cap \tilde E_2 \cap \eqref{eq:stuff-finite-3}] > 0$.

\begin{equation}
\tilde h = h + \frac{\phi}{\xi} \log(\frac{C_2}{C_1} \wedge 1) .
\end{equation}

Recall from (<ref>) that $B_{\delta_0}(U) \subset V$. Hence, on $\tilde E_1 \cap \tilde E_2 \cap \eqref{eq:stuff-finite-3}$,

\begin{align*}
\diam^{\delta_0}_{\tilde h}(B_{3 \delta_0}(W) \backslash V) &\leq \frac{C_2}{C_1} \diam^{\delta_0}_{h}(B_{3 \delta_0}(W) \backslash V) \quad \mbox{(Weyl scaling)} \\
&\leq D_h(\partial B_{2 \delta_0}, \partial B_{3\delta_0}) \quad \mbox{(the event \eqref{eq:stuff-finite-3})} \\
&= D_{\tilde h}(\partial B_{2 \delta_0}, \partial B_{3\delta_0}) \quad \mbox{($\phi \equiv 0$ on $\overline B_{3 \delta_0}$)},
\end{align*}

which is the event $\tilde E_3$ with $\tilde h$ in place of $h$. Hence, this event has positive probability.

As we will see below, by Weyl scaling,  the fact $\log(\frac{C_2}{C_1} \wedge 1) \leq 0$, and since $\phi \equiv 0$ on $U \cup \A_{3,6} \cup B_{3 \delta_0}$,
on the event $\tilde E_1 \cap \tilde E_2(M) \cap \eqref{eq:stuff-finite-3}$ the event $\tilde E_1 \cap \tilde E_2(M)$ occurs with $\tilde h$ in place of $h$. By <cit.>, the laws of $h$ and $\tilde h$, viewed modulo additive constant, are mutually absolutely continuous. Since the occurrence of the event $\tilde E_1 \cap \tilde E_2(M) \cap \tilde E_3$ is unaffected by adding a constant to $h$, we conclude that $\P[\tilde E_1\cap \tilde E_2(M) \cap \tilde E_3] > 0$, as required.

that adding the bump function did not change the occurrence of the events $\tilde E_1$ and $\tilde E_2$,
defined in (<ref>) and (<ref>) respectively.
The first inequality in $\tilde E_1$ is

\begin{align*}
D_{\tilde h}(Q', \partial Q) &= D_h(Q', \partial Q) \quad \mbox{($\phi \equiv 0$ on $\overline Q$)} \\
&> \sup_{x \in \partial B_{5}} D_h(0, x; (B_{2 \delta_0}(Q \cup W))^c) \quad \mbox{(event $\tilde E_1$ for $h$)} \\
&\geq e^{- \log(\frac{C_2}{C_1} \wedge 1)} \sup_{x \in \partial B_{5}} D_{\tilde h}(0, x; (B_{2 \delta_0}(Q \cup W))^c) \quad \mbox{(Weyl scaling)} \\
&\geq  \sup_{x \in \partial B_{5}} D_{\tilde h}(0, x; (B_{2 \delta_0}(Q \cup W))^c) \quad \mbox{($\log(\frac{C_2}{C_1} \wedge 1) \leq 0$)}.
\end{align*}

The second inequality in $\tilde E_1$ and the first inequality in $\tilde E_2$ are checked in a similar fashion, using
$\log(\frac{C_2}{C_1} \wedge 1) \leq 0$, and $\phi \equiv 0$ on $\overline{B_{2 \delta_0}}$.

Since $\phi \equiv 0$ on $U \cup \A_{3,6}$, the last inequality in $\tilde E_2$ is preserved.
Hence, on the event $\tilde E_1 \cap \tilde E_2(M) \cap \eqref{eq:stuff-finite-3}$ the event $\tilde E_1 \cap \tilde E_2(M)$ occurs with $\tilde h$ in place of $h$.

§ OBSTACLE PROBLEM FOR RADON MEASURES

In this appendix we provide the proofs which were omitted in Section <ref>.
For clarity, we prove these results for any Radon measure $\mu$ satisfying, for some $R > 0$,

\begin{equation} \label{eq:volume-growth}
r^{\beta^-} \leq \mu(B_r(z)) \leq r^{\beta^-} \quad \mbox{for all $z \in B_R$},
\end{equation}

for some exponents $\beta^+, \beta^- > 0$, for all $r$ sufficiently small. This implies the results in Section <ref> as the Liouville measure is a.s. a Radon measure which satisfies (<ref>). Indeed, this follows from Lemma <ref> and the scaling properties of $h$ and $\mu_h$, namely (<ref>) and (<ref>).

For $R > 0$, let

\begin{equation} \label{eq:expected-exit-time}
q_{B_R}(\cdot) = \int_{B_R} G_{B_R}(y,\cdot) \mu(dy),
\end{equation}

where $G_{B_R}$ is the Green's function for $B_R$.
Under the condition (<ref>), the function $q_{B_R}$ satisfies the following properties:

* Continuous: $q_{B_R}$ is Hölder continuous in $\overline{B_R}$ and finite;

* Potential: $q_{B_R}$ is superharmonic and $\Delta q_{B_R} = -\mu$  in $B_R$;

* Zero boundary: $q_{B_R}(z) = 0$ for $z \in \partial B_R$;

* Positive: $q_{B_R}(x) > 0$ for $x \in B_R$.

The first property follows by the same argument outlined in the proof of Proposition <ref>;
the second, by, for example,  <cit.>; the third as $G_{B_R}(0, \cdot) = 0$ on $\partial B_R$;
and the fourth by the strong maximum principle.

For notational simplicity, we consider $R = 1$ in all but the last subsection.

§.§ Definition

For $t > 0$,  denote the obstacle $\beta_t: \overline{B_1} \to \R \cup \{\infty\}$ as

\begin{equation} \label{eq:obstacle}
\beta_t(x) = -t G_{B_1}(0,x) + q_{B_1}(x).
\end{equation}

The set of  supersolutions

\begin{equation} \label{eq:shifted-supersolutions}
\mclSshift{}{t} = \{ w \in C(\overline{B_1}) :  \Delta w \leq 0 \mbox{ in $B_1$}  \mbox{ and }  w \geq   \beta_t \mbox{ in $\overline{B_1}$}\},
\end{equation}

where $C(\overline{B_1})$ denotes the set of continuous functions on the closed unit ball.

Consider the least supersolution or least superharmonic majorant
as the pointwise infimum of all functions in $\mclSshift{}{t}$

\begin{equation} \label{eq:appendix-least-supersolution}
\lssshift{}{t} = \inf\{ w \in \mclSshift{}{t} \}
\end{equation}

and the odometer

\begin{equation} \label{eq:appendix-limit-odometer}
v_t = \lssshift{}{t}  - \beta_t.
\end{equation}

Note that $\mclSshift{}{t}$ is non-empty as it contains $q_{B_1}$ — thus $\lssshift{}{t}$ always exists.
Denote the non-coincidence set by

\begin{equation} \label{eq:appendix-non-co-set}
\mathit{\Lambda_t} = \{ x \in B_1 : \lssshift{}{t} > \beta_t \}.
\end{equation}

Note that the least supersolution in (<ref>) is related
to (<ref>) by $\lss{}{t}= \lssshift{}{t} - q_{B_1}$. In particular, $\mathit{\Lambda_t}$
coincides with ${\cluster{}{t}}$ and $v_t$ with $\odometer{}{t}$  — each of the lemmas in Section <ref> will follow via this substitution.
We choose to work with $\mclSshift{}{t}$ as this allows us to directly cite results concerning superharmonic functions.

§.§ Existence

We first verify that the solution to the obstacle problem is non-degenerate in the following sense, this implies
Lemma <ref>.

For all $t > 0$, $\lssshift{}{t}$ is finite, continuous, and an element of $\mclSshift{}{t}$.

\begin{equation} \label{eq:lsc-leastsuperharmonic}
f_t  := \inf \{ g: g \mbox{ is superharmonic in $B_1$ and $g \geq \beta_t$ in $\overline{B}_1$} \},
\end{equation}

where, as before, the infimum is pointwise.
Note that this definition differs from $\lssshift{}{t}$ in that admissible superharmonic functions need only be lower semicontinuous.
It suffices to show that $f_t$ satisfies the desired properties. Indeed, by definition $f_t \leq  \lssshift{}{t}$
and the reverse inequality follows from $f_t \in \mclSshift{}{t}$.

Step 1: Finiteness.

If $g \geq \beta_t$ is superharmonic, then, by the minimum principle, $g \geq 0$ on $B_1$ as $g \geq \beta_t = 0$ on $\partial B_1$.
As this holds for all such $g$, $f_t \geq 0$. This together with $\infty > q_{B_1} \geq f_t$ ($q_{B_1}$ is admissible in (<ref>)) shows finiteness.

Step 2: Superharmonicity.

We use  <cit.>
which we recall for the reader's convenience. Let $O$ be a bounded open set and let $f:O \to [-\infty, \infty]$.
The lower semicontinuous regularization of $f$ is defined by

\begin{equation} \label{eq:lower-semicontinuous}
\hat{f}(x) = \min \{ f(x), \liminf_{y \to x} f(y) \}.
\end{equation}

<cit.> states that if $f > -\infty$ is the infimum of a family of superharmonic functions on $O$
then $\hat{f}$ is superharmonic on $O$ and $\hat{f}(x) = \liminf_{y \to x} f(y)$.

Since $f_t$ is finite, we may use this to see that its lower semicontinuous regularization, $\hat{f}_t$, is superharmonic on $B_1$ and satisfies $\hat{f}_t(x) = \liminf_{y \to x} f_t(y) \leq f_t(x)$.
In fact, (<ref>) implies $f_t$ is equal to its lower semicontinuous regularization. Indeed, $\hat{f}_t$ is superharmonic and

\[
\hat{f}_t(x) = \liminf_{y \to x} f_t(y) \geq \liminf_{ y \to x} \beta_t(y) =  \beta_t(x)
\]

as the obstacle, $\beta_t$, is continuous, implying that $\hat{f}_t \geq f_t$ by (<ref>).

Step 3: Continuity.

By Step 2, $f_t$ is lower semicontinuous. It remains to verify upper semicontinuity.
Let $\epsilon > 0$ and $x_0 \in B_1$ be given. By continuity of $\beta_t$, there exists $\delta > 0$ sufficiently small so that

\[
\beta_t(x) +\epsilon/2 \geq \beta_t(y), \quad \forall x,y \in B_{\delta}(x_0).
\]

and hence, since $f_t(x) \geq \beta_t(x)$,

\begin{equation} \label{eq:lower-bound-continuity}
\epsilon + f_t(x) \geq \beta_t(x_0) + \epsilon/2 \geq \beta_t(x), \quad \forall x \in B_{\delta}(x_0).
\end{equation}

Let $g_1$ be the unique function which is harmonic in $B_{\delta}(x_0)$ and coincides with $\epsilon + f_t$ on $\partial B_{\delta}(x_0)$.
Note that since $\epsilon + f_t - g_1$ is superharmonic in $B_{\delta}(x_0)$ and equal to 0 on $\partial B_{\delta}(x_0)$,

\begin{equation} \label{eq:superharmonic-poisson-inequality}
\epsilon + f_t \geq g_1, \quad \mbox{on $B_{\delta}(x_0)$}.
\end{equation}

Define the function $g$ to be $g_1$ on $B_{\delta}(x_0)$ and $\epsilon + f_t$ on $B_1 \backslash B_{\delta}(x_0)$.
One may check, using the super-mean-value property and (<ref>), that $g$ is superharmonic in $B_1$. Hence, by (<ref>), $g$ is admissible in (<ref>) and, in turn, $g \geq f_t$.

\begin{align*}
\limsup_{x \to x_0} f_t(x) &\leq \limsup_{x \to x_0} g(x) \quad \mbox{($g \geq f_t$ on $B_1$)} \\
&= g(x_0)\quad \mbox{($g$ is harmonic and hence continuous in a neighborhood of $x_0$)} \\
&\leq \epsilon + f_t(x_0) \quad \mbox{(by \eqref{eq:superharmonic-poisson-inequality} and definition of $g$)}.
\end{align*}

We conclude by observing the prior inequality holds for any $\epsilon > 0$.

We next check that the least supersolution is harmonic on the non-coincidence set, this together with Lemma <ref> implies
Lemma <ref>.

The non-coincidence set, ${\cluster{}{t}}$, is open and connected and

\[
\Delta \lssshift{}{t} = 0 \quad \mbox{on ${\cluster{}{t}}$}.
\]

Step 1: ${\cluster{}{t}}$ is open.

As $\lssshift{}{t}$ and $\beta_t$ are continuous and the disk $B_1$ is open, the set ${\cluster{}{t}}$
is open (this is the topological definition of a continuous function).

Step 2: $\lssshift{}{t}$ is harmonic on ${\cluster{}{t}}$.

If ${\cluster{}{t}}$ is empty, we are done, so suppose not. (This never happens but is proved later in Proposition <ref>.)
Further suppose for sake of contradiction that $\lssshift{}{t}$ is not harmonic on ${\cluster{}{t}}$. Since we know that $\lssshift{}{t}$ is superharmonic (Lemma <ref>), this means that $\lssshift{}{t}$ is not subharmonic on ${\cluster{}{t}}$.

The idea of the rest of the proof is the following. Since $\lssshift{}{t} > \beta_t$ on ${\cluster{}{t}}$, there is some extra room to `lower' $\lssshift{}{t}$. If $\Delta \lssshift{}{t}(z) < 0$ at some $z \in {\cluster{}{t}}$, then we can decrease $\lssshift{}{t}$ around $z$ by bending it up just enough to not break superharmonicity. This contradicts the minimality of $\lssshift{}{t}$.
We can't carry out this strategy literally since $\lssshift{}{t}$ is a priori not differentiable, so we instead use one of the equivalent definitions of subharmonic.

Here are the details. Since we are assuming that $\lssshift{}{t}$ is not subharmonic, by, ,
<cit.>, there is some $z \in {\cluster{}{t}}$ such that for every $R > 0$, there is a closed ball
$\overline{B_r}(z) \subset {\cluster{}{t}}$ of radius $r < R$ and a function $H:\overline{B_r}(z) \to \R$ which is continuous in $\overline{B_r}(z)$ and harmonic in $B_r(z)$ such that

\begin{equation}  \label{eq:smaller}
H \geq \lssshift{}{t} \quad \mbox{on $\partial B_r(z)$}
\end{equation}

\begin{equation} \label{eq:below-w}
\lssshift{}{t}(x_0) > H(x_0)
\end{equation}

for some $x_0 \in B_r(z)$.

Next, since $\lssshift{}{t}$ and $\beta_t$ are continuous and $\overline{B_R}(z) \subset {\cluster{}{t}}$, for $R$ sufficiently small,

\begin{equation} \label{eq:lies-way-above}
\inf_{x \in \partial B_r(z)} \lssshift{}{t}(x) > \sup_{y \in \overline{B_r}(z)} \beta_t(y), \quad\forall r < R .
\end{equation}

Fix $r > 0$ small so that (<ref>), (<ref>), and (<ref>) hold.  As $H$ is harmonic,

\[
\inf_{x \in B_r(z)} H(x) \geq \inf_{y \in \partial B_r(z)} H(y) \geq  \inf_{x \in \partial B_r(z)} \lssshift{}{t}(x) > \sup_{y \in \overline{B_r}(z)} \beta_t(y),
\]

in particular,

\begin{equation} \label{eq:above-obstacle}
H > \beta_t \quad \mbox{in $\overline{B_r}(z)$}.
\end{equation}

The above inequalities allow us to `lower' $\lssshift{}{t}$ using $H$.  Indeed, take the function $\psi: \overline{B_r}(z) \to \R$ defined by

\begin{equation}
\psi := \min(H, \lssshift{}{t})
\end{equation}

and note that by continuity and (<ref>), $\psi = \lssshift{}{t}$ in a neighborhood of $\partial B_r(z)$. In particular,
we may continuously extend $\psi$ to all of $\overline{B_1}$ by defining $\psi = \lssshift{}{t}$ on $\overline{B_1} \backslash B_r(z)$.
As $H$ is harmonic in $B_r(z)$ and $\lssshift{}{t}$ is superharmonic, this extension $\psi$ is superharmonic. Also, by (<ref>) $\psi \geq \beta_t$.
This shows that $\psi \in \mclSshift{}{t}$. However, by (<ref>),  $\lssshift{}{t}(x_0) > \psi(x_0)$,
contradicting the minimality of $\lssshift{}{t}$.

Step 3: ${\cluster{}{t}}$ is connected.

Otherwise there is a connected component of ${\cluster{}{t}}$ not containing the origin
upon which $\odometer{}{t}$ is non-zero, subharmonic, and 0 on its boundary — this violates the strong maximum principle.

§.§ Monotonicity

We now check monotonicity, this proves Lemma <ref>.

If $t_1 \leq t_2$, then $ \cluster{}{t_1} \subseteq  \cluster{}{t_2}$.

Recall that the odometer can be expressed as $\odometer{}{t} = \lssshift{}{t} - \beta_t$.
Showing monotonicity is equivalent to verifying $\odometer{}{t_1} \leq \odometer{}{t_2}$.
Unpack the difference to see that

\begin{align*}
\odometer{}{t_2} -  \odometer{}{t_1} &= \lssshift{}{t_2} - \lssshift{}{t_1} + \beta_{t_1} - \beta_{t_2} \\
&= \lssshift{}{t_2} - \lssshift{}{t_1} + (-t_1+t_2) G_{B_1}(0,\cdot).
\end{align*}

This motivates considering the superharmonic function

\[
\tilde{s} := \lssshift{}{t_2} + (t_2 - t_1) G_{B_1}(0,\cdot).
\]

In particular, we have

\[
\odometer{}{t_2} - \odometer{}{t_1} = \tilde{s}-\lssshift{}{t_1},
\]

thus it suffices to show

\begin{equation} \label{eq:lower-bound}
\tilde{s} \geq \lssshift{}{t_1}.
\end{equation}

This inequality follows from the obstacle problem. Indeed,

\[
\lssshift{}{t_2} \geq \beta_{t_2} =  q_{B_1} - t_2 G_{B_1}(0,\cdot)
\]

we have, after plugging in the definition of $\tilde{s}$,

\[
\tilde{s} \geq q_{B_1} - t_1 G_{B_1}(0,\cdot) = \beta_{t_1}.
\]

Therefore, $\tilde{s} \in \mclSshift{}{t_1}$ and by minimality of $\lssshift{}{t_1}$
we have (<ref>).

§.§ Conservation of mass

In this section we prove that no mass comes in from the boundary, that is, $\mu_h({\cluster{}{t}}) \leq t$.
This establishes Lemma <ref>.
To that end, we observe that the odometer is 0 on the boundary of the domain.

For all $t > 0$, $\odometer{}{t} = 0$ on $\partial B_1$.

This is immediate from $q_{B_1} \in  \mclSshift{}{t}$ and $q_{B_1} = G_{B_1}(0,\cdot) = 0$ on $\partial B_1$

We then use this together with the definition of weak normal derivative to prove the desired claim.

For all $t > 0$, $\mu({\cluster{}{t}}) \leq t$. Moreover, if $\overline{{{\cluster{}{t}}}} \subset B_1$ and $\mu(\partial {\cluster{}{t}}) = 0$, then $\mu_h({\cluster{}{t}}) = t$.

Fix $t > 0$ and recall that ${\cluster{}{t}}$ is an open set. By Lemma <ref>, Lemma <ref>,
the superharmonicity of $\lssshift{}{t}$, and the definition of $\odometer{}{t}$,

\begin{equation} \label{eq:laplacian-of-v}
\begin{cases}
\odometer{}{t} = 0  &\mbox{on $\partial B_1$} \\
\Delta \odometer{}{t} = -t \delta_0 + \mu |_{_{{{\cluster{}{t}}}}} + \nu |_{_{\partial {\cluster{}{t}}}} &\mbox{on $B_1$}
\end{cases}
\end{equation}

where $0 \leq \nu \leq \mu$ is a Radon measure which is absolutely continuous with respect to $\mu$. Hence,
$\odometer{}{t}$ solves a linear Dirichlet problem on $B_1$, and so, by  <cit.>,
there exists a weak normal derivative $\frac{ \partial \odometer{}{t}}{\partial n}$ so that

\begin{equation} \label{eq:weak-integration-by-parts}
-t + \mu({\cluster{}{t}}) + \nu(\partial {\cluster{}{t}}) = \int_{\partial B_1} \frac{ \partial \odometer{}{t}}{\partial n} d \sigma,
\end{equation}

where $d \sigma$ denotes integration with respect to surface measure.
Since $\odometer{}{t} \geq 0$ on $B_1$ and $\odometer{}{t} = 0$ on $\partial B_1$, by  <cit.>,

\begin{equation}
\frac{ \partial \odometer{}{t}}{\partial n} \leq 0
\end{equation}

almost everywhere with respect to the surface measure. Moreover,

\begin{equation}
\overline{{{\cluster{}{t}}}} \subset B_1 \implies \frac{ \partial \odometer{}{t}}{\partial n} = 0,
\end{equation}

as the weak normal derivative coincides with the classical normal derivative if it exists.
In particular, by (<ref>),

\begin{equation}
-t + \mu({\cluster{}{t}}) + \nu(\partial {\cluster{}{t}}) \leq 0,
\end{equation}

\begin{equation}
\overline{{{\cluster{}{t}}}} \subset B_1 \implies -t + \mu({\cluster{}{t}}) + \nu(\partial {\cluster{}{t}}) = 0,
\end{equation}

completing the proof as $0 \leq \nu \leq \mu$.

§.§ Compatibility

We prove Lemma <ref> in this section.

As previously mentioned, the results proved so far in this appendix apply to $\lss{B_R}{t}, \cluster{B_R}{t}, \odometer{B_R}{t}$ as long as (<ref>)
is satisfied for $\mu$ in $B_R$. That is, if (<ref>) is satisfied, then $\lss{B_R}{t} \in \mclS{B_R}{t}$ and

\begin{equation} \label{eq:laplacian-of-v-general}
\begin{cases}
\odometer{B_R}{t} = 0  &\mbox{on $\partial B_R$} \\
\Delta \odometer{B_R}{t} = -t \delta_0 + \mu |_{_{\cluster{B_R}{t}}} + \nu |_{\partial \cluster{B_R}{t}} &\mbox{on $B_R$}
\end{cases}
\end{equation}

for a Radon measure $0 \leq \nu \leq \mu$ which is absolutely continuous with respect to $\mu$.

Suppose (<ref>) is satisfied for fixed $R > 0$.
For all $s_1 \leq R$, if, for some $s_2 \in [s_1, R]$, we have $\overline{\cluster{B_{s_2}}{t}} \subset B_{s_1}$,
then  $\cluster{B_{s}}{t} = \cluster{B_{s_2}}{t}$ for all $s \in [s_1, R]$.

Fix $R > 0$. Note that if (<ref>) is satisfied for $R$, then it is satisfied for all $s \leq R$.
meaning $\lss{B_s}{t} \in \mclS{B_s}{t}$ and (<ref>) holds for $\odometer{B_s}{t}$ for all $s \leq R$.  Let $s_1 \leq R$ be given and fix $s_2 \in [s_1, R]$ for which we have $\overline{\cluster{B_{s_2}}{t}} \subset B_{s_1}$.

We first claim that

\begin{equation} \label{eq:monotonicity-domains}
\odometer{B_{r_1}}{t} \leq \odometer{B_{r_2}}{t} \quad \mbox{in $B_{r_1}$}, \quad \forall r_1 \leq r_2 \leq R.
\end{equation}

To prove (<ref>), fix $r_1 \leq r_2 \leq R$ and write

\begin{equation} \label{eq:intermediate-inequality-monotonicity}
\odometer{B_{r_1}}{t} - \odometer{B_{r_2}}{t} =  \lss{B_{r_1}}{t} -  \lss{B_{r_2}}{t} + t G_{B_{r_1}}(0,\cdot) - t G_{B_{r_2}}(0,\cdot) = \lss{B_{r_1}}{t} - \hat{w}_t
\end{equation}

\begin{equation}
\hat{w}_t := \lss{B_{r_2}}{t} - t (G_{B_{r_1}}(0,\cdot) - G_{B_{r_2}}(0,\cdot))  .
\end{equation}

Since $\lss{B_{r_2}}{t} \in \mclS{B_{r_2}}{t}$ and $\Delta G_{B}(0,\cdot) = -\delta_0$ in $B$ for any ball $B$,

\begin{equation}
\Delta \hat{w}_t = \Delta \lss{B_{r_2}}{t} - t (\Delta G_{B_{r_1}}(0,\cdot) - \Delta G_{B_{r_2}}(0,\cdot)) \leq \mu  \quad \mbox{in $B_{r_1}$}
\end{equation}

\begin{equation}
\hat{w}_t = (\lss{B_{r_2}}{t}  + t G_{B_{r_2}}(0,\cdot)) - t G_{B_{r_1}}(0,\cdot)  \geq - t G_{B_{r_1}}(0,\cdot) \quad \mbox{in $B_{r_1}$}.
\end{equation}

Therefore, $\hat{w}_t \in \mclS{B_{r_1}}{t}$, which shows $\lss{B_{r_1}}{t} \leq \hat{w}_t$ and hence (<ref>) by (<ref>).

For the other direction, we use the hypothesis $\overline{\cluster{B_{s_2}}{t}} \subset B_{s_1}$. This together with (<ref>)
implies $\odometer{B_{s_1}}{t}$ is identically zero in a neighborhood of $\partial B_{s_1}$ and so can be extended
by 0 to be harmonic in $B_R \backslash B_{s_1}$. Also, observe that (<ref>) implies $\cluster{B_{s_1}}{t} \subset \cluster{B_R}{t}$ by definition.  Therefore, by (<ref>) for $B_{s_1}$ and $B_R$, $\odometer{B_R}{t} - \odometer{B_{s_1}}{t}$ is subharmonic in $\cluster{B_R}{t}$ and 0 on its boundary
which shows

\begin{equation} \label{eq:monotonicity-domains-two}
\odometer{B_R}{t} \leq \odometer{B_{s_1}}{t}.
\end{equation}

Combining (<ref>) and (<ref>) shows that

\begin{equation}
\odometer{B_{R}}{t} = \odometer{B_{s_1}}{t} \leq \odometer{B_{s}}{t} \leq \odometer{B_{r}}{t} \leq \odometer{B_R}{t}, \quad \forall s_1 \leq s \leq r \leq R,
\end{equation}

completing the proof by the definition of ${\cluster{}{t}}$.