The random variable $L$ is not a sum of independent random variables, but each walk which contributes to the sum in $L$ can be mapped to the point at which it exits the current cluster. Therefore,
by the strong Markov property,

\begin{equation} \label{eq:strong-markov-reduction}
L \leq \hat{L} := \sum_{a \in O^{{\eps}}_2} 1^a\{\tau_b < T\}
\end{equation}

where $1^a\{\tau_b < T\}$ denotes the event that an independent walker started at $a \in {\VGeps}$ hits site $b$ before it leaves $O^{{\eps}}_2$.
Hence, by (<ref>) we have that

\begin{equation} \label{eq:expected-value-of-l}
\E[\hat{L} | h , \eta] = \sum_{a \in O^{\eps}_2} \P^a[\tau_b < T | h, \eta] =  \frac{ \sum_{a \in O^{{\eps}}_2} \Gr{{\eps}}{O_2^{{\eps}}}(a, b)}{ {\Gr{{\eps}}{O_2^{{\eps}}}(b, b)}}.
\end{equation}

By the assumption (<ref>), we have that (<ref>) and  (<ref>) imply that
there exists a deterministic $q > 0$ and $\sigma(h,\eta)$-measurable finite random variable $\mathfrak{s} > 0$ so that with probability approaching one as ${\eps} \to 0$,

\begin{equation} \label{eq:m-bounded-below-by-l}
\E[M | h ,\eta] \geq (1 + \mathfrak{s}) \E[\hat{L} | h , \eta] \geq (1 + \mathfrak{s}) {\eps}^{-q},
\end{equation}

with the latter inequality following by the assumption (<ref>).

Recall that if $S$ is a finite sum of independent indicators with mean $\mu$, then

\begin{equation} \label{eq:concentration-inequality}
\P\left[ |S - \mu| \geq \mu^{3/4} \right] \leq 2 \exp\left( -\frac{1}{4} \mu^{1/2} \right)
\end{equation}

for all sufficiently large $\mu$ (see, , <cit.>).  Now, set $k = (1 + \mathfrak{s}/4) \E[\hat{L} |h,\eta]$ in (<ref>)
and compute

\begin{align*}
&\P\left[L \geq (1 + \mathfrak{s}/4) \E[\hat{L} | h , \eta] | h , \eta \right] \\
&\leq \P \left[ \hat{L} \geq (1 + \mathfrak{s}/4) \E[\hat{L} | h ,\eta] | h , \eta \right] \quad \mbox{(by \eqref{eq:strong-markov-reduction})} \\
&\leq \P \left[ \hat{L} \geq \E[\hat{L} | h ,\eta] + \E[\hat{L} | h ,\eta]^{3/4} | h , \eta \right]
\quad\mbox{(by \eqref{eq:m-bounded-below-by-l} for all ${\eps}$ small depending on $\mathfrak{s}$)} \\
&\leq 2 \exp \left(-\frac{1}{4} \E[\hat{L} | h , \eta]^{1/2}\right)  \quad \mbox{(by \eqref{eq:concentration-inequality} with $\mu = \E[\hat{L} | h, \eta])$} \\
&\leq 2 \exp\left(-\frac{1}{4} {\eps}^{-q/2} \right) \quad \mbox{(by \eqref{eq:m-bounded-below-by-l})}.
\end{align*}

\begin{align*}
&\P \left[ M \leq (1 + \mathfrak{s}/4) \E[\hat{L} | h , \eta] | h , \eta \right]  \\
&\leq  \P \left[ M \leq \E[M | h , \eta] - \E[M | h , \eta]^{3/4} | h , \eta \right]
\quad\mbox{(by \eqref{eq:m-bounded-below-by-l} for all ${\eps}$ small depending on $\mathfrak{s}$)} \\
&\leq 2 \exp\left(-\frac{1}{4} {\eps}^{-q/2} \right)  \quad \mbox{(by \eqref{eq:m-bounded-below-by-l} and \eqref{eq:concentration-inequality} with $\mu = \E[M | h, \eta])$}.
\end{align*}

Therefore, if we set $k = (1 + \mathfrak{s}/4) \E[\hat{L} | h, \eta]$ and use (<ref>), we have (<ref>), completing the proof.

§ IDLA UPPER BOUND

As in the previous section we fix a $\rho \in (0,1)$ so that $B_{\rho} \Subset B_1$.
Recall the notation for the subset of $\C$ corresponding to a subset of ${\VGeps}$, (<ref>).
Here we prove that the IDLA cluster is contained in a harmonic ball.

Recall the time $\TT$ from (<ref>).
For each $\delta \in (0, \rho)$ and $t > 0$, on the event $\{t < \TT\}$,
it holds except on an event of probability tending to 0 as ${\eps} \to 0$  that

\[
\overline{A}_{{\eps}}(\lfloor t {\eps}^{-1} \rfloor) \subset \Bdeltap(\Lambda_t),
\]

where $\Lambda_t$ is as in Theorem <ref>.

Our proof of the upper bound combines ideas of Lawler <cit.> and Duminil-Copin-Lucas-Yadin-Yehudayoff <cit.>.
The idea is utilize the diffusive, `smoothing' nature of IDLA. That is, it is extremely unlikely that a random walk travels a
long distance without hitting many sites while doing so. Since the IDLA cluster is built by running independent random walks,
the cluster itself has this same property.

In our setting, we would like to use this smoothing to assert the following: given the current cluster $\Theta \subset {\VGeps}(B_{\rho})$, after adding $\delta {\eps}^{-1}$ walkers to the origin, the resulting cluster must be contained in ${\VGeps}(\Bdeltapp{C \delta^c}(\overline{\Theta}))$, (for some constants $c, C$) with extremely high probability.

Duminil-Copin-Lucas-Yadin-Yehudayoff <cit.> showed that IDLA does indeed have this property if the underlying graph
is regular enough. Roughly, they required the graph to obey volume doubling and have an intrinsic metric which can be controlled by the Euclidean metric.
Unfortunately, these required properties do not hold on the mated-CRT map. In particular, with positive probability, it is possible for random walkers
to travel a long Euclidean distance in the mated-CRT map without hitting many cells. Fortunately, this is not the case
for `most' of the mated-CRT map.

We formalize `most' by considering the behavior of random walk on a Euclidean annulus intersected with the mated-CRT map (under the usual SLE/LQG embedding).
Roughly, for a `good' annulus, for all sufficiently small ${\eps}$, it is difficult for a random walk on ${\VGeps}$
to travel across the annulus without hitting many cells along the way.

We start in Section <ref> by defining what it means for an annulus to be good.
We then use the definition of good annulus in Section <ref> to provide
estimates for the hitting measure of random walk on the mated-CRT map. We then use these estimates
together with an iterative argument to prove the upper bound in Section <ref>.

Below we frequently make use of the Abelian nature of IDLA. This Abelian property was first observed by Diaconis-Fulton in <cit.>
and states that the distribution of the final IDLA cluster is unaffected by the order in which walkers are sent out. In particular,
we may use this to `start' and `stop' walkers as they go through certain sets.  To that end, we introduce the following definitions.
For continuity of literature, we use the same notation as <cit.>.

Given sets $\Theta \subset \Theta' \subset {\VGeps}$, and vertices $a_1, \ldots, a_k \in \Theta$,
denote by

\[
A(\Theta; a_1, \ldots, a_k \to \Theta') \subset {\VGeps}
\]

the IDLA aggregate with the following initial condition: the set $\Theta$ is completely occupied, $k$ independent simple random walks are started at $a_1, \ldots, a_k$, and the walkers are stopped upon exiting $\Theta'$. Let

\[
P(\Theta; a_1, \ldots, a_k \to \Theta') \subset \partial \Theta'
\]

be the final positions of those walkers among $a_1, \ldots, a_k$, which are stopped before being absorbed into the aggregate.
We write

\[
A(\Theta; a_1, \ldots, a_k) := A(\Theta; a_1, \ldots, a_k \to {\VGeps}),
\]

when the walkers are not stopped, , $\Theta' = {\VGeps}$.

In this notation, the Abelian property of IDLA is the following.

For every $\sigma(h,\eta)$-measurable choice of $\Theta \subset \Theta' \subset {\VGeps}$ and $a_1, \ldots, a_k \subset \Theta$, the conditional laws given $(h,\eta)$ of

\[
\begin{aligned}
A(\Theta; a_1, \ldots, a_k) \quad \text{and} \quad
A( A(\Theta; a_1, \ldots, a_k \to \Theta'); P(\Theta; a_1, \ldots, a_k \to \Theta'))
\end{aligned}
\]

agree. That is, the IDLA aggregate has the same (conditional) distribution as any restarted, stopped IDLA aggregate.

§.§ Good annuli

Let $\boldsymbol c_0 =\boldsymbol c_0 (\rho, \gamma) \in (0,1/5)$ be a deterministic constant chosen so that

\begin{equation} \label{eq:choice-of-constant-annulus}
C_0^{-1} \log \left(\frac{1}{\boldsymbol c_0} \right) -1 \geq \frac{1}{2}.
\end{equation}

where $C_0 = C_0(\rho, \gamma)$ is the constant from Lemma <ref>. This will be used in Lemma <ref> below.

For $z \in \C$, $r > 0$, and parameter $C >0$, let $E_r(z) := E_r(z; C)$ be the event that the following
is true:

\begin{equation} \label{eq:good-annulus-event}
\sup_{v \in \A_{r, 2r}(z)} \int_{\A_{r,2r}(z)} \left(\log \frac{1}{|u-v|} + 1 \right) d \mu_h(u) \leq C \min_{w \in \A_{r, 2 r}(z)} \mu_h(B_{\boldsymbol{c_0} r}(w)).
\end{equation}

An annulus $\A_{r, 2 r}(z)$ for which $E_r(z)$ holds it said to be good. In this subsection we prove that there are many good annuli in $B_{\rho}$.

There exists a constant $C$ depending only on $\gamma$ such that the following holds with polynomially high
probability as $\delta \to 0$.
For each $z \in ( B_{\rho - 3\sqrt{\delta}} \setminus B_{10 \sqrt{\delta}} ) \cap \frac{\delta}{100} \Z^2$ there is at least one radius $r \in [\delta,\delta^{1/2}] \cap \{7^{-n}\}_{n \in \N}$ for which $E_r(z)$ occurs,
where $E_r(z) = E_r(z;C)$ is as in (<ref>).

The argument is similar to <cit.>.  We start by showing that for each $z \in \C$ and $r > 0$, the event $E_r(z)$ occurs
with high probability provided $C > 0$ is chosen to be large (Lemma <ref>).
We will then use the near-independence of the GFF across disjoint concentric annuli (Lemma <ref>)
to show that for each fixed $z \in \C$, it holds with very high probability when $\delta$ is small that there is at least one radius $r \in [\delta, \delta^{1/2}] \cap \{7^{-n}\}_{n \in \N}$ for which $E_r(z)$ occurs.
Finally, we will take a union bound over all $z \in B_{\rho - 3\sqrt{\delta}} \setminus B_{10 \sqrt{\delta}} \cap \frac{\delta}{100} \Z^2$.

Throughout this section, we use the fact, recalled in (<ref>),  that in $B_1$ the $\gamma$-quantum cone field $h$ agrees in law with
the whole-plane GFF plus $-\gamma \log|\cdot|$ normalized so that its circle average over $B_1$ is 0. Also recall
the notation for annuli from (<ref>).

For each $p  \in (0,1)$, there exists $C  = C(p) > 0$ such that the event $E_r(z) := E_r(z; C)$ of (<ref>) satisfies

\[
\begin{aligned}
\P\left[ E_r(z) \right] \geq p,&\quad \forall r > 0, \\
&\quad \forall z \in\C \mbox{ such that $\A_{r/3, 3r}(z) \subset B_{\rho}$ and $\dist(\A_{r/3, 3r}(z), \{0\}) \geq r/100$}.
\end{aligned}
\]

Denote the left and right sides of the inequality in the event (<ref>)
by $\mathcal{L}^h_r(z)$ and $\mathcal{R}^h_r(z)$ respectively.  We first show that it suffices to prove the lemma with $h$ replaced by $h^{\C}$, , the GFF without a log-singularity ($\gamma = 0$ in (<ref>)). We then prove the lemma for $h^{\C}$.
Step 1: Reduction to $h^{\C}$.

Recall that $h$ restricted to $B_1$ agrees in law with $h^\C -\gamma \log |\cdot|$ where $h^\C$ is a whole-plane GFF.
So, we can couple $h^{\C}$ and $h$ so that $h|_{B_1} = h^\C|_{B_1} - \gamma\log|\cdot|$.
Suppose the statement of the lemma holds for $h^{\C}$ with $C_1 > 0$ in place of $C$.

Fix $r > 0$ and then $z$ such that the annulus $\A_{r/3, 3 r}(z)$ lies at Euclidean distance at least $r/100$ from the origin.
Note that $E_r(z)$ depends only on $h|_{\A_{r/3,3r}(z)}$. Thus, the log function is bounded above and below in the region of dependence of $E_r(z)$.
Hence, by Weyl scaling (Fact <ref>), there is a constant $A>0$ depending only on $\gamma$ such that

\begin{equation}
\frac{\mathcal L^{h}_{r}(z)}{\mathcal{R}^h_r(z)} \leq A  \frac{\mathcal L^{h^\C}_{r}(z)}{\mathcal{R}^{h^\C}_r(z)}.
\end{equation}

\[
\frac{\mathcal L^{h^\C}_{r}(z)}{\mathcal{R}^{h^\C}_r(z)}  \leq C_1 \implies \frac{\mathcal L^{h}_{r}(z)}{\mathcal{R}^h_r(z)} \leq C_1 A
\]

and hence (<ref>) occurs for $C := A \times C_1$ if it occurs for $h^{\C}$.
Step 2: Whole-plane GFF.

We use the fact that the law of $  h^{\C}$ is both scale and translation invariant modulo additive constant (<ref>). By the Weyl scaling property of the measure $\mu_h$ (Fact <ref>), the event $E_r(z)$ is a.s. determined by $h$ viewed modulo additive constant. From this and the LQG coordinate change formula for $\mu_h$, we infer that the probability of $E_r(z)$ does not depend on $r$ or $z$. Hence, it suffices to find $C > 0$ as in the lemma statement such that $\P[E_1(0)] \geq p$. This however, follows immediately from the fact that $\frac{\mathcal{L}^h_1(0)}{\mathcal{R}^h_1(0)}$ is a positive, finite random variable.

The following lemma is a special case of <cit.>.

Fix $0 < s_1<s_2 < 1$. Let $\{r_k\}_{k\in\N}$ be a decreasing sequence of positive numbers such that $r_{k+1} / r_k \leq s_1$ for each $k\in\N$ and let $\{E_{r_k} \}_{k\in\N}$ be events such that $E_{r_k}$ is a.s. determined by $h |_{\A_{s_1 r_k , s_2 r_k}(0)  } $, viewed modulo additive constant, for each $k\in\N$.
For $K\in\N$, let $N(K)$ be the number of $k\in [1,K] \cap \Z$ for which $E_{r_k}$ occurs.
For each $\alpha > 0$ and $p = p(\alpha,s_1,s_2) \in (0,1)$ and $C = C(\alpha,s_1,s_2) > 0$ (independent of the particular choice of $\{r_k\}$ and $\{E_{r_k}\}$) such that if

\begin{equation} \label{eqn-annulus-iterate-prob}
\P\left[ E_{r_k}  \right] \geq p , \quad \forall k\in\N  ,
\end{equation}

\begin{equation} \label{eqn-annulus-iterate}
\P\left[ N(K)  = 0 \right] \leq C e^{-\alpha K} ,\quad\forall K \in \N.
\end{equation}

We now prove the desired claim.

The event $E_r(z)$ depends only on the measure $\mu_h|_{\A_{r/3,3r}(z)}$
Moreover, multiplying this measure by a constant does not change whether $E_r(z)$ occurs.
Therefore, $E_r(z)$ is a.s. determined by $h|_{\A_{r/3,3r}(z)}$ viewed modulo additive constant.

We now apply Lemma <ref> with $K = \lfloor \log_7  \epsilon^{-1/2} \rfloor$, the radii $r_1,\dots,r_K \in [ \delta, \delta^{1/2}] \cap \{7^{-n}\}_{n\in\N}$, the events $E_{r_k} = E_{r_k}(z)$, and an appropriate universal constant choice of $\alpha$. We find that there exist universal constants $p\in (0,1)$ and $c>0$ such that if $\P[E_r(z)] \geq p$
for each $r > 0$ and each $z \in B_{\rho} \setminus B_{10 r}$,  then for all $z \in B_{\rho} \setminus B_{10 \sqrt{\delta}}$,

\begin{equation} \label{eqn-annulus-hit-iterate}
\P\left[ \text{$E_r(z)$ occurs at least once for $r \in [ \delta, \delta^{1/2}] \cap \{7^{-n}\}_{n\in\N}$} \right] \geq 1 - O_\delta(\delta^3)
\end{equation}

with a universal implicit constant in the $O_\delta(\cdot)$.

By Lemma <ref>, there exists $C = C(\gamma) > 0$ such that for this choice of $C$, one has $\P[E_r(z)] \geq p$ for each $r > 0$ and each $z \in B_{\rho} \setminus B_{10 r}$. Therefore, the estimate (<ref>) holds for this choice of $C$. We then conclude by means of a union bound over all $z\in  (B_{\rho-3 \sqrt{\delta}} \setminus B_{10 \sqrt{\delta}}) \cap \frac{\delta}{100} \Z^2$.

§.§ Harmonic measure estimates

Equipped with the prevalence of good annuli, we are now ready to prove the following random walk estimate.
Roughly, it states that with probability bounded away from zero, random walk cannot travel through an annulus of constant size without hitting at least a mesoscopic number of vertices.	See Figure <ref> for an illustration.

Much stronger versions of this estimate have been proved on $\Z^d$ <cit.> and the supercritical percolation cluster <cit.>.

There exists deterministic $\omega = \omega(\gamma) > 0$ so that for each $s_0 \in (0,1)$, there exists $s_1, s_2 \in (0,1)$
and a random  $\sigma(h)$-measurable $\boldsymbol{\delta}>0$ such that with probability approaching one as ${\eps}$ goes to zero, the following is true for all $r \in [10 {\eps}^{\omega}, \boldsymbol{\delta}]$ and all sets $\Theta \subset {\VGeps}$ such that $\overline{\Theta} \Subset B_{\rho/2}$ and $B_{s_0} \subset \overline{\Theta}$ with $\overline{\Theta}$ as in (<ref>).

Let $a \in \cl(\Theta)$ and let $S \subset {\VGeps}(\overline{\Theta} + B_r)$ be such that $|S \setminus \Theta| < s_1 r^{2 \beta^-} {\eps}^{-1}$
where $\beta^-$ is as in Lemma <ref>.
Let $\Xi$ denote the trace of a random walk started at $a$ and stopped upon exiting
${\VGeps}(\overline{\Theta} + B_r)$. Then,

\[
\P[\Xi \cap {\VGeps}(\overline{\Theta} + B_r) \setminus (S \cup \Theta) \neq \emptyset | h, \eta] \geq s_2.
\]

A law of large numbers version of this estimate is Lemma <ref> below and this will be used to show that the IDLA cluster
cannot develop long tentacles in Lemma <ref>.

The set $\Theta$ in Lemma <ref> may be thought of as the current IDLA cluster
and $S$ as the cluster after releasing an additional, small number of walkers (fewer than $s_1 r^{2 \beta^-} {\eps}^{-1}$).
The lemma states that with (conditional) probability bounded from below by a constant, $s_2$, which is independent
of $r$ and ${\eps}$, a random walk started at an arbitrary vertex in $\cl(\Theta)$
will be absorbed into the intermediate cluster (, hit a vertex not in $S \cup \Theta$) before exiting the $B_r$ neighborhood
of the initial cluster, ${\VGeps}(\overline{\Theta} + B_r)$. In the case when $\Theta$ is a large IDLA cluster, the condition $B_{s_0} \subset \overline{\Theta}$
is satisfied for some small $s_0 \in (0,\rho)$ for all small ${\eps}$ due the lower bound, Proposition <ref>
and the corresponding property for harmonic balls, Theorem <ref>.

To prove the divisible sandpile upper bound in Section <ref>, we require the following stronger version of Lemma <ref>.

There exists deterministic $\omega = \omega(\gamma) > 0$ so that for each $s_0 \in (0,1)$, there exists $s_1, s_2 \in (0,1)$
and a random  $\sigma(h)$-measurable $\boldsymbol{\delta}>0$ such that with probability approaching one as ${\eps}$ goes to zero, the following is true for all $r \in [10 {\eps}^{\omega}, \boldsymbol{\delta}]$ and all sets $\Theta \subset {\VGeps}$ such that $\overline{\Theta} \Subset B_{\rho/2}$ and $B_{s_0} \subset \overline{\Theta}$ with $\overline{\Theta}$ as in (<ref>).

For all $A \subset {\VGeps}(\overline{\Theta} + B_{r}) \setminus \Theta$
such that, for all $z \in \C$,  $|A \cap {\VGeps}(B_r(z))| < s_1 r^{2 \beta^-} {\eps}^{-1}$, where $\beta^-$ is as in Lemma <ref>, we have

\[
\sup_{a \in {\VGeps}(\partial (\overline{\Theta} + B_{r/2}))} \P[\mbox{$X^{a, {\eps}}$ exits $ {\VGeps}(\overline{\Theta} + B_{r}) \setminus \Theta $ before exiting $A$} | h , \eta] \leq  1 - s_2 ,
\]

where $X^{a, {\eps}}$ is simple random walk started at $a \in {\VGeps}$.

By the strong Markov property and Lemma <ref>, the trace of a random walk started at $a \in \cl(\Theta)$
contains the trace of a random walk started at some site $b \in {\VGeps}(\partial (\overline{\Theta} + B_{r/2}))$ with probability approaching one as ${\eps} \to 0$. Moreover, if $S \subset {\VGeps}(\overline{\Theta} + B_r)$ satisfies $|S \setminus \Theta| < s_1 r^{2 \beta^-} {\eps}^{-1}$
then $A := S   \setminus \Theta$ satisfies  $|A| < s_1 r^{2 \beta^-} {\eps}^{-1}$.
Therefore, for all $z \in \C$,  $|A \cap {\VGeps}(B_r(z))| < s_1 r^{2 \beta^-} {\eps}^{-1}$.

The previous three sentences imply, by Lemma <ref> (under the conditions of the lemma)

\[
\P[\Xi \cap {\VGeps}(\overline{\Theta} + B_r) \setminus (S \cup \Theta) = \emptyset | h, \eta] \leq  1-s_2 ,
\]

which implies Lemma <ref>.

An illustration of the statement in Lemma <ref>. The filled in cells corresponding to vertices in the cluster $\overline{\Theta}$ are shown in light blue and a neighborhood, $\overline{\Theta} + B_r$ is outlined by a dashed black line.  The set $\overline{S \setminus \Theta}$ is shown in red. Random walk started at a vertex in $\cl(\Theta)$
is unlikely to exit ${\VGeps}(\overline{\Theta} + B_r)$ via paths which stay in $S$.

Let $\omega = \omega(\gamma) > 0$ be as in Lemma <ref>.
We may assume, by possibly taking $\omega$ smaller, that $2 \beta^{-} \omega \leq \alpha$, where $\beta^-$ and $\alpha$ are as in
Lemmas <ref> and <ref>. This will be used in (<ref>) below.

Let $E_{r}(z) = E_r(z; C)$
and $C > 0$ be the event and parameter from Lemma <ref>.
By Lemma <ref>, it holds with polynomially high probability as $\delta \to 0$ that for each $z \in (B_{\rho - 3 \sqrt{\delta}} \setminus B_{10 \sqrt{\delta}}) \cap \frac{\delta}{100} \Z^2$, there exists $\mathbf{r}_z = \mathbf{r}_z(\delta) \in [\delta,\delta^{1/2}]$ such that $E_{\mathbf{r}_z}(z)$ occurs.
By Lemma <ref>, it also holds with polynomially high probability as $\delta \to 0$ that

\begin{equation} \label{eq:volume-growth-lower-bound}
\mu_h\left( B_{\delta}(x) \right)  \geq \delta^{\beta^-}, \quad \forall x \in B_{1 - \delta}.
\end{equation}

By the Borel-Cantelli lemma, a.s. there exists a random $\boldsymbol{\delta}' > 0$ such that the preceding two sentences hold for all dyadic $\delta < \boldsymbol{\delta}'$.
We restrict to the event that the statements in Lemmas <ref>, <ref>, and <ref> all hold, which happens with probability tending to 1 as ${\eps}\to 0$.  (For Lemma <ref> we suppose that convergence holds for some countable family of polynomials which are dense in the uniform topology in $B_{\rho}$.) Our calculations below will require ${\eps}$ to be small in a deterministic way, so we fix a small deterministic $\boldsymbol{\pi}$ (which we will determine below) and an ${\eps} < \boldsymbol{\pi}$.

Step 1: Reduction to random walk in a good annulus.

We surround $\Theta$ by good annuli. By our assumptions on $\Theta$, by possibly taking $\boldsymbol{\pi}$ smaller, depending only on $s_0$, we can arrange that

\begin{equation} \label{eq:boundary-inside}
(\overline{\Theta} + B_{20 \sqrt{\delta }}) \setminus (\overline{\Theta} + B_{\sqrt{\delta }}) \Subset  B_\rho \setminus B_{10 \sqrt{\delta }}, \quad \forall \delta < \boldsymbol{\pi} \wedge \boldsymbol{\delta}'.
\end{equation}

By (<ref>) and the fact that $\mathbf{r}_z \in [\delta , \delta^{1/2}]$, we obtain that for each $\delta < \boldsymbol{\pi} \wedge \boldsymbol{\delta}'$ there exists $\mathbf{Z} = \mathbf{Z}(\delta) \subset (B_{\rho - \delta } \setminus B_{10 \sqrt{\delta }})  \cap \frac{\delta }{100} \Z^2$ for which

\begin{equation} \label{eq:cover-continuous}
(\overline{\Theta} + B_{4 \sqrt{\delta }}) \setminus (\overline{\Theta} + B_{3 \sqrt{\delta }}) \Subset \bigcup_{{z} \in \mathbf{Z}} B_{\mathbf{r}_{{z}}/2}({z}) \Subset B_{\rho- 10 \sqrt{\delta}}
\end{equation}

\begin{equation} \label{eq:cover-in-complement}
\bigcup_{{z} \in \mathbf{Z}} B_{3 \mathbf{r}_{{z}}}({z}) \Subset (\overline \Theta + B_{2 \sqrt{\delta}})^c.
\end{equation}

We now fix a dyadic $\delta \in (10 {\eps}^{\omega}, \boldsymbol{\pi} \wedge \boldsymbol{\delta}')$
and let $r \in [\delta, \sqrt{\delta}] < \rho/4$. We continue with this choice of $r$ as the lemma statement follows once we choose
$\boldsymbol{\delta} = \sqrt{\boldsymbol{\pi} \wedge \boldsymbol{\delta}'}$.

By (<ref>) and the strong Markov property of simple random walk, it suffices to prove the lemma
with $a$ replaced by $b \in {\VGeps}(\partial B_{{(4/3)} \mathbf{r}_{z}}({z}))$ for ${z} \in \mathbf{Z}$. We in fact prove the following stronger statement:
a simple random walk $X^{b, {\eps}}_t$ started at such $b$ hits

\[
\mathfrak{E} := {\VGeps}(B_{\mathbf{r}_{z/4}}(\eta(b))) \setminus A
\]

before exiting ${\VGeps}(B_{\mathbf{r}_{z/4}}(\eta(b)))$ with conditional probability given $(h,\eta)$ bounded from below by a constant.
See Figure <ref>.
Step 2: Random walk in a good annulus.

Let $\xi_{\mathbf{r}_{z}}$
be the first time that $X^{b, {\eps}}_t$ hits ${\VGeps}(\partial B_{\mathbf{r}_{z/4}}(\eta(b)))$ and let

\begin{equation} \label{eq:visit-count}
V = \sum_{t=0}^{\xi_{\mathbf{r}_{z}}} 1 \{ X^{b, {\eps}}_t \in \mathfrak{E}\}  = \left( \text{number of visits of $\mathfrak{E}$ by $X^{b, {\eps}}$ before time $\xi_{\mathbf{r}_{z}}$} \right) .
\end{equation}

By the second moment method,

\begin{equation} \label{eq:second-moment-lower-bound}
\P[V \geq 1 | h, \eta] \geq \frac{(\E[V | h, \eta])^2}{\E[V^2 | h , \eta]}.
\end{equation}

In the next two steps we show that

\begin{equation} \label{eq:lower-bound-on-V}
\E[ V | h ,\eta] \geq  {\eps}^{-1} C^{-1} \times \mu_h(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b)))
\end{equation}

\begin{equation} \label{eq:upper-bound-on-V}
\E[ V^2 | h ,\eta]   \leq  C {\eps}^{-2} \left(\sup_{v \in \A_{\mathbf{r}_{z}, 2\mathbf{r}_{z}}(z)} \int_{\A_{\mathbf{r}_{z},2 \mathbf{r}_{z}}(z)} \left( \log \frac{1}{|u-v|} + 1  \right) d \mu_h(u)\right)^2,
\end{equation}

for some large deterministic constant $C = C(\rho, \gamma) > 0$.
Combining (<ref>) and (<ref>)
together with the definition of the event $E_r(z)$ in (<ref>), leads to
a constant order lower bound in (<ref>). This completes the proof by the definition (<ref>) of $V$.
Step 3: Lower bound on the numerator.

We lower bound the numerator:

\begin{align*}
\E[ V | h ,\eta] &= \sum_{b' \in \mathfrak{E}} \Gr{{\eps}}{B_{\mathbf{r}_{z}/4}(\eta(b))}(b, b') \qquad \mbox{(definition of $V$ and $\Gr{{\eps}}{B_{\mathbf{r}_{z}}(\eta(b))}$)} \\
&\geq \sum_{b' \in {\VGeps}(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b))) \cap \mathfrak{E}} \Gr{{\eps}}{B_{\mathbf{r}_{z}/4}(\eta(b))}(b, b') \quad \mbox{(where $\boldsymbol{c_0}$
is as in \eqref{eq:choice-of-constant-annulus})} \\
&\geq \left( \inf_{b' \in {\VGeps}(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b)))} \Gr{{\eps}}{B_{\mathbf{r}_{z}/4}(\eta(b))}(b, b') \right) (|{\VGeps}(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b))| - |A \cap {\VGeps}(B_{\mathbf{r}_{z}}(\eta(b))| ) \\
&\geq \left( \inf_{b' \in {\VGeps}(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b)))} \Gr{{\eps}}{B_{\mathbf{r}_{z}/4}(\eta(b))}(b, b') \right) {\eps}^{-1} (\mu_h(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b))) -  s_1 r^{2 \beta^-}
- {\eps}^{\alpha}) 		\stepcounter{equation}\tag{\theequation}\label{eq:lower-bound-on-V-intermediate} \\
&\qquad \mbox{(by the assumption on $A$ and Lemma \ref{lemma:mu_h-convergence})}.
\end{align*}

We claim that for a sufficiently small, deterministic choice of $\boldsymbol{\pi} > 0$ and $s_1 > 0$,

\begin{equation}\label{eq:use-of-assumption-on-omega}
\mu_h(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b))) -  s_1 r^{2 \beta^-}
- {\eps}^{\alpha} \geq  C'\mu_h(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b)))
\end{equation}

for some deterministic constant $C' = C'(\rho, \gamma)$.
Indeed, for a deterministic constant $c = c(\rho, \gamma)$ which will change from line to line we have

\begin{align*}
&\mu_h(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b))) -  s_1 r^{2 \beta^-}
- {\eps}^{\alpha} \\
&= \frac{1}{2} \mu_h(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b)))  + \frac{1}{2} \mu_h(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b))) -
s_1 r^{2 \beta^-}
- {\eps}^{\alpha}  \\
&\geq \frac{1}{2} \mu_h(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b)))  + c \times \mathbf{r}_{z}^{\beta^-} -
s_1 r^{2 \beta^-}
- {\eps}^{\alpha}  \quad \mbox{(by Lemma \ref{lemma:volume-growth})} \\
&\geq \frac{1}{2} \mu_h(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b)))  + c \times \delta^{\beta^-} -
s_1 \times \delta^{\beta^-}
- {\eps}^{\alpha}  \quad \mbox{(since $\mathbf{r}_{z} > \delta$ and $r < \sqrt{\delta}$)} \\
&\geq \frac{1}{2} \mu_h(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b)))  + c \times \delta^{\beta^-}
- {\eps}^{\alpha}  \quad \mbox{(for small $s_1 < c$)}  \\
&\geq \frac{1}{2} \mu_h(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b)))  + c \times {\eps}^{\omega \beta^-}
- {\eps}^{2 \beta^- \omega}  \quad \mbox{($\delta > {\eps}^{\omega}$ and $\alpha \geq 2 \beta^- \omega$ by assumption)}  \\
&\geq \frac{1}{2} \mu_h(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b))) \quad \mbox{(for $\boldsymbol{\pi}$ small depending only on $c$).}
\end{align*}

\begin{align*}
&\E[ V | h ,\eta] \\
&\geq \left( \inf_{b' \in {\VGeps}(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b)))} \Gr{{\eps}}{B_{\mathbf{r}_{z}/4}(\eta(b))}(b, b') \right) {\eps}^{-1} (\mu_h(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b))) -  s_1 r^{2 \beta^-}
- {\eps}^{\alpha})
\quad \mbox{(by \eqref{eq:lower-bound-on-V-intermediate})} \\
&\geq \frac{1}{2} {\eps}^{-1} (\mu_h(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b))) -  s_1 r^{2 \beta^-}
- {\eps}^{\alpha})
\quad \mbox{(by \eqref{eq:choice-of-constant-annulus} and Lemma \ref{lemma:greens-kernel-lower-bound})} \\
&\geq \frac{1}{2} {\eps}^{-1} \times  C'\mu_h(B_{\boldsymbol{c_0} \mathbf{r}_{z}}(\eta(b)))
\quad \mbox{(by \eqref{eq:use-of-assumption-on-omega})},
\end{align*}

which implies (<ref>) with $C^{-1} := C'/2$.
Step 4: Upper bound on the denominator

We finally upper bound the denominator in (<ref>); here the deterministic constant $C = C(\rho, \gamma) > 0$ also changes from line to-line:

\begin{align*}
&\E[ V^2 | h ,\eta] \\
&\leq \E[ \xi_{\mathbf{r}_{z}}^2 | h ,\eta] \\
&\leq C {\eps}^{-2} \left(\sup_{v \in B_{\mathbf{r}_{z/4}}(\eta(b))} \int_{B_{\mathbf{r}_{z/4}+{\eps}^{\alpha}}(\eta(b))} \left( \log \frac{1}{|u-v|} + 1  \right) d \mu_h(u) \right)^2  \quad \mbox{(by Lemma \ref{lemma:exit-time-upper-bound})} \\
&\leq C {\eps}^{-2} \left(\sup_{v \in \A_{\mathbf{r}_{z}, 2\mathbf{r}_{z}}(z)} \int_{\A_{\mathbf{r}_{z},2 \mathbf{r}_{z}}(z)} \left( \log \frac{1}{|u-v|} + 1  \right) d \mu_h(u) \right)^2 \\
&\qquad \mbox{(since $r_{z} \geq \delta \geq 10 {\eps}^{\alpha}$ and $b \in {\VGeps}(\partial B_{(4/3) \mathbf{r}_z}(z)$)} .
\end{align*}

This shows (<ref>).

Illustration of part of the proof of Lemma <ref>; only parts of the sets are shown and aspect ratios are not drawn to scale.
The set $\overline{\Theta} + B_{r/2}$ is shown in blue with a solid black boundary,
the boundary of $\overline{\Theta} + B_{r}$ is shown as a dashed black line.
A good annulus $\A_{r_z, 2 r_z}(z)$ is shown in gray, an interior ball is outlined in black, and the set $A$ is shown
in light-red. A random walk started at $a$ must hit ${\VGeps}(B_{{(4/3)} r_{z}}(z))$ for some ${z} \in \mathbf{Z}$
at a point $b \in {\VGeps}(\partial B_{{(4/3)} r_{z}}(z))$. The proof then reduces to showing a random walk started at $b$ must exit
$A$ before exiting ${\VGeps}(B_{r_{z}/4}(b))$.

We next show that with exponentially high probability in $r$, if one starts $k \leq r^{2 \beta^-} {\eps}^{-1}$ walkers somewhere
in the closure of the current cluster $\cl(\Theta)$, then the number of walkers which are absorbed
before exiting ${\VGeps}(\overline\Theta + B_r)$ is at least a constant fraction of $k$.
The proof uses the constant-order lower bound given by Lemma <ref> together with a concentration inequality for negative binomials.
The statement is a modification of <cit.>. Note that <cit.> is much stronger than what we have below,
but that stronger statement is not needed to complete the proof of the IDLA upper bound. For the statement, we recall the definition of the modified IDLA aggregate from Definition <ref>.

There exists deterministic $\omega = \omega(\gamma) > 0$ and $s_3 > 0$ so that for each $s_0 \in (0,1)$, there exists $s_4, s_5 \in (0,1)$
and a random $\sigma(h)$-measurable $\boldsymbol{\delta} >0$ such that with probability approaching one as ${\eps}$ goes to zero, the following holds for $r \in [10 {\eps}^{\omega}, \boldsymbol{\delta}]$
for all sets $\Theta \subset {\VGeps}$
such that $\overline{\Theta} \Subset B_{\rho/2}$ and $B_{s_0} \subset \overline{\Theta}$ with $\overline{\Theta}$ as in (<ref>):
for each $k \leq s_5 r^{2 \beta^-} {\eps}^{-1}$ and $a_1, \ldots, a_{k} \in \cl(\Theta)$,

\[
\P[ |A(\Theta; a_1, \ldots, a_{k} \to {\VGeps}(\overline{\Theta} + B_r)) \setminus \Theta| \leq s_4 k | h ,\eta] \leq \exp(-s_3 k).
\]

For $s_0 \in (0,1)$, let $\omega$, $s_1, s_2$ be as in Lemma <ref>
and fix ${\eps}$ and $r \geq 10 {\eps}^{\omega}$ for which the event of Lemma <ref>  occurs. Set $s_5 = \min(\frac{s_1}{s^{-1}_2 + C}, s_1)$ for a
deterministic constant $C > 0$ from (<ref>) below.  Let $k \leq s_5 r^{2 \beta^-} {\eps}^{-1}$ and $a_1, \ldots, a_{k} \in \Theta$
be given.

Start with the set $\Theta$ completely occupied and consider an infinite number of walkers each started at an arbitrary site in $\cl(\Theta)$ and stopped upon exiting ${\VGeps}(\overline\Theta + B_r)$. Let $\mathcal{N}(k_1)$ denote the number of such walkers
needed until $k_1$ walkers have been absorbed into the cluster. That is, $\mathcal N(k_1)$ counts the number of walkers needed
until the aggregate has an additional $k_1$ walkers in ${\VGeps}(\overline\Theta + B_r) \setminus \Theta$.

We seek to iteratively apply Lemma <ref>. Define $S_0 = \Theta$ and for each $k' \in \{1, \ldots, k_1\}$
condition on the first $(k'-1)$ walkers to determine the current aggregate $S_{k'-1}$.
We then apply Lemma <ref> with $S := S_{k'-1}$ to the $k'$-th walker under the conditional law given the first $(k'-1)$
walkers. By the Markov property of IDLA this implies that

\[
\mathcal{N}(k_1) \leq \mathrm{NegBinom}(k_1, s_2), \quad \forall k_1 \leq s_1 r^{2 \beta^-} {\eps}^{-1}
\]

where $\mathrm{NegBinom}(k_1,s_2)$ denotes a random variable drawn from a negative binomial distribution with $n = k_1$ and $p = s_2$, sampled independently from $(h,\eta)$.
Therefore, by a concentration inequality for negative binomial random variables, for each $k_1 \leq s_1 r^{2 \beta^-} {\eps}^{-1}$,

\begin{equation} \label{eq:number-of-walkers-nb-bound}
\P[ \mathcal{N}(k_1) \leq (s_2^{-1} + C) k_1 | h, \eta] \leq \exp(-c k_1), \quad \mbox{for deterministic constants $C, c > 0$}.
\end{equation}

Thus, if we choose $k_1 = (s_2^{-1} + C)^{-1} \times k$, we have by our choice of $s_5$ and

\[
\P\left[ \mathcal{N}((s_2^{-1} + C)^{-1} \times k) \leq k | h , \eta \right]  \leq \exp(-c k),
\]

where the deterministic constant $c$ has changed compared to (<ref>).
By definition of $\mathcal{N}$, this implies the desired conclusion with $s_3 = c$ and $s_4 = (s_2^{-1} + C)^{-1}$.

§.§ Iteration

We now prove a general bound on how far walkers can spread by iterating Lemma <ref>.
The iteration uses the Abelian property of IDLA and involves starting and stopping walkers which reach a certain distance
of the current cluster. The proof is similar to that of <cit.>.

There exists a deterministic exponent $\beta = \beta(\rho, \gamma) > 0$ and a constant $C = C(\rho, \gamma)>0$ so that
for each $s_0 \in (0,1)$ there exists a $\sigma(h)$-measurable random variable $\boldsymbol{\delta}_0$
such that with probability approaching one as ${\eps} \to 0$,  for each $\overline{\Theta} \Subset B_{\rho/2}$ with $B_{s_0} \subset \Theta$
and each $\delta_0 < \boldsymbol{\delta}_0$ the following is true.
The IDLA cluster started with  $\Theta$ completely occupied and
$j \leq \lceil \delta_0 {\eps}^{-1} \rceil$ walkers at sites $\{a_1, \ldots, a_{j}\} \subset \cl(\Theta)$
is contained in a $C \delta_0^\beta$ neighborhood of $\Theta$:

\[
A(\Theta; a_1, \ldots, a_{j}) \subset {\VGeps}(\Bdeltapp{C \delta_0^\beta}(\overline{\Theta})) \subset {\VGeps}(B_{\rho}).
\]

For $s_0 \in (0,1)$, let the parameters $\omega, s_3, s_4, s_5$ and random variable $\boldsymbol{\delta}$ be as in  Lemma <ref>.
We truncate on the event of Lemma <ref> and fix an ${\eps}$.
Below we use the fact that Lemma <ref> is stated for all  choices of $\Theta$ satisfying the hypotheses and IDLA is a Markov process,
Lemma <ref>. 	Throughout the proof we introduce several other events
which occur with probability approaching one as ${\eps} \to 0$ and will truncate on those events.

Let $\beta^-$ be as in Lemma <ref>.
Let $\boldsymbol{\delta_0} = \boldsymbol{\delta_0}(\boldsymbol{\delta}, \rho, \gamma) < \boldsymbol{\delta}$ be such that $(\boldsymbol{\delta_0} s_5^{-1})^{1/2 \beta^-} \leq \boldsymbol{\delta}$, so that Lemma <ref> holds with $r \in ( 10 {\eps}^{\omega}, (\boldsymbol{\delta_0} s_5^{-1})^{1/2 \beta^-})$.
At the end of the proof, we reveal, in a deterministic fashion,  how small  $\boldsymbol{\delta_0}$ must be
so that the final cluster is contained in ${\VGeps}(B_{\rho})$. Fix $\delta_0 < \boldsymbol{\delta_0}$.

Fix $\omega_0 \in (0,\omega)$ which we will choose in (<ref>) below.
Set $\Theta^0 = \Theta$, $P_0 =\{a_1, \ldots, a_j\}$,
and for $k \geq 1$, inductively define (using the notation of Definition <ref>)

\[
\begin{cases}
\mbox{Current radius: } & r_k :=  \max\left( (|P_{k-1}| \times {\eps} \times s_5^{-1})^{1/{2 \beta^-}}, 10 {\eps}^{\omega_0} \right) \\
\mbox{Current cluster: } & \Theta^{k} := A(\Theta^{k-1}; P_{k-1} \to {\VGeps}(\overline{\Theta^{k-1}} + B_{r_k})) \\
\mbox{Stopped walkers: } & P_k := P(\Theta^{k-1}; P_{k-1} \to {\VGeps}(\overline{\Theta^{k-1}} + B_{r_k})).
\end{cases}
\]

Define the stopping time

\begin{equation} \label{eq:iterate-stop-time3}
\tau_{{\eps},3} = \min \{ k \geq 0 : \overline{\Theta^{k}} \not \in B_{\rho}\}.
\end{equation}

In the analysis below, we will truncate on a high probability event which ensures that $\tau_{{\eps},3}  =\infty$.

Denote the first time there are fewer than
${\eps}^{-\omega_0}$ stopped walkers by

\begin{equation} \label{eq:iterate-stop-time2}
\tau_{{\eps},2} = \min \left( \min \{ k \geq 0 : |P_k| \leq {\eps}^{-\omega_0}\}, \tau_{{\eps},3} \right)
\end{equation}

and the first time (before $\tau_{{\eps},2}$) that the current radius is less than $10 {\eps}^{\omega_0}$ by

\begin{equation}
\tau_{{\eps},1} = \min \left( \min \{ k \geq 0 : r_k \leq 10 {\eps}^{\omega_0} \}, \tau_{{\eps},2} \right).
\end{equation}

By the Abelian property of IDLA, Lemma <ref>, the distribution of the final IDLA cluster $A(\Theta; a_1, \ldots, a_{\lceil \delta_0 {\eps}^{-1} \rceil})$
coincides with that of $\Theta^{\infty}$. Thus, it suffices to iteratively control how far the clusters $\Theta^{k}$ spread. In the first step, we control how far the aggregate spreads until
time $\tau_{{\eps},1}$, in the second, until time $\tau_{{\eps},2}$. In the third step, we control how far ${\eps}^{-\omega_0}$ walkers can spread.
We show in the final step, after making $\boldsymbol{\delta_0}$ smaller in a deterministic way, that in fact we have $\tau_{{\eps},2} < \tau_{{\eps},3}$ and
$\tau_{{\eps},3} = \infty$.

Before proceeding, we note that by Lemma <ref> (after truncating on another high probability event),  for each $\zeta > 0$,

\begin{equation} \label{eq:cell-size-estimate-local}
\diam(H^{{\eps}}_a) \leq {\eps}^{2/(2+\gamma)^2 - \zeta} \quad \forall a \in {\VGeps}(B_{\rho}).
\end{equation}

Choose $\zeta$ in a way which depends only on $\gamma$ and then choose $\omega_0$ small enough so that

\begin{equation} \label{eq:choice-of-omega0}
{\eps}_0^{-\omega_0}  \times {\eps}_0^{2/(2+\gamma)^2 - \zeta} \leq {\eps}_0^{\omega_0/2} ,\quad \forall {\eps}_0 \in (0,1) .
\end{equation}

Step 1: $t < \tau_{{\eps},1}$.

All of our parameter choices so far have been so that we can iteratively apply
Lemma <ref> for each $k \in \{1, \ldots, \tau_{{\eps},1}-1\}$ with $\Theta = \Theta^{k}$,
$r = r_k$, and $k = |P_{k-1}| \geq {\eps}^{-\omega_0}$.

By (<ref>) and (<ref>) and the fact $r_k \geq 10 {\eps}^{\omega_0}$, there is a `shell' of additional vertices in
$\Theta^{k}$ that are not in $\Theta^{k-1}$; that is, $\partial \Theta^{k} \cup \Theta^{k-1}$ is not connected.
Therefore, at least one walker is absorbed in each step and hence we have $\tau_{{\eps},1} \leq \tau_{{\eps},2} \leq {\eps}^{-1}$.
Thus, by Lemma <ref> (with $s_3 > 0$ and $s_4 \in (0,1)$ from there)  and a union bound over all $k=1,\dots,\lfloor{\eps}^{-1} \rfloor$,

\begin{equation} \label{eqn:use-absorb}
\P \left[ \exists k \in \{1, \ldots, \tau_{{\eps},1}-1\} : |P_k| \geq (1-s_4) |P_{k-1}| | h, \eta \right] \leq {\eps}^{-1} \exp(- s_3 {\eps}^{-\omega_0}).
\end{equation}

We henceforth truncate on the event in (<ref>), which happens with probability tending to 1 as ${\eps}\to 0$
to see that

\begin{equation} \label{eq:good-event}
|P_k| \leq (1-s_4) |P_{k-1}|, \quad \forall k \in \{1, \ldots, \tau_{{\eps},1}-1\} .
\end{equation}

This implies that $|P_k| \leq (1-s_4)^k |P_0| \leq (1-s_4)^k \delta_0 {\eps}^{-1}$ for each $k\leq \tau_{{\eps},1}-1$. By the definition of $r_k$, this, in turn, implies that

\begin{equation} \label{eq:good-event2}
r_k \leq (\delta_0\times (1-s_4)^k \times s_5^{-1})^{1/2\beta^-}, \quad \forall k \in \{1, \ldots, \tau_{{\eps},1}-1\} .
\end{equation}

\begin{equation} \label{eq:bound-on-radii-sum}
\sum_{k=0}^{\tau_{{\eps},1}-1} r_k \leq \sum_{k=0}^{\tau_{{\eps},1}-1} (\delta_0\times (1-s_4)^k \times s_5^{-1})^{1/2\beta^-} \leq C \times \delta_0^{1/2\beta^-} \times \frac{1}{1 - (1-s_4)^{1/2\beta^-}},
\end{equation}

where $C = C(s_5) > 0$ is deterministic.
Step 2: $k \in [\tau_{{\eps},1}, \tau_{{\eps},2})$.

In this step we again apply Lemma <ref> for each $k \in \{\tau_{{\eps},1}, \ldots, \tau_{{\eps},2}\}$
with the same choice of parameters as in Step 1 except now the radius is fixed $r_k = 10 {\eps}^{\omega_0}$.
Note that $r_k \geq (|P_{k-1}| \times {\eps} \times s_5^{-1})^{1/{2 \beta^-}}$ by definition and hence the conditions of Lemma <ref>
are satisfied.
We show that

\begin{equation} \label{eq:time-upper-bound}
\tau_{{\eps},2} \leq C \log {\eps}^{-1} + \tau_{{\eps},1}, \quad \mbox{for deterministic $C = C(s_5, \gamma) > 0$}
\end{equation}

and hence

\begin{equation}
\sum_{k=\tau_{{\eps},1}}^{\tau_{{\eps},2}} r_k \leq C \times {\eps}^{\omega_0} \times \log {\eps}^{-1}, \quad \mbox{for deterministic $C = C(s_5, \gamma) > 0$}.
\end{equation}

To see (<ref>), we first use exactly the same argument as in Step 1 (and truncate on another event) to get that

\[
|P_k| \leq (1-s_4) |P_{k-1}|, \quad \forall k \in \{\tau_{{\eps},1}, \ldots, \tau_{{\eps},2} \} .
\]

Hence, by the definition (<ref>) of $\tau_{{\eps},2}$,

\[
{\eps}^{-\omega_0} \leq |P_{\tau_{{\eps},2}-1}| \leq (1-s_5)^{\tau_{{\eps},2}- \tau_{{\eps},1}-1} {\eps}^{-1}
\]

which implies (<ref>) upon re-arranging.
Step 3: $k \geq \tau_{{\eps},2}$.

It remains to control how far the remaining ${\eps}^{-\omega_0}$ walkers can go before being absorbed.
We do this crudely by appealing to the upper bound on the Euclidean diameter of cells in the mated-CRT map:
by (<ref>) and (<ref>)
we have that ${\eps}^{-\omega_0}$ walkers cannot go farther than Euclidean distance ${\eps}^{\omega_0/2}$ before being absorbed into the cluster,

\begin{equation} \label{eq:distance-of-final-cluster}
\dist(\overline{\Theta^{\infty}}, \overline{\Theta^{\tau_{{\eps},2}}}) < {\eps}^{\omega_0/2},
\end{equation}

since each walker occupies at least one cell.
Step 4: Reduce ${\eps}$ and $\boldsymbol{\delta_0}$ and conclude.

By combining Steps 1-3, the final aggregate $\overline{\Theta^{\infty}}$ is contained in $\overline{\Theta} + B_R$, where

\[
\begin{aligned}
R &=  C \times \delta_0^{1/2\beta^-}  \quad \mbox{(by \eqref{eq:bound-on-radii-sum})} \\
&+  C \times {\eps}^{\omega} \times \log {\eps}^{-1}  \quad \mbox{(by \eqref{eq:time-upper-bound})} \\
&+ {\eps}^{\omega_0/2} \quad \mbox{(by \eqref{eq:distance-of-final-cluster})}
\end{aligned}
\]

for deterministic constants $C = C(s_4, s_5, \beta^-, \gamma)$.
The first term in $R$ dominates for small ${\eps}$.
This completes the proof after we decrease $\boldsymbol{\delta_0}$ in a deterministic fashion and then ${\eps}$ depending on $\delta_0$ so that $R < \rho/4$ and hence the time $\tau_{{\eps},3} $ of (<ref>) is infinite, as $\overline{\Theta} \Subset B_{\rho/2}$ by assumption.

§.§ Proof of upper bound

We now prove the upper bound by combining the asymptotically correct lower bound together with the general bound of the previous subsection.
The idea is the following. By the Abelian property, we may construct the IDLA cluster by first letting walkers evolve until they exit ${\VGeps}(\Lambda_t)$,
where $\Lambda_t$ is as in Theorem <ref>.
By the lower bound, Proposition <ref>,
most of the walkers will have been absorbed into the aggregate at this point. In fact, as the LQG measure of the boundary of harmonic balls is zero (Theorem <ref>),
the number of remaining walkers can be made arbitrarily small. Thus, we may apply Lemma  <ref> to see that the remaining walkers do not spread too far.

Let the parameters $\beta, C$ be as in Lemma <ref>.
Let $\delta \in (0, \rho)$ be given. Truncate on the event that $t < \TT$ and let $\Lambda_t$ be the harmonic ball satisfying the conditions in Theorem <ref>.
Since $\Lambda_t$ is open and contains the origin, we have that there exists a random $\sigma(h)$-measurable $s_0 \in (0,1)$ so that $B_{s_0} \Subset \Lambda_t$.
Recall the notation $B_{\delta}^{\pm}$ from (<ref>).
Since $\mu_h(\partial \Lambda_t) = 0$ (Theorem <ref>),

\begin{equation} \label{eq:zero-measure-boundary}
\lim_{\delta \downarrow 0} \left( \mu_h(\Bdeltap(\Lambda_t)) - \mu_h(\Bdeltam(\Lambda_t)) \right) = 0.
\end{equation}

Denote by

\[
A_{{\eps}}(\lfloor t {\eps}^{-1} \rfloor \to {\VGeps}(\Lambda_t)) := A(\emptyset ; 0,\dots,0 \to {\VGeps}(\Lambda_t))
\]

the IDLA aggregate formed by $\lfloor t {\eps}^{-1} \rfloor$ walkers started at the origin
stopped upon exiting ${\VGeps}(\Lambda_t)$  and let $P_{{\eps}}(\lfloor t {\eps}^{-1} \rfloor \to {\VGeps}(\Lambda_t))$ denote the positions of the stopped (not absorbed) walkers
on the boundary of ${\VGeps}(\Lambda_t)$.
We have by Proposition <ref>
that for each $\delta_0 \in (0, \rho)$,

\begin{equation} \label{eq:asymptotic-lower-bound}
\Bdeltamm{\delta_0}(\Lambda_t)  \subset \overline{A}_{{\eps}}(\lfloor t {\eps}^{-1} \rfloor \to {\VGeps}(\Lambda_t))
\end{equation}

except on an event of probability tending to 0 as ${\eps} \to 0$.

Choose, as in the statement of Lemma <ref>, a $\sigma(h)$-measurable $\boldsymbol{\delta_0}$ depending on $s_0$. Assume that $C \boldsymbol{\delta_0}^\beta < \delta$.
By (<ref>), we may choose a $\delta_0 < \boldsymbol{\delta_0}$ sufficiently small so that

\begin{equation} \label{eq:small-boundary}
\mu_h\left(	\Bdeltapp{\delta_0}(\Lambda_t) \setminus	\Bdeltamm{\delta_0}(\Lambda_t) \right) \leq  \boldsymbol{\delta_0}/2.
\end{equation}

The aggregate lower bound (<ref>) implies

\[
|P_{{\eps}}(\lfloor t {\eps}^{-1} \rfloor \to {\VGeps}(\Lambda_t))| \leq t {\eps}^{-1} - |{\VGeps}(\Bdeltamm{\delta_0}(\Lambda_t))|
\]

except on an event of probability tending to 0 as ${\eps} \to 0$.
Let $\alpha = \alpha(\gamma) > 0$ be the parameter from Lemma <ref>.
Since by Theorem <ref>, $\mu_h(\Lambda_t) = t$, we have by Lemma <ref>,
except on an event of probability tending to 0 as ${\eps} \to 0$,

\[
{\eps}^{-1} \left( t - {\eps}^{\alpha}  \right) \leq |{\VGeps}(\Bdeltapp{\delta_0}(\Lambda_t))|  \leq {\eps}^{-1} \left( \mu_h(\Bdeltapp{\delta_0}(\Lambda_t)) + {\eps}^{\alpha} \right)
\]

\[
|{\VGeps}(\Bdeltamm{\delta_0}(\Lambda_t))| \geq {\eps}^{-1} \left( \mu_h(\Bdeltamm{\delta_0}(\Lambda_t)) - {\eps}^{\alpha} \right).
\]

The previous three indented equations imply that except on an event of probability tending to 0 as ${\eps} \to 0$,

\[
|P_{{\eps}}(\lfloor t {\eps}^{-1} \rfloor \to {\VGeps}(\Lambda_t))| \leq {\eps}^{-1} \left( \mu_h(\Bdeltapp{\delta_0}(\Lambda_t) \setminus \Bdeltamm{\delta_0}(\Lambda_t))
+ 3 {\eps}^{\alpha} \right).
\]

This implies by (<ref>) that

\begin{equation}
|P_{{\eps}}(\lfloor t {\eps}^{-1} \rfloor \to {\VGeps}(\Lambda_t))|
\leq \boldsymbol{\delta}_0 {\eps}^{-1}
\end{equation}

except on an event of probability tending to 0 as ${\eps} \to 0$.

By the Abelian property of IDLA (Lemma <ref>), the distribution of the final aggregate
coincides with that created after releasing the paused walkers:

\[
A_{{\eps}}(\lfloor t {\eps}^{-1}\rfloor ) \overset{d}{=} A( A_{{\eps}}(\lfloor t {\eps}^{-1} \rfloor \to {\VGeps}(\Lambda_t));P_{{\eps}}(\lfloor t {\eps}^{-1} \rfloor \to {\VGeps}(\Lambda_t))),
\]

where the equality in distribution is conditional on $h$ and $\eta$.
Since IDLA is a Markov process and $\overline{\Lambda}_t \Subset B_{\rho/2}$  ($t < \TT$), we may apply Lemma <ref> with $\Theta = A_{{\eps}}(\lfloor t {\eps}^{-1} \rfloor \to {\VGeps}(\Lambda_t))$ and at most $\boldsymbol{\delta}_0 {\eps}^{-1}$ walkers started at $P_{{\eps}}(\lfloor t {\eps}^{-1} \rfloor \to {\VGeps}(\Lambda_t))$ to see that
except on an event of probability approaching zero as ${\eps} \to 0$,

\[
\overline{A}_{{\eps}}(\lfloor t {\eps}^{-1} \rfloor )
\subset
\Bdeltapp{C \boldsymbol{\delta_0}^\beta}(\overline{A_{{\eps}}}(\lfloor t {\eps}^{-1} \rfloor \to {\VGeps}(\Lambda_t))
\subset
\Bdeltapp{C \boldsymbol{\delta_0}^\beta}(\Lambda_t).
\]

This concludes the proof upon recalling $\delta > C \boldsymbol{\delta_0}^\beta$.

§ DIVISIBLE SANDPILE UPPER BOUND

We combine Lemma <ref> together with ideas from <cit.>
to prove an upper bound on the divisible sandpile cluster. For the statement, recall the notation of the cluster $D_{{\eps}}$ from (<ref>)
and $\overline{D_{{\eps}}}$ from (<ref>).

Recall the time $\TT$ from (<ref>).
For each $\delta \in (0, \rho)$ and $t > 0$, on the event $\{t < \TT\}$,
it holds except on an event of probability tending to 0 as ${\eps} \to 0$  that

\[
\overline{D}_{{\eps}}(t {\eps}^{-1}) \subset \Bdeltap(\Lambda_t),
\]

where $\Lambda_t$ is as in Theorem <ref>.

Recall from (<ref>) the least action principle for the divisible sandpile odometer,

\[
\frac{v^{{\eps}}_t}{\deg^{{\eps}}}  = \min \{ w:{\VGeps} \to [0,\infty) : \deg^{{\eps}} \times \Delta^{{\eps}} w + t {\eps}^{-1} \delta_0 \leq 1 \},
\]

and that

\begin{equation} \label{eq:cluster-contained-in}
\begin{aligned}
\Lambdateps &= \{a \in \VGeps: v^{{\eps}}_t > 0 \} \\
D_{{\eps}}(t {\eps}^{-1}) &=  \cl(\Lambdateps).
\end{aligned}
\end{equation}

In this section we mainly work with the stopped odometer, defined by

\begin{equation} \label{eq:stopped-odometer-de}
u_t^{{\eps}} = \min \{ w: {\VGeps}(B_{\rho}) \to [0,\infty) : \Delta^{{\eps}} w + {\deg^{{\eps}}(0)}^{-1} t {\eps}^{-1} \delta_0 \leq (\deg^{{\eps}})^{-1}\},
\end{equation}

which, as discussed in the beginning of Section <ref>, relates to $\wteps$ via  $u_t^{{\eps}} = \wteps + (t {\eps}^{-1}) \times  \gr{{\eps}}{B_\rho}(0, \cdot)$.
The reason for doing so is that the results of Section <ref>, a priori, apply just to the stopped odometer.  However, as we will show, for small $t$, the odometer
$\frac{v^{{\eps}}_t}{\deg^{{\eps}}}$ agrees with the stopped odometer. In particular,
it is straightforward to see that

\begin{equation} \label{eq:consistency}
u_t^{{\eps}} = 0 \quad \mbox{on $\cl({\VGeps}(\partial B_{\rho}))$}
\implies
u_t^{{\eps}} = \frac{v^{{\eps}}_t}{\deg^{{\eps}}} \quad  \mbox{in ${\VGeps}(\cl(B_{\rho}))$}.
\end{equation}

Therefore, the upper bound for the divisible sandpile cluster will follow from (<ref>) once we show that $u_t^{{\eps}}$ is zero near the boundary of $\Lambda_t$ for $t < \TT$.

Our strategy for the upper bound is similar to the argument in <cit.>, but with some simplifications.
We start by establishing harmonic comparison lemmas in Section <ref> which allow us to compare the mass
of the cluster to the size of the stopped odometer.  We then use these lemmas together with an iteration to show the upper bound in Section <ref> as follows. By convergence of the stopped odometer away from the origin, we have that $u_t^{{\eps}}$ is small near ${\VGeps}(\partial \Lambda_t)$. This implies that the amount of mass in the shell ${\VGeps}(\Lambda_t + B_r) \setminus {\VGeps}(\Lambda_t)$ is small.
Hence, by Lemma <ref> the probability that a random walk exits the cluster
before exiting this shell is bounded from below. By comparison, this implies the size of the odometer must be small on ${\VGeps}(\partial (\Lambda_t + B_r))$. Iterate this argument with shells of decreasing size to see that the odometer must be zero outside of
${\VGeps}(\Lambda_t + B_R)$ for some small $R$.

This is simpler than <cit.> as we do not have to redo <cit.> since we have already proved a harmonic measure estimate in Lemma <ref>. In particular, we do not consider `very good annuli' explicitly. Further, since we have an a priori strong upper bound on the size of the odometer on ${\VGeps}(\partial \Lambda_t)$ due to its convergence (Theorem <ref>), the inductive counterpart of <cit.> is shorter.

§.§ Harmonic comparison

In this subsection, we prove discrete analogues of the harmonic comparison lemmas for the Hele-Shaw odometer from <cit.>. First note that since  $\Delta^{\eps} u_t^{\eps} = \Delta^{\eps} \wteps + (t {\eps}^{-1}) \times \Delta^{\eps} \gr{{\eps}}{B_{\rho}}$,  by combining (<ref>) and Lemma <ref> we have that

\begin{equation} \label{eq:laplacian-of-uteps}
\begin{aligned}
0 \leq \Delta^{{\eps}} u_t^{{\eps}}  + {\deg^{{\eps}}(0)}^{-1} t {\eps}^{-1} \delta_0 &\leq (\deg^{{\eps}})^{-1}
\quad \mbox{on ${\VGeps}(B_{\rho})$} \\
u_t^{{\eps}} &\geq 0  	\quad \mbox{on ${\VGeps}(B_{\rho})$} \\
\Delta^{{\eps}} u_t^{{\eps}}  + {\deg^{{\eps}}(0)}^{-1} t {\eps}^{-1} \delta_0 &= (\deg^{{\eps}})^{-1}
\quad \mbox{on $\Lambdateps$} \\
u_t^{{\eps}} &= 0 \quad \mbox{on ${\VGeps}(B_{\rho}) \setminus \Lambdateps$}.
\end{aligned}
\end{equation}

We use this to show the following.

There exists deterministic constants $\omega = \omega(\gamma) > 0$,  $C = C(\rho, \gamma) > 0$, $\boldsymbol{c}_0 = \boldsymbol{c}_0(\rho, \gamma) \in (0, 1)$ such that with polynomially high probability as ${\eps} \to 0$, the following holds. For all $t > 0$,
all connected sets $\Theta \subset {\VGeps}(B_{\rho})$ containing the origin,
and all $s > 0$, $z \in \C$ such that ${\VGeps}(B_{s}(z)) \subset {\VGeps}(B_{\rho}) \setminus \Theta$,

\[
\left| {\VGeps}(B_{\boldsymbol{c}_0 s}(z)) \cap \Lambdateps \right| \leq C \times \sup_{\partial \Theta} u_t^{{\eps}}.
\]

Truncate on the events of Lemmas <ref> and <ref>.
By (<ref>) we have that $\Delta^{{\eps}} u_t^{{\eps}} \geq 0$ away from the origin and $\Delta^{{\eps}} u_t^{{\eps}} = (\deg^{{\eps}})^{-1}$ on $\Lambdateps \setminus \{0\}$.
Hence, by the maximum principle,  as $u_t^{{\eps}} = 0$ on $\partial B_{\rho}$ and $u_t^{{\eps}}$ is subharmonic away from the origin,

\begin{equation} \label{eq:upper-bound-on-ut}
\sup_{\Lambdateps \cap {\VGeps}(B_s(z))} u_t^{{\eps}} \leq  \sup_{\partial \Theta} u_t^{{\eps}}.
\end{equation}

We will prove the desired inequality by comparing $u_t^{{\eps}}$ to the  following function

\[
\mathfrak{f}^{{\eps}}(\cdot) =  \sum_{b \in {\VGeps}(B_s(z)) \cap \Lambdateps \setminus  {\VGeps}(\partial B_s(z))} \gr{{\eps}}{B_s(z)}(\cdot,b)
\]

which, for the same reasoning leading to (<ref>), satisfies

\[
\begin{cases}
\Delta^{{\eps}} \mathfrak{f}^{{\eps}}(\cdot) = -\deg^{{\eps}}(\cdot)^{-1} 1\{ \cdot \in \Lambdateps\} \quad &\mbox{in ${\VGeps}(B_s(z)) \setminus {\VGeps}(\partial B_s(z))$} \\
\mathfrak{f}^{{\eps}}(\cdot) = 0 \quad &\mbox{otherwise}.
\end{cases}
\]

We then compute:

\begin{align*}
\sup_{{\VGeps}(B_s(z))} \mathfrak{f}^{{\eps}}  &\leq 	\sup_{{\VGeps}(B_s(z))} ( u_t^{{\eps}} + \mathfrak{f}^{{\eps}} ) \quad \mbox{($u_t^{{\eps}} \geq 0$)} \\
&\leq \sup_{{\VGeps}(\partial B_s(z))} (	u_t^{{\eps}} + \mathfrak{f}^{{\eps}} )  \quad \mbox{($u_t^{{\eps}} + \mathfrak{f}^{{\eps}}$ is subharmonic)} \\
&= \sup_{{\VGeps}(\partial B_s(z))} 	u_t^{{\eps}} \quad \mbox{($\mathfrak{f}^{{\eps}} = 0$ on ${\VGeps}(\partial B_s(z))$)} \\
&\leq \sup_{\partial \Theta} u_t^{{\eps}} \quad \mbox{(by \eqref{eq:upper-bound-on-ut})}.
\end{align*}

Let $\boldsymbol{c}_0$ be as in (<ref>) so that
by Lemma <ref>

\[
\mathfrak{f}^{{\eps}} \geq \sum_{b \in {\VGeps}(B_{\boldsymbol{c_0} s}(z)) \cap \Lambdateps}  \gr{{\eps}}{B_s(z)}(\cdot,b)
\geq \frac{1}{2} 	 \left| {\VGeps}(B_{\boldsymbol{c}_0 s}(z)) \cap \Lambdateps \right|.
\]

Combining the previous two indented equations completes the proof.

The following holds for all $t > 0$ and all sets $B \subset {\VGeps}(B_{\rho}) \setminus \{0\}$:

\[
u_t^{{\eps}}(b) \leq \left( \sup_{\partial B} u_t^{{\eps}} \right) \P[\mbox{$X^{b, {\eps}}$ exits $B$ before hitting $B \setminus \Lambdateps$} | h, \eta], \quad \forall b \in B,
\]

where $X^{b, {\eps}}$	is a simple random walk on ${\Geps}$ started from the vertex $b \in {\VGeps}$.

The function

\[
f(b) := \left( \sup_{\partial B} u_t^{{\eps}} \right) \P[\mbox{$X^{b, {\eps}}$ exits $B$ before hitting $B \setminus \Lambdateps$} | h, \eta]
\]

\[
\begin{cases}
\Delta^{{\eps}} f = 0 \quad \mbox{on $B \cap \Lambdateps$} \\
f = 0 \quad \mbox{on $B \cap (\Lambdateps)^c$} \\
f = \left( \sup_{\partial B} u_t^{{\eps}} \right) \quad \mbox{on $B^c \cap \Lambdateps$}
\end{cases}
\]

Therefore, by (<ref>), we have

\[
\begin{cases}
\Delta^{{\eps}} (u_t^{{\eps}} - f) \geq 0 \quad \mbox{on $B \cap \Lambdateps$} \\
(u_t^{{\eps}} - f) = 0 \quad \mbox{on $B \cap (\Lambdateps)^c$} \\
(u_t^{{\eps}} - f) \leq 0 \quad \mbox{on $B^c \cap \Lambdateps$},
\end{cases}
\]

which implies the claim by the maximum principle and the fact $u_t^{{\eps}} = 0$ on $(\Lambdateps)^c$.

§.§ Iteration

The proof of Proposition <ref> proceeds by a similar iteration to the proof of Proposition <ref>
but with the odometer replacing the role of the random walks.  We show that the maximum of the odometer decreases across shells of
increasing radii.
Eventually the odometer will be so small the upper bound we derive from the following estimate will suffice.

The following is true for all $t > 0$ with polynomially high probability as ${\eps} \to 0$.
For every point $a \in D_{{\eps}}(t {\eps}^{-1}) \setminus \{0\}$  there is a path $a = a_0 \sim a_1 \sim \cdots \sim a_m = 0$ in $D_{{\eps}}(t {\eps}^{-1})$

\[
u^{{\eps}}_t(a_{i+1}) \geq u^{{\eps}}_t(a_i) + \log({\eps}^{-1})^{-2}.
\]

By <cit.> it holds with polynomially high probability as ${\eps} \to 0$ that

\begin{equation} \label{eq:degree-bound}
\max_{a \in {\VGeps}(B_{\rho})} \deg^{{\eps}}(a) \leq (\log {\eps}^{-1})^2.
\end{equation}

Let $a_0 = a$. Inductively, assume that $i\geq 0$ and $a_i \in  D_{{\eps}}(t {\eps}^{-1}) $ has been defined. If $a_i = \{0\}$, set $m=i$ and conclude the induction. If $a_i \not=0$, let $a_{i+1}$ be a neighbor of $a_i$ maximizing $u^{{\eps}}_t(a_{i+1})$.
Then $a_{i+1} \in  D_{{\eps}}(t {\eps}^{-1})$. Since $\Delta^{{\eps}} u^{{\eps}}_t = (\deg^{{\eps}})^{-1}$ on $D_{{\eps}}(t {\eps}^{-1})$,
we have that

\begin{align*}
u^{{\eps}}_t(a_{i+1}) &\geq \frac{1}{\deg^{{\eps}}(a_i)} \sum_{y \sim a_i} u^{{\eps}}_t(y) \quad \text{(choice of $a_{i+1}$)} \\
&=  u^{{\eps}}_t(a_i) + \Delta^{{\eps}} u^{{\eps}}_t(a_i) \quad \text{(definition of $\Delta^{\eps}$)} \\
&= u^{{\eps}}_t(a_i) + \deg^{{\eps}}(a_i)^{-1} \\
&\geq u^{{\eps}}_t(a_i) + \log({\eps}^{-1})^{-2} \quad\text{(by~\eqref{eq:degree-bound})}.
\end{align*}

As the number of vertices in $D_{{\eps}}(t {\eps}^{-1})$ is finite and contains the origin, Lemma <ref>, there must be a finite $i$ for which $a_i = 0$.

Illustration of the objects appearing in Step 1 of the proof of Proposition <ref>.
Part of the current cluster $\overline{\Theta_{k-1}}$ is shown in light blue
and parts of the boundaries of the intermediate clusters $\overline{\Theta}$ and $\overline{\Theta} + B_r$, defined in (<ref>)
and the next cluster $\overline{\Theta_k}$ are outlined by dashed lines. The ball $B_r(z')$ is shown in light gray
and part of the set $A$, defined in (<ref>), is shown in red.

Let $\delta \in (0, \rho)$ be given. Truncate on the event that $t < \TT$ and let $\Lambda_t$ be the harmonic ball satisfying the conditions in Theorem <ref>.
Since $\Lambda_t$ is open and contains the origin, we have that there exists a random $\sigma(h)$-measurable $s_0 \in (0,1)$ so that $B_{s_0} \Subset \Lambda_t$.

For this $s_0$,  let the parameters $\omega, s_1, s_2, \boldsymbol{\delta}$ be as in Lemma <ref>. Let $\beta^-$ be as in Lemma <ref> and

\begin{equation} \label{eq:s6}
s_6 := s_1^{-1} \boldsymbol{c_0}^{-2 \beta^-} 4^{2 \beta^-} \times C
\end{equation}

where $C$ and $\boldsymbol{c_0}$ are the deterministic constants from Lemma <ref>. Let $\boldsymbol{\delta_0} = \boldsymbol{\delta_0}(\boldsymbol{\delta}, \rho, \gamma) < \boldsymbol{\delta}$ be such that $(\boldsymbol{\delta_0} s_6)^{1/2 \beta^-} \leq \boldsymbol \delta$, so that Lemma <ref> holds with $r \in ( 10 {\eps}^{\omega}, (\boldsymbol{\delta_0} s_6)^{1/2 \beta^-})$.
At the end of the proof we state how small $\boldsymbol{\delta_0}$ must be so that the final cluster is contained in ${\VGeps}(B_{\rho})$. Fix $\delta_0 <
\boldsymbol{\delta_0}$ and decrease $\boldsymbol{\delta_0}$ so that $2 \times C_1 \times \boldsymbol{\delta_0}^\beta < \delta$ for a random constant $C_1$ to be determined below.

By the uniform convergence of the scaled version $\overline u_t^{\eps}$ of $u_t^{{\eps}}$ away from the origin (Theorem <ref> and Lemma <ref>) and the fact that $\Lambda_t$ a.s. contains a neighborhood of 0, on the event $\{t < \TT\}$,

\begin{equation} \label{eq:start-iteration}
\sup_{{\VGeps}(\partial \Lambda_t)} u_t^{{\eps}} < \delta_0 {\eps}^{-1},
\end{equation}

except on an event of probability tending to 0 as ${\eps} \to 0$.
Fix $\omega_0 \in (0,\omega)$ which we will choose in (<ref>) below. Set
$\Theta_0 = {\VGeps}(\Lambda_t)$, $r_0 = \delta_0$, $O_0 = \max_{\partial \Theta_0} u^{{\eps}}_t$
and for $k \geq 1$, inductively define

\begin{equation} \label{eq:div-sandpile-cluster-iteration}
\begin{cases}
\mbox{Current cluster: } & \Theta_k := {\VGeps}(\overline{\Theta_{k-1}} + B_{r_{k-1}}) \\
\mbox{Current radius: } & r_k :=  \max\left( (|O_{k-1}| \times {\eps} \times s_6)^{1/{2 \beta^-}}, 40 \boldsymbol{c_0}^{-1} {\eps}^{\omega_0} \right) \\
\mbox{Current maximum odometer: } & O_k := \max_{\partial \Theta_k} u^{{\eps}}_t.
\end{cases}
\end{equation}

Denote the first time the odometer is smaller than
${\eps}^{-\omega_0}$ by

\begin{equation} \label{eq:iterate-stop-time2-div}
\tau_{{\eps},2} = \min\{ k \geq 0 : |O_k| \leq {\eps}^{-\omega_0} \}
\end{equation}

and the first time (before $\tau_{{\eps},2}$) that the current radius is less than $40 \boldsymbol{c_0}^{-1} {\eps}^{\omega_0}$ by

\begin{equation}
\tau_{{\eps},1} = \min \left( \min\{ k \geq 0 : r_k \leq 40 \boldsymbol{c_0}^{-1} {\eps}^{\omega_0} \}, \tau_{{\eps},2} \right).
\end{equation}

By Lemma <ref>, for each $\zeta > 0$,

\begin{equation} \label{eq:cell-size-estimate-local-div}
\diam(H^{{\eps}}_a) \leq {\eps}^{2/(2+\gamma)^2 - \zeta} \quad \forall a \in {\VGeps}(B_{\rho}),
\end{equation}

w.p. tending to 1 as ${\eps} \to 0$.
Choose $\zeta$ in a way which depends only on $\gamma$ and then choose $\omega_0$ small enough so that

\begin{equation} \label{eq:choice-of-omega0-div}
{\eps}_0^{-\omega_0}  \times {\eps}_0^{2/(2+\gamma)^2 - \zeta} \leq {\eps}_0^{\omega_0/2} ,\quad \forall {\eps}_0 \in (0,1) .
\end{equation}

Step 1: $k < \tau_{{\eps},1}$.

We first show that for each $t \in \{1, \ldots, \tau_{{\eps},1}-1\}$,

\begin{equation} \label{eq:iterative-bound-of-o}
O_k \leq (1-s_2) O_{k-1},
\end{equation}

except on an event of probability tending to 0 as ${\eps} \to 0$.
By Lemma <ref>, for all $z \in \C$ and $s > 0$ such that $B_{s}(z) \subset B_{\rho} \setminus \overline{\Theta_{k-1}}$:

\begin{equation} \label{eq:mass-bound-in-proof}
\left| {\VGeps}(B_{\boldsymbol c_0 s}(z)) \cap \Lambdateps \right| \leq C \times \sup_{\partial \Theta_{k-1}} u_{t}^{{\eps}},
\end{equation}

where $\boldsymbol{c_0}$ is as in Lemma <ref>, except on an event of probability tending to 0 as ${\eps} \to 0$.
As we justify just below, this implies we can apply Lemma <ref> with parameters

\begin{equation} \label{eq:parameter-choices}
\begin{aligned}
r &= \boldsymbol{c_0} r_k/4 \\
\Theta &= {\VGeps}(\overline{\Theta_{k-1}} + B_{r_k/4}) \\
A &= \Lambdateps \cap \{ {\VGeps}(\overline{\Theta} + B_r) \setminus \Theta \}.
\end{aligned}
\end{equation}

See Figure <ref> for an illustration of these sets. Indeed, we have that $A \subset {\VGeps}(\overline{\Theta} + B_r) \setminus \Theta$ and, by (<ref>), for $z \in \C$:

\begin{align*}
|A \cap {\VGeps}(B_r(z))|  &=
|A \cap {\VGeps}(B_{\boldsymbol{c_0} r_k/4}(z))|  \\
&\leq |\Lambdateps \cap {\VGeps}(B_{\boldsymbol{c_0} r_k/4}(z'))|  \\
&\quad \mbox{(for some $z' \in \C$ s.t.\ $B_{r_k/4}(z') \subset B_{\rho} \setminus \overline{\Theta_{k-1}}$  by definition of $A$ \eqref{eq:parameter-choices})} \\
&\leq C \times \sup_{\partial \Theta_{k-1}} u_t^{{\eps}} \quad \mbox{(by \eqref{eq:mass-bound-in-proof})} \\
&= C \times O_{k-1}   \quad \mbox{(definition of $O_{k-1}$ \eqref{eq:div-sandpile-cluster-iteration})} \\
&=  r_k^{2 \beta^-} \times {\eps}^{-1} \times s_6^{-1} \times C  \quad \mbox{(definition of $r_k$ \eqref{eq:div-sandpile-cluster-iteration})} \\
&=  s_1  r^{2 \beta^-} {\eps}^{-1} \quad \mbox{(definition of $r$ and $s_6$  \eqref{eq:parameter-choices} and \eqref{eq:s6})}.
\end{align*}

Thus, as $r \geq 10 {\eps}^{\omega}$ the hypotheses of Lemma <ref> are satisfied and we may apply it to get

\begin{equation} \label{eq:probability-bound-in-proof}
\begin{aligned}
\sup_{a \in {\VGeps}(\partial (\overline{\Theta_{k-1}} + B_{r_k/4 + \boldsymbol{c_0} r_k/8}))}
\P[&\mbox{$X^{a, {\eps}}$ exits ${\VGeps}(\overline{\Theta_{k-1}} + B_{r_k/4 + \boldsymbol{c_0} r_k/4} )\setminus
{\VGeps}(\overline{\Theta_{k-1}} + B_{r_k/4}))$} \\
&\mbox{before exiting $A$} | h , \eta]
\leq (1 - s_2)
\end{aligned}
\end{equation}

except on an event of probability tending to 0 as ${\eps} \to 0$.
Next, by Lemma <ref> applied to the set

\[
B := {\VGeps}(\overline{\Theta_{k-1}} + B_{r_k/4 + \boldsymbol{c_0} r_k/4} )\setminus
{\VGeps}(\overline{\Theta_{k-1}} + B_{r_k/4})
\]

together with the maximum principle for $u_t^{{\eps}}$ ($u_t^{{\eps}} = 0$ on ${\VGeps}(\partial B_{\rho})$),

\[
u_t^{{\eps}}(b) \leq O_{k-1} \times \P[\mbox{$X^{b, {\eps}}$ exits $B$ before exiting $A$} | h , \eta], \quad \forall b \in B.
\]

Another application of the maximum principle shows

\[
O_{k} \leq  \sup_{b \in {\VGeps}(\partial (\overline{\Theta_{k-1}} + B_{r_k/4 + \boldsymbol{c_0} r_k/8}))} u_{t-1}^{{\eps}}(b).
\]

The previous two indented equations together with (<ref>) imply (<ref>).

We now use (<ref>), the definition of $r_k$,  and the initial bound given by (<ref>) to see that

\begin{equation} \label{eq:good-event3}
r_k \leq (\delta_0\times (1-s_2)^k \times s_6)^{1/2\beta^-}, \quad \forall k \in \{1, \ldots, \tau_{{\eps},1}-1\}
\end{equation}

and consequently,

\begin{equation} \label{eq:bound-on-radii-sum-div}
\sum_{k=0}^{\tau_{{\eps},1}-1} r_k \leq \sum_{k=0}^{\tau_{{\eps},1}-1} (\delta_0\times (1-s_2)^k \times s_6)^{1/2\beta^-} \leq C \times \delta_0^{1/2\beta^-} \times \frac{1}{1 - (1-s_2)^{1/2\beta^-}},
\end{equation}

except on an event of probability tending to 0 as ${\eps} \to 0$ where $C = C(s_6) > 0$ is a random constant.
Step 2: $t \in [\tau_{{\eps},1}, \tau_{{\eps},2})$.

We again apply Lemma <ref> for each $k \in \{\tau_{{\eps},1}, \ldots, \tau_{{\eps},2}\}$ with the same choice of parameters as in Step 1.
Reasoning similar to the proof of Step 1 shows that

\[
O_k \leq (1-s_2) O_{k-1}, \quad \forall k \in \{\tau_{{\eps},1}, \ldots, \tau_{{\eps},2} \}
\]

except on an event of probability tending to 0 as ${\eps} \to 0$.
Hence, by the definition (<ref>) of $\tau_{{\eps},2}$,

\[
{\eps}^{-\omega_0} \leq |O_{\tau_{{\eps},2}-1}| \leq C (1-s_2)^{\tau_{{\eps},2}- \tau_{{\eps},1}-1} {\eps}^{-1}, \quad \mbox{for $C = C(s_6) > 0$}
\]

which, upon re-arranging, implies

\begin{equation} \label{eq:time-upper-bound-div}
\tau_{{\eps},2} \leq C \log {\eps}^{-1} + \tau_{{\eps},1}, \quad \mbox{for random $C = C(s_2, s_6) > 0$}
\end{equation}

and hence

\begin{equation}
\sum_{k=\tau_{{\eps},1}}^{\tau_{{\eps},2}} r_k \leq C \times {\eps}^{\omega_0} \times \log {\eps}^{-1}, \quad \mbox{for random $C = C(s_2, s_6) > 0$}
\end{equation}

except on an event of probability tending to 0 as ${\eps} \to 0$.
Step 3: $k \geq \tau_{{\eps},2}$.

We control the size of the odometer by Lemma <ref>
and the upper bound on the Euclidean diameter of cells in the mated-CRT map.
By Lemma <ref>,  w.p. tending to 1 as ${\eps} \to 0$, there is a path
of at most $\log({\eps}^{-1})^2 {\eps}^{-\omega_0}$ cells
from the boundary of the support of $u_t^{{\eps}}$
to $\partial \Theta_{\tau_{{\eps},2}}$ along which the odometer increases by ${\eps}^{-\omega_0}$.
Thus, by the definition of $\tau_{{\eps},2}$, (<ref>), and (<ref>)

\begin{equation} \label{eq:distance-of-final-cluster-div-sandpile}
u_t^{{\eps}}(a) = 0  \quad \mbox{for $a \in {\VGeps}(B_{\rho})$ such that $\dist(\eta(a), \eta(\Theta_{\tau_{{\eps},2}})) \geq  {\eps}^{\omega_0/2} \log({\eps}^{-1})^2$}
\end{equation}

except on an event of probability tending to 0 as ${\eps} \to 0$.
Step 4: Reduce ${\eps}$ and $\boldsymbol{\delta_0}$ and conclude.

Recall that we have set $\Theta_0 = {\VGeps}(\Lambda_t)$, so that $\overline\Theta_0$ is the union of the ${\eps}$-mated-CRT map cells which intersect $\Lambda_t$.
By combining Steps 1-3, (<ref>), and (<ref>),
the divisible sandpile cluster, $\overline{D}_{{\eps}}(t {\eps}^{-1})$ is contained in $\overline{\Theta_0} + B_R$, where

\[
\begin{aligned}
R &=  C_1 \times \delta_0^{1/2\beta^-}  \quad \mbox{(by \eqref{eq:bound-on-radii-sum-div})} \\
&+  C_1 \times {\eps}^{\omega} \times \log {\eps}^{-1}  \quad \mbox{(by \eqref{eq:time-upper-bound-div})} \\
&+ {\eps}^{\omega_0/2} \log({\eps}^{-1})^2 \quad \mbox{(by \eqref{eq:distance-of-final-cluster-div-sandpile})}
\end{aligned}
\]

for random constants $C_1 = C_1(s_2, s_6, \beta^-, \gamma)$, provided that $R < \rho/2$,
except on an event of probability tending to 0 as ${\eps} \to 0$.
The first term in the definition of $R$ dominates for small ${\eps}$.
Decrease $\boldsymbol{\delta_0}$ in a deterministic fashion and then decrease ${\eps}$ depending on $\delta_0$ so that $R < \rho/2$ and hence $u_t^{{\eps}} = v_t^{{\eps}}$.
This completes the proof after recalling that $\delta > 2 \times C_1 \boldsymbol{\delta_0}^\beta$.

§ CONVERGENCE

We start by combining Propositions <ref>, <ref>, <ref>, and  <ref>
into the following statement.

Recall the time $\TT$ from (<ref>)
and the notation $B_{\delta}^{\pm}$ from (<ref>).  For each $\delta \in (0, \rho)$ and $t > 0$, on the event $\{t < \TT\}$,
it holds except on an event of probability tending to 0 as ${\eps} \to 0$  that

\[
\Bdeltam(\Lambda_t) \subset \overline{\mathcal{X}}_{{\eps}}(t {\eps}^{-1}) \subset \Bdeltap(\Lambda_t)
\]

for $\mathcal{X} \in \{A, D\}$,
where $\Lambda_t$ is as in Theorem <ref>,  , both the IDLA cluster and divisible sandpile cluster converge.

To prove Theorems <ref> and <ref>, we will combine the previous proposition with a scaling argument to transfer from $t\in (0,\TT)$ to an arbitrary $t>0$.

§.§ Proof of Theorems <ref> and <ref>

We use the scale-invariance property of the $\gamma$-quantum cone. Fix $b > 0$ and for $R_b$ as in
(<ref>), let

\[
h^b := h(R_b \cdot) + Q \log R_b - \frac{1}{\gamma} \log b \quad \text{and} \quad  \eta^b = R^{-1}_b \eta(b \cdot).
\]

By Lemma <ref>, $(h, \{\Lambda_t\}_{t > 0} ) \overset{d}{=} (h^b, \{ R_b^{-1} \Lambda_{b t} \}_{t>0})$.
Hence, by the scale-invariance of the law of $\eta$, viewed modulo time parametrization, and the independence of $\eta$ and $(h, \{\Lambda_t\}_{t>0})$,

\begin{equation} \label{eq:scale-invariance-of-field-and-eta}
(h,\eta,\{\Lambda_t\}_{t > 0})  \overset{d}{=} (h^b,\eta^b , \{R_b^{-1} \Lambda_{b t}\}_{t > 0} ) .
\end{equation}

Hence, if we let

\[
T^b := \sup \{ t > 0 : R_b^{-1} \Lambda_{b t} \subset B_{\rho/3} \}
\]

then $T^b$ has the same law as $\TT$, defined in (<ref>).
By Proposition <ref>, for each $\delta \in (0, \rho)$ and $t > 0$, on the event $\{t < \TT\}$,
it holds except on an event of probability tending to 0 as ${\eps} \to 0$  that

\[
\Bdeltam(\Lambda_t) \subset \overline{\mathcal{X}}_{{\eps}}( t {\eps}^{-1} ) \subset \Bdeltap(\Lambda_t),
\]

for $\mathcal{X} \in \{A, D\}$, , both the IDLA cluster and divisible sandpile cluster converge.
Let $\mathcal{X} \in \{A, D\}$ be given.
By (<ref>),
this implies that for each $\delta \in (0, \rho)$ and $t > 0$, on the event $\{t < T^b\}$,
it holds except on an event of probability tending to 0 as ${\eps} \to 0$  that

\[
\Bdeltam(R^{-1}_b \Lambda_{b t}) \subset  R^{-1}_b \overline{\mathcal{X}}_{{\eps}}( b t {\eps}^{-1} ) \subset \Bdeltap(R^{-1}_b \Lambda_{b t}).
\]

By applying the above statement with $t/b$ in place of $t$ and $\delta / R_b$ in place of $\delta$, then re-scaling space by $R_b$, we get the following.
For each $\delta \in (0, \rho)$ and $t > 0$, on the event $\{t < b \times T^b\}$,
it holds except on an event of probability tending to 0 as ${\eps} \to 0$  that

\[
\Bdeltam(\Lambda_t) \subset \overline{\mathcal{X}}_{{\eps}}( t {\eps}^{-1} ) \subset \Bdeltap(\Lambda_t).
\]

As $b > 0$ was arbitrary, it remains to show that in probability,

\begin{equation} \label{eq:btb-converge-to-inf}
\lim_{b \to \infty} b \times T^b = \infty.
\end{equation}

This, however, follows immediately from the fact that $\TT$ is strictly positive and $\TT \overset{d}{=} T^b$.

§ DISCRETE OBSTACLE PROBLEM

We provide the proofs which were omitted in Section <ref> in the general setting of an undirected, locally finite, and infinite graph $G$. We consider a finite, connected set of vertices of $G$, denoted by $V$, which contain a marked vertex which we call the origin, denoted by $0$.
The graph Laplacian, $\Delta$, is defined by,

\[
\Delta u(a) =  \frac{1}{\deg(a)} \sum_{b \sim a}(u(b) - u(a)), \quad \mbox{for all vertices $a$ of $G$}
\]

where the sum $b \sim a$ is over the vertices $b$ which share an edge with $a$ and $\deg(a)$ denotes the number of such vertices.  Observe that the analogue of the maximum principle, Lemma <ref>, and the following discrete divergence theorem are satisfied on $G$.

Let $VG$ denote all the vertices of $G$ and suppose that $f,g: VG \to \R$ are functions which are zero outside of a finite subset of $VG$.

\begin{equation} \label{eq:gauss-green}
\sum_{a \in VG} f(a) \Delta g(a) \deg(a) = \sum_{a \in VG} g(a) \Delta f(a) \deg(a).
\end{equation}

This is well known for general, undirected graphs — see for example <cit.> or <cit.>.

Let $\mathfrak{r}:V \to (0, \infty)$ denote a threshold function
and $\phi : \cl(V) \to (-\infty, 0]$ the obstacle
which satisfies the following equation

\begin{equation} \label{eq:laplacian-of-phi}
\begin{cases}
\Delta \phi  = \mathfrak t \delta_0 \quad \mbox{on $V$} \\
\phi = 0 \quad \mbox{on $\partial V$}
\end{cases}
\end{equation}

where $\mathfrak t > 0$. Consider the following discrete obstacle problem

\begin{equation} \label{eq:discrete-least-supersolution-general}
\begin{aligned}
w := &\min \{ u: \cl(V) \to \R : \Delta u \leq \mathfrak{r} \mbox{ on $V$ and $u \geq \phi$}\}.
\end{aligned}
\end{equation}

We will make use of the following function $\mathfrak{q}: \cl(V) \to \R$ which is defined by the unique solution to

\begin{equation} \label{eq:definition-of-exit-time-general}
\begin{cases}
\Delta \mathfrak q  = \mathfrak{r} \quad \mbox{on $V$} \\
\mathfrak{q} = 0 \quad \mbox{on $\partial V$}.
\end{cases}
\end{equation}

We also denote the discrete cluster by

\begin{equation} \label{eq:discrete-cluster-general}
\Lambda = \{ a \in \cl(V) :  w(a) > \phi(a) \}.
\end{equation}

The results of Section <ref> may be achieved from those of this section
via the following map, for ${\eps} > 0$, $\rho \in (0,1)$,

\begin{align*}
\mathfrak{t} &\to {\eps}^{-1} t \deg^{{\eps}}(0)^{-1} \\
w &\to \wteps \\
G &\to {\Geps} \\
V &\to {\VGeps}(B_{\rho}) \setminus {\VGeps}(\partial B_{\rho}) \\
\mathfrak{r} &\to (\deg^{{\eps}})^{-1} \\
\phi &\to -(t {\eps}^{-1}) \times  \gr{{\eps}}{B_\rho}(0, \cdot) \\
\mathfrak q &\to  -\qeps{B_{\rho}} \\
\Lambda &\to \Lambdateps.
\end{align*}

Lemma <ref> is a special case of the following.

We have that

\[
\max(\mathfrak q,\phi) \leq w \leq 0
\]

and  $\Delta w \leq \mathfrak{r}$ on $V$.

Step 1: $w \leq 0$.

As $0 \geq \phi$ and $\mathfrak{r} > 0$ the function which is identically 0 is admissible in (<ref>).
Step 2: $w \geq \phi$ and $\Delta w \leq \mathfrak{r}$.

Observe that in general, for two functions $u_1, u_2$ defined on the vertices
of the graph $G$, we have

\[
\Delta \min(u_1, u_2) \leq \min(\Delta u_1 , \Delta u_2),
\]

where the minimum is pointwise.
Indeed, write $u = \min(u_1,u_2)$ and suppose for a vertex $a$ that $u(a) = u_1(a)$. Then, by definition,

\[
\Delta u(a) = \frac{1}{\deg(a)} \sum_{b \sim a}(u(b) - u_1(a))
\leq \frac{1}{\deg(a)} \sum_{b \sim a}( u_1(b) - u_1(a)) = \Delta u_1(b).
\]

Therefore, by compactness, the inequalities in the definition of (<ref>)
are also satisfied by $w$.
Step 3: $w \geq \mathfrak q$.

By Steps 1 and 2, since $\phi$ is zero on $\partial V$, we have that $w$ is zero on $\partial V$.
Thus, by (<ref>) and Step 2, 	$(w - \mathfrak{q})$ is superharmonic on $V$ and zero on
$\partial V$, and hence the claim follows by the maximum principle.

Lemma <ref> is a special case of the following.

We have that $\Delta w = \mathfrak{r}$ on $\Lambda$
and $\mathfrak{r} \geq \Delta w \geq 0$ on $V$.
Moreover,  $\Lambda$ is connected and if $\mathfrak{t} > \mathfrak{r}(0)$, contains the origin.

Step 1: $\Delta w = \mathfrak{r}$ on $\Lambda$.

Note that the inequality $\Delta w \leq \mathfrak{r}$ was established in Lemma <ref>.
Suppose for sake of contradiction $w > \phi$
but $\Delta w(a) < \mathfrak{r}(a)$ for some $a \in V$.  Let

\[
\delta = \min \left( \mathfrak{r}(a) - \Delta w(a), w(a) - \phi(a)  \right) > 0
\]

and consider the function $u: V \to \R$ defined by,

\[
u := w - \delta \times 1\{ \cdot = a\}.
\]

We will see that the function $u$ is admissible in (<ref>), but strictly less than $w$,  a contradiction.
By definition

\[
\Delta u(b) \leq \Delta w(b) \quad \forall b \neq a
\]

\begin{align*}
\Delta u(a) &= \frac{1}{\deg(a)} \sum_{b \sim a} (w(b) - u(a))
=  \Delta w(a) + \delta
\leq \mathfrak{r}(a).
\end{align*}

\[
u(b) = w(b) \geq \phi(b) \quad \forall b \neq a
\]

\begin{align*}
= w(a) - \delta
\geq  w(a) - w(a) + \phi(a)
\end{align*}

completing the proof.
Step 2: $\Delta w \geq 0$.

If $a \in V \cap \Lambda^c$, we have

\begin{align*}
\Delta w(a) &= \frac{1}{\deg(a)} \sum_{b \sim a} (w(b) -  w(a)) \\
&= \frac{1}{\deg(a)} \sum_{b \sim a} (w(b) - \phi(a)) \quad \mbox{(assumption on $a$)} \\
&\geq \Delta \phi(a)  \quad \mbox{(Lemma \ref{lemma:discrete-basic-properties-general})} \\
&\geq 0 \quad \mbox{(since $\phi$ is subharmonic)}.
\end{align*}

In light of Step 1, this completes the proof.
Step 3:  $\Lambda$ is connected.

Consider the function $v =  w - \phi$.
If $\Lambda$ is disconnected then there is a connected component of $\Lambda$ not containing the origin;
call this component $A$.
By Step 1 and (<ref>), we have that

\[
\begin{cases}
\Delta v = \mathfrak{r} \quad \mbox{on $A$} \\
v > 0 \quad \mbox{on $A$} \\
v = 0 \quad \mbox{on $\partial A$},
\end{cases}
\]

which contradicts the maximum principle.
Step 4: If $t > \mathfrak{r}(0)$,  $\Lambda$ contains the origin.

We demonstrate the contrapositive. Observe that the argument in Step 3 shows that $\Lambda$ is either empty or contains the origin.
If $\Lambda$ is empty, then

\begin{align*}
\mathfrak{t} &= \Delta \phi(0) \qquad \mbox{(by \eqref{eq:laplacian-of-phi})} \\
&= 	\Delta  w(0) \qquad \mbox{(as $\Lambda$ is empty)} \\
&\leq \mathfrak{r}(0) \qquad \mbox{(by Lemma \ref{lemma:discrete-basic-properties-general})},
\end{align*}

completing the proof.

We conclude with a discrete conservation of mass lemma; this is a general version
of Lemma <ref>.

If $\mathfrak{r} = \deg^{-1}$, we have that $|\Lambda| \leq  \mathfrak{t} \times \deg(0)$. Moreover,
if $\Lambda \subset \inte(V)$,

\[
\sum_{a \in V} \Delta w(a) \deg(a) =  \mathfrak{t} \times \deg(0).
\]

Let	$v = w - \phi$ and extend $v$ to a function on $VG$, the set of all vertices of $G$, by defining it to be zero on $\cl(V)^c$.
We compute,

\begin{align*}
-\mathfrak{t} \times  \deg(0) + |\Lambda| &\leq \sum_{a \in V} \Delta v(a) \deg(a)
\quad \mbox{(by Lemma \ref{lemma:discrete-laplacian-bound-general}, \eqref{eq:laplacian-of-phi}, and $\mathfrak{r} = \deg^{-1}$)} \\
&= \sum_{a \in VG} 1\{ a \in V\}  \Delta v(a) \deg(a) \\
&= \sum_{a \in VG}   v(a) \deg(a) \Delta 1\{ a \in V\}
\quad \mbox{(by Lemma \ref{lemma:discrete-divergence-theorem})} \\
&= \sum_{a \in V}   v(a) \deg(a) \Delta 1\{ a \in V\}
\quad \mbox{($v = 0$ on $V^c$)} \\
&= -\sum_{a \in V}  v(a)  \left( \mbox{\# neighbors of $a$ not in $V$} \right) \\
&\leq 0 \quad \mbox{(since $v \geq 0$ by Lemma \ref{lemma:discrete-basic-properties-general})}.
\end{align*}

If $\Lambda \subset \inte(V)$, then $v = 0$ on all vertices which share an edge with a vertex in $\partial V$.
Thus, an identical computation, where the first line is replaced by

\[
-\mathfrak{t} \times  \deg(0) + \sum_{a \in V} \Delta w(a) \deg(a) = \sum_{a \in V} \Delta v(a) \deg(a)
\]

and the last inequality by an equality, shows the moreover clause.