Independent connected-set envelopes and a greedy lattice-animal bound

Every supremum defining $M_z$ is measurable because the roots are countable and each $S_y$ has countably many possible finite values. Translation of the common set law and [F2] give ${\sum }_{y}P(z\in S_y,\mathrm{\mid }{S}_y\mathrm{\mid }>u)=\mathbb{E}[\mathrm{\mid }{S}_0\mathrm{\mid }{\mathbf{1}}_{\{\mathrm{\mid }{S}_0\mathrm{\mid }>u\}}]$. This tail tends to zero: for example it is at most $u^{-1}\mathbb{E}\mathrm{\mid }{S}_0{\mathrm{\mid }}^2$. The union bound therefore shows $P(M_z>u)\to 0$, hence $M_z<\mathrm{\infty }$ almost surely. At least $S_z$ contains $z$, and every size is a positive integer, so a finite supremum is attained. Taking the countable union of the null exceptional events proves this simultaneously for all $z$. No independence of the $M_z$ is asserted.

We first prove the assertion for the independent field $X$. Suppose sites are independently marked with probability $r$. If an animal of size at most $m$ contains $k$ distinct marked sites, a depth-first traversal of a spanning tree of that animal, rooted at $0$, visits them in some order using at most $2m$ steps. The successive displacement vectors $v_1,\dots ,v_k\in {\mathbb{Z}}^2$, starting at $0$, have total ${\mathrm{\ell }}^1$ length at most $2m$. There are at most $2^{2k}\left(\genfrac{}{}{0px}{}{2m+2k}{2k}\right)\le (Cm\mathrm{/}k{)}^{2k}$ such ordered lists for $1\le k\le m$: choose signs of their two coordinates and distribute at most $2m$ units among $2k$ nonnegative coordinate magnitudes. Repeated sites and zero coordinates only cause overcounting. For a list of distinct visited sites, all its sites are marked with probability $r^k$. Thus $P(\text{some animal of size at most }m\text{ has }k\text{ marked sites})\le (C^{\mathrm{\prime }}r{m}^2\mathrm{/}{k}^2{)}^k$. This counts marked-site lists rather than all animals and is uniform over all their shapes.

Put $m=2^l$, $r_j=P(X_0\ge 2^j)$ and fix $ϵ=1\mathrm{/}4$. Choose a deterministic $a$ such that $C^{\mathrm{\prime }}\mathrm{/}{a}^2\le e^{-2}$, and a fixed $B\ge 2$. For $0\le j\le \lfloor ϵ{\log }_{2}m\rfloor$ set $k_j=\lceil \max \{am\sqrt{r_j},B\log m\}\rceil$. If $k_j>m$ the corresponding forbidden event is empty; otherwise step 1.2 bounds its probability by $e^{-2k_j}\le m^{-2B}$. The union over these $O(\log m)$ values of $j$ is summable along dyadic $m$. For a moment order $Q>2\mathrm{/}ϵ$, the elementary inequality $P(X_0>v)\le v^{-Q}\mathbb{E}{X}_0^Q$ and the fact that every size-at-most-$m$ animal lies in ${\mathrm{\Lambda }}_m=[-m,m{]}^2\cap {\mathbb{Z}}^2$ give $P({\max }_{{\mathrm{\Lambda }}_m}{X}_y>m^{ϵ})\le (2m+1{)}^2\mathbb{E}{X}_0^Q{m}^{-ϵQ}$, also summable in $l$.

If events $E_l$ have summable probabilities, then $P({\bigcup }_{l\ge L}{E}_l)\le {\sum }_{l\ge L}P(E_l)\to 0$; countable additivity implies that only finitely many $E_l$ occur almost surely. Apply this elementary argument to the failures in step 2.1. Almost surely, for every sufficiently large dyadic $m$, all weights in ${\mathrm{\Lambda }}_m$ are at most $m^{ϵ}$, and every size-at-most-$m$ animal has fewer than $k_j$ sites with $X_y\ge 2^j$ at each relevant level. The pointwise layer bound $X_y\le 1+{\sum }_{j\ge 0}{2}^{j+1}{\mathbf{1}}_{\{X_y\ge 2^j\}}$ then gives, for every animal with $m\mathrm{/}2<\mathrm{\mid }\mathrm{\Gamma }\mathrm{\mid }\le m$, $\mathrm{\mid }\mathrm{\Gamma }{\mathrm{\mid }}^{-1}{\sum }_{y\in \mathrm{\Gamma }}{X}_y\le 1+4a{\sum }_{j\ge 0}{2}^j\sqrt{r_j}+O(m^{ϵ-1}\log m)$. Levels above $ϵ{\log }_{2}m$ contribute zero; the error comes from the additive $B\log m+1$ in $k_j$. The constants are deterministic.

A fixed moment order $Q>2$ gives $\sqrt{r_j}\le (\mathbb{E}{X}_0^Q{)}^{1\mathrm{/}2}{2}^{-Qj\mathrm{/}2}$, so the series in step 3.1 converges. Its value is deterministic, and the error tends to zero. Choose $K$ larger than $2+4a{\sum }_{j\ge 0}{2}^j\sqrt{r_j}$. Every sufficiently large animal belongs to one of the dyadic size ranges in step 3.1. Thus almost surely its average $X$-weight is at most this same $K$, proving the general independent-field assertion.

Apply step 4.1 to $X_y=\mathrm{\mid }{S}_y{\mathrm{\mid }}^2$, which is an independent identically distributed field with all moments. Work on the probability-one event where its large-animal bound holds and all $M_z$ are finite. Let $\mathrm{\Gamma }$ be any sufficiently large animal, $m=\mathrm{\mid }\mathrm{\Gamma }\mathrm{\mid }$, $A={\sum }_{z\in \mathrm{\Gamma }}{M}_z$, and $t=A\mathrm{/}(2m)>0$. For each of the finitely many $z\in \mathrm{\Gamma }$ with $M_z>t$, choose a root $y(z)$ attaining its maximum; to avoid even a selection convention take the first such root in a fixed enumeration of ${\mathbb{Z}}^2$. Let $Y$ be the distinct chosen roots, $s_y=\mathrm{\mid }{S}_y\mathrm{\mid }$, $U={\sum }_{y\in Y}{s}_y^2$, and $V={\sum }_{y\in Y}{s}_y$. Low-weight sites contribute at most $A\mathrm{/}2$, and at most $s_y$ sites can be assigned to a given $S_y$. Hence $U\ge A\mathrm{/}2=mt$. Since $s_y>t$ for chosen roots, $V\le U\mathrm{/}t$.

The set $\mathrm{\Delta }=\mathrm{\Gamma }\cup {\bigcup }_{y\in Y}{S}_y$ is a finite connected animal containing every root in $Y$, because every chosen set meets $\mathrm{\Gamma }$ and is connected. Its size is at least $m$ and at most $m+V$. Therefore the same large-animal bound gives $U\le {\sum }_{y\in \mathrm{\Delta }}\mathrm{\mid }{S}_y{\mathrm{\mid }}^2\le K\mathrm{\mid }\mathrm{\Delta }\mathrm{\mid }\le K(m+U\mathrm{/}t)$. If $t>2K$, this would imply $U<2Km$ while step 5.1 gives $U\ge mt>2Km$, a contradiction. Thus $t\le 2K$ and $A\le 4Km$. Taking the deterministic constant $K_0=4K$ proves the envelope assertion uniformly over every sufficiently large animal. Only the given probability laws and countable additivity were used, without constructing any new infinite family.