Percolation and number of phases in the 2D Ising model

Question

Update. As my previous figure had conceptual mistakes I decided to change the picture to another, more instructive.

After a long time I came back to try to understand an article on the Ising model. The review article is Percolation and number of phases in the 2D Ising model by Hans-Otto Georgii and Yasunari Higuchi (published in 2000 in the Journal of Mathematical Physics as Percolation and number of phases in the two-dimensional Ising model).

I confess that since after the first half of the statement of the first theme got lost. For an expert my doubts about the second half of the proof of the lemma below is certainly primitive.

I reproduce below the statement of the lemma and its proof together as some figures to illustrate what I did the first half of the demonstration of the lemma.

Lema 2.1(Existence of infinite clusters) If $\mu\in\mathcal{G}$ is different from $\mu^-$, there exists with positive probability an infinite $+$cluster. That is, $\mu(E^+)>0$ when $\mu\ne\mu^-$.

[Here, $\mathcal{G}$ is the set of Gibbs measures of the Ising model of first neighbors in the net $\mathbb{Z}^2$. And $\mu^-$ is the extremal Gibbs measure, which is obtained by thermodynamic limit of finite volume measures whit borders fixed in negative sign. $E^+$ denotes the event que there exists an infinite cluster of spins in state $+$.]

Proof. Suppose that $\mu(E^+)=0$. Then any given square $\Delta$ is almost surely surrounded by a $-*$circuit, and with probability close to $1$ such a circuit can already be found within a square $\Lambda\supset\Delta$ provided $\Lambda$ is large enough. If this occurs, we let $\Gamma$ be the largest random subset of $\Lambda$ which is the interior of such a $-*$circuit. (A largest such set exists because the union of such sets is again the interior of a $-*$circuit.) In the alternative case we set $\Gamma=\emptyset$. By maximality, $\Gamma$ is determined from outside [ see figure below].

So far I understand the argument. And in any section of this theorem repeats this argument in several demonstrations. But what comes next does not seem intelligible to a non specialist. I have no idea how the properties below to get the desired equality through the instruction below:

Continuation of proof: The strong Markov property together with the stochastic monotonicity $\mu^-_\Gamma\preceq \mu^-$ therefore implies (in the limit $\Lambda\uparrow\mathbb{Z}^2$) that $\mu\preceq\mu^-$ on $\mathcal{F}_\Delta$. Since $\Delta$ was arbitrary and $\mu^-$ is minimal we find that $\mu=\mu^-$, and the lemma is proved. $\Box$

Question: How to use the strong Markov property and the stochastic monotonicity to finish the proof of lemma? Below the properties used as set out in Article.

$\bullet$ the strong Markov property of Gibbs measures, stating that $\mu(\cdot\,|\mathcal{F}_{\Gamma^c})(\omega) =\mu_{\Gamma(\omega)}^\omega$ for $\mu$-almost all $\omega$ when $\Gamma$ is any finite random subset of $\mathbb{Z}^2$ which is determined from outside, in that $\{\Gamma=\Lambda\}\in\mathcal{F}_{\Lambda^c}$ for all finite $\Lambda$, and $\mathcal{F}_{\Gamma^c}$ is the $\sigma$-algebra of all events $A$ outside $\Gamma$, in the sense that $A\cap \{\Gamma=\Lambda\}\in\mathcal{F}_{\Lambda^c}$ for all finite $\Lambda$. (Using the conventions $\mu_{\emptyset}^\omega=\delta_\omega$ and $\mathcal{F}_{\emptyset^c}=\mathcal{F}$ we can in fact allow that $\Gamma$ takes the value $\emptyset$.) For a proof
one simply splits $\Omega$ into the disjoint sets $\{\Gamma=\Lambda\}$ for finite $\Lambda$.

.

$\bullet$ the stochastic monotonicity (or FKG order) of Gibbs distributions; writing $\mu\preceq\nu$ when $\mu(f)\leq\nu(f)$ for all increasing local (or, equivalently, all increasing bounded measurable) real functions $f$ on $\Omega$, we have $\mu_{\Lambda}^\omega\preceq\mu_{\Lambda}^{\omega'}$ when $\omega\leq\omega'$, and
$\mu_{\Lambda}^\omega\preceq\mu_{\Delta}^\omega$ when $\Delta\subset\Lambda$ and $\omega\equiv +1$ on $\Lambda\setminus\Delta$ (the opposite relation holds when $\omega\equiv -1$ on $\Lambda\setminus\Delta$).

Answer 1

Assume, as in the first part of the proof, that there exists a set $\Gamma$, determined from the outside, such that $\Delta\subset\Gamma\subset\Lambda$. Moreover, by construction, the exterior boundary of $\Gamma$ is entirely composed of $-$ spins (not as on your picture! $\Gamma$ is exactly the interior of this circuit).

Therefore, by the strong Markov property, we have, on this event, $$ \mu(\cdot|\mathcal{F}_{\Gamma^c}) = \mu^-_{\Gamma}. $$ Let $f$ be an increasing local function with support inside $\Delta$. By FKG, $$ \mu^-_{\Gamma}(f) \leq \mu^-_{\Lambda}(f) \leq \mu_{\Lambda'}^-(f), $$ for all $\Lambda'\supset\Lambda$. Let $\Lambda'\uparrow\mathbb{Z}^2$ to conclude that $$ \mu^-_{\Gamma}(f)\leq \mu^-(f). $$ Let $\mathcal{A}_{\Lambda,\Gamma}$ be the event that there is a $-\star$-circuit inside $\Lambda$, surrounding $\Delta$, and that $\Gamma$ is the interior of the largest such circuit. Let also $\mathcal{A}_\Lambda=\bigcup_\Gamma \mathcal{A}_{\Lambda,\Gamma}$. Then, $$ \mu(f) \leq \sum_{\substack{\Gamma:\\\Delta\subset\Gamma\subset\Lambda}} \mu( \mu(f | \mathcal{F}_{\Gamma^c}) 1_{\mathcal{A}_{\Lambda,\Gamma}}) + \mu(\mathcal{A}_{\Lambda}^c) \leq \mu^-(f) + \mu(\mathcal{A}_{\Lambda}^c). $$ Since $\lim_{\Lambda\uparrow\mathbb{Z}^2}\mu(\mathcal{A}_\Lambda^c) \leq \mu(E^+)=0$, we conclude that $$ \mu(f) \leq \mu^-(f), $$ for all increasing local functions $f$. This implies that $\mu\preccurlyeq\mu^-$. But, since $\mu^-$ is stochastically minimal, this means that $\mu=\mu^-$.