Proof: Limit measure $$\mu^x$$ exists

We follow an argument given by Adams, \cite{A_12}. We note that the measure $$\mu^x$$ can also be seen as the limit of a sequence of image measures $$\mu_{N}^{\nabla,\psi_{x}}=\mu_{N}^{\psi_{x}}\circ\nabla^{-1}$$, where $$\mu_{N}^{\psi_{x}}$$ measures height configurations on $$\Lambda_N$$ with boundary condition $$\psi_{x}(y)=xy$$, and $$\nabla:\Lambda_{N}\rightarrow \mathcal B ( \Lambda_N )$$ maps height configurations to bond configurations, \cite{F_03} pp.152-153. Note that $$\nabla$$ is not invertible, however $$\mu_{\Lambda_{N}}^{\psi_{u}}\circ\nabla^{-1}$$ is still well defined because the preimage $$\nabla^{-1}(\eta)$$ consists of height configurations that all have the same mass under $$\mu_{N}^{\psi_{x}}$$, thus there is no ambiguity.

Denoting $$b_i = (i, i+1) \in \mathcal{B}(\Lambda)$$, we define the height function of a configuration $$\eta \in \Omega_N$$ to be

$\phi_\eta(y)\coloneqq\sum_{i=1}^{y}\eta(b_i), \qquad y \in \Lambda_N$

so that $$\nabla(\phi_\eta) = \eta$$. Now if $$(\eta(b_i) )_{i=0}^N$$ are i.i.d.\@ with distribution $$\nu(\cdot) \propto \exp(- V( \cdot ) )$$, we see that the measure obtained by conditioning on the final height satisfies $$\nu^{\otimes N}( \cdot \, | \phi(N) = xN ) = \mu_N^{\psi_x} \circ \nabla ^{-1}$$, where equality is in law. Hence the limit $$\mu^x$$ is the weak limit of this sequence of conditioned measures.

In Theorem 3.5 of \cite{GPV_88}, Guo et al.\@ show that in the limit, the conditioned probability measure above can be expressed as a product measure of the relevant Cram\'er transformed measures; in particular letting $$\lambda = \lambda(x)$$, they show that this limit is an infinite product of $$\hat \nu_\lambda$$. Now we are done since

$\mu^x(\cdot) = \lim_{N \rightarrow \infty} \mu_N^{\nabla, \psi_x}(\cdot) = \lim_{N\rightarrow \infty} \nu^{\otimes N}( \, \cdot | \phi(N) = xN ) = \prod_{i \in \mathbb Z} \hat \nu_\lambda(\cdot) \qedhere$