It is possible to prove the existence of Gibbs measures using the Kolmogorov extension theorem? If yes how? If the proof is too long to write here is there any reference?
Thank you.
Edit.
Let $S$ be a countable set and $\mathscr{S}=\mathscr{S}(S)=\{\Lambda: \Lambda \subset S, \quad 0 <| \Lambda | <\infty \} $. To keep in mind we can take $S$ as $ \mathbb{N}, \mathbb{Z}$ or $\mathbb{Z}^2$.
Let $ \mathscr{E}$ a $\sigma$-algebra on the set $\mathbb{E}$. If $(\mathbb{E}, \mathscr{E}) = (\mathbb{E}^i, \mathscr{E}^i), \quad \forall i \in S $ define:$(\mathbb{E}^\Lambda, \mathscr{E}^\Lambda)=\bigotimes_{i\in \Lambda}(\mathbb{E}^i, \mathscr{E}^i).$
If $\Omega \triangleq \mathbb{E}^S = \{\omega=(\omega_i)_{i\in S}: \omega_i \in \mathbb{E}^i, \forall i\in S \} $ then for $ \Lambda, \Gamma \in \mathscr{S} $ with $ \Lambda\subset \Gamma $ define:
$ \Pi_i: \Omega \to \mathbb{E}^i $ as the natural projection of $\Omega \triangleq \mathbb{E}^S $ in $\mathbb{E}^i $,
$ \Pi_\Lambda: \Omega \to \mathbb{E}^\Lambda $ as the natural projection of $ \Omega \triangleq \mathbb{E}^S $ in $ \mathbb{E}^\Lambda $,
$ \Pi_{\Gamma, \Lambda}: \mathbb{E}^\Gamma \to\mathbb{E}^\Lambda$ as the natural projection of $ \Omega \triangleq \mathbb{E}^S$ in $\mathbb{E}^\Lambda $.
Now consider the following $ \sigma $-algebras defined on $\Omega $. $ \mathcal{F}_\Lambda = \sigma \big(\Pi_\Lambda\big) $, $ \mathcal{J}_{\Lambda}=\sigma \big(\Pi_{S/\Lambda}\big)$, $\mathcal{F}=\sigma \big (\{\Pi_\Lambda \}_{\Lambda \in \mathscr{S}}\big)$.
And also consider the $ \sigma$-algebras $ \mathcal{F}_{\Gamma, \Lambda} $ and $ \mathcal{J}_{\Gamma, \Lambda} $ defined on $ \Omega_\Gamma = \mathbb{E}^\Gamma $ respectively by $ \sigma \big(\Pi_{\Gamma, \Lambda} \big) $ and $ \sigma \big(\Pi_{\Gamma/\Lambda, \Lambda} \big)$. In this notation the Kolmogorov extension theorem can be stated as follows.
Definition Given a family of probability measures $\{\mu^{\Gamma}\}_{\Lambda \in \mathscr{S}}$ on $(\mathbb{E}^\Lambda, \mathscr{E}^\Lambda)$ the equations
\begin{equation}\mu^\Lambda(\quad)=\mu^\Gamma\big(\Pi^{-1}_{\Gamma \Lambda} (\;\cdot\;) \big) \quad \forall ,\Gamma, \Lambda \in \mathscr{S} \text{ with } \Lambda \subset \Gamma \end{equation}
are called Kolmogorov consistency condition.
and
Theorem [Kolmogorov extension] If $(\mu_\Gamma)_{\Gamma \in\mathscr{S}} $ is a family of probability measures on $(\mathbb{E}^\Gamma,\mathscr{E}^\Gamma)$, meeting the consistency condition Kolmogorov then there exists a unique probability measure on $(\mathbb{E}^S, \mathscr{E}^S = \mathcal{F})$ such that
\begin{equation} \mu^\Lambda = \mu \big(\Pi_{\Lambda}^{-1}(\; \cdot \;)\big) \quad \forall \;\Lambda \end{equation}
In a brief term as a mean Gibbs measure $ \mu $ on the $(\Omega,\mathcal{F})$ satisfies that the condition Dobrushin-Lanford-Ruelle equivalently that is the same as $ \mu \Big(\Pi_\Lambda(A) \times \{\Pi_{S/\Lambda}(\omega)\}\Big)=\mu\Big(A|\mathcal{J}_\Lambda \Big)(\omega) $ for $A\in \mathcal{F}$ and $\mu|_{\mathcal{J}_\Lambda}\mbox{-a.e. }\omega\in\Omega$. Here $\mu|_{\mathcal{J}_\Lambda}$ is the restriction of the measure $\mu$ to $\mathcal{J}_\Lambda$.
In other words it's like $\mu$ to be specified by the probability of kernels $(\Omega,\mathcal{J}_\Lambda)$ to $(\Omega,\mathcal{F})$ given by $ \mathscr{\mathcal{F}}\times\Omega\ni(A,\omega) \longmapsto \mu\Big(\Pi_\Lambda(A)\!\times\!\{\Pi_{S/\Lambda}(\omega)\}\Big) $