# It is possible to prove the existence of Gibbs measures using the Kolmogorov extension theorem?

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 $\mathbb{Z}^2$.

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

$$\mu^\Lambda(\quad)=\mu^\Gamma\big(\Pi^{-1}_{\Gamma \Lambda} (\;\cdot\;) \big) \quad \forall ,\Gamma, \Lambda \in \mathscr{S} \text{ with } \Lambda \subset \Gamma$$

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

$$\mu^\Lambda = \mu \big(\Pi_{\Lambda}^{-1}(\; \cdot \;)\big)$$

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 $$\mathscr{\mathcal{F}}\times\Omega\ni(A,\omega) \longmapsto \mu\Big(\Pi_\Lambda(A)\!\times\!\{\Pi_{S/\Lambda}(\omega)\}\Big)$$

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Can you say precisely what you mean by "Gibbs measures"? –  Nate Eldredge Jun 16 '12 at 20:16
@ Nate Eldredge. I add explanations on Gibbs measure and Kolmogorov's extension theorem. Thank's. –  Elias Jun 17 '12 at 14:52
Perhaps you should try mathoverflow.net ? –  leonbloy Jun 19 '12 at 14:43
Need not be frightened by the notation. I believe that one need not resort to MathOverflow. But I will consider the suggestion! –  Elias Jun 20 '12 at 14:33

You can, at least for finite state spaces $\mathbb{E}^i$. The link to Kolmogorov's extension theorem is used explicitly in Theorem 5 of these notes: