Thermodynamic limit and Gaussian measures

Question

Let $\Lambda \subset \mathbb{Z}^{d}$ be finite and fixed and consider $\mathbb{R}^{|\Lambda|}$ be the vector space of all sequences $\varphi = (\varphi_{x})_{x\in \Lambda}$. We equip $\mathbb{R}^{|\Lambda|}$ with its Borel $\sigma$-algebra $\mathbb{B}(\mathbb{R}^{|\Lambda|})$. We denote by $\nu$ the Lebesgue measure restricted to $\mathbb{B}(\mathbb{R}^{|\Lambda|})$.

Now, in statistical mechanics, one usually consider an action $S_{\Lambda}: \mathbb{R}^{|\Lambda|}\to \mathbb{R}$ which is assumed to be a measurable function. This action gives rise to a finite volume Gibbs measure $\mu_{\Lambda}$, defined by means of its density with respect to $\nu$: \begin{eqnarray} d\mu_{\Lambda}(\varphi) := \mbox{const.} e^{-S_{\Lambda}(\varphi)}d\nu(\varphi) \tag{1}\label{1} \end{eqnarray} where the $\mbox{const.}$ term in (\ref{1}) is a normalization factor so that $\mu_{\Lambda}$ is a probability measure on $\mathbb{R}^{|\Lambda|}$.

From the physics point of view, we are interested in studying the behavior of the system in the infinite volume limit $\Lambda \nearrow \mathbb{Z}^{d}$. Thus, we have to first ensure that the infinite volume measure $\mu_{\mathbb{Z}^{d}} \equiv \mu$ exists in some sense. We usually take this limit as the weak-limit: \begin{eqnarray} \int d\mu e^{i\langle f, \varphi\rangle} := \lim_{n\to \infty}\int d\mu_{\Lambda_{n}}e^{i\langle f, \varphi\rangle} \tag{2}\label{2} \end{eqnarray} provided this limit exists. Here, $\Lambda_{n}$ is a sequence of increasing sets converging to $\mathbb{Z}^{d}$.

A particular case of the above scenario is when the action $S_{\Lambda}$ comes from a strictly positive quadratic form $Q_{\Lambda}: \mathbb{R}^{|\Lambda|}\times \mathbb{R}^{|\Lambda|} \to \mathbb{R}$. If we set $S_{\Lambda}(\varphi) = Q_{\Lambda}(\varphi, \varphi)$, the measure (\ref{1}) becomes Gaussian.

Now, it is a known fact that Gaussian measures are consistent in the sense of Kolmogorov Theorem, so that this very same Theorem implies the existence of a measure $\mu$ on $\mathbb{R}^{\mathbb{Z}^{d}}$. (*)

My Question is whether we can interpret this measure $\mu$ obtained using Kolmogorov's Theorem as a weak limit of the form (\ref{2}) in some sense. My point here is: it seems legit to consider this $\mu$ as a infinite volume measure, but the path taken to define this limit was not by means of a limit such as (\ref{2}). Besides, the inner product $\langle f,\varphi \rangle$ only makes sense on $\mathbb{R}^{\mathbb{Z}^{d}}$ if we restrict it to some subspace, say $\mathcal{l}^{2}(\mathbb{Z}^{d})$. How to connect these two scenarios?

EDIT: Let me elaborate more on (*). Simon's book states Kolmogorov's Theorem as follows. Theorem [Kolmogorov]: Let $\mathcal{I}$ be a countable set and let a probability measure $\mu_{|I|}$ on $\mathbb{R}^{|I|}$ be given for each finite set $I\subset \mathcal{I}$, so that the family of $\mu_{I}$'s is consistent (i.e. $\mu_{I}(A) = \mu_{I'}(A\times \mathbb{R}^{|I'|-|I|})$ if $|I'|\ge |I|$). Then there is a probability measure space $(X, \mathcal{F}, \mu)$ and random variables $\{f_{\alpha}\}_{\alpha \in \mathcal{I}}$ so that $\mu_{I}$ is the joint probability distribution of $\{f_{\alpha}\}_{\alpha \in \mathcal{I}}$.

Now, the proof of this theorem shows that $X$ is actually $\dot{\mathbb{R}}^{|\mathcal{I}|}$ where $\dot{\mathbb{R}}:=\mathbb{R}\cup \{+\infty\}$ is a compactification of $\mathbb{R}$. Thus, if we take $\mathcal{I}=\mathbb{Z}^{d}$, and for each finite $\Lambda \subset \mathbb{Z}^{d}$ the associate $\mu_{\Lambda}$ to be a Gaussian measure (say, nondegenerate and associated to some positive-definite matrix $C_{\Lambda}$), the above Theorem proves the existence of a (Gaussian) measure on $\mathbb{R}^{\mathbb{Z}^{d}}$.

EDIT 2: The measures $\mu_{\Lambda_{n}}$ are, in fact, defined in $\mathbb{R}^{\mathbb{Z}^{d}}$ with the product topology. It can be constructed from (\ref{1}) by taking, instead of $S_{\Lambda}$, an action $S:\mathbb{R}^{\mathbb{Z}^{d}}\to \mathbb{R}$ with free boundary conditions outside each $\Lambda_{n}$.

Answer 1

Not quite an answer (because I still don't understand the question) but too long for a comment.

Take the following very simple example where the $\varphi_{x}$ are iid $\mathscr{N}(0,1)$ random variables. There is no problem constructing their joint law $\mu$ as a probability measure on $\mathbb{R}^{\mathbb{Z}^d}$ with the product topology. The way to do that is using the Daniell-Kolmogorov Extension Theorem. Now what about the physicists' beloved notion of action? Well, it just does not make sense.

In this case, a physicist may want to write formally $$ d\mu(\varphi)=\frac{1}{Z}\ e^{-S(\varphi)}\ \prod_{x\in\mathbb{Z}^{d}} d\varphi_x $$ with the action $S(\varphi)=\frac{1}{2}Q(\varphi,\varphi)$, (BTW please put the 1/2) where $$ Q(\varphi,\varphi)=\sum_{x\in\mathbb{Z}^d} \phi_x^2\ . $$ Now if the random variable $\varphi\in\mathbb{R}^{\mathbb{Z}^d}$ is sampled according to the (well defined) probability measure $\mu$, then almost surely $$ Q(\varphi,\varphi)=\infty $$ as can be shown say by Kolmogorov's Three Series Theorem.

This is all to say the question would make more sense if, rather than talking about actions and the matrix $Q$, it talked instead about the covariance matrix/bilinear form $C=Q^{-1}$.