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 $\Lambda_{n}$.