Kolmogorov Extension Theorem for Gaussian measures

Question

Let $Y$ be a non-empty finite set and $\mu_{Y}$ be a (non-degenerate) Gaussian measure on $\mathbb{R}^{Y}$, that is, $\mu_{Y}$ has density $\frac{1}{(2\pi)^{\frac{1}{2}|Y|}}e^{-\frac{1}{2}\langle x, C x\rangle}$ with respect to the Lebesgue measure. Here, $|Y|$ denotes the cardinality of $Y$, $\langle \cdot, \cdot \rangle$ is the usual inner product on $\mathbb{R}^{Y}$ and $C = (C_{ij})$, $i,j \in Y$ is a positive-definite matrix.

If $\emptyset \neq X \subset Y$, let $C_{X} = (C_{ij})$, $i,j \in X$ be a submatrix of $C$ and $\mu_{X}$ the associate Gaussian measure.

Question 1: I want to prove the compatibility condition: for any Borel set $E \in \mathbb{R}^{X}$, the following identity holds: $$\mu_{X}(E) = \mu_{Y}(E\times \mathbb{R}^{Y\setminus X}). \tag{1}\label{1}$$

My proof: Write: $$\mu_{Y}(E\times \mathbb{R}^{Y\setminus X}) = \frac{1}{(2\pi)^{\frac{1}{2}|Y|}}\int_{E\times \mathbb{R}^{Y\setminus X}}e^{-\frac{1}{2}\langle x, C x\rangle}dx = \frac{1}{(2\pi)^{\frac{1}{2}|Y|}}\int_{E\times \mathbb{R}^{Y\setminus X}}e^{-\frac{1}{2}\langle Tx, C_{X}Tx\rangle}e^{-\frac{1}{2}\langle x-Tx, C_{Y\setminus X}(x-Tx)\rangle}dx \\ = \frac{1}{(2\pi)^{\frac{1}{2}|X|}}\int_{E}e^{-\frac{1}{2}\langle y, C_{X} y\rangle}dy \frac{1}{(2\pi)^{\frac{1}{2}|Y\setminus X|}}\int_{\mathbb{R}^{Y\setminus X}}e^{-\frac{1}{2}\langle x-Tx, C_{Y\setminus X}(x-Tx)\rangle}dx \tag{2}\label{2}$$ Where $T$ is an orthogonal projection $\mathbb{R}^{Y} \ni x \mapsto Tx := x|_{X}$ and $y = Tx$ is just a change of variables, with $\det T = 1$. Because: $$\int_{\mathbb{R}^{Y\setminus X}}e^{-\frac{1}{2}\langle x-Tx, C_{Y\setminus X}(x-Tx)\rangle}dx = 1,$$ the result follows.

Is my proof correct?

Question 2: We can consider $X$ and $Y$ to be finite subsets of $\mathbb{Z}^{d}$. By Kolmogorov Extension Theorem, this implies the existence of a unique measure $\mu$ on $\mathbb{R}^{\mathbb{Z}^{d}}$. I have a problem understanding what is the proper $\sigma$-algebra which $\mathbb{R}^{\mathbb{Z}^{d}}$ is equipped with. Is it just the usual Borel $\sigma$-algebra, that is, the $\sigma$-algebra generated by the open sets of $\mathbb{R}^{\mathbb{Z}^{d}}$, when the latter is equipped with the product topology?

Answer 1

1. No, this argument is wrong. In the second equality of (2), you claim that $$\langle x, Cx \rangle = \langle Tx, C_X Tx \rangle + \langle x - Tx, C_{Y\setminus X} (x - Tx) \rangle,$$ where $T$ denotes orthogonal projection onto $X$. In general, though, the RHS will also contain cross terms $$\langle Tx, C(x - Tx)\rangle,$$ which do not appear in your expression (2). (To argue this correctly, you need to e.g. write the density $f_Y$ as a product $f_Xf_{Y \mid X}$, using the conditional distribution on $Y$ given evaluation on $X$.)

2. The $\sigma$-algebra on $\mathbb R^{\mathbb Z^d}$ is the product $\sigma$-algebra, $$\sigma(\{\pi_i^{-1}(B):i \in \mathbb Z^d, B \in \mathcal B\}),$$ where $\pi_i(x) = x_i$ are the coordinate projections. In this case, as $\mathbb Z^d$ is countable, the product $\sigma$-algebra is equal to the $\sigma$-algebra generated by the product topology, but in general they may be different.