Imprecise Definition of a $\sigma$-algebra

Question

I'm reading some works on the hierarchical model in statistical mechanics and I came across an strange definition, which I need to clarify. Consider a finite set $\Lambda \subset \mathbb{Z}^{d}$. The set of all functions $\varphi : \Lambda \to \mathbb{R}$ is isomorphic to $\mathbb{R}^{|\Lambda|}$, so that these functions are represented as vectors $\varphi = (\varphi_{x})_{x\in \Lambda} \in \mathbb{R}^{|\Lambda|}$, and are called fields. Given $\xi_{1},...,\xi_{N} \in \mathbb{R}^{|\Lambda|}$, we set $\varphi_{j} := \sum_{k> j}\xi_{k}$ for each $j=0,...,N-1$ and $\varphi_{N} = 0$.

Now the author states the following: "Given $X\subset \Lambda$, let $\mathcal{N}_{j}(X)$ be the algebra of functions measurable with respect to the $\sigma$-algebra generated by $\{\varphi_{j}(x), \hspace{0.1cm} x \in X\}$. In more down to earth terms, an element of $\mathcal{N}_{j}(X)$ is a function only of fields at points $x \in X$."

I don't understand what is this $\sigma$-algebra. It seems confusing to me. What does it mean?

Edit: The text can be found in math.ubc.ca/~db5d/Seminars/PCMILectures/lectures.pdf

Answer 1

$\newcommand\vpi{\varphi}$ $\newcommand\La{\Lambda}$ $\newcommand\R{\mathbb R}$ You misunderstood the notes: For each $j\in\{0,\dots,N\}$ and each $x\in\La$, $\vpi_j(x)$ is (not a real number but) a real-valued random variable (r.v.).

Indeed, (i) formula (2.17) on page 25 in the notes implies that $\vpi_j=\sum_{k=j+1}^N\zeta_j$ (with $\vpi_N=0$), (ii) line 2 on page 22 there tells us that $\zeta_j\sim N(C_j)$, and (iii) by formula (1.20) on page 10, $\zeta_j\sim N(C_j)$ means that $\zeta_j=(\zeta_j(x)\colon x\in\La)$ is a random zero-mean Gaussian vector (with values in $\R^\La$ and) with the covariance matrix $C_j=(C_j(x,y))_{(x,y)\in\La^2}$.

So, for each $j\in\{0,\dots,N\}$ and each $X\subseteq\La$, the $\sigma$-algebra in question is the $\sigma$-algebra generated by the set $\{\varphi_j(x)\colon x\in X\}$ of real-valued r.v.'s, that is, the smallest $\sigma$-algebra with respect to which all the r.v.'s $\varphi_j(x)$ with $x\in X$ are measurable.

Answer 2

Disclaimer: maybe this would be more appropriate as a comment since I do not know the cited text, but I am posting this as an answer because I do cannot yet comment:

My understanding would be that it is meaning the $\sigma$-algebra generated by the restriction $\phi_j|_{X}$, that is the $\sigma$-algebra generated by all the sets $C \subseteq \Lambda$ such that $C=\{x\,: \, \phi_j|_X(x) \in C\}$.

By a standard result, a real function $g$ is measurable with respect to a $\sigma$-algebra generated by another real function $f$ if and only if there exist a measurable function $h:\mathbb{R} \to \mathbb{R}$ such that $g=h \circ f$.

So in this case, if the definition is the one I conjectured, then $f \in \mathcal{N}_j$ if and only if there is a measurable $h$ such that $f=h \circ \phi_j|_X$, that is, $f$ would depend only on the values of $\phi_j$ on the set $X$, as the text says. Cannot help with the physics though :)