The following content was based on Minlos' book on statistical physics. Let $\Lambda \subset \mathbb{R}^{d}$ be fixed (Minlos takes $d=3$ but I think the ideas follow without change to $d \ge 1$). We denote by $\Lambda^{N}$ the $N$-fold cartesian product of $\Lambda$ with itself and $(\Lambda^{N})'$ $N$-uples $(x_{1},...,x_{N})$ in $\Lambda$ with different entries, i. e. $x_{i}\neq x_{j}$ if $i\neq j$. Also $\Gamma_{\Lambda, N}:=\{\omega \subset \Lambda, \hspace{0.1cm} \mbox{card}(\omega) = N\}$, where $\mbox{card}(\omega)$ is the cardinality of the set $\omega$. Define $\Pi: (\Lambda^{N})' \to \Gamma_{\Lambda, N}$ by $(x_{1},...,x_{N}) \mapsto \{x_{1},...,x_{N}\}$. For every subset $A$ of $\Gamma_{\Lambda, N}$, Minlos set: $$ \mu_{\Lambda}^{(N)}(A) := \frac{\mbox{Vol}(\Pi^{-1}(A))}{N!} $$ Then, he states that $\mu_{\Lambda}(\Gamma_{\Lambda, N}) = \frac{|\Lambda|^{N}}{N!}$. The problem is that he doesn't seem to define $\mbox{Vol}$ or $|\cdot|$ anywhere and it is getting me a little confused. At first, I thought $\mbox{Vol}$ was just the Lebesgue measure on $\mathbb{R}^{dN}$. But it would be a little odd because if I take $A = \{x_{1},...,x_{N}\}$ it seems that $\mu_{\Lambda}^{(N)}(\{x_{1},...,x_{N}\}) = 0$. Besides, how come does the second statement about $\mu_{\Lambda}(\Gamma_{\Lambda, N})$ follow? If $|\Lambda|$ is the cardinality of $\Lambda$ (which I don't know for sure), does it follow from de definition of $\mu_{\Lambda}$?