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 $\mu_{\Lambda}$?
1 Answer
If $\text{Vol}$ denotes the Lebesgue measure on $(\mathbb R^d)^N$, if $\cdot$ denotes the Lebesgue measure on $\mathbb R^d$, and if $\mu_\Lambda=\mu_\Lambda^{(N)}$, then indeed $\mu_\Lambda(\Gamma_{\Lambda,N})=\frac{\Lambda^N}{N!}$. This follows because (i) $\Pi^{1}(\Gamma_{\Lambda,N})=(\Lambda^N)'$ and (ii) (say, by the FubiniTonelli theorem) $\text{Vol}((\Lambda^N)')=\text{Vol}(\Lambda^N)=\Lambda^N$.
As for $\mu_\Lambda^{(N)}(A)$ with $A=\{x_{1},...,x_{N}\}$, it is undefined. Indeed, for $\mu_\Lambda^{(N)}(A)$ be defined, $A$ must be a subset of $\Gamma_{\Lambda,N}$, whereas $\{x_{1},...,x_{N}\}$ is a subset of $\mathbb R^d$ but not of $\Gamma_{\Lambda,N}$. On the other hand, if $A$ is the singleton set $\{(x_{1},...,x_{N})\}$ with pairwise distinct $x_j$'s in $\mathbb R^d$, then indeed $\mu_\Lambda^{(N)}(A)=0$, and there is nothing odd about that.

$\begingroup$ This is a very good answer! In summary, it looks like $\cdot$ is not the cardinality of the set, but it $\cdot$ and $\mbox{Vol}$ seem to be two Lebesgue measures on different spaces. It makes sense to me! $\endgroup$– MathMathCommented Feb 9, 2020 at 13:52