The following content was based on [Minlos'](https://rads.stackoverflow.com/amzn/click/com/0821813374) 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 $\mu\_{\Lambda}$?



Measure, volume and cardinality on Minlos' book on statistical physics
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 Fubini--Tonelli 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. 


