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Let's consider a bounded (maybe compact) set $\Lambda \subset \mathbb{R}^{d}$ with particles interacting on it. Suppose, for each $N \in \mathbb{N}$, $U_{N}: (\mathbb{R}^{d})^{N} \to \mathbb{R}\cup \{+\infty\}$ is function which represents the potential energy of a system consisting in $N$ interacting particles. Now, the partition function of such a system in the grand-canonical ensemble is given by: \begin{eqnarray} \Xi_{\Lambda}(\beta, z) := \sum_{N=0}^{\infty}\frac{z^{N}}{N!} \int_{\Lambda}d\mu(x_{1})\cdots \int_{\Lambda}d\mu(x_{N}) e^{-\beta U_{N}(x_{1},...,x_{N})} \tag{1}\label{1} \end{eqnarray}

My Question is: What is the $\sigma$-algebra in which the product measures $\mu(x_{1})\times\cdots\times \mu(x_{N})$ are defined? I mean, for each $N$, the $N$-th term of the sum (\ref{1}) involves a product of $N$ measures and I'm having trouble understanding what is happening here. It seems to me that this is just a weak-limit in the sense that: \begin{eqnarray} \int d\mu_{n} f \to \int d\mu f \tag{2}\label{2} \end{eqnarray} but this implies that each $\mu_{n} = \mu(x_{1})\times \cdots \times \mu(x_{N}) $ is defined, for every $N$, on a "bigger" $\sigma$-algebra. What is this $\sigma$-algebra?

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    $\begingroup$ woh you know it's serious when capital xi comes out :-) $\endgroup$
    – nomen
    Commented Feb 7, 2020 at 22:56

2 Answers 2

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Looking at your formula (1), it appears that $\mu$ must be a measure defined on a $\sigma$-algebra $\mathscr F$ over the finite set $\Lambda$. The natural $\sigma$-algebra over the finite set $\Lambda$ is the largest $\sigma$-algebra over $\Lambda$, which is the (power) set $2^\Lambda$ of all subsets of $\Lambda$. By definition, the product measure $\mu^{\otimes N}$ is defined on the product $\sigma$-algebra $\mathscr F^{\otimes N}$. If $\mathscr F=2^\Lambda$, then $\mathscr F^{\otimes N}=(2^\Lambda)^{\otimes N}=2^{\Lambda^N}$, the set of all subsets of $\Lambda^N$.

Your formula (1) can then be rewritten simply as $$ \Xi_{\Lambda}(\beta, z) := \sum_{N=0}^{\infty}\frac{z^{N}}{N!} \int_{\Lambda^N}e^{-\beta U_{N}(x_{1},...,x_{N})}\,\mu^{\otimes N}\Big(\prod_{j=1}^N dx_j\Big) \\ =\sum_{N=0}^{\infty}\frac{z^{N}}{N!} \sum_{(x_{1},...,x_{N})\in\Lambda^N}e^{-\beta U_{N}(x_{1},...,x_{N})}\,\prod_{j=1}^N \mu(\{x_j\}). $$

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  • $\begingroup$ I just realized I mistyped my question. Actually, for continuous systems, $\Lambda$ must be a bounded set, not finite. I'll edit it. How does your answer change in this case? $\endgroup$ Commented Feb 7, 2020 at 14:57
  • $\begingroup$ The case $\Lambda$ finite is assumed when we're dealing with $\mathbb{Z}^{d}$ rather than $\mathbb{R}^{d}$. Your answer seems to address the $\mathbb{Z}^{d}$ case and it is still very useful to me because I get in troble understanding the problem in $\mathbb{Z}^{d}$ a well. But my original post should have considered $\Lambda$ bounded and I didn't notice my mistake. $\endgroup$ Commented Feb 7, 2020 at 14:59
  • $\begingroup$ @Willy.K: Iosif's answer applies also in $\mathbb{R}^d$. Looks to me you are trying to follow the lectures by Brydges which are rather advanced stuff without knowing measure theory and in particular the definition of product $\sigma$-algebras and product of measures, etc. You need to review that first. To do it quickly, you can for example read the first chapter of the book "Probability: Theory and Examples" by Rick Durrett. $\endgroup$ Commented Feb 7, 2020 at 15:05
  • $\begingroup$ @AbdelmalekAbdesselam this time I'm not following Brydges notes. But I would expect $\mu(x_{1})\times \cdots \times \mu(x_{N})$ to be a measure on, I don't know, $\mathbb{R}^{\infty}$ and $\int_{\Lambda}$ to be just the integral restriced to $\Lambda$. Adapting the answer would lead to a $\sigma$-algebra consisting on $2^{\Lambda}$, that is, every subset of $\Lambda$ is measurable, right? $\endgroup$ Commented Feb 7, 2020 at 15:11
  • $\begingroup$ In some references, the $\sigma$-algebra is assumed to be $\cup_{N}\Gamma_{N}(\Lambda)$ where $\Gamma_{N}(\Lambda) := \{(x_{1},...,x_{N})\in (\mathbb{R}^{d})^{\Lambda}, x_{i}\in \Lambda\}$ but I don't know if this is the case. $\endgroup$ Commented Feb 7, 2020 at 15:17
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The configuration space is the disjoint union of $\Lambda^N$ for each nonnegative integer $N$. You can take the Borel $\sigma$-algebra on each of these (or Lebesgue if you prefer, but you're unlikely to encounter non-Borel sets in real life).

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  • $\begingroup$ Thanks for your answer! Just to clarify: I set $\Omega = \cup_{N}\Lambda^{N}$ to be my configuration space and I can take the Borel $\sigma$-algebra on each $\Lambda^{N} \subset \mathbb{R}^{dN}$, right? With what $\sigma$-algebra should I equip $\Omega$? $\endgroup$ Commented Feb 7, 2020 at 19:38

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