# Grand-canonical Gibbs measure for continuous systems

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?

• woh you know it's serious when capital xi comes out :-) Feb 7, 2020 at 22:56

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

• 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? Feb 7, 2020 at 14:57
• 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. Feb 7, 2020 at 14:59
• @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. Feb 7, 2020 at 15:05
• @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? Feb 7, 2020 at 15:11
• 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. Feb 7, 2020 at 15:17

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).

• 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 $\Omega$? Feb 7, 2020 at 19:38