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After my first post here, I have one more doubt which is bothering me. It concerns Minlos's book Introduction to mathematical statistical physics again. To fix the notation, we have $\Lambda \subset \mathbb{R}^{d}$, and we set $C_{\Lambda}^{(N)}:=\{\omega \subset \Lambda, \lvert\omega\rvert = N\}$ where $\lvert\omega\rvert$ denotes the cardinality of $\omega$. Suppose $\mu_{\Lambda}^{(N)}$ is a measure on $\mathbb{P}(C_{\Lambda}^{(N)})$, the set of all subsets of $C_{\Lambda}^{(N)}$. As you can see here, Minlos introduces $C_{\Lambda}:=\bigcup_{N=0}^{\infty}C_{\Lambda}^{(N)}$ (with $C_{N}^{(0)} = \{\emptyset\}$) and his aim is to define a measure $\mu_{\Lambda}$ on $C_{\Lambda}$. But he states $\mu_{\Lambda}(\emptyset) = 1$. So, I have two questions concerning this definition. The first is: what does it mean to define a measure on a set? Should I assume the underlying $\sigma$-algebra is $\mathbb{P}(C_{\Lambda})$? Second, how can it be that $\mu_{\Lambda}(\emptyset) = 1$ if a measure has to satisfy $\mu_{\Lambda}(\emptyset) = 0$?

EDIT: I think I understood my second question. The point is that, in this case, $\emptyset$ is viewed as an element not a subset of the $C_{\Lambda}$.

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    $\begingroup$ Yeah, it seems like this is just slightly sloppy notation, and Minlos really means $\mu_{\Lambda}(\{\emptyset\})$. $\endgroup$ Commented Feb 9, 2020 at 18:59
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    $\begingroup$ Thank you!! I'm new here! $\endgroup$
    – MathMath
    Commented Feb 9, 2020 at 19:11
  • $\begingroup$ Perhaps you copied something wrong ...$C_{\Lambda}^{(N)}:=\{\omega \subset \Lambda, \lvert\omega\rvert<+\infty\}$ does not depend on $N$. $\endgroup$ Commented Feb 9, 2020 at 20:56
  • $\begingroup$ Yes. It should be $C_{\Lambda}^{(N)}:=\{\omega \subset \Lambda, |\omega| = N\}$. I'll edit it. $\endgroup$
    – MathMath
    Commented Feb 9, 2020 at 21:20

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