consider the space $M = \left\{ \mu : \mathscr{B}(\mathbb{R}) \to \mathbb{R} \cup \left\{ -\infty, +\infty \right\} \ | \ \mu \text{ signed Radon measure} \right\}$ which is not a vector space, since we allow infinite Radon measures, e.g. the Lebesgue measure. In Kallenberg: Random Measures, he considers only positive Radon measures $M_+ \subseteq M$ and defines a $\sigma$-algebra on $M_+$ as the initial $\sigma$-algebra of the maps $\mu \to \mu B$ for bounded (i.e. relatively compact) sets $B \in \mathscr{B}(\mathbb{R})$. It turns out, that this $\sigma$-algebra coincides with the $\sigma$-algebra induced by the vague topology on $M_+$, which is defined by the convergence $\mu_n \to \mu$ by testing on functions with compact support $C_c(\mathbb{R})$, i.e. $\int f d\mu_n \to \int f d\mu$ for all $f \in C_c(\mathbb{R})$.

Now my questions:

  1. We can consider the extended positive real line $[0, \infty]$ as a topological space by a one-point compactification, i.e. $A \subseteq [0,\infty]$ is open if $A \cap [0,\infty)$ is open in $[0,\infty)$. Therefore, we could also consider the maps $\mu \to \mu B \in [0,\infty] $ for $\mu \in M_+$ but $B \in \mathscr{B}(\mathbb{R})$ not necessarily bounded. Is the induced $\sigma$-algebra on $M_+$ the same as for bounded sets $B$?

  2. We can also consider the construction by Kallenberg of a $\sigma$-algebra on all Radon measures $M$, i.e. considering the maps $\mu \to \mu B$ for bounded $B$ but $\mu \in M$. Is this in some way natural and does this correspond to gaining a $\sigma$-algebra by representing $\mu = \mu^+ - \mu^-$ as the difference of two positive measures, where we already can generate the $\sigma$-algebras for both parts and then merge these $\sigma$-algebras together. Does this also correspond to the $\sigma$-algebra generated by the vague topology on $M$?

  3. Merging questions 1. and 2. together: Is the $\sigma$-algebra on $M$ as in 2. the same as the initial $\sigma$-algebra of the evaluation maps $\mu \to \mu B$ for $\mu \in M$ and all $B \in \mathscr{B}(\mathbb{R})$ (not necessarily bounded)?

Best regards.


I think the statements are all correct and the proof is straightforward.

In detail (for question 3), take on $\mathbb{R}$ the 2-point compactification to $[-\infty, \infty]$ and the corresponding Borel $\sigma$-algebra $\mathcal{B}([-\infty, \infty])$ which is an extension of the Borel $\sigma$-algebra on $\mathbb{R}$, since the trace $\sigma$-algebra on $\mathbb{R}$ is $\mathcal{B}(\mathbb{R})$. Consider the $\sigma$-algebra $\mathcal{M}$ on $M$ generated by evaluation maps $\mu \to \mu B$ for bounded sets $B \in \mathcal{B}(\mathbb{R})$ and the $\sigma$-algebra $\mathcal{M}'$ on $M$ generated by evaluation maps for all sets $B$. By definition it holds that $\mathcal{M} \subseteq \mathcal{M}'$. In order to show that $\mathcal{M}' \subseteq \mathcal{M}$ it is enough to show that for all $B \in \mathcal{B}(\mathbb{R})$ the evaluation map $\mu \to \mu B$ is $\mathcal{M}/\mathcal{B}([-\infty,\infty])$-measurable. Define $B_n := B \cap [-n,n]$ such that $B_n$ is a nondecreasing sequence with $B = \bigcup_{n \in \mathbb{N}} B_n$. Split $\mu = \mu^+ - \mu^-$ according to the Jordan decomposition such that $\mu^\pm$ are positive Borel measures. By continuity from below of $\mu^\pm$ it follows that $\mu B = \lim_{n \to \infty} \mu B_n$. Since $\mu \to \mu B_n$ is $\mathcal{M}/\mathcal{B}([-\infty,\infty])$-measurable it follows that $\mu \to \mu B$ is also $\mathcal{M}/\mathcal{B}([-\infty,\infty])$-measurable as a limit of measurable functions.


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