1
$\begingroup$

Let $(E,\mathcal{E})$ be any measurable space and denote by $M_{\mathrm{atom}}(E)$ the set of all measures $\mu$ of the form $$\mu=\sum_{i \in I} \delta_{x_i}$$ with $I$ at most countable and $x_i \in E$ for every $i \in I$ (the $x_i$ are not necessarily distinct). Equip $M_{\mathrm{atom}}(E)$ with the smallest $\sigma$-algebra for which all the functions $$\begin{array}{ccccc} & & M_{\mathrm{atom}}(E) & \to & \{0,1,2,\ldots,\infty\} \\ & & \mu & \mapsto & \mu(A) \\ \end{array}$$ for $A \in \mathcal{E}$ are measurable.

For $\mu=\sum_{i \in I} \delta_{x_i} \in M_{\mathrm{atom}}(E)$, set $$\mu_2=\sum_{i \neq j \in I} \delta_{(x_i,x_j)} \in M_{\mathrm{atom}}(E^2)$$ with $E^2$ equipped with the product $\sigma$-algebra.

Question 1. Is it true that $$\begin{array}{ccccc} & & M_{\mathrm{atom}}(E) & \to & M_{\mathrm{atom}}(E^2) \\ & & \mu & \mapsto & \mu_2 \\ \end{array}$$ is measurable?

Edit. Here is some motivation. Assume that the diagonal $D= \{(x,x): x \in E\}$ is measurable in $E \otimes E$ and consider a random measure $M$ with values in $M_{\mathrm{atom}}(E)$. Then "$M$ is simple" is an event (that is $\{ \forall x \in E: M(\{x\}) \leq 1 \}$ is an event) when the mapping $\mu \mapsto \mu_2$ is measurable, since $$\{ \mu \in M_{\mathrm{atom}}(E) : \forall x \in E: \mu(\{x\}) \leq 1 \}= \{ \mu \in M_{\mathrm{atom}}(E) : \mu_2(D)=0\}.$$

Edit 2. Reformulation of the motivation in measure-theoretical terms. Let $M_{\mathrm{atom}}^{\mathrm{simple}}(E)$ be the subset of $M_{\mathrm{atom}}(E)$ made of measures $\mu$ of the form $$\mu=\sum_{i \in I} \delta_{x_i}$$ with $I$ at most countable and $x_i \in E$ with $x_i \neq x_i$ for $i \neq j$.

Question 2. Assume that the diagonal $D= \{(x,x): x \in E\}$ is measurable in $E \otimes E$. Is it true that $M_{\mathrm{atom}}^{\mathrm{simple}}(E)$ is a measurable subset of $M_{\mathrm{atom}}(E)$?

If $\mu \mapsto \mu_2$ is measurable, then $M_{\mathrm{atom}}^{\mathrm{simple}}(E)= \{ \mu \in M_{\mathrm{atom}}(E): \mu_2(D)=0\}$, so the answer would be yes. But perhaps it is simpler to answer directly Question 2 than Question 1.

Partial answer for a separable metric space. If $(E,d)$ is a separable metric space equipped with its Borel $\sigma$-algebra, then the answer to both questions is affirmative: writing $D^{(1/n)}= \{z \in E^2: d(z,D)<1/n\}$, since $D= \cap_{n \geq 1} D^{(1/n)}$, it suffices to check that $\mu \mapsto \mu_2(O)$ is measurable for every open set $O \subset E^2$ [Edit: this is flawed: the measure $\mu_2$ is not necessarily finite...]. By separability, $O$ can be written as a countable disjoint union of measurable products $A \times B$, so it suffices in turn to check that $\mu \mapsto \mu_2(A \times B)$ is measurable for $A,B$ measurable. This readily follows from the fact that for every integer $k \geq 0$, $$\{\mu \in M_{\mathrm{atom}}(E): \mu_2(A\times B)=k\}=\{\mu \in M_{\mathrm{atom}}(E): \mu(A \cap B)<\infty, \mu(A)\mu(B)-\mu(A \times B)=k\}.$$

$\endgroup$
5
  • $\begingroup$ Think of $E=\mathbb{R}$ and the rationals as $x_i$. $\endgroup$ – user64494 Dec 4 '20 at 19:57
  • $\begingroup$ Do you assume that $\mathcal{E}$ contains all one-point sets? In this case $M_{atom}(E)$ is the set of $\sigma$-finite $\bar{\mathbb{N}}_0$-valued measures on $\mathcal{E}$ and you can assume that $\mathcal{E} = \sigma(\{\{x\} \colon x \in E\})$. $\endgroup$ – Dieter Kadelka Dec 4 '20 at 20:24
  • $\begingroup$ No, there are no assumptions on $(E,\mathcal{E})$. Roughly speaking, this would be an analog of Proposition 4.3 in math.kit.edu/stoch/~last/seite/lectures_on_the_poisson_process/… in the case of so-called "proper" measures (at first glance I guess the proof carries through in this case, but it is quite involved and I am wondering if there is a simpler argument). $\endgroup$ – Gagar Dec 4 '20 at 21:40
  • $\begingroup$ @Gagar: I still have problems with the original formulation. If f.i. $\mathcal{E} = \{\emptyset,E\}$, then as measure all $\delta_x$ are identical. The problem then is the definition of $\mu \to \mu_2$. Is this map well defined and does this matter. (I for myself have not investigated this.) $\endgroup$ – Dieter Kadelka Dec 6 '20 at 12:13
  • $\begingroup$ @DieterKadelka Oh, I see! Indeed there can be an issue of the uniqueness of the representation of $\mu$ which raises the question of the definition of $\mu_2$. So let's say that $ \mathcal{E}$ contains all one-point sets (which is indeed the case in the context of Question 2 where the diagonal is measurable). However measures in $M_{atom}(E)$ are not necessarily $\sigma$-finite (take a same atom infinitely many times), and also in general $\mathbb{N}_{0}$-valued measures are not necessarily sums of Dirac measures. $\endgroup$ – Gagar Dec 6 '20 at 13:15
1
$\begingroup$

It's been a long time for me without using $\sigma$-algebra : I am feeling a bit rusty so please forgive (by order of gravity) any lengthy < naive < wrong remarks of mine. This is only a partial answer, unfortunately.

I assume you're using the discrete $\sigma$-algebra on $\overline{\mathbf{N}}:=\mathbf{N}\cup\{\infty\}$.

For any measurable space $(E,\mathcal{E})$ let's call $\mathfrak{B}(E,\mathcal{E})$ the $\sigma$-algebra that you use on $M_{atom}(E)$, that is \begin{align*} \mathfrak{B}(E,\mathcal{E}) = \sigma\big\{\chi_A^{-1}(\{n\}) \,:\, A\in \mathcal{E}, n\in\overline{\mathbf{N}}\big\}, \end{align*} where for $A\in\mathcal{E}$, $\chi_A$ denotes the evaluation map defined on $M_{atom}(E)$ by $\mu\mapsto \mu(A)$.

With the same notation as above, you choose to equip $M_{atom}(E^2)$ with $\mathfrak{B}(E^2,\mathcal{E}\otimes \mathcal{E})$. I claim that \begin{align*} \mathfrak{B}(E^2,\mathcal{E}\otimes \mathcal{E}) = \sigma\big\{\chi_{A_1\times A_2}^{-1}(\{n\}) \,:\, (A_1,A_2)\in \mathcal{E}^2. n\in\overline{\mathbf{N}}\big\},\qquad (\star) \end{align*} One inclusion is a consequence (by minimality) of the definition of $\mathfrak{B}(E^2,\mathcal{E}\otimes \mathcal{E})$ given above. The second inclusion which is needed to be proven is (the first equality is simply the definition) \begin{align*} \mathfrak{B}(E^2,\mathcal{E}\otimes \mathcal{E}) = \sigma\big\{\chi_B^{-1}(\{n\}) \,:\, B\in \mathcal{E}\otimes\mathcal{E}, n\in\overline{\mathbf{N}}\big\}\subseteq \sigma\big\{\chi_{A_1\times A_2}^{-1}(\{n\}) \,:\, (A_1,A_2)\in \mathcal{E}^2, n\in\overline{\mathbf{N}}\big\}. \end{align*} To establish this inclusion, introduce the subset $\mathfrak{S}\subseteq\mathscr{P}(E\times E)$ of all $B\in\mathcal{E}\otimes\mathcal{E}$ such that for all $k\in\overline{\mathbf{N}}$ \begin{align} \chi_B^{-1}(\{k\})\in \sigma\big\{\chi_{A_1\times A_2}^{-1}(\{n\}) \,:\, (A_1,A_2)\in \mathcal{E}^2, n\in\overline{\mathbf{N}}\big\}. \end{align} One checks that $\mathfrak{S}$ forms a $\sigma$-algebra, containing obviously all $A_1\times A_2$ for $(A_1,A_2) \in \mathcal{E}^2$. Using that $\mathcal{E}\otimes \mathcal{E}$ is precisely the $\sigma$-algebra generated by these " tensor " elements we recover $\mathcal{E}\otimes\mathcal{E}\subseteq \mathfrak{S}$, which implies the desired inclusion, again by a minimality argument. $(\star)$ is thus proved.

Now, my hope was to obtain something stronger than $(\star)$, namely :

\begin{align*} \mathfrak{B}(E^2,\mathcal{E}\otimes \mathcal{E}) = \sigma\big\{\chi_{A_1\times A_2}^{-1}(\{n\}) \,:\, (A_1,A_2)\in \mathcal{E}^2,\, A_1\times A_2 \cap \Delta = \emptyset,\,n\in\overline{\mathbf{N}}\big\},\qquad (\star\star) \end{align*} where $\Delta:=\{(x,x)\,:\, x\in E\}$ is the diagonal of $E\times E$. To establish $(\star\star)$ the most natural thing would be to prove \begin{align*} \mathcal{E}\otimes\mathcal{E} = \sigma\big\{A_1\times A_2\,:\,(A_1,A_2)\in \mathcal{E}^2,\, A_1\times A_2 \cap \Delta = \emptyset\big\},\qquad(\star\star\star) \end{align*} and simply reproduce the above argument of minimality.

PROBLEM : It's not hard to check that a necessary condition for $(\star\star\star)$ to hold is that $\Delta$ itself belongs to $\mathcal{E}\otimes\mathcal{E}$. This is not an empty assumption (see Nedoma's pathology), but if you're working in a not-so-fat-and-ugly space, maybe you could cope with it. I have a strong feeling that if $\Delta$ does belong to the $\sigma$-algebra, then $(\star\star\star)$ holds -- and so does $(\star\star)$ in that case -- but I did not manage to write it rigorously. I expect a somehow general statement about generated $\sigma$-algebras, when you just erase a part of the generating system and yet recover all the $\sigma$-algebra ...


What follows is done under assumption $(\star\star)$.

With this characterization of the measurable sets at the arrival of your map $T:\mu\mapsto \mu_2$, the measurability of the latter boils down to establish, for any $(A_1,A_2)\in\mathcal{E}^2$ for which $A_1\times A_2\cap \Delta = \emptyset$ and $n\in\overline{\mathbf{N}}$, \begin{align*} T^{-1}\big(\chi_{A_1\times A_2}^{-1}(\{n\})\big) \in \mathfrak{B}(E,\mathcal{E}). \end{align*} $T^{-1}\big(\chi_{A_1\times A_2}^{-1}(\left\{n\right\})\big)$ is composed of measures $\mu \in M_{atom}(E)$ such that $\mu_2(A_1\times A_2)=n$. Now $\delta_{(x_i,x_j)}(A_1\times A_2) = \delta_{x_i}(A_1)\delta_{x_j}(A_2)$ and we are therefore asking (since $A_1\times A_2$ does not meet $\Delta$) \begin{align*} \sum_{i\neq j \in I} \delta_{x_i}(A_1)\delta_{x_j}(A_2)&=\sum_{i,j \in I} \delta_{x_i}(A_1)\delta_{x_j}(A_2)\\ &= \chi_{A_1}(\mu)\chi_{A_2} (\mu) \\ &= n. \end{align*} The set of such measures $\mu$ is indeed in $\mathfrak{B}(E,\mathcal{E})$ as it can be written for finite $n$ as \begin{align*} \bigcup_{d|n} \chi_{A_1}^{-1}(\{d\})\cap \chi_{A_2}^{-1}\left(\left\{\frac{n}{d}\right\}\right) \end{align*} and for infinite $n$ as \begin{align*} \chi_{A_1}^{-1}(\{\infty\})\cup \chi_{A_2}^{-1}(\{\infty\}). \end{align*}

$\endgroup$
1
  • $\begingroup$ Great! I have edited the original post to add Question 2: proving $(\star \star \star)$ assuming that $\Delta \in \mathcal{E} \otimes \mathcal{E}$ would do the job! $\endgroup$ – Gagar Dec 6 '20 at 11:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.