# Measurability for countable sums of Dirac measures

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 $$k \geq 0$$, $$\{\mu \in M_{\mathrm{atom}}(E): \mu_2(A\times B)=k\}=\{\mu \in M_{\mathrm{atom}}(E): \mu(A \cap B)<\infty, \mu(A)\mu(B)-\mu(A \times B)=k\}.$$

• Think of $E=\mathbb{R}$ and the rationals as $x_i$. – user64494 Dec 4 '20 at 19:57
• 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 $\mathcal{E} = \sigma(\{\{x\} \colon x \in E\})$. – Dieter Kadelka Dec 4 '20 at 20:24
• 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). – Gagar Dec 4 '20 at 21:40
• @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.) – Dieter Kadelka Dec 6 '20 at 12:13
• @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. – Gagar Dec 6 '20 at 13:15

## 1 Answer

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*}

• 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! – Gagar Dec 6 '20 at 11:41