Number of atoms of a probability measure

Question

Let $P\mathbb{R}$ be the space of probability measures on $\mathbb{R}$. Is the function \begin{align*} P\mathbb{R} &\to \mathbb{N} \cup \{\infty\}\\ \mu &\mapsto \#\{ x \in \mathbb{R} \mid \mu\{x\} > 0 \} \end{align*} measurable?

Answer 1

Yes.

First, one can show the $\sigma$-algebra in question is equal to the Borel $\sigma$-algebra induced by the weak topology. (We could do without this fact, but I find it makes the proof simpler.)

Now for any $n \in \mathbb{N}$ and any $\epsilon, R, r > 0$, let $A_{n, \epsilon, R, r}$ be the set of all probability measures $\mu$ for which there exist $n$ distinct points $x_1, \dots, x_n \in [-R,R]$, each of which is an atom of mass at least $\epsilon$ (i.e. $\mu\{x_i\} \ge \epsilon$ for each $i$), and with $|x_i - x_j| \ge r$ for $i \ne j$. I claim that $A_{n, \epsilon, R, r}$ is closed in the weak topology. To see this, suppose $(\mu_k)$ is a sequence in $A_{n, \epsilon, R, r}$ and $\mu_k \to \mu$ weakly. For each $k$, let $x_1^k, \dots, x_n^k \in [-R,R]$ be atoms for $\mu_k$ satisfying the above conditions. By the compactness of $[-R,R]^n$, we can pass to a subsequence so that for each $i$, the sequence $x_i^k$ converges to some $x_i \in [-R,R]$. By continuity of distance, we have $|x_i - x_j| \ge r$ for all $i \ne j$, and by standard properties of weak convergence, $\mu\{x_i\} \ge \epsilon$ for each $i$. So we indeed have $\mu \in A_{n, \epsilon, R, r}$.

Now the set $$ A_n = \bigcup_{\epsilon > 0} \bigcup_{R=1}^\infty \bigcup_{r > 0} A_{n, \epsilon, R, r}$$ is Borel, where the unions over $\epsilon$ and $r$ are taken over a countable sequence tending to 0. But $A_n$ is exactly the set of probability measures $\mu$ having at least $n$ atoms. For if $\mu$ admits $n$ atoms $x_1, \dots, x_n$, then $\mu \in A_{n, \epsilon, R, r}$ where $\epsilon < \min_i \mu\{x_i\}$, $R > \max_i |x_i|$, and $r < \min_{i \ne j} |x_i - x_j|$ are all positive / finite.

Finally, the set of all $\mu$ having exactly $n$ atoms is just $A_n \setminus A_{n+1}$, so it too is Borel. (For $n=0$ we use $A_1^c$, and for $n=\infty$ we use $\bigcap_{n=1}^\infty A_n$.)

Answer 2

$\lim_{N\to\infty}\lim_{\epsilon\to0}\lim_{k\to\infty}\sum_{j=-Nk}^{Nk}\mu[j/k,(j+1)/k)^\epsilon$.