An integer valued and $\sigma$-finite measure without atoms is costantly zero? Consider an integer valued ($\mu(A) \in \mathbb{N}$) and $\sigma$-finite measure $\mu$ on $S$, a complete and separable metric space.
Is it true that if $\mu (\{x\})=0$ for all $x \in S$, then $\mu(A)=0$ for all Borel sets?
This question comes from an attempt to show that a.e. realization of a point process has the form $\sum_{i \in \mathbb{N}} k_i \delta _{x_i}$ with $k_i \in \mathbb{N}, x_i \in S$.
Indeed, this last statement would be easily proved if the answer at my question is Yes.
 A: Let $\mu$ be  a nonzero, integer valued   measure   on a complete   separable metric space $S$ with metric $\rho$. We will show there exists $x \in S$ such that $\mu\{x\} >0$.
Let $D$ be a countable dense set in $S$. Since $S$ can be written as a union  $\cup_{d \in D} B(d,1)$ of balls with radius $1/2$, there must exist  $d_1 \in D$ such that $\mu [B(d_1,1/2)] \ge 1$.
We continue inductively: Suppose that we have already found some $d_k \in D$ such that  $\mu [B(d_k,2^{-k})] \ge 1$.
Define $D_k:=D \cap B(d_k,2^{-k})$.
Since $$B(d_k,2^{-k}) \subset \bigcup_{d \in D_k} B(d, 2^{-k-1}) \,$$ there  must exist  some $d_{k+1} \in D_k$ such that $\mu [B(d_{k+1},2^{-k-1})] \ge 1$. This completes the inductive step.
Since $\rho(d_k,d_{k+1})<2^{-k}$, the sequence $\{d_k\}_{k \ge 1}$ is a Cauchy sequence in $S$. Therefore this sequence converges to some $x \in S$.
Given $\epsilon>0$, if $k$ is large enough, then $\rho(d_k,x) <\epsilon$ and $2^{-k}<\epsilon$, so $B(x,2\epsilon) \supset B(d_k,2^{-k})$. Thus $\mu [B(x,2\epsilon)] \ge 1$. Since this holds for every  $\epsilon>0$, we conclude (by intersecting over rational $\epsilon>0$) that $\mu\{x\}  \ge 1$.
