I'm working through the proof of Theorem VI.2.17 in the book "Probability and Stochastics" by Cinlar, and am having some problems understanding it (or maybe there are some conditions missing?)
Let $(E,\mathcal{E})$ be a measurable space, and let N be a Poisson random measure on it. Suppose that its mean (measure) $\nu$ is $\Sigma$-finite. Then, N is a random counting measure if and only if $\nu is diffuse.
I'll recall some of the definitions here, in case the reader is not familiar with it (since it might not be standard definitions).
A measure $\nu$ is diffuse if $\nu(\{x\}) =0$ for every $x \in E$.
A measure $\nu$ is $\Sigma$-finite if there exists a sequence of finite measures $\nu_n$ such that $\nu = \sum_n \nu_n$.
A random measure N is called a random counting measure if, for almost every $\omega$, the measure $A \mapsto M(\omega,A)$ is purely atomic and its every atom has weight one.
The necessity part of the proof is quite simple, we just show that $\nu(\{x\}) = 0$ for fixed (arbitrary) $x \in E$, which follows from some simple calculation for the random counting measure N.
However, in the sufficiency part I'm not sure if every step is correct. The proof goes as follows:
We assume that $\nu$ is diffuse (and $\Sigma$-finite). Because the probability law of a Poisson random measure is determined by its mean measure, we may assume that N is constructed by $$N = \sum_{i=1}^\infty N_n, \qquad N_n = \sum_{i=1}^{K_n(\omega)} 1_A (X_{n,i}),$$ where the $X_i$ are i.i.d and K is a Poisson random variable with mean $\nu$. (I'm not actually sure that this holds without any further assumptions, I have a feeling we may need to restrict it to countably generated $\mathcal{E}$). Note that the collection $X = \{X_{n,i}: n \geq 1, i \geq 1\}$ is independent, and every member has a diffuse distribution.
Take some pair of distinct indices $(n,i)$ and $(m,j)$. Then $P = (X_{ni} = X_{mj}) = 0$ (I'm also not sure this part holds without any further assumptions). Then, the countable union of all these events is also negligible, and hence N is a counting measure.
I'm looking for explanations why these parts of the proof are correct - or that further conditions are needed.