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Assume $(S,\mathcal{S},\eta)$ is an arbitrary $\sigma$-finite measure space. Let $N:\mathcal S\rightarrow\{0,1,2,\ldots\}\cup\{\infty\}$ in a way such that $\{N(A):A\in\mathcal S\}$ are random variables defined on the probability space $(\Omega,\mathcal{F},\mathbb{P})$. Then $N$ is called a Poisson random measure on $(S,\mathcal{S},\eta)$ if

(i) for mutually disjoint $A_1,\ldots,A_n$ in $\mathcal{S}$, the variables $N(A_i),\ldots,N(A_n)$ are independent,

(ii) for each $A\in\mathcal{S}$, $N(A)$ is Poisson variable distributed with parameter $\eta(A)$.

(iii) $\mathbb{P}$-almost surely, $N$ is a measure.

In the context of a Levy process, I can see that $\eta$ corresponds to the product measure of the jump measure in a Levy process and Lebesgue measure on $t$, and ($S$, $\mathcal{S}$) is the measure space generated by the Borel $\sigma$-algebra on $\mathbb{R}\times[0,\infty)$.

now here is a colloary to a proof such measure exists

Suppose that $N$ is a Poisson random measure on $(S,\mathcal{S}, \eta)$., the the support of $N$ is $\mathbb{P}$ almost surely countable.

I don't quite understand what is the support of $N$ here means and would appreciate an answer.

Lost1
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2 Answers2

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To say "the support of $N$ is countable" means there is a countable set $A \subset S$ with $N(A) = 1$.

More precisely, in this context, for $\mathbb{P}$-almost every $\omega$, there is a countable set $A(\omega) \subset S$ such that $N(\omega)(A(\omega)) = 1$.

Nate Eldredge
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To say that the support of $N$ is countable means that there exists a countable random set $T$ in $\mathcal S$ such that, $P$-almost surely, $N(S\setminus T)=0$ and $N(\{t\})=1$ for every $t$ in $T$. Thus, $N=\sum\limits_{t\in T}\delta_t$.

Equivalently, for every (deterministic) $A$ in $\mathcal S$, one asks that $N(A)(\omega)=\# (A\cap T(\omega))$ for $P$-almost every $\omega$ in $\Omega$.

Did
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