Suppose that $v$ is a Radon measure on $(0,\infty)$ and let $X$ be a Poisson Point Process on $(0,\infty)$ with intensity measure $v$. Let $Y:=\int xX(dx)$.
In Theorem 24.17 of Probability Theory by A. Klenke (3rd version), the author, in showing that
$$ P[Y<\infty]>0 \Leftrightarrow P[Y<\infty]=1 \Leftrightarrow \int v(dx) \min(1,x) <\infty $$
writes $Y = Y_0+Y_\infty$ where $Y_0:=\int_{(0,1)} x X(dx)$ and $Y_\infty:= \int_{[1,\infty)} x X(dx)$.
I do not understand why it should be "clear" that
$$P[Y_\infty<\infty]>0 \Leftrightarrow P[Y_\infty<\infty]=1 \Leftrightarrow v([1,\infty))<\infty$$
My guess is that this last equivalence chain is linked to the fact that $Y_\infty$ is calculating the expectation of Poisson random variable on $[1,\infty)$ with parameter $v([1,\infty])$.
Any help in understanding this is very appreciated. Let me know if more details are needed. Thanks.
If $v([1,\infty))=\infty$ then denoting $v_n=v([n,n+1))$ we have $\sum_{n=1}^{\infty}v_n=\infty.$ Denote by $Y_n$ the number of points in $[n,n+1)$ of the Poisson process. By definition the $Y_n$ are Poisson independent and $E(Y_n)=v_n.$ Therefore $$Y_{\infty}=\sum_{n=1}^{\infty} Y_n=\infty.$$ A quick proof of this last fact goes as follows. Denote $w_n=\Pr(Y_n>0)=1-e^{-v_n}.$ If $\sum_{n=1}^{\infty}w_n<\infty$ this implies that $v_n\to 0$ and therefore $w_n\sim v_n$ implying the contradiction $\sum_{n=1}^{\infty}v_n<\infty.$ Hence $\sum_{n=1}^{\infty}\Pr(Y_n>0)=\infty$ and from the Borel Cantelli lemma we deduce that an infinite number of $Y_n$ are not $0.$