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Let $\xi$ be a Poisson Random Measure of intensity $\mu$ (informally $\mathbb E\xi = \mu$).

(For $f \ge 0$, say) when does $\xi f = \infty?$ Kallenberg (Foundations of Modern Probabilility) claims the following:

• $\xi f$ exists iff $\mu(|f| \wedge 1) < \infty$.

It's certainly a basic result on Poisson Random Measure that $\mu(|f| \wedge 1) < \infty$ implies $\xi f < \infty$ a.s.. However, he doesn't actually seem to prove the converse, and I'm not sure what he means by the converse either. It could be either of

• $\mathbb P(\xi f < \infty) = 0$ 1$implies$\mu(|f| \wedge 1) < \infty$•$\mathbb P(\xi f < \infty) < 1$> 0$ implies $\mu(|f| \wedge 1) < \infty$

although perhaps these two are equivalent by Kolmogorov's 0-1 law.

Can anyone shed any light on how to prove the converse, or point me to a better reference for Poisson Random Measure?

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# Non-existence of integral with respect to Poisson Random Measure

Let $\xi$ be a Poisson Random Measure of intensity $\mu$ (informally $\mathbb E\xi = \mu$).

(For $f \ge 0$, say) when does $\xi f = \infty?$ Kallenberg (Foundations of Modern Probabilility) claims the following:

• $\xi f$ exists iff $\mu(|f| \wedge 1) < \infty$.

It's certainly a basic result on Poisson Random Measure that $\mu(|f| \wedge 1) < \infty$ implies $\xi f < \infty$ a.s.. However, he doesn't actually seem to prove the converse, and I'm not sure what he means by the converse either. It could be either of

• $\mathbb P(\xi f < \infty) = 0$ implies $\mu(|f| \wedge 1) < \infty$
• $\mathbb P(\xi f < \infty) < 1$ implies $\mu(|f| \wedge 1) < \infty$

although perhaps these two are equivalent by Kolmogorov's 0-1 law.

Can anyone shed any light on how to prove the converse, or point me to a better reference for Poisson Random Measure?