How to Show RV Related to Poisson Random Measure is a.s. Finite? I'm really new to this area of random measures, and I'm a bit confused on how to get started on this problem.

Let $\mu$ be a measure on $\mathbb{R}$ with
  $\mu(\left\{0\right\}) = 0$ and $\int_\mathbb{R} |x| \wedge 1 \mu(dx)
 < \infty$.
Define $\nu = \lambda \times \mu$.
Let $M$ be a Poisson Random Measure with mean $\nu$. Let $Z_t =
 \int_{[0,t] \times \mathbb{R}} M(ds,dx)$.
Prove that $Z_t < \infty$ a.s., i.e. $\mathbb{P}(Z_t<\infty)=1$.

I'm not really sure how to approach this problem. I managed to show that $\nu$ must be $\sigma$-finite. I also know the following:
$$P(Z_t < \infty) = lim_{\alpha \rightarrow 0} E[e^{\alpha Z_t}]\ \text{and}\ Z_t = lim_{m \rightarrow \infty} \int_{[0,t] \times (\frac{1}{m}, \infty]} M(ds,dx).$$
Normally the way I usually show something is finite a.s. is to show it has finite expectation a.s., but I don't see a way to make that work in this case.
 A: This statement is false.
If $\mu$ is inifnite (e.g., $\mu(dx)=|x|^{-1}1_{\{|x|<1\}}dx$) then $Z_t=M([0,t]\times\mathbb{R})$ is a Poisson random variable with mean 
$\mathbb{E}[M([0,t]\times\mathbb{R})]=\nu([0,t]\times\mathbb{R})=t\mu(\mathbb{R})=\infty$ and is therefore infinite with probability 1.
I think you meant $Z_t=\int_{[0,t]\times\mathbb{R}}xM(ds,dx)$. The proof would go as follows: 


*

*Find the Laplace transform of $Z_t^\pm=\int_{[0,t]\times\mathbb{R}_\pm}|x|M(ds,dx)$: Let $\xi$ be a Poisson random measure with mean $m$ (on the space, say $S$). Using simple functions first (i.e., functions of the form $f(x)=\sum_{k=1}^n c_k 1_{B_k}(x)$ 
for disjoint $B_1,\ldots,B_n$), one may prove that for nonnegative functions 
$f:S\to(0,\infty)$, we have 
$$
\mathbb{E}\left[\exp\left(-\int_S f(x)\xi(dx)\right)\right]
=\exp\left(-\int_S \big(1-e^{f(x)}\big)m(dx)\right).
$$
Take $\xi=M$ and $f(x)=ax1_{\{x>0\}}$, then
$$
\mathbb{E}[\exp(-aZ_t^\pm)]
=\exp\left(-\int_{[0,t]\times\mathbb{R}_\pm} \big(1-e^{a|x|}\big)\nu(ds,dx)\right)
=\exp\left(-t\int_{\mathbb{R}_\pm} \big(1-e^{a|x|}\big)\mu(dx)\right).
$$

*Use the inequality $e^{y}\geq (1+y)\vee 0$, or equivalently, 
$y\wedge 1\geq 1-e^{-y}$, to deduce that 
$\int_{\mathbb{R}_\pm} \big(1-e^{a|x|}\big)\mu(dx)\leq
a\int_{\mathbb{R}_\pm} |x|\wedge 1\mu(dx)=aC_\pm$ for a finite $C_\pm$, by hypothesis.

*Conclude that $1\geq\mathbb{E}[\exp(-aZ_t^\pm)]\geq\exp(-aC_\pm)\to 1$
as $a\to0$.

*Conclude that $Z_t^\pm<\infty$ a.s. and thus, so is $Z_t=Z_t^+-Z_t^-$. 
