Integration with respect to a Poisson random measure Let $N$ be a Poisson random measure (PRM) on a Polish space, $\left(X,\mathcal{B}(X)\right)$,
and let $\tilde{\nu}$ be its mean measure. Then, let $f$ be any non
negative and bounded function on $X$. The following are equivalent:
$$
1. \int_{X}f(x)N(dx)<+\infty,\ a.s.;
$$
$$
2. \int_{X}\left(1-e^{-f(x)}\right)\tilde{\nu}(dx)<+\infty;
$$
$$
3. \int_{X}\min\left\{ \left|f(x)\right|,1\right\} \tilde{\nu}(dx)<+\infty.
$$
Why is it so?
 A: (1) $\iff$ (2)
Since $f \geq 0$ is measurable, there exists a sequence $(f_n)_{n \in \mathbb{N}}$ of simple functions (aka elemtary functions/step functions) such that $0 \leq f_n \uparrow f$. Each $f_n$ can be written in the form
$$f_n(x) = \sum_{j=1}^{N_n} c_{j,n} 1_{B_{j,n}}(x).$$
for suitable constants $c_{j,n} \geq 0$ and $B_{j,n} \in \mathcal{B}(X)$. Then
$$\int f_n(x) \, N(dx) = \sum_{j=1}^{N_n} c_{j,n} N(B_{j,n})$$
and using that $N$ is a Poisson random measure, we find
$$\begin{align*} \mathbb{E} \exp \left( - \lambda \int f_n(x) \, N(dx) \right) &= \prod_{j=1}^{N_n} \exp \left(- \bar{\nu}(B_{j,n}) (1-e^{-\lambda c_{j,n}}) \right) \\ &= \exp \left( - \int (1-e^{-\lambda f_n(x)}) \, \bar{\nu}(dx) \right) \end{align*}$$
Since $f \geq 0$ is non-negative we may apply the dominated convergence theorem to conclude
$$\begin{align*} \mathbb{E} \exp \left( - \lambda \int f(x) \, N(dx) \right) &= \lim_{n \to \infty} \mathbb{E} \exp \left( -\lambda \int f_n(x) \, N(dx) \right) \\ &= \lim_{n \to \infty} \exp \left( - \int (1-e^{-\lambda f_n(x)}) \, \bar{\nu}(dx) \right) \\ &= \exp \left( - \int (1-e^{-\lambda f(x)}) \, \bar{\nu}(dx) \right). \end{align*}$$
(Just as a side remark: This identity is also known as Campbell's formula.)
If $\int f(x) \, N(dx)<\infty$ almost surely, we know that the left-hand side is strictly positive and this, in turn, implies $\int (1-e^{-\lambda f(x)}) \, \bar{\nu}(dx)<\infty$. On the other hand, if $\int (1-e^{-\lambda f(x)}) \, \bar{\nu}(dx)<\infty$ for some $\lambda>0$, this implies that the right-hand side is strictly positive and so $\int f(x) \, N(dx)<\infty$ almost surely.
(3) $\implies$ (2)
Obviously, $0 \leq 1-e^{-y} \leq 1$ for any $y \geq 0$. On the other hand, it is not difficult to see (e..g from Taylor's formula) that $1-e^{-y} \leq y$ for all $y \geq 0$. Thus, $$1-e^{-y} \leq \min\{1,y\}$$ and using this estimate for $y=f(x)$ proves the assertion.
(2) $\implies$ (3)
Since $y \leq 2(1-e^{-y})$ for any $0 \leq  y \leq 1$, we have
$$\int_{|f(x)| \leq 1} |f(x)| \, \bar{\nu}(dx) \leq 2 \int_X (1-e^{-f(x)}) \, \bar{\nu}(dx)<\infty.$$
On the other hand, by the monotonicity of the mapping $y \mapsto 1-e^{-y}$,
$$\int_{|f(x)|>1} 1\, \bar{\nu}(dx) \leq \int_{|f(x)|>1} \frac{1-e^{-f(x)}}{1-e^{-1}} \, \bar{\nu}(dx) \leq \frac{1}{1-e^{-1}} \int_X(1-e^{-f(x)}) \, \bar{\nu}(dx)<\infty.$$
Consequently,
$$\int_X \min\{|f(x)|,1\} \, \bar{\nu}(dx)<\infty.$$
