[proofplan] The moment formulae and Laplace/characteristic [function](/page/Function) identities are established by a three-stage approximation: first for indicators $f = \mathbb{1}_A$ (where $M(A) \sim \operatorname{Po}(\mu(A))$ and the formulae are standard Poisson computations), then for [simple functions](/page/Simple%20Function) (by linearity and independence), and finally for general $f$ by monotone convergence or dominated convergence. [/proofplan] [step:Verify the formulae for indicator functions $f = \mathbb{1}_A$] For $A \in \mathcal{E}$ with $\mu(A) < \infty$, $M(\mathbb{1}_A) = M(A) \sim \operatorname{Po}(\mu(A))$. The Poisson [distribution](/page/Distribution) with parameter $\lambda = \mu(A)$ has mean $\lambda$ and variance $\lambda$, giving \begin{align*} \mathbb{E}[M(\mathbb{1}_A)] = \mu(A) = \int_E \mathbb{1}_A \, d\mu, \qquad \operatorname{Var}(M(\mathbb{1}_A)) = \mu(A) = \int_E \mathbb{1}_A^2 \, d\mu. \end{align*} For the [Laplace transform](/page/Laplace%20Transform) with $u > 0$: the moment generating function of $\operatorname{Po}(\lambda)$ gives \begin{align*} \mathbb{E}[e^{-u M(A)}] = \sum_{k=0}^\infty e^{-uk} \frac{\lambda^k e^{-\lambda}}{k!} = e^{-\lambda} \sum_{k=0}^\infty \frac{(\lambda e^{-u})^k}{k!} = e^{-\lambda + \lambda e^{-u}} = \exp\!\left(-\lambda(1 - e^{-u})\right). \end{align*} Setting $\lambda = \mu(A)$ and $f = \mathbb{1}_A$: $\mathbb{E}[e^{-u M(f)}] = \exp(-\int_E (1 - e^{-u f}) \, d\mu)$. For the characteristic function: the same computation with $iu$ replacing $-u$ gives $\mathbb{E}[e^{iu M(A)}] = \exp(\lambda(e^{iu} - 1)) = \exp(\int_E (e^{iu f} - 1) \, d\mu)$. [/step] [step:Extend to non-negative simple functions by independence] Let $f = \sum_{j=1}^m c_j \mathbb{1}_{A_j}$ with $c_j > 0$ and $A_1, \ldots, A_m$ disjoint. Then $M(f) = \sum_{j=1}^m c_j M(A_j)$, where $M(A_1), \ldots, M(A_m)$ are independent Poisson random variables. For the moments: \begin{align*} \mathbb{E}[M(f)] = \sum_{j=1}^m c_j \mathbb{E}[M(A_j)] = \sum_{j=1}^m c_j \mu(A_j) = \int_E f \, d\mu, \end{align*} and by independence, $\operatorname{Var}(M(f)) = \sum_{j=1}^m c_j^2 \operatorname{Var}(M(A_j)) = \sum_{j=1}^m c_j^2 \mu(A_j) = \int_E f^2 \, d\mu$. For the Laplace transform, independence gives the product form: \begin{align*} \mathbb{E}[e^{-u M(f)}] = \prod_{j=1}^m \mathbb{E}[e^{-u c_j M(A_j)}] = \prod_{j=1}^m \exp\!\left(-\mu(A_j)(1 - e^{-u c_j})\right) = \exp\!\left(-\sum_{j=1}^m \mu(A_j)(1 - e^{-u c_j})\right). \end{align*} Since $\sum_{j=1}^m \mu(A_j)(1 - e^{-u c_j}) = \int_E (1 - e^{-u f(y)}) \, d\mu(y)$, the Laplace transform formula holds for simple functions. The characteristic function formula follows by the identical argument with $iu$ replacing $-u$. [/step] [step:Extend to general measurable $f$ by monotone and dominated convergence] **Laplace transform (part (ii)):** Let $f : E \to \mathbb{R}_+$ be measurable. Choose non-negative simple functions $f_n \uparrow f$ pointwise. Then $M(f_n) = \int_E f_n \, dM \uparrow \int_E f \, dM = M(f)$ (by the [Monotone Convergence Theorem](/theorems/509) applied to the counting measure $M(\omega, \cdot)$). Since $e^{-u M(f_n)} \downarrow e^{-u M(f)}$ (the exponential is decreasing), the [Monotone Convergence Theorem](/theorems/509) applied to $(e^{-u M(f_n)})$ (which are bounded by $1$, so dominated convergence applies directly) gives \begin{align*} \mathbb{E}[e^{-u M(f)}] = \lim_{n \to \infty} \mathbb{E}[e^{-u M(f_n)}] = \lim_{n \to \infty} \exp\!\left(-\int_E (1 - e^{-u f_n}) \, d\mu\right). \end{align*} Since $0 \leq 1 - e^{-u f_n} \uparrow 1 - e^{-u f}$ pointwise, the [Monotone Convergence Theorem](/theorems/509) gives $\int_E (1 - e^{-u f_n}) \, d\mu \to \int_E (1 - e^{-u f}) \, d\mu$. Therefore $\mathbb{E}[e^{-u M(f)}] = \exp(-\int_E (1 - e^{-u f}) \, d\mu)$. **Moment formulae (part (i)):** For $f \in L^1(\mu) \cap L^2(\mu)$, write $f = f^+ - f^-$ and use the Laplace transform to identify $\mathbb{E}[M(f)]$ and $\operatorname{Var}(M(f))$. The mean follows by differentiating the characteristic function at $u = 0$: $\frac{d}{du} \mathbb{E}[e^{iu M(f)}]\big|_{u=0} = i \mathbb{E}[M(f)]$, and the right-hand side gives $i \int_E f \, d\mu$. The variance follows from the second [derivative](/page/Derivative). **Characteristic function (part (iii)):** For $f \in L^1(\mu)$, approximate $f$ by simple functions $f_n$ with $|f_n| \leq |f|$. The bound $|e^{iuf_n(y)} - 1| \leq |u| |f_n(y)| \leq |u| |f(y)|$ ensures $\int_E |e^{iuf_n} - 1| \, d\mu \leq |u| \int_E |f| \, d\mu < \infty$. By the [Dominated Convergence Theorem](/theorems/4), $\int_E (e^{iuf_n} - 1) \, d\mu \to \int_E (e^{iuf} - 1) \, d\mu$. The convergence $M(f_n) \to M(f)$ almost surely and $|e^{iuM(f_n)}| = 1$ give $\mathbb{E}[e^{iuM(f_n)}] \to \mathbb{E}[e^{iuM(f)}]$ by dominated convergence. Combining: $\mathbb{E}[e^{iuM(f)}] = \exp(\int_E (e^{iuf} - 1) \, d\mu)$. [/step]