[proofplan] Uniqueness follows from the fact that the joint [distributions](/page/Distribution) of $(M(A_1), \ldots, M(A_k))$ for disjoint [sets](/page/Set) form a $\pi$-system that determines the law of $M$. For existence, we first handle the finite case $\mu(E) < \infty$ by scattering $N \sim \operatorname{Po}(\mu(E))$ independent points with distribution $\mu / \mu(E)$, then extend to the $\sigma$-finite case by partitioning $E$ into sets of finite measure and taking independent copies. [/proofplan] [step:Prove uniqueness via the $\pi$-system characterisation] For any finite collection of disjoint sets $A_1, \ldots, A_k \in \mathcal{E}$, properties (ii) and (iii) determine the joint distribution of $(M(A_1), \ldots, M(A_k))$: independence specifies the joint law as the product of individual Poisson distributions. The collection of events $\{M(A_1) = n_1, \ldots, M(A_k) = n_k\}$ over all choices of disjoint $A_1, \ldots, A_k$ and non-negative integers $n_1, \ldots, n_k$ forms a $\pi$-system (it is closed under finite intersections, by refining the partition). This $\pi$-system generates the $\sigma$-algebra governing the law of $M$. By Dynkin's $\pi$-$\lambda$ lemma, the law of $M$ is uniquely determined. [/step] [step:Construct a Poisson random measure when $\mu(E) < \infty$] Suppose $\lambda = \mu(E) < \infty$. Let $N \sim \operatorname{Po}(\lambda)$ and let $(Y_n)_{n \geq 1}$ be an independent [sequence](/page/Sequence) of i.i.d. random variables with common distribution $\mu / \lambda$ (the normalised measure), all defined on a common probability space and independent of $N$. Define \begin{align*} M(A) = \sum_{n=1}^{N} \mathbb{1}_{Y_n \in A}, \qquad A \in \mathcal{E}. \end{align*} Countable additivity (i) holds since indicator [functions](/page/Function) are additive over disjoint sets. For properties (ii) and (iii): let $A_1, \ldots, A_k$ be disjoint sets with $p_j = \mu(A_j)/\lambda$. Conditional on $N = m$, the vector $(M(A_1), \ldots, M(A_k))$ has a multinomial distribution $\operatorname{Mult}(m; p_1, \ldots, p_k)$. By the [Splitting Property of the Poisson Distribution](/theorems/1193), if $N \sim \operatorname{Po}(\lambda)$ and the counts are multinomially distributed given $N$, then $M(A_1), \ldots, M(A_k)$ are independent with $M(A_j) \sim \operatorname{Po}(\lambda p_j) = \operatorname{Po}(\mu(A_j))$. [/step] [step:Extend to the $\sigma$-finite case by partition and independent superposition] Since $(E, \mathcal{E}, \mu)$ is $\sigma$-finite, there exists a partition $E = \bigcup_{k=1}^\infty E_k$ with $\mu(E_k) < \infty$ for each $k$. By the previous step, for each $k$ there exists a Poisson random measure $M_k$ on $(E_k, \mathcal{E}|_{E_k}, \mu|_{E_k})$. Construct these on independent probability spaces and define \begin{align*} M(A) = \sum_{k=1}^\infty M_k(A \cap E_k), \qquad A \in \mathcal{E}. \end{align*} We verify the three properties. **(i) Countable additivity:** For disjoint $(A_j)_{j \geq 1}$, $M(\bigcup_j A_j) = \sum_k M_k((\bigcup_j A_j) \cap E_k) = \sum_k \sum_j M_k(A_j \cap E_k) = \sum_j \sum_k M_k(A_j \cap E_k) = \sum_j M(A_j)$, where the exchange of summation is justified because all terms are non-negative. **(iii) Poisson marginals:** For a fixed $A \in \mathcal{E}$, the random variables $M_k(A \cap E_k)$ are independent (by independence of the $M_k$) with $M_k(A \cap E_k) \sim \operatorname{Po}(\mu(A \cap E_k))$. The sum of independent Poisson random variables is Poisson: $M(A) = \sum_k M_k(A \cap E_k) \sim \operatorname{Po}(\sum_k \mu(A \cap E_k)) = \operatorname{Po}(\mu(A))$. **(ii) Independence:** For disjoint $A_1, \ldots, A_m$, the random variables $M(A_j) = \sum_k M_k(A_j \cap E_k)$ are independent because, for each fixed $k$, the variables $M_k(A_1 \cap E_k), \ldots, M_k(A_m \cap E_k)$ are independent (by the finite-case construction), and the families are independent across different values of $k$ (by independence of the $M_k$). Therefore the sums $M(A_j) = \sum_k M_k(A_j \cap E_k)$ are independent. [/step]