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For finitely many disjoint bounded intervals $J_1,\dots ,J_r$, all dyadic cells used in their level-$n$ sums are disjoint. The independent occupied indicators proved in part (a) form independent groups, so $M_n(J_1),\dots ,M_n(J_r)$ are independent. Their almost-sure limits are also independent: for $0\le z_j\le 1$, bounded convergence givesThis joint probability generating function determines the joint law as that of independent Poisson counts. If an interval is unbounded, truncate all intervals by $[0,R)$ and let $R\uparrow \infty$ in the joint Laplace transforms. An unbounded interval has infinite count almost surely, since its Poisson means tend to infinity; this extends the independence assertion to such intervals as well.

For an interval-step function $s={\sum }_j{a}_j{1}_{J_j}$ with $a_j\ge 0$ and disjoint bounded intervals, independent Poisson distributions yieldWe now extend this to arbitrary nonnegative Borel functions, rather than stopping at interval-step functions. The mean measure $m(A)=\mathbb{E}M(A)$ is countably additive by Tonelli theorem, is finite on bounded intervals, and agrees with $\lambda$ on intervals by the preceding Poisson count calculation. Uniqueness of finite measures on each bounded Borel space, then exhaustion by bounded spaces, gives $m=\lambda$ on every Borel set. Simple-function approximation therefore givesFor bounded $f$ supported in a bounded interval, choose nonnegative interval-step functions $s_k$ with $\int |s_k-f|d\lambda \to 0$. Their existence follows by approximating measurable level sets in finite Lebesgue measure by finite unions of intervals, then approximating $f$ by simple functions. The mean-measure identity impliesThe function $e^{-u}$ is one-Lipschitz on $[0,\infty )$, so the expectations of the exponentials converge. The deterministic integrals of $1-e^{-s_k}$ converge by the same Lipschitz bound. Hence the Laplace formula holds for this bounded Borel $f$.

For general $f\ge 0$, put $f_k=(f\wedge k)1_{[0,k)}$. These functions increase to $f$, giving $M(f_k)\uparrow M(f)$ and $\int (1-e^{-f_k})d\lambda \uparrow \int (1-e^{-f})d\lambda$. Bounded convergence for the exponentials and monotone convergence for the integrals prove the Laplace functional of a Poisson random measure:with $e^{-\infty }=0$.

Finally take $f={\sum }_j{u}_j{1}_{A_j}$ for disjoint Borel sets of finite Lebesgue measure. The formula gives the product joint Laplace transform of independent Poisson counts with means $\lambda (A_j)$. Uniqueness of that transform proves the required count laws on all Borel sets. For infinite-measure sets the formula gives $\mathbb{E}{e}^{-uM(A)}=0$ for $u>0$, so their counts are infinite almost surely. Together with the given local finiteness, these facts show that $M$ is a Poisson random measure with intensity $\lambda$. This completes the exponential void probabilities characterize a Poisson random measure argument; simplicity was used to recover counts from fine-cell occupancy.

The void probability functional assigns to a measurable set the probability that it contains no atom of a random point measure. For a Poisson random measure of intensity $\mu$, it is $e^{-\mu (A)}$. The exponential void probabilities characterize a Poisson random measure result uses simplicity; doubling every atom of a Poisson measure preserves void probabilities but changes count distributions.