Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 33 / 6

Let $E_i=\{M(I_i)=0\}$. For any subset $K\subseteq \{1,\dots ,n\}$, simultaneous voidness is voidness of the union. Since the intervals are disjoint, the assumed void probability functional givesThis includes the empty subset. To check all event patterns explicitly, apply the inclusion-exclusion principle to the complements in a chosen pattern:Thus the void events are mutually independent, and their complementary occupied-cell indicators are mutually independent too. This is joint independence, not merely pairwise independence.

At level $n$, the $N_n$ dyadic cells fully contained in $J$ are disjoint. Their occupied indicators are independent by the preceding part, and every cell has length $2^{-n}$, so each has success probabilityTheir sum therefore satisfiesIf there are no fully contained cells, $N_n=0$ and this means the deterministic zero count. The binomial law follows directly from independence of occupation indicators, without assuming that the full counts $M(D_k^n)$ are already Poisson.

Refining an occupied cell replaces it by two child cells, at least one of which is occupied. All children of a cell contained in $J$ remain contained in $J$, and additional cells can enter the sum. Hence $M_n(J)$ is nondecreasing, pathwise, and is at most $M(J)$.

We must also address boundary points for a fixed interval $J$. For every fixed deterministic $x$, take bounded intervals $I_r$ containing $x$ with lengths tending to zero. The assumed void law givesThus the fixed endpoints of $J$ are almost surely not atoms. This is the exponential void probabilities forbid fixed atoms observation; it is not a simultaneous assertion for all possible real endpoints.

In a fixed bounded interval containing $J$, local finiteness gives only finitely many atoms. Simplicity makes their locations distinct, so there is a positive minimum distance between any two, unless there is at most one. Every atom in the interior of $J$ also has positive distance from its endpoints. For sufficiently fine dyadic cells each such atom lies alone in a cell fully contained in $J$, including if the atom itself is at a dyadic boundary because the cells are half-open. Thus the dyadic occupancy approximation of a simple point measure eventually equals $M(J)$, almost surely. In particular,

The union of fully contained level-$n$ cells misses at most two boundary portions of total length at most $2^{1-n}$. Consequently $N_n{2}^{-n}\to \lambda (J)$, and $p_n/2^{-n}\to 1$. For $0\le z\le 1$, the probability generating function from the binomial calculation givesThe convergence follows by expanding the logarithm, since $N_n{p}_n\to \lambda (J)$ and $N_n{p}_n^2\to 0$. Since the counts converge almost surely, bounded convergence identifies the same expression as $\mathbb{E}[z^{M(J)}]$. It is the generating function of the Poisson distribution, whose power-series coefficients determine its law. Thus

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.