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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