= Solution
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 gives
$$
\mathbb P(M(\{x\})>0)\leq\mathbb P(M(I_r)>0)=1-e^{-\lambda(I_r)}\longrightarrow0.
$$
Thus 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,
$$
\boxed{M_n(J)\uparrow M(J)\quad\text{almost surely}.}
$$
The union of fully contained level-$n$ cells misses at most two boundary portions of total length at most $2^{1-n}$. Consequently $N_n2^{-n}\to\lambda(J)$, and $p_n/2^{-n}\to1$. For $0\leq z\leq1$, the <probability generating function> from the binomial calculation gives
$$
\mathbb E[z^{M_n(J)}]=(1+p_n(z-1))^{N_n}\longrightarrow\exp(\lambda(J)(z-1)).
$$
The convergence follows by expanding the logarithm, since $N_np_n\to\lambda(J)$ and $N_np_n^2\to0$. 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
$$
\boxed{M(J)\sim\operatorname{Poisson}(\lambda(J)).}
$$
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