= Solution
For finitely many disjoint bounded intervals $J_1,\ldots,J_r$, all dyadic cells used in their level-$n$ sums are disjoint. The independent occupied indicators proved in part (a) form independent groups, so $M_n(J_1),\ldots,M_n(J_r)$ are independent. Their almost-sure limits are also independent: for $0\leq z_j\leq1$, bounded convergence gives
$$
\mathbb E\prod_j z_j^{M(J_j)}
=\lim_n\prod_j\mathbb E z_j^{M_n(J_j)}
=\prod_j\exp(\lambda(J_j)(z_j-1)).
$$
This 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 $s=\sum_j a_j\mathbf1_{J_j}$ with $a_j\geq0$ and disjoint bounded intervals, independent <Poisson distributions> yield
$$
\mathbb E e^{-M(s)}=\prod_j\exp\left(\lambda(J_j)(e^{-a_j}-1)\right)
=\exp\left(-\int_0^\infty(1-e^{-s(x)})\,dx\right).
$$
We 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 gives
$$
\mathbb E M(g)=\int g\,d\lambda\qquad(g\geq0\text{ Borel}).
$$
For bounded $f$ supported in a bounded interval, choose nonnegative interval-step functions $s_k$ with $\int|s_k-f|\,d\lambda\to0$. 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 implies
$$
\mathbb E|M(s_k)-M(f)|\leq\int|s_k-f|\,d\lambda\longrightarrow0.
$$
The 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 $f\geq0$, put $f_k=(f\wedge k)\mathbf1_{[0,k)}$. 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>:
$$
\boxed{\mathbb E[e^{-M(f)}]=\exp\left(-\int_0^\infty(1-e^{-f(x)})\,dx\right),}
$$
with $e^{-\infty}=0$.
Finally take $f=\sum_j u_j\mathbf1_{A_j}$ 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 \b[$M$ 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.
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