= Solution
For a measurable $A$ with $(\mu+\nu)(A)<\infty$, independence of the two <Poisson random measures> gives
$$
\mathbb P((\Phi+\Psi)(A)=k)=\sum_{j=0}^k e^{-\mu(A)}\frac{\mu(A)^j}{j!}e^{-\nu(A)}\frac{\nu(A)^{k-j}}{(k-j)!}
=e^{-(\mu+\nu)(A)}\frac{((\mu+\nu)(A))^k}{k!}.
$$
Thus its count has the <Poisson distribution> with the summed parameter. If $A_1,\ldots,A_m$ are disjoint, the variables within each family of counts are <independent>, and independence of $\Phi$ and $\Psi$ makes the two families <independent> of each other. Hence the pairs $(\Phi(A_j),\Psi(A_j))$, and therefore their sums, are <independent> over $j$.
Adding sample counting measures preserves countable additivity. The definition in (a) therefore proves
$$
\boxed{\Phi+\Psi\text{ is a Poisson random measure of intensity }\mu+\nu.}
$$
This proves the required <Superposition theorem for Poisson point processes> directly from count distributions.
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