= Solution
Let $E_i=\{M(I_i)=0\}$. For any subset $K\subseteq\{1,\ldots,n\}$, simultaneous voidness is voidness of the union. Since the intervals are disjoint, the assumed <void probability functional> gives
$$
\mathbb P\left(\bigcap_{i\in K}E_i\right)
=e^{-\lambda(\bigcup_{i\in K}I_i)}
=\prod_{i\in K}e^{-\lambda(I_i)}
=\prod_{i\in K}\mathbb P(E_i).
$$
This includes the empty subset. To check all event patterns explicitly, apply the <inclusion-exclusion principle> to the complements in a chosen pattern:
$$
\mathbb P\left(\bigcap_{i\in K}E_i\cap\bigcap_{j\notin K}E_j^c\right)
=\prod_{i\in K}\mathbb P(E_i)\prod_{j\notin K}(1-\mathbb P(E_j)).
$$
Thus \b[the void events are mutually independent], and their complementary occupied-cell indicators are mutually independent too. This is joint <independence>, not merely pairwise <independence>.
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