= Solution
Assume the zero set $A=\{s\in[0,t]:g(s)=0\}$ is measurable, as is ensured by measurability of $g$. This is an implicit necessary hypothesis: for an arbitrary real function the printed <integral> and event need not be measurable. Put $a=\int_0^t\mathbf1_A(s)\,ds$. Conditional on $N_t=n$, the <uniform order statistics> description makes the <probability> of no jump in $A$ equal to $(1-a/t)^n$. Averaging over the <Poisson distribution> gives
$$
\mathbb P(0\notin\mathcal R(g)[0,t])
=e^{-\lambda t}\sum_{n=0}^\infty\frac{[\lambda t(1-a/t)]^n}{n!}=e^{-\lambda a}.
$$
Consequently
$$
\boxed{\mathbb P(0\in\mathcal R(g)[0,t])=1-\exp\left(-\lambda\int_0^t\mathbf1_{\{g(s)=0\}}\,ds\right).}
$$
In particular a measure-zero zero set is almost surely missed, despite possibly containing infinitely many points.
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