 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 $a={\int }_0^t{1}_A(s)ds$. 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 givesConsequentlyIn particular a measure-zero zero set is almost surely missed, despite possibly containing infinitely many points.