Let $Z_t = 1\{X\geq t\}$, where $X$ is continuously distributed.
The trajectories of $Z$, i.e. $t\mapsto Z_t(\omega)$ have derivative zero everywhere, except for a single point $t_\omega$. Can I conclude that since $X$ is continuously distributed, the derivative of $Z_t$ is zero a.s (or in $L_p$)?
The trajectories are a.e. constant: $$ Z_t(\omega) = \begin{cases} 1\quad t \leq X(\omega), \\ 0 \quad t > X(\omega) \end{cases}. $$ So we have that $$ \Bbb P\Bigl( \omega \in \Omega : Z_t'(\omega) = 0 \Bigr) = \Bbb P\Bigl( \omega \in \Omega :X(\omega) \neq t \Bigr) , \quad \text{for all} \ t \geq 0. $$ which is equal to $1$ if $X$ has a continuous distribution. So for fixed $t \geq 0$ the paths of $Z$ are a.s. differentiable at $t$. But $$ \Bbb P\Bigl( \omega \in \Omega : t \mapsto Z_t(\omega) \ \text{is differentiable for all} \ t \geq 0 \Bigr) = 0 $$ So no path is differentiable as a whole.