We introduce a probability space $(\Omega,\mathscr{F},\mathbb{P})$ equipped with a filtration $\mathbb{F}$. We define a Poisson process $N$ with intensity $\lambda$ adapted to $\mathbb{F}$, and we designate by $\tau$ the $\mathbb{F}$-stopping time representing the time of the first jump. Let $Y$ be a jump-diffusion process whose jump component is $N$, while $X$ is a simple jump process defined as: \begin{align} X_t&:=1_{\{t\leq \tau\}} \end{align} By definition, $Y$ is right-continuous with left limits (RCLL). On the other hand, $X$ is left-continuous with right limits (LCRL). We introduce a third process $Z$ defined as the product of $X$ and $Y$.
What can be said about the continuity and the evaluation of $Z$ when $t=\tau$?
Limit: let us consider the simple case where $Y=N$, then we have: \begin{align} &\lim_{t\nearrow\tau}Z_t =\lim_{t\nearrow\tau}\{X_tY_t\} =X_\tau\lim_{t\nearrow\tau}\{Y_t\} =\lim_{t\nearrow\tau}\{N_t\} =0 \\ &\lim_{t\searrow\tau}Z_t =\lim_{t\searrow\tau}\{X_tY_t\} =\lim_{t\searrow\tau}\{X_t\}Y_\tau =\lim_{t\searrow\tau}\{1_{\{t\leq \tau\}}\}N_\tau =0 \end{align} As expected the limits are well-defined.
Evaluation: using the fact that $Z_t=X_{t^-}Y_{t^+}$, I get: \begin{align} Z_\tau=X_{\tau^-}Y_{\tau^+}=1\times N_\tau=1 \end{align} So $Z$ has both well-defined left and right limits at $\tau$, but it is discontinuous at this point from both sides because its value at the stopping time is different from both limits. Moreover, it will be identically equal to $0$ for all $t\in\mathbb{R}_{\geq0}\backslash\{\tau\}$ and equal to $1$ for $t\in\{\tau\}$.
Is this reasoning sensible?