# How to argue $t \mapsto Y_t(\omega)$ is cadlag for every $\omega \in \Omega$ if the index is changed?

I am trying to understand a proof in my book where the setup is:

Let $$(Y_t)_{t \geq 0}$$ be a $$(\mathcal{F}_t)_{t \geq 0}$$-Lévy-proces and $$\tau$$ an extended stopping time wrt. $$(\mathcal{F}_t)_{t \geq 0}$$. Assume that $$\tau$$ is finite almost surely and put $$D_\tau Y_t = (Y_{\tau + t} - Y_\tau) \textbf{1}_{\{\tau < \infty\}}$$. Then I have to argue that $$(D_\tau Y_t)_{t \geq 0}$$ is a Lévy-proces wrt. $$(\mathcal{F}_{(\tau + t)})_{t \geq 0}$$

Therefore, for one of the conditions, I have to prove that $$t \mapsto D_\tau Y_t(\omega)$$ is cadlag for every $$\omega \in \Omega$$.

Since $$(Y_t)_{t \geq 0}$$ is a Lévy-process, I know that $$t \mapsto Y_t(\omega)$$ is cadlag for every $$\omega \in \Omega$$. However, what is the argument then that $$Y_{\tau + t}$$ is cadlag (I know that cadlag is simply right-continuity with left limits)? Is there some basic properties that I need to know of? For continuity I know that if both $$f,g$$ is continuous then their sum is continuous. Is this also what is used here? Or do we use that $$\tau$$ is an extended stopping time which is finite almost surely?

This does not involve Probability Theory at all. Fix $$\omega$$ with $$\tau (\omega)<\infty$$ and note that $$t \to Y_{\tau (\omega)+t}(\omega)-Y_{\tau (\omega)}(\omega)$$ is right continuous with left limits. [If $$f$$ is cadlag so is $$f(x+y)-c$$ for any fixed $$y,c$$].