Let $(\Omega, \mathcal{F}, \{\mathcal{F_n}\}, P)$ be a filtered probability space, let $X = \{X_n : n = 0,1,2,...\}$ be an $\{\mathcal{F}_n\}$-martingale, let $\tau$ be an $\{\mathcal{F}_n\}$-stopping time, and let $$\mathcal{F_\tau} = \{F \in \mathcal{F} : F \} .$$
Is it then true that $\{X_{\tau + k} : k = 0, 1, 2, ...\}$ is an $\mathcal{F}_{\tau + k}$-martingale?
I have shown that each $X_{\tau + k}$ is $\mathcal{F}_{\tau + k}$-measurable (and I think the integrability of each $X_{\tau + k}$ is clear... although I'm actually not too sure about this if $\tau$ is not bounded), but I can't seem to prove that $$ E[X_{\tau + k} \mid \mathcal{F}_{\tau + k-1}] = X_{\tau+k-1}.$$ I take it that (since $\tau$ is a random variable) I cannot just "reindex" $\tau + k$ (i.e., let "$\tau + k = n$") and just use the fact that $X$ is an $\{\mathcal{F}_n\}$-martingale...
(Note that the result may not be true.)
You have to assume that $\tau$ is finite valued (so that $X_{\tau+k}$ makes sense).
What you have to show is $$\int_A X_{\tau+k}dP= \int_A X_{\tau+k-1}dP$$ for all $A \in \mathcal F_{\tau+k-1}$.
It is enough to show that $$\int_{A\cap (\tau=j)} X_{j+k}dP= \int_{A\cap (\tau=j)} X_{j+k-1}dP$$ for all $A \in \mathcal F_{\tau+k-1}$ for each $j$.
For this note that $B\equiv A\cap (\tau=j)=A\cap (\tau+k-1=j+k-1)\in \mathcal F_{j+k-1} $ by the definition of $\mathcal F_{j+k-1} $. Now the result follows from the definition of a martingale.