Example of a random variable $X$ that is an $\mathscr{F}_t$-local martingale, but not an $\mathscr{F}_t^X$-local martingale. This is a problem from Ethier and Kurtz' Markov Processes. The book introduces some theorems on local martingales but they all involve the process being right continuous. I think this problem must be solved using the definitions of a local martingale, that is, there is $\mathscr{F}_t$ stopping times $\tau_1\le \tau_2 \le \cdots$ with $\tau_n \to \infty$ a.s. such that $X^{\tau_n}$ is a martingale. However, I am new to this concept and am lost on how to solve this problem. I would greatly appreciate some help.
Let $\eta$ and $\xi$ be independent random variables with $P(\eta=1)=P(\eta=-1)=\frac{1}{2}$ and $E|\xi|=\infty$. Define 
$$X(t) =
\begin{cases}
0,  & 0\le t<1, \\
\eta \xi, & t\ge 1,
\end{cases}$$
and
$$ \mathscr{F}_t =
\begin{cases}
\sigma(\xi),  & 0\le t<1 \\
\sigma(\xi,\eta), & t\ge 1.
\end{cases}$$
Show that $X$ is an $\mathscr{F}_t$-local martingale, but that $X$ is not an $\mathscr{F}_t^X$-local martingale. 

Because $E|X_t|=E|\xi|=\infty$ for $t\ge 1$, $\{X_t\}$ is not a martingale (with respect to any filtration). 
Define $\tau_n:=n$ on $\{|\xi|\le n\}$ and $\tau_n=0$ on $\{|\xi|>n\}$. Then $\{\tau_n\}$ is an increasing sequence of $(\mathcal F_t)$ stopping times (note that $\xi$ is $\mathcal F_0$-measurable) with limit $\infty$ on $\{|\xi|<\infty\}$. I assume that $P(|\xi|<\infty)=1$. Because, for $t\ge 1$,  $X_{t\wedge \tau_n} = \eta\xi1_{\{|\xi|\le n\}}$, the stopped process $X^{\tau_n}$ is an $(\mathcal F_t)$-martingale. This shows that $X$ is an $(\mathcal F_t)$ local martingale. 
But, for $n$ sufficiently large, the random variable $\tau_n$ is not an $({\mathcal F}^X_t)$ stopping time, because ${\mathcal F}^X_0=\{\emptyset,\Omega\}$ but $\{\tau_n=0\}=\{|\xi|\le n\}$. 
As @d.k.o shows, $(X_t)$ does not admit a localizing sequence of $({\mathcal F}^X_t)$ stopping times, 
and so $(X_t)$ cannot be an $({\mathcal F}^X_t)$ local martingale.
