Assume that $X$ is $L^p$ integrable for $1\leq p\leq \infty$ (i.e., for all $t$, $X_t\in L^p$) and is also a (Cadlag) local martingale. If $T_n$ is a localizing sequence of stopping times for $X$. Is it true that the stopped process $X^{T_n}$ remains $L^p$ integrable?
The answer is clearly yes for the discrete time case and also in the continuous case for the cases $p=1,p=\infty$. So I'm guessing it should be yes in general but I don't know how to prove it.
Well I am not sure this is true. I give you an argument to be discussed about which I'm not totally comfortable.
Let's fix $T_n=T$, let's suppose it has a law $P_T(ds)$ and that $T$ is independent of $(M_t)_{t>0}$ and finally that $M_t$ positive for all $t>0$.
Now for any $t>0$ we have a fixed $p\in (1,\infty)$ such that $M_t\in L^p$, and we have :
$$E[(M^T_t)^p]=E[(M_t.1[T>t]+M_T.1[T\leq t])^p]\geq E[(M_T.1[T\leq t])^p]= E[M_T^p.1[T\leq t]]$$
Now using the law of $T$ :
$$E[M_T^p.1[T\leq t]]=E[\int_0^t M_s^pdP_T(ds)]$$
and independence :
$$E[M_T^p.1[T\leq t]]=\int_0^t E[M_s^p]dP_T(ds)$$
Noting $g_{(p,M)}(s)=E[M_s^p]$, we see that if $g_{(p,M)}(s).1[0,t](s)$ is not in $L^1(P_T(ds),\mathbb{R}^+)$ the process is not $L^p$-integrable in t.
The truth is that I am not fully sure about this and especially about the fact that such conditions can be met in practice.
Best regards