TITLE: Stopped $L^2$ - martingale is again $L^2$ integrable
QUESTION [6 upvotes]: Given a $L^2$ martingale $X_t$, for a finite stopping time $T$, is it true that $X_T$ is again $L^2$ integrable?
If $X_t$ is additionally uniform integrable, then we can use Jensen-inequality:$$E[X_T^2]=E[E[X_\infty\mid F_T]^2]\leq E[X_\infty^2]$$ to obtain the $L^2$ integrability of $X_T$ provided $X_\infty$ is square integrable.
However, I don't know if the statement is still true without the assumption on uniform integrability and $X_\infty^2$ is integrable.

REPLY [2 votes]: There is also Doob's example:  Let $F$ be any distribution on the real line. Choose the appropriate function $g$ so that $g(B_1)$ has distribution $F$, and then define $T:=\inf\{t>1: B_t=g(B_1)\}$. Clearly $T$ is a.s. finite by the recurrence of Brownian motion, and $B_T$ has distribution $F$. And this $T$ is a stopping time of the filtration of the given Brownian motion.