Let $ (M_t)_{t \ge 0}$ be a continuous local martingale (as defined in LeGall). Let $\mathbb{E}$ denote expectation.
Is $\mathbb{E}[M_t^2] < \infty$ for all $t \ge 0$? If yes, then how does one prove it? If no, then how does one construct a counterexample?
My attempt: Fix $\tau > 0$. The sequence of stopping times $T_n = n$ reduces $M$ (from LeGall). Define a constant stopping time $T=\tau$. By the same argument, $M^T$, which is the martingale $M_{t \wedge T}$, is uniformly integrable, and therefore, bounded in $L^1$, which gives that $M_{\tau} \in L^1$.
Perhaps, I might be missing something very basic here, but I do not see the $L^2$ conclusion I am seeking appear naturally from here. Any hints or suggestions are appreciated.
$M_t=B_tX$ where $X$ is an integrable but not square-integrable random variable, and $(B_t)_{t\ge0}$ is a brownian motion independent of $X$.
If you allow $M_t$ not to start from $0$ then it is even more simple: $M_t=X$.