Square Integrable local martingale or locally square integrable martingale?

Question

I have a question about martingales. What is the difference between "locally square integrable martingale" and "square integrable local martingale"? In particular, which set does $M_{loc}^2$ represent?

Answer 1

Protter gives the following definition:

Let $X$ be a stochastic process. A property $\pi$ is said to hold locally if there exists a sequence of stopping times $(T_n)_{n \geq 1}$ increasing to $\infty$ almost surely such that $X^{T_n} 1_{\{T_n>0\}}$ has property $\pi$, each $n \geq 1$.

If we speak of a square integrable local martingale $(X_t)_{t \geq 0}$, then

1. $(X_t)_{t \geq 0}$ is a local martingale, i.e. there exists stopping times $(T_n)$, $T_n \uparrow \infty$ such that the stopped process $(X_t^{T_n} 1_{\{T_n>0\}})_{t \geq 0}$ is a martingale.
2. $X_t \in L^2(\mathbb{P})$ for all $t \geq 0$.

In contrast, a locally square integrable martingale $(X_t)_{t \geq 0}$ satisfies

1. There exists a sequence of stopping times $(T_n)$, $T_n \uparrow \infty$, such that $X_{t}^{T_n} \in L^2$ for all $n \geq 1$, $t \geq 0$.
2. $(X_t)_{t \geq 0}$ is a martingale.

Sometimes, the notion locally square integrable martingale is also used for processes for which there exist stopping times $(T_n)$ such that $(X_t^{T_n})_{t \geq 0}$ is a martingale and $X_t^{T_n} \in L^2$ - in this case, "local" refers both to integrability and martingale property. This highly depends on the book (or paper) you are reading.