In the context of stochastic integration, we showed how it's possible to define the stochastic integral $\int H dM$ for $H \in L^2(M)$ and $M \in \mathcal M^2_0$ (martingales null at $0$ such that $\sup_{t} E[|M_t|^2] < \infty$) Then we want to expand this, and we introduce the "local" version of the spaces, and we say that $M \in \mathcal M^2_{0,loc}$ if exists a sequence of stopping times $\tau_n \nearrow \infty$ such that the stopped process $M^{\tau_n} = M_{t \wedge \tau_n} \in \mathcal M^2_0$ for every $n$.
Similarly we say that a predictable process $H \in L^2_{loc}(M)$ if there exists a sequence of stopping times $\tau_n \nearrow \infty$ such that $HI_{]0, \tau_n]} \in L^2(M)$ for each $n$, where $]0, \tau_n] = \{(\omega, t) \in \bar \Omega: 0 < t \le \tau_n(\omega)\}$
Now the professor's notes say that for
$M \in \mathcal M^2_{0,loc}$ and $H \in L^2_{loc}(M)$, defining the stochastic integral is straightforward: set $$\int H dM = \int HI_{]0, \tau_n]} dM^{\tau_n}$$ on $]0, \tau_n]$.
Question (finally):
It seems to assume that it's possible to find a sequence of stopping times $\tau_n$ that localise simultaneously both $H$ and $M$, which seems surprising to me. Is it true that it's always possible to find such a sequence or am I misunderstanding?
Yes, it is possible to find a common localizing sequence.
By the very definition, there exist localizing stopping times $(\sigma_n)_{n \in \mathbb{N}}$ and $(\tau_n)_{n \in \mathbb{N}}$ for $M$ and $H$, respectively. If we set
$$\varrho_n := \min\{\tau_n, \sigma_n\}$$
then $\varrho_n$ is a stopping time satisfying $\varrho_n \uparrow \infty$. Since $\varrho_n \leq \tau_n$, it is obvious that $H 1_{]0,\varrho_n]} \in L^2(M)$. Moreover, the optional stopping theorem (applied for the martingale $M^{\sigma_n}$) gives that
$$M^{\varrho_n}_t = M^{\sigma_n}_{t \wedge \tau_n}$$
is a martingale. Hence, $M^{\varrho_n} \in \mathcal{M}_0^2$.
This shows that $(\varrho_n)_{n \in \mathbb{N}}$ is a localizing sequence for both $M$ and $H$.