Let $\mathfrak{H}_{0}^{2,c}$ denote the space of square-integrable continuous martingales starting from zero. Let $\mathfrak{A}$ be a closed linear subspace of $\mathfrak{H}_{0}^{2,c}$. Show that
$\mathfrak{A}$ is stable under stopping, i.e., $M_{\cdot\wedge\tau} \in \mathfrak{A}$ for any $M\in \mathfrak{A}$ and stopping time $\tau$
implies that
$\mathfrak{A}$ is stable under integration, i.e., $\int_0^\cdot H_tdM_t\in \mathfrak{A}$ for any $M\in \mathfrak{A}$ and $H$ predictable with $E[\int_0^t H_sd<M>_s] <\infty$.
I tried to approach the question via Monotone Class Theorem. Let $$F:= \{H: \int_0^\cdot H_sdM_s \text{ for any } M\in \mathfrak{A}, H \text{ predictable}, E[\int_0^t H_sd<M>_s] <\infty\}.$$
Here, we use the result in The closure of the elementary processes are the predictable processes.
It is easy to verify that $F$ contains constant functions (as $\mathfrak{A}$ is a linear subspace), is closed under uniform convergence (by Ito's Isometry), and is closed under uniformly bounded convergence (by DCT). It remains to show that $F$ includes all elementary processes. Consider $$ H_t = \sum_{i=0}^{n-1} h_i \mathbb{I}_{(t_i, t_{i+1}]}. $$ for $h_i$ bounded, $F_{t_i}$-measureable. Since $\mathfrak{A}$ is closed under stopping, $M_{t_i\wedge \cdot} \in \mathfrak{A}$ for all $i$. Yet, how can we extend this to $$ \int_0^\cdot H_sdM_s = \sum_{i=0}^{n-1} h_i(M_{t_{i+1}\wedge \cdot} - M_{t_i\wedge \cdot})? $$ A prior, we do not know whether $h_i(M_{t_{i+1}\wedge \cdot} - M_{t_i\wedge \cdot}) \in \mathfrak{A}$.