Assume we have a possibly multidimensional Brownian motion on a probability space $(\Omega,\mathcal F,\mathbb P)$ where $(\mathcal F_t)_{t\in[0;T]}$ is the Brownian standard filtration. Let $\Vert X\Vert_{\mathcal M} := \mathbb E(\int^T_0 X_s^2 \mathrm ds )^{1/2}$ and let $$ \mathcal M := \big\{ X\colon \Omega \times [0;T] \to \mathbb R \vert \text{$X$ is a semimartingale}, \Vert X\Vert_{\mathcal M} < \infty \big\}. $$ Now we define the subset $$ \mathcal S := \bigg\{ \bigg\{ \int^t_0 Y_s \mathrm ds \bigg\}_{t\in[0;T]} \bigg\vert Y \colon \Omega \times [0;T] \to \mathbb R \text{ is adapted, continuous and bounded} \bigg\} $$
Is the $\Vert X\Vert_{\mathcal M}$-closure of $\mathcal S$ equal to $\mathcal M$? If not, can we say something about this closure like that every semimartingale of the form $\ldots$ or with the property $\ldots$ is inside this closure? References welcome!
My approach so far: I have proven already that every process $X$ with the following property is in the closure: For all $\delta,\beta>0$, there is a $\nu>0$ such that $$ \mathbb P \bigg[ \sup_{s,t\in[0;T],|s-t| \le \nu} \vert X_s - X_t \vert \ge \beta \bigg] \le \delta. $$ My proof is pathwise, so I hope that one can get weaker conditions. Moreover, this property seems too technical and too uncommon for me. I have also proven that Ito processes of the form $$ \int^\cdot_0 \mu_s \mathrm ds + \int^\cdot_0 \sigma_s \mathrm dW_s $$ with $\mu \in L^2$ and $\sigma \in L^{2+\varepsilon}$ satisfy this property. If I knew that this property is satisfied even if $\sigma \in L^2$, this would also be quite helpful.