Oksendal. Stohastic differential equations. Problem 4.7

Question

I am stacked at problem 4.7 (b) in the book Oksendal. Stohastic Differential equations.

The conditions of the problem are the next:

Let $X_{t}$ be an Ito integral.

$dX_{t}=v\left(t,\omega\right)dB_{t}\left(\omega\right)$ where $v\in\mathbb{R}^n$, $v\in\mathcal{V}\left(0,T\right)$, $B_{t}\in\mathbb{R}^{n}$ and $0\leq t\leq T$

Prove that if $v$ is bounded then $M_{t}=X_{t}^{2}-\int\left|v_{s}\right|^{2}ds$ is a martingale, where $\left\langle X,X\right\rangle _{t}:=\int\left|v_{s}\right|^{2}ds$ quadratic variation of the martingale process $X_{t}$ .

Part of my solution is the next:

1. Let's consider the process $M_{t}=X_{t}^{2}-\int_{0}^{t}\left|v_{s}\right|^{2}ds$.
2. According to Corollary 3.2.6 we have to show that $M_{t}$ can be represented in the form of $M_{t}\left(\omega\right)=\int_{0}^{t}f\left(s,\omega\right)dB_{s}$, because this process is a martingale. $(f\left(t,\omega\right)\in\mathcal{V}\left(0,T\right))$
3. According to Ito formula we have $dM_{t}=-\left|v_{t}\right|^{2}dt+2X_{t}dX_{t}+\left(dX_{t}\right)^{2}$
4. Using the definition of $dX_{t}$ and substituting it at (3) we have that $dM_{t}=-\left|v_{t}\right|^{2}dt+2X_{t}v\left(t,\omega\right)dB_{t}\left(\omega\right)+\left(v\left(t,\omega\right)dB_{t}\left(\omega\right)\right)^{2}$
5. Simplifying terms in (4) the equation can be simplifyed to $dM_{t}=2X_{t}v\left(t,\omega\right)dB_{t}\left(\omega\right)\implies M_{t}=M_{0}+\int_{0}^{t}2X_{s}v\left(s,\omega\right)dB_{s}$
6. Let's assume that $M_{0}=0$ and we next have to prove that $2X_{t}v\left(t,\omega\right)\in\mathcal{V}\left(0,T\right)$

I have to prove the statement in point (6) to finish the proof. What do I need to do next, to get the needed result?

For convenience, I add the definition of $\mathcal{V}\left(0,T\right)$ space in the picture:

$\mathcal{V}\left(0,T\right)$ space in the picture" />