$\mathbb{E}[(\int_{0}^{t} V_s^2: d\langle M^{c}\rangle_{s})^{\frac{1}{2}}]<\infty$ implies $N_t:= \int_0^t V_s \cdot d M_{s}^{c}$ is a martingale.

Question

Suppose that $M$ is a continuous martingale and $V_s$ is a progressive process with $\mathbb{E}\left[\left(\int_{0}^{t} Tr(V_sV_s^T d\left\langle M\right\rangle_{s})\right)^{\frac{1}{2}}\right]<\infty$.

Show that $N_t:= \int_0^t V_s \cdot d M_{s}$ is a martingale.

I found a similar statement here, but I have no idea how to show this. Holder's inequality doesn't seem to work.

Any help would be appreciated.