Given $M_t$ a martingale, $\lim_{t \to \infty}E[M_t]^2=0$, and $t \le 1$, show $$E\left[\left( \left| \int_0^t M_s dB_s \right|^2 \right) \right] \le E[M_t^2] t$$
The first step is that I want to use Itô's isometry to show (I am a little unsure about the absolute value here but I think it works like this)
$$E\left[\left( \left| \int_0^t M_s dB_s \right|^2 \right) \right] = E\left[\left( \int_0^t \left| M_s \right|^2 ds \right) \right]$$
Now since we have the traditional integral we can move the expected value inside the integral.
$$=\left( \int_0^t E\left[\left| M_s \right|^2\right] ds \right) $$
If this has been correct until here, my biggest confusion is now how to evaluate this integral. Considering the desired answer it seems like we treat the integrand here as a "constant" in the traditional sense so I would just have $\int_0^t E\left[\left| M_s \right|^2\right] ds = tE[\left| M_t \right|^2] - 0E[\left| M_0 \right|^2]$ but I doubt it. Can someone help me here? Thanks.
We have \begin{align} E\left[\left( \int_0^t M_s dB_s \right)^2 \right] &= E\left[\int_0^t M_s^2 ds\right] \\ &=\int_0^t E\left[ M_s^2 \right] ds \le \int_0^t E\left[ M_t^2 \right] ds =E\left[ M_t^2 \right]t \\ \end{align}
Here, we use the fact that $E[M_s^2] \le E[M_t^2]$, effectively, applying the Jensen's inequality $$M_s^2 =E[M_t|\mathcal{F}_s]^2 \le E[M_t^2|\mathcal{F}_s]$$ $$\implies E[M_s^2] \le E[E[M_t^2|\mathcal{F}_s]] = E[M_t^2]$$