Integral with respect to Brownian motion, Variance

Question

Good day. Imagene we have a martingale $M(t)=\int_0^t f(s)dB(s)$ which satisfies Dambis-Dubins-Schwarz Theorem. At the same time $M(t)^2 - <M>(t)$ is a Martingale starting in $0$ as well. If i am not mistaken, then from Dambis-Dubins-Schwarz Theorem i could conclude that \begin{align*} M(t) \sim N(0,\int_0^t f(s)^2ds) \end{align*} On another side $\mathbb{V}(M(t))=\mathbb{E}(M(t)^2)=\mathbb{E}(<M>(t))$. It follows that \begin{align*} \mathbb{E}(\int_0^t f(s)^2ds)=\int_0^t f(s)^2ds \end{align*} If $f(s)$ is deterministic, then it is not a problem, but what if not? I am sure there is a mistake somewhere in my thoughts. Good if someone could fix it :)