Ito integral's zero mean My Sto Cal prof gave a long proof for the fact that $E[\int_{0}^{t} f_s dW_s] = 0$ where W is Brownian and f is Borel x $\mathscr{F}$-measurable, adapted and satisfies some integrability condition. Apparently it follows from the fact that is is true for simple functions (the proof was actually for simple functions) and then we take the limit for general functions or something like that.

However, it was pointed out to me on another question that the zero-mean property actually follows from the fact that

if we define $I(t) \equiv \int_{0}^{t} f_s dW_s$,

then $I(0) = \int_{0}^{0} f_s dW_s = 0$

and $I(t) = \int_{0}^{t} f_s dW_s$ is a martingale.

So, I guess it then follows that $E[I(t)] = E[I(0)] = E[0] = 0$?

I suspect that my prof may not have done this because proof of the martingale property was not discussed and might come out in our exam, but anyway does the fact that $I(0)=0$ and that $I(t)$ is a martingale really imply that $E[I(t)]=0$ ?
The answer is certainly: yes. By definition of a martingale, $\mathsf E M_t = \mathsf E M_0$ (just using the tower property) and since $M_0 = 0$ in your case, all follows. Note that Ito integral may not be a martingale unless some integrability conditions are satisfied, in general it may be only a 'local martingale' - so you need to see which integrability conditions do you have.
The answer is certainly: yes. By definition of a martingale, $\mathsf E M_t = \mathsf E M_0$ (just using the tower property) and since $M_0 = 0$ in your case, all follows. Note that Ito integral may not be a martingale unless some integrability conditions are satisfied, in general it may be only a 'local martingale' - so you need to see which integrability conditions do you have.