A reference stated : "Because $I(t)$ is a martingale and $I(0) = 0$, we have $E[I(t)]=0$ for all $t\geq0$"
What are the role of the martingale and initial value to determine that $E[I(t)] = 0$?
Because we know that $I(t)=\int\Delta(u)dW(u)$ already explained that the the expectation of Brownian motion's increment here is zero, so $E[I(t)]$ must be zero.
By the conditional expectation property, relatively to the $\sigma$-algebra $\mathcal{F}_0$ generated by $I(0)$, we have $\mathbf{E}[I(t)] = \mathbf{E}\big[\mathbf{E}[I(t)|\mathcal{F_0}]\big] = \mathbf{E}[I(0)] = 0$, the second equality being the martingale property for $I(t)$. It is a general fact in martingale theory that the expectation of a martingale is a constant.