In Oksendal's Stochastic Differential Equations, the set $\mathcal{V}(S,T)$ is defined to be all functions $f:[0,\infty)\times \Omega \to\mathbb{R}$ such that
- $(t,\omega)\to f(t,\omega)$ is $\mathcal{B}([0,\infty))\otimes \mathcal{F}$-measurable
- $f(t,\omega)$ is $\mathcal{F}_t-$adapted
- $\mathbb{E}[\int_S^T f^2(t,\omega)dt]<\infty$
A function $\phi\in \mathcal{V}(S,T)$ is called elementary if $\phi$ has the form
$$\phi(t,\omega) = \sum_{j\geq 0}e_j(\omega)\chi_{[t_j, t_{j+1})}(t) $$
Then $e_j$ must be $\mathcal{F}_{t_j}$ measurable and thus independent from $B_{t_{j+1}}-B_{t_j}$ where $\{B_t\}$ is Brownian motion centered anywhere. The Ito integral is then defined for elementary functions to be $$\int_S^T \phi(t,\omega)dB_t(\omega) = \sum_{j\geq 0}e_j(\omega)[B_{t_{j+1}}-B_{t_{j}}](\omega)$$
Since $e_j$ is independent from $B_{t_{j+1}}-B_{t_{j}}$ then we have that
$$\begin{align*}\mathbb{E}\Big[\int_S^T \phi(t,\omega)dB_t(\omega)\Big] &= \sum_{j\geq 0}\mathbb{E}[e_j(\omega)[B_{t_{j+1}}-B_{t_{j}}](\omega)]\\
&= \sum_{j\geq 0}\mathbb{E}[e_j]\mathbb{E}[B_{t_{j+1}}-B_{t_{j}}] \\
&= \sum_{j\geq 0}\mathbb{E}[e_j]\cdot 0 = 0\end{align*}$$
Where we use that $B_{t_{j+1}}-B_{t_j}\sim\mathcal{N}(0, t_{j+1}-t_j)$.
Edit: Ending up answering my own question and keeping this up to avoid confusion when people look up this post Itō Integral has expectation zero when they are asking about a property in Oksendal.
Best Answer
The key observation is in the assumption that $\phi\in\mathcal{V(S,T)}$ because this ensures that $\phi(t,\omega) = \sum_{j\geq} e_j(\omega)\chi_{[t_j, t_{j+1})}(t)$ where $e_j$ is $\mathcal{F}_{t_j}-$measurable and thus independent of $B_{t_{j+1}}-B_{t_j}$ which is a normally distributed random variable with mean $0$. Further because $B_{t_{j+1}}-B_{t_j}$ is independent of $\mathcal{F}_{t_j}$ we get this result. These are the two fact that caused me enough problems that I tried posting it here.