Suppose $\{H(s,\omega): s\ge 0 , \omega \in \Omega \}$ is progressively measurable and $\{ B(t): t \ge 0\}$ a linear Brownian motion. Show that for any stopping time T with: $\mathbb{E} \left[ \int_{0}^{T} H(s)^2 ds \right] < \infty$ we have:
(a) $\mathbb{E} \left[ \int_{0}^{T} H(s) dB(s) \right]=0$
(b) $\mathbb{E} \left[ \left( \int_{0}^{T} H(s) dB(s) \right)^2 \right] = \mathbb{E} \left[ \int_{0}^{T} H(s)^2 ds \right]$.
I have tried to use several lemmas such as approximations by progressively measurable processes. I don't really understand why T should be a stopping time. If I apply the optional stopping theorem for (a) would it work? Also we could define the process $H^T(s, \omega)$ such that $\int_{0}^{T} H(s) dB(s)= \int_{0}^{\infty} H^T(s)dB(s)$ and try to work with that. I am a bit confused of what should work in this case. Any help would be appreciated!
Let $M(t) = \int_0^{t \wedge T} H(s) dB(s)$. Then $M(t)$ is a stopped local martingale and hence is also a local martingale.
Further, $\langle M \rangle_t = \int_0^{t \wedge T} H(s)^2 ds$ and so your assumption gives $\mathbb{E}[\langle M \rangle_\infty] < \infty$. It is a standard result that this implies that $M$ is an $L^2$-bounded martingale and $\mathbb{E}[M(\infty)^2] = \mathbb{E}[M(0)^2] + \mathbb{E}[\langle M \rangle_\infty]$. Hence $$\mathbb{E}[\int_0^T H(s)dB(s)] = \mathbb{E}[M(\infty)] = \mathbb{E}[M(0)] = 0$$ and similarly $$\mathbb{E}[\bigg(\int_0^T H(s)dB(s)\bigg)^2] = \mathbb{E}[\langle M \rangle_\infty] = \mathbb{E}[\int_0^T H(s)^2 ds]$$ as desired.