Show that $\mathbb{E}\{\int_U^T g(s)dB(s)\mid \mathcal{F}_U\} = 0$

Question

Suppose that $g$ is a simple process in the class $\mathcal{V}=\mathcal{V}[U,T]$. Using the notations

$g_k=g(t_k)$, $\Delta B_k = B(t_{k+1})-B(t_k)$, and $\mathcal{F}_k=\mathcal{F}_{t_k}$, with the partition being $U=t_0<t_1<\ldots <t_{n-1}< t_n=T$,

I want to show that

$\mathbb{E}\{\int_U^T g(s)dB(s)\mid \mathcal{F}_U\} = 0$. My reasoning is as follows:

$\mathbb{E}\{\int_U^T g(s)dB(s)\mid \mathcal{F}_U\} = \sum_k\mathbb{E}[g_k\Delta B_k\mid\mathcal{F}_U] = \{g_k,\Delta B_k \text{ are independent of }\mathcal{F}_U\}=\Sigma_k\mathbb{E}[g_k\Delta B_k]=\Sigma_k \mathbb{E}[\mathbb{E}[g_k\Delta B_k\mid \mathcal{F}_k]]=\{g_k\in\mathcal{F}_k,\Delta B_k\text{ is independent of }\mathcal{F}_k\}=\sum_kg_k\mathbb{E}[\Delta B_k]=0. $

My question is if these steps are correct. In particular, I am unsure of the steps following the second equality sign.

Any comment or suggestion is most welcome. Thanks in advance :)

Answer 1

I would only change a little detail at the end: \begin{eqnarray*} \mathbb{E}\{\int_U^T g(s)dB(s)\mid \mathcal{F}_U\} &=& \sum_k\mathbb{E}[g_k\Delta B_k\mid\mathcal{F}_U] \\ &=&\Sigma_k\mathbb{E}[g_k\Delta B_k]\\ &=&\Sigma_k \mathbb{E}[\mathbb{E}[g_k\Delta B_k\mid \mathcal{F}_k]]\\ &=&\sum_k\mathbb{E}[g_k \mathbb E[\Delta B_k]]=0 \end{eqnarray*}