Prove that $\mathbb{E} \left[ \int_0^t f_s \, dW_s \right] = 0$, for any continuous stochastic process $f_s$ I am currently trying to prove that
$$
\mathbb{E} \left[ \int_0^t f_s \, dW_s \right] = 0
$$
where $f_s = f(W_s)$ is some continuous bounded stochastic process.
My attempt to do this is as follows:
$$
\mathbb{E} \left[ \int_0^t f_s \, dW_s \right] = \mathbb{E} \left[ \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} f_{s_i} \cdot \Delta W_{s_i} \right]
$$
where $\Delta W_{s_i} := W_{s_{i+1}} - W_{s_i}$. This is then equal to
$$
\lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \mathbb{E} \left[f_{s_i} \cdot \Delta W_{s_i} \right]
$$
which I believe is then equal to
$$
\lim_{n \rightarrow \infty} \left( n \cdot \mathbb{E} \left[f_{s_i} \right] \cdot \mathbb{E} \left[ \Delta W_{s_i} \right] \right)
$$
Thus, since $\mathbb{E} \left[ \Delta W_{s_i} \right] = 0$, we have our result.
So my question is as follows:


*

*Is this a correct proof (i.e. Have I made any false assumptions)?

*Could this have been done better?

 A: You are on the right track. Intuitively, your proposal works quite well, especially if you do not require mathematical rigor. If you require so, two comments are:
(1) Specific reasons are necessary for the commutativity of $\mathbb{E}$ and $\lim_{n\to\infty}$, and
(2) Conditional expectation could be adopted to clarify the expectation of each term of the sum.
Here is a sketch of proof.
Recall the definition of Ito's integral
$$
Y_t=\int_0^tf_s{\rm d}W_s:=\lim_{n\to\infty}\sum_{j=0}^{n-1}f_{s_j}\left(W_{s_{j+1}}-W_{s_j}\right),
$$
where $0=s_0<s_1<\cdots<s_n=t$ forms a partition of $\left[0,t\right]$. Note that the limit hereinabove is in the sense of the existence of some square-integrable $Y_t$ that satisfies
$$
\lim_{n\to\infty}\mathbb{E}\left(Y_t-\sum_{j=0}^{n-1}f_{s_j}^n\left(W_{s_{j+1}}-W_{s_j}\right)\right)^2=0,
$$
where $\left\{f_t^n\right\}_{n\in\mathbb{N}}$ is a sequence of random simple function that approximates $f_t$, i.e., $f_t^n\to f_t$ almost surely (guaranteed by the boundedness of $f_t$, or more generally, square-integrability of $f_t$). Denote
$$
Y_t^n=\sum_{j=0}^{n-1}f_{s_j}^n\left(W_{s_{j+1}}-W_{s_j}\right).
$$
With these understandings, our target could be figured out by
$$
\mathbb{E}Y_t=\mathbb{E}Y_t^n+\mathbb{E}\left(Y_t-Y_t^n\right).
$$
Thanks to the vanishing of the last term, i.e. (using Cauchy-Schwarz inequality and the definition of $Y_t$),
$$
\left|\mathbb{E}\left(Y_t-Y_t^n\right)\right|\le\mathbb{E}\left|Y_t-Y_t^n\right|\le\sqrt{\mathbb{E}\left(Y_t-Y_t^n\right)^2}\to 0,
$$
it suffices to figure out the limit of $\mathbb{E}Y_t^n$.
$Y_t^n$ is a finite sum, for which
\begin{align}
\mathbb{E}Y_t^n&=\mathbb{E}\left(\sum_{j=0}^{n-1}f_{s_j}^n\left(W_{s_{j+1}}-W_{s_j}\right)\right)\\
&=\sum_{j=0}^{n-1}\mathbb{E}\left(f_{s_j}^n\left(W_{s_{j+1}}-W_{s_j}\right)\right)\\
&=\sum_{j=0}^{n-1}\mathbb{E}\left(\mathbb{E}\left(f_{s_j}^n\left(W_{s_{j+1}}-W_{s_j}\right)|\mathcal{F}_{s_j}\right)\right)\\
&=\sum_{j=0}^{n-1}\mathbb{E}\left(f_{s_j}^n\mathbb{E}\left(W_{s_{j+1}}-W_{s_j}|\mathcal{F}_{s_j}\right)\right)\\
&=\sum_{j=0}^{n-1}\mathbb{E}\left(f_{s_j}^n\cdot 0\right)\\
&=0.
\end{align}
Therefore, $\mathbb{E}Y_t^n\to 0$, for which $\mathbb{E}Y_t=0$ follows immediately as per the reasoning from above.
You may refer to here for more detailed steps with regards to this topic.
