Expectation of Product of Ito Integrals wrt Independent Brownian Motions Let $W_1(t)$, $W_2(t)$, $W_3(t)$ be independent Brownian motions and $f$, $g$ smooth functions. I want to know if the following is true:
$$
   \mathbb{E}\left[
      \left(
         \int\limits_0^t f(W_3(s),s) \,dW_1(s)
      \right)
      \left(
         \int\limits_0^t g(W_3(s),s) \,dW_2(s)
      \right)
   \right] = 0
$$
I think it is, because, as an argument from intuition:
$$
\begin{align}
   \mathbb{E}&\left[
      \left(
         \int\limits_0^t  f(W_3(s),s) \,dW_1(s)
      \right)
      \left(
         \int\limits_0^t g(W_3(s),s) \,dW_2(s)
      \right)
   \right]\\ 
&=
\mathbb{E}\left[
      \left(
         \lim_{n\rightarrow\infty}\sum\limits_{i=0}^{n-1} 
                  f(W_3(t_i),t_i) [W_1(t_{i+1})-W_1(t_{i})]
      \right)
      \left(
         \lim_{n\rightarrow\infty}\sum\limits_{j=0}^{n-1} 
                  g(W_3(t_j),t_j) [W_2(t_{j+1})-W_2(t_{j})]
      \right)
   \right]\\
&=
\lim_{n\rightarrow\infty}
\sum\limits_{i=0}^{n-1}
\sum\limits_{j=0}^{n-1}
\mathbb{E}\left[
      f(W_3(t_i),t_i) g(W_3(t_j),t_j)
   \right]
\underbrace{
\mathbb{E}\left[ 
   W_1(t_{i+1})-W_1(t_{i})
\right]
}_0
\underbrace{
\mathbb{E}\left[ 
   W_2(t_{j+1})-W_2(t_{j})
\right]
}_0\\
&=0
\end{align}
$$
since increments of $W$ are Gaussian. But I am not sure if there are some stochastic analysis concepts I am missing. Thanks!
 A: As Did mentioned, $[W_1,W_2](t)=0$. Let  $\{t_i\}_{i=1}^{n}$ be a partition of $[0,t]$. We want to consider 
$$A_n=\sum_{i=0}^{n-1}(\,W_1(t_{i+1})-W_1(t_{i})\,)(\,W_2(t_{i+1})-W_2(t_{i})\,)$$
Using independent of $W_1$ and $W_2$ , thus $\mathbb{E}[A_n]=0$. Since increment of Wiener process are independent, the variance of sum is sum of variance , and we have
$$\operatorname{Var}(A_n)=\sum_{i=0}^{n-1}\mathbb{E}[(\,W_1(t_{i+1})-W_1(t_{i})\,)^2]\,\,\mathbb{E}[(\,W_2(t_{i+1})-W_2(t_{i}))^2\,]$$
$$\operatorname{Var}(A_n)=\sum_{i=0}^{n-1}(t_{i+1}-t_i)^2 $$
$$\operatorname{Var}(A_n)\le\sum_{i=0}^{n-1}(t_{i+1}-t_i)\max\{t_{i+1}-t_i\}_{i=0}^{n-1}$$
$$\operatorname{Var}(A_n)\le\max\{t_{i+1}-t_i\}_{i=0}^{n-1}\sum_{i=0}^{n-1}(t_{i+1}-t_i)$$
$$\operatorname{Var}(A_n)\le\max\{t_{i+1}-t_i\}_{i=0}^{n-1}\,\,t$$
therefore $\operatorname{Var}(A_n)=\mathbb{E}[A_n^2]\to 0$ as $\delta_n=\max\{t_{i+1}-t_i\}_{i=0}^{n-1}\to 0$. This implies that $A_n\to 0$, in probability.
Now set
$$X_t=\int\limits_0^t f(W_3(s),s) \,dW_1(s) $$
in other words 
$$dX_t=f(W_3(t),t) \,dW_1(t)\quad,\quad X_0=0$$
and
$$Y_t=\int\limits_0^t g(W_3(s),s) \,dW_2(s) $$
in other words 
$$dY_t=g(W_3(t),t) \,dW_2(t)\quad,\quad Y_0=0$$
By application of Ito's lemma
$$d(X_tY_t)=Y_t dX_t+X_tdY_t+\underbrace{d[X_t,Y_t]}_{0}$$
as a result
$$X_tY_t=\underbrace{X_0Y_0}_{0}+\int_{0}^{t}Y_sdX_s+\int_{0}^{t}X_sdY_s$$
finally, we have
$$\mathbb{E}\left[X_tY_t\right]=\mathbb{E}\left[\int_{0}^{t}Y_sdX_s\right]+\mathbb{E}\left[\int_{0}^{t}X_sdY_s\right]=0$$
