Ito integral property 
Let $X$ be adapted integrand, for which we know that $E[\int_0^T X_s^2 \, ds]< \infty $. Let $0≤s<t≤T$ and we have some $W∈F_s$ ($F_s$ is $\sigma$-algebra) and we know $E[\int_s^t W^2X_u^2 \, du]<\infty$.
I have to show  that $\int_s^t W X_u \, dB_u=W \int_s^t X_u dBu$.

I know that for simple integrands this follows from definition.
We also know that for $X$ there exists sequence of elementary integrands $X^n$ such that $E[ \int_0^t (X^n_u-X_u)^2 \, du]$ converges to $0$, when $n \to \infty$. From this one can see that it also holds $E[ (\int_0^t (X^n_u-X_u) \, dB_u)^2]$.
I wonder if I can show that it  holds $E[ |(\int_s^t W (X^n_u-X_u) \, dB_u)|] \to 0$ as $n \to \infty$ . (I think I should use Cauchy- Schwarz inequality: $E(XY)^2 \leq E(X^2)E(Y^2)$, but not sure where. If I could show this convergence in $L^1$, it would be enough for my proof.
If I rewrite it as $E[E[|(\int_s^t W(X^n_u-X_u) \, dB_u)|  \mid F_s]]$ I don't know if I can put $W$ out of integral. Would it be correct?
Any help would be really very appreciated. Thanks.
 A: Let us first assume that $W$ is bounded, i.e. that there exists $k \geq 0$ such that $\mathbb{P}(|W| \leq k)=1$. For given $X$ choose a sequence of approximating elemtary integrands $X^n$ such that $X_n \to X$ in $L^2(\mathbb{P} \otimes \lambda|_{[0,t]})$. By Itô's isometry and the boundedness of $W$, we have
$$\begin{align*} \mathbb{E} \left( \left| \int_s^t W X_u^n \, dB_u - \int_s^t W X_u \, dB_u \right|^2 \right) &= \mathbb{E} \left( \int_s^t W^2 (X_u^n-X_u)^2 \, du \right) \\ &\leq k^2 \mathbb{E} \left( \int_s^t (X_u-X_u^n)^2 \, du \right) \xrightarrow[]{n \to \infty} 0. \tag{1} \end{align*}$$
Similarly, we find that
$$\mathbb{E} \left( \left| W \int_s^t X_u \, dB_u - W \int_s^t X_u^n \, dB_u \right|^2 \right) \xrightarrow[]{n \to \infty} 0. \tag{2}$$
Since we know that the assertion holds for $X^n$, we get
$$\begin{align*} \int_s^t W X_u \, dB_u &\stackrel{(1)}{=} L^2-\lim_{n \to \infty} \int_s^t W X_u \, dB_u \\\ &= L^2-\lim_{n \to \infty} W \int_s^t X_u^n \, dB_u \\ &\stackrel{(2)}{=} W \int_s^t X_u \,dB_u. \end{align*}$$
Now we can consider the general case, i.e. that $W$ is not necessarily bounded. By the first part of the proof, we have
$$\int_s^t \min\{W,k\} X_u \, dB_u = \min\{W,k\} \int_s^t X_u dB_u. \tag{3}$$
By Itô's isometry,
$$\begin{align*} \mathbb{E} \left( \left| \int_s^t W X_u \, dB_u - \int_s^t \min\{W,k\} X_u \, dB_u \right|^2 \right) &= \mathbb{E} \left( \int_s^t (W-\min\{W,k\})^2 X_u^2 \, du \right) \\ &\leq 4 \mathbb{E} \left( \int_s^t W^2 1_{\{|W|>k\}} X_u^2 \, du \right). \end{align*}$$
Since $\mathbb{E}(\int_s^t W^2 X_u^2 \, du)<\infty$, it follows from the dominated convergence theorem that the right-hand side converges to $0$ as $k \to \infty$. A very similar reasoning shows that
$$\mathbb{E} \left( \left| W \int_s^t X_u \, dB_u - \min\{W,k\} \int_s^t X_u \, dB_u \right|^2 \right) \xrightarrow[]{k \to \infty} 0.$$
Letting $k \to \infty$ in $(3)$ we get
$$\int_s^t W X_u \, dB_u = W \int_s^t X_u \, dB_u.$$
