Compute $\mathbb{E}[\tilde{X}_t]$, where $\tilde{X}_t=X_t=(1-t)\int_0^t\frac{1}{1-s}dW_s$ for $0\le t<1$ and $\tilde{X}_t=0$ for $t=1$ I have the following exercise and I don't really understand the answer. I am going to write my professor's answer first, then a question about what I don't understand about my professor's answer and my attempt at the end. Any help would be greatly appreciated, thanks a lot in advance!
QUESTION: Compute $\mathbb{E}[\tilde{X}_t]$, where $\tilde{X}_t=X_t=(1-t)\int_0^t\frac{1}{1-s}dW_s$ for $0\le t<1$ and $\tilde{X}_t=0$ for $t=1$.
PROFESSOR'S ANSWER:
Note that
$$\int_0^t\left(\frac{1}{1-s}\right)^2 ds=\frac{1}{1-t}<\infty \quad\text{for all }t\in[0,1)$$
Thus,
$$\mathbb{E}\left[\int_0^t\frac{1}{1-s} dW_s\right]=0\implies \mathbb{E}[\tilde{X}_t]=0\quad\text{for all }t\in[0,1)$$
and since $\mathbb{E}[X_1]=0$, we have that $\mathbb{E}[\tilde{X}_t]=0$ for all $t\in[0,1]$.
MY QUESTION:
1. Why does the first integral imply that $\mathbb{E}\left[\int_0^t\frac{1}{1-s} dW_s\right]=0$?
MY ATTEMPT:
$$\mathbb{E}[\tilde{X}_t]=(1-t)\mathbb{E}\int_0^t\frac{1}{1-s}dW_s=(1-t)\mathbb{E}\frac{W_t}{1-t}=\mathbb{E}W_t=0\quad\text{for all }t\in\mathbb{R}$$
Hence, $\mathbb{E}[\tilde{X}_t]=0$ for $t\in[0,1].$
2. Would this be correct?
 A: *

*Your attempt: How do you conclude that $$\mathbb{E} \left( \int_0^t \frac{1}{1-s}  \, dW_s \right) = \mathbb{E} \left( \frac{W_t}{1-t} \right)$$ ...? To me it looks as you used something of the form $$\int_0^t f(s) \, dW_s = f(t) W_t,$$ but this identity is, in general, not correct.

*Your professor's answer: Your professor uses the following well-known theorem:

Let $f : [0,\infty) \times \Omega \to \mathbb{R}$ be a progressively-measurable function such that $$\mathbb{E} \left( \int_0^t f(s)^2 \, ds \right) <\infty$$ for all $t \geq 0$. Then the stochastic integral $$M_t := \int_0^t f(s) \, dW_s$$ is well-defined for all $t \geq 0$. Moreover, $(M_t)_{t \geq 0}$ is a martingale; in particular, $$\mathbb{E}(M_t) = \mathbb{E}(M_0)=0.$$

In your setting the function $f$ is determinstic, i.e. $f$ does not depend on $\omega$. Therefore, it suffices to check that $f$ is Borel-measurable (which it is) and  that the integrability condition $$\mathbb{E} \left( \int_0^t f(s)^2 \, ds \right)  = \int_0^t f(s)^2 \, ds < \infty$$ is satisfied. That's exactly what your professor did. 
