Showing that Ito's integral of deterministic function follows a normal distribution Let $f(t, \omega)$ be a stochastic process on $0 \leq t \leq T$ where $\omega$ indicates that $f$ depends on a Brownian motion. Assume $f(t, \omega)$ is independent of the increments of the Brownian motion in the future (i.e., $f(t, \omega)$ is independent of $w(t + s, \omega) - w(t, \omega)$ for all $s > 0$). Furthermore, suppose
$$\int_0^T E[f^2(s, \omega)] ds < \infty$$
I would like to show the following three properties:
1)
$$\int_{0}^{t} f(s) dw(s, \omega) \sim N\left(0, \int_0^t f^2(s) ds\right)$$


*For any $0 \leq \tau \leq t \leq T$,

$$ E\left[\int_\tau^t f(s, \omega) dw(s,\omega)\right] = 0$$

I found lots of resources for the first property online, but I cannot find any definite proofs about them. Here is one post on quant.SE that proves properties of the covariance: https://quant.stackexchange.com/questions/53212/integration-of-a-deterministic-function-w-r-t-a-brownian-motion
However, I'm not entirely sure how to show that it follows a normal distribution with varince equal to the provided quantity. I think that I need to utilize the assumption that the integral of the expectation of the square of the stochastic process is finite.
Any help is appreciated. This is an exercise from my book, and I've been stuck on it for a couple of days.
 A: Let $t_{n,k}=\frac{k}{2^n}t$ and note that
$$
\int_0^t f(s)\textrm{d}W_s=\lim_{n\to\infty} \sum_{k=0}^{2^n-1} f(t_{n,k})(W_{t_{n,k+1}}-W_{t_{n,k}}),
$$
where the limit is taken in probability. Now, the increments of Brownian motion are independent and Gaussian, so we have
$$
 \sum_{k=0}^{2^n-1} f(t_{n,k})(W_{t_{n,k+1}}-W_{t_{n,k}})\sim N\left(0, \sum_{k=0}^{2^n-1} Var(f(t_{n,k}) (W_{t_{n,k+1}}-W_{t_{n,k}}))\right)=N\left(0,\sum_{k=0}^{2^n-1} f(t_{n,k})^2 2^{-n}t\right)
$$
where we've used that $f$ is deterministic and the variance of the Brownian increments.
Now, $\lim_{n\to\infty}\sum_{k=0}^{2^n-1} f(t_{n,k})^2 2^{-n}t=\int_0^t (f(s))^2\textrm{d}s$ and it's a classic result that Gaussian variables converge weakly if and only if their means and variances converge and their limit is the corresponding (possibly degenerate) Gaussian variable. If you don't know this, it can be proven by considering characteristic functions.
Since convergence in probablity implies weak convergence, this establishes the desired.
