How can I show that $\int_0^T f(s)dW_s\sim \mathcal N\left(0, \int_0^T f(s)^2ds\right)$? Let $f$ a determinisitic functio in $L^2(0,T)$. How can I show that $$\int_0^T f(s)dW_s\sim \mathcal N\left(0,\int_0^T f(s)^2ds\right) \ \ ?$$
First of all, it's clear that $$\mathbb E\int_0^T f(s)dW_s=0$$
and 
$$\mathbb E\left(\int_0^T f(s)dW_s\right)^2=\int_0^T f(s)^2ds.$$
So, if $\int_0^T f(s)dW_s$ is normally distributed, then the claim follow. To prove it's normally distributed, I tried to use Itô formula with $g(x,t)=xf(t)$. This gives (as far as $f$ derivable)$$W_tf(t)=\int_0^t W_s f'(s)ds+\int_0^t f(s)dW_s,$$
but unfortunately, I can't conclude. Any idea ?
 A: 
Step 1: If the assertion holds for all continuous deterministic $f \in L^2(0,T)$, then it holds for all deterministic $f \in L^2(0,T)$.

Proof: Fix a deterministic function $f \in L^2(0,T)$. By the denseness of the continuous functions, there exists a sequence $(f_n)_{n \in \mathbb{N}}$ of continuous functions such that $f_n \to f$ in $L^2(0,T)$, i.e. $$\lim_{n \to \infty} \int_0^T |f_n(t)-f(t)|^2 \, dt = 0. \tag{1}$$ By Itô's isometry, this implies that the stochastic integral $X_n := \int_0^T f_n(s) \, dW_s$ converges in $L^2(\mathbb{P})$ to $X:=\int_0^T f(s) \, dW_s$. Since $f_n$ is continuous, we know that $X_n$ is Gaussian with mean $\mu_n =0$ and variance $\sigma_n^2 = \int_0^T f_n(s)^2 \, ds$. As $\lim_{n \to \infty} \mu_n=0$ and $\lim_{n \to \infty} \sigma_n^2 = \int_0^T f(s)^2 \, ds$, it follows that the limit $X=\lim_{n \to \infty} X_n$ is Gaussian with mean zero and variance $\int_0^T f(s)^2 \, ds$ (this can be shown e.g. by using that the characteristic function of $X_n$ converges pointwise to the characteristic function of $X$).

Step 2: Prove the assertion for continuous deterministic $f$.

For continuous deterministic functions $f$, the stochastic integral can be obtained as ($L^2$-)limit of Riemann sums: $$\int_0^T f(s) \, dW_s = \lim_{n \to \infty} \sum_{i=0}^{n-1} f \left( T\frac{i}{n} \right) (W_{T(i+1)/n}-W_{Ti/n}). \tag{2}$$ Using that the increments $W_{T(i+1)/n}-W_{Ti/n}$ are independent and distributed like $N(0,T/n)$, we find that
\begin{align*} \mathbb{E}\exp \left( i \xi \int_0^T f(s) \, dW_s \right) &= \lim_{n \to \infty} \prod_{i=0}^{n-1} \mathbb{E}\exp \left( i \xi f(Ti/n) (W_{T(i+1)/n}-W_{Ti/n}) \right) \\ &= \lim_{n \to \infty} \prod_{i=0}^{n-1} \exp \left( - \xi^2 \frac{f(Ti/n)^2}{2} \frac{T}{n} \right) \\ &= \lim_{n \to \infty} \exp \left(-\xi^2 \sum_{i=0}^{n-1} f(Ti/n)^2 \left[ \frac{T(i+1)}{n} - \frac{Ti}{n} \right] \right) \\ &= \exp \left( - \frac{\xi^2}{2} \int_0^T f(s)^2 \,d s \right). \end{align*}
This shows that $\int_0^T f(s) \, dW_s$ is Gaussian with mean zero and variance $\int_0^T f(s)^2 \, ds$.
