I feel like this might be really simple, but I have tried for some time and not been able to make any headway. If $f \in L^{2}( [0, \infty))$ and $\{B_{t},\{\mathcal{F}_{t}\}, t \in [0,\infty)\}$ is the Brownian motion, the Wiener integral $\int_{0}^{t}f(s)dB_{s}$ is defined as the $L^{2}$-limit of the Riemann sums $\sum_{i=0}^{n}f(t_{i})(B_{t_{i+1}}-B_{t_{i}})$ over a partition $\pi_{n}$ of $[0,t]$, as the mesh size goes to zero. Since the Riemann sums themselves are mean zero Normal random variables, so is $\int_{0}^{t}f(s)dB_{s}$ with variance $\int_{0}^{t}f(s)^{2}ds$.
I was trying to derive the same as a special case of Ito's integral by considering $f(t):\equiv f(t)1_{\Omega}(\omega)$. This function is clearly measurable with respect to $\mathcal{B}([0,\infty))\times \mathcal{F}$. Since $E\int_{0}^{\infty}f(s)^{2}ds < \infty$ and the quadratic variation of $B_{t}$ is $t$, which is absolutely continuous, the integral $I_{t}(f)=\int_{0}^{t}f(s)dB_{s}$ is defined. Since this is a martingale and $I_{0}=0$ a.s, $E(I_{t}(f))=0$ and by the Ito isometry, the variance is $\int_{0}^{t}f(s)^{2}ds$. But how do I show the distribution of $I_{t}(f)$ is Normal? Any help would be appreciated.