Intuition Behind this Theorem About Brownian Motion I am having a hard time with the intuition behind some of the representation theorems dealing with Brownian Motion. I think if someone can simply explain the intuition behind this theorem then everything will fall into place:
Theoreom:
For any $0\le a <b$ and any finite random variable $X\in\mathscr{F}_a$, there is a stopping time $\tau$ with $a\le\tau<b$ such that 
\begin{align*}X=\int_a^{\tau}\frac{1}{b-t}dB_t\end{align*}
That is the theorem. One thing I am having a massive problem with is understanding how this can be true since the right side of the above equality is a Gaussian process, but I don't see why $X$ needs to be a Gaussian Process. Can anyone explain this?

For any deterministic $T>0$ the random variable
$$Y_T := \int_a^T \frac{1}{b-t} \, dB_t$$
is Gaussian, but for a stopping time $\tau$ the integral
$$Y_{\tau} = \int_a^{\tau} \frac{1}{b-t} \, dB_t$$
does not need to be Gaussian. Just consider Brownian motion itself: For each fixed $T>0$ we know that $B_T$ is Gaussian, but this does not imply that $B_{\tau}$ is Gaussian for any stopping time $\tau$ (consider e.g. $\tau := \inf\{t>0; B_t = 1\}$, then $B_{\tau}=1$ almost surely).
