Q:

Expectation of Exponential of Stochastic Integral

Let $z$ be the standard Brownian motion, $\omega$ an element of the sample space.  Is it true that
$$
\mathbf E\bigg[\exp\Big(\int_0^t f(\omega,s)\,\mathrm dz(s)\Big)\bigg] = \mathbf E\bigg[\exp\Big(\frac{1}{2}\int_0^t f(\omega,s)^2 \,\mathrm ds\Big)\bigg]
$$
I can prove it is true for $f$ depending not on $\omega$ but only on $t$ by looking at the Riemann sum of the integral and taking conditional expectations.  However, when $f$ depends on $\omega$, how should one treat the expectations at successive time points?  

Edit:
The above is incorrect.  It should be rather:
$$
\mathbf E\bigg[\exp\Big(\int_0^t f(\omega,s)\,\mathrm dz(s)-\frac{1}{2}\int_0^t f(\omega,s)^2 \,\mathrm ds\Big)\bigg] = 1
$$
I can prove this with telescoping sum, Taylor expansion and $(dz)^2 = dt$. The following proof is not completely rigorous.  We shall improve on that later. 
Specifically, let
$$g(t) = \exp\Big(\int_0^t f(\omega,s)\,\mathrm dz(s)-\frac{1}{2}\int_0^t f(\omega,s)^2 \,\mathrm ds\Big)-1.$$ 
$$g(t) = \sum_{i=0}^n \big(g(t_{i+1})-g(t_i)\big)= \sum_{i=0}^n g(t_i)\Big(\frac{g(t_{i+1})}{g(t_i)}-1\Big)$$
\begin{align}
\frac{g(t_{i+1})}{g(t_i)}-1 &= \int_{t_i}^{t_{i+1}} f(\omega,s)\,\mathrm dz(s)-\frac{1}{2}\int_{t_i}^{t_{i+1}} f(\omega,s)^2 \,\mathrm ds+\frac{1}{2}\Big(\int_{t_i}^{t_{i+1}} f(\omega,s)\,\mathrm dz(s)\Big)^2+O\big(\|t_{i+1}-t_i\|^2\big) \\
&= f(\omega,t_i)(z(t_{i+1})-z(t_i))-\frac{1}{2} f(\omega,t_i)^2(t_{i+1}-t_i)+\frac{1}{2}\big(f(\omega,t_i)(z(t_{i+1})-z(t_i))\big)^2+O\big(\|t_{i+1}-t_i\|^2\big) \\
&= f(\omega,t_i)(z(t_{i+1})-z(t_i))+O\big(\|t_{i+1}-t_i\|^2\big)
\end{align}
where the last line results from 
$$(z(t_{i+1})-z(t_i))^2 = t_{i+1}-t_i+O\big(\|t_{i+1}-t_i\|^2\big).$$
Taking the limit of $n\rightarrow\infty$ such that $\max\limits_i\|t_{i+1}-t_i\|\rightarrow 0$,
$$g(t) = \int_0^t g(s)f(\omega,s)dz(s).$$
Therefore
$$\mathbf E[g(t)]=0.$$

In fact the proposition is a simple application of the Ito's Lemma. 

A:

$$X_t = exp(\int_0^t {f(\omega,s)dB_s}-\frac 12 \int_0^t{f(\omega,s)^2 ds})$$ and  $Y_t = \ln(X_t)$
then $$Y_t = \int_0^t {f(\omega,s)dB_s}-\frac 12 \int_0^t{f(\omega,s)^2 ds}$$ then we have $dY_t=f(\omega,t)dB_t - \frac 12 f^2(\omega,t)dt$ .  so by Ito formula we have 
$$dX_t = d(e^{Y_t}) = e^{Y_t}dY_t + \frac 12 e^{Y_t}(dY_t)^2=X_t f(\omega,t)dB_t - \frac 12 X_t f^2(\omega,t)dt + \frac 12 X_t f^2(\omega,t)dt =  X_t f(\omega,t)dB_t $$
so we have 
$$X_t = X_0 + \int_0^t {X_s f(\omega,s)dB_s}$$
since ito integral is martingale and its expectation is zero, we have $$ E[X_t]=E[X_0]=1 $$

