How can I use the moment generating function to prove independence between $X_s$ and $X_t-X_s$ Consider a process $X_t$ with the dynamics $$dX_t=f(t)dW_t$$
where $W_t$ is Brownian motion and $X_0=0$. Define $\sigma^2(t)=\int_0^tf^2(u)du$
Clearly $X_s$ and $X_t-X_s$ for $t>s$ are independent (due to the properties of Brownian motion) and $X_t\sim (0,\sigma^2(t))$ 
My question is: Can I use the moment generating function to prove the independence between $X_s$ and $X_t-X_s$?
Hence computing $E[\exp( a_1X_s+a_2(X_t-X_s) )]$ and use the result to prove independence
 A: Let $\{f^n\}$ be a sequence of simple functions converging to $f$ and $X^n_t\triangleq \int_0^tf^n(s)dW_s$. Then for each $n$, $X^n_s$ and $X^n_t-X^n_s$ are clearly independent; and we get the independence of the limits $X_s$ and $X_t-X_s$ from this result. Proof of the last result, as you can see, is likely to rely on characteristic functions as you desire.
Or you can repeat a step in Levy's characterization of BM (e.g., Karatzas and Shreve, Theorem 3.3.16). It suffices to show
$$
E(e^{iu(X_t-X_s)}|\mathcal F_s)=e^{-\frac{1}{2}u^2\int_s^tf^2(\tau)d\tau}
$$
which is equivalent to the statement that for all $A\in\mathcal F_s$,
$$
E(1_Ae^{iu(X_t-X_s)})=P(A)e^{-\frac{1}{2}u^2\int_s^tf^2(\tau)d\tau}.
$$
Now, apply Ito's lemma to $x\mapsto e^{iux}$ to obtain
$$
e^{iuX_t}=e^{iuX_s}+iu\int_s^te^{iuX_\tau}f(\tau)dW_\tau-\frac{1}{2}u^2\int_s^te^{iuX_\tau}f^2(\tau)d\tau.
$$
Under a mild regularity condition on $f$ (e.g., $\int_0^tf^2(\tau)d\tau<\infty$ for all $t$), the stochastic integral is a martingale, so
$$
E(e^{iuX_t}|\mathcal F_s)=e^{iuX_s}-\frac{1}{2}u^2\int_s^tE(e^{iuX_\tau}|\mathcal F_s)f^2(\tau)d\tau.
$$
Multiply $1_Ae^{-iuX_s}$ and take expectations:
$$
E(1_Ae^{iu(X_t-X_s)})
=P(A)-\frac{1}{2}u^2\int_s^tE(1_Ae^{iu(X_\tau-X_s)})f^2(\tau)d\tau.
$$
Let $z(t)=E(1_Ae^{iu(X_t-X_s)})$. We then have
$$
z(t)=P(A)-\frac{1}{2}u^2\int_s^tz(\tau)f^2(\tau)d\tau,\quad t\ge s,
$$
whose unique solution is
$$
z(t)=P(A)e^{-\frac{1}{2}u^2\int_s^tf(\tau)^2d\tau}.
$$
