I'm working on Drifted Brownian Motion at the moment and I'm having some difficulty with understanding Girsanov's Theorem (from Shreve):
Let $W(t) \ (0 \leq t \leq T)$ be a BM on probability space $(\Omega, F, \Pr)$. Let $\Theta(t) \ (0 \leq t \leq T)$, be an adapted process. Define
$$Z(t) = \exp \left( -\int_0^t\Theta(u)\mathrm{d}W(u) - \frac{1}{2}\int_0^t \Theta^2(u)\mathrm{d}u \right),$$
$$\widetilde{W}(t) = W(t) + \int_0^t\Theta(u)\mathrm{d}u,$$
and assume that $\displaystyle\mathbb{E}\left[\int_0^T \Theta^2(u)Z^2(u)\mathrm{d}u\right] < \infty$. Set $Z=Z(T)$. Then $\mathbb{E}[Z]=1$ and >under probability measure $\widetilde P$ given by
$$\widetilde \Pr(A) = \int_A Z(\omega)\mathrm{d}\Pr(\omega)\qquad\forall A \in F$$
the process $\widetilde W(t) \ (0 \leq t \leq T)$ is a Brownian Motion.
Say I want to apply this thorem to $W(t) + \mu t$, then, $Z(t) = e^{-\mu W(t) - \frac{1}{2}\mu^2}$. Would you then have
$$\widetilde P(A) = e^{-\mu W(T) - \frac{T}{2}\mu^2}\int_A\mathrm{d}\Pr(\omega)\qquad\forall A \in F$$
as a function for the probability under $\widetilde \Pr$? I dont really understand what you would need to do with the $\omega$ in $Z(\omega)$.
The expression $$\widetilde{P}(A) = e^{-\mu W(T)-\frac{T}{2}\mu^2} \int_A dP(\omega)$$ doesn't make sense because $\widetilde{P}(A)$ is just a number (the probability of something), but $W(T)$ is random.
For the question about what to do with $\omega$, remember that $\omega$ is what keeps track of the randomness in the random variables. When we write $W(T)$, we really mean $W(T,\omega)$. So the correct expression would be $$\widetilde{P}(A) = \int_A e^{-\mu W(T)-\frac{T}{2}\mu^2} dP(\omega) = \int_A e^{-\mu W(T,\omega)-\frac{T}{2}\mu^2} dP(\omega).$$