If $\tilde W_t=W_t-\int_0^t f(s)ds$ why $\mathbb P(\tilde W_t\in A)=\int_{\{W\in A\}}\exp(-\int_0^tf(s)dW_s-\frac{1}{2}\int_0^tf(s)^2ds)d\mathbb P$? Let $(\Omega ,\mathcal F,\mathbb P)$ a Brownian motion and $f:\mathbb R\to \mathbb R$ a continuous function. Set $$\tilde W_t=W_t-\int_0^t f(s)\,\mathrm d s.$$
It's well known (Girsanov) that $(\tilde W_t)$ is a Bronwian motion under the measure $$\mathbb Q(d\omega )=\exp\left\{\int_0^t f(s)dW_s-\frac{1}{2}\int_0^tf(s)^2ds\right\}\mathbb P(d\omega ).\tag{1}$$
Now, in my lecture, it's written that
$$\frac{d\mathbb P\tilde W_t^{-1}}{d\mathbb PW_t^{-1}}=\exp\left\{-\int_0^t f(s)dW_s-\frac{1}{2}\int_0^tf(s)^2ds\right\},\tag{2}$$
and thus 
$$\mathbb P\{\tilde W_t\in A\}=\int_\Omega \mathbb 1_{\{W\in A\}}\exp\left\{-\int_0^t f(s)dW_s-\frac{1}{2}\int_0^tf(s)^2ds\right\}\mathbb P(d\omega ).\tag{3}$$

Questions
1) How can I get (2) ? does it comes from (1) ?
2) From (2), for me 
$$\mathbb P\{\tilde W_t\in A\}=\int_A\exp\left\{\int_0^t f(s)dW_s-\frac{1}{2}\int_0^tf(s)^2ds\right\}\mathbb PW_t^{-1}(dx),$$
but this looks strange because the integrand in the RHS is defined on $\Omega $, not on $\mathbb R$. So, how can I get (2) ? does it follow from (1) ?
3) In fact, (3) says that $$\mathbb P\{\tilde W_t\in A\}=\int_{\{W_t\in A\}}\frac{d\mathbb PW_t^{-1}}{d\mathbb PW_t}\mathbb P(d\omega ).\tag{4}$$
I don't really understand how to get this formula (4) from (2). 
 A: For brevity, set
$$q_t := \exp \left( \int_0^t f(s) \,d W_s - \frac{1}{2} \int_0^t f(s)^2 \, ds \right),$$
i.e. $d\mathbb{Q} = q_t \, d\mathbb{P}$. Since
$$W_t = \tilde{W}_t + \int_0^t f(s),$$
we have
\begin{align*} q_t &= \exp \left( \int_0^t f(s) dW_s + \int_0^t f(s) (f(s) \, ds) - \frac{1}{2} \int_0^t f(s)^2 \,ds \right) \\ &= \exp \left( \int_0^t f(s) \, d\tilde{W}_s + \frac{1}{2} \int_0^t f(s)^2 \, ds \right). \tag{1} \end{align*}
From
$$\mathbb{P}(\tilde{W}_t \in A) = \int 1_{\{\tilde{W}_t \in A\}} \underbrace{d\mathbb{P}}_{=1/q_t \, d\mathbb{Q}} = \int 1_{\{\tilde{W}_t \in A\}} \frac{1}{q_t} \, d\mathbb{Q}$$
we see that
$$\mathbb{P}(\tilde{W}_t \in A) \stackrel{(1)}{=} \int 1_{\{\tilde{W}_t \in A\}} \exp \left( -\int_0^t f(s) \, d\tilde{W}_s - \frac{1}{2} \int_0^t f(s)^2 \, ds \right) \, d\mathbb{Q}.$$
By Girsanov' theorem, $(\tilde{W}_s)_{s \leq t}$ is a Brownian motion with respect to $\mathbb{Q}$, i.e. it has the same distribution as $(W_s)_{s \leq t}$ with respect to $\mathbb{P}$. Consequently,
$$\mathbb{P}(\tilde{W}_t \in A) = \int 1_{\{W_t \in A\}} \exp \left( - \int_0^t f(s) \, dW_s - \frac{1}{2} \int_0^t f(s)^2 \, ds \right) \, d\mathbb{P}.$$
