Girsanov Theorem and Different Probability Measures Can it be that $W(t)$ induces $\mathbb{P}$ (and therefore $\mathbb{P}$ is defined by the density of $W(t)$): and then, you have $X(t):=W(t)−rt$ (and therefore $X(t)$ has a different density) under the same measure $\mathbb{P}$ ? Wouldn't $X(t)$ define its own measure via its own density?
Specifically, we can define $\mathbb{P}$ for any $A:a \epsilon \mathbb{R}$ as follows: $\mathbb{P}(A):=\int^{a}_{-\infty}f_{W_t}(h)dh$
We could then define $\mathbb{Q}$ for any $A:a \epsilon \mathbb{R}$ as follows: $\mathbb{Q}(A):=\int^{a}_{-\infty}f_{X_t}(h)dh$.
In the above, $f_{W_t}()$ is Normal density with variance $t$, whilst  $f_{X_t}()$ is Normal density with variance $t$ and mean $-rt$. 
The R-N Derivative to get from $\mathbb{P}$ to $\mathbb{Q}$ would then be $\frac{\mathrm{d}\mathbb{Q}}{\mathrm{d}\mathbb{P}}=\exp(-\int_0^t rdW_h -\frac{1}{2}\int_0^tr^2 dh)$
My confusion arises from the fact that one could also write: $\mathbb{P}(B):=\mathbb{P}(X(t)\leq b)=\mathbb{P}(W(t)-r\leq b)=\mathbb{P}(W(t)\leq b+r)=\int^{b-r}_{-\infty}f_{W_t}(h)dh$. 
So that makes me think that we can discuss probabilistic events related to $X(t)$ and $W(t)$ under the same measure $\mathbb{P}$. However the fact that we can define a measure via a process-specific density then makes me think that it doesn't make sense to have two different processes with different densities under the same probability measure.
 A: There is a big confusion about notation and naming in your post....
For clarification let's start by define the probability space $(\Omega,\mathcal{F},\mathbb{S})$ where all of your random variables live on.
Then you assume that $W$ induces a probability measure $\mathbb{P}$ which means nothing else that $$\mathbb{P}(A) := \mathbb{S}(W \in A)$$
This definition is always well defined and now, if there exists a measurable function $f_W \ge 0$ s.t. $$\mathbb{P}(A) = \int_A f_W(h) dh$$ then we call $f_W$ the densitiy  of $W$.
Consider that your "definition" $$\mathbb{P}(A):=\int^{a}_{-\infty}f_{W}(h)dh$$ only is valid for sets $A = (-\infty,a)$ and not in general.
Luckily on $\mathcal{B}(\mathbb{R})$  is enough to show/have this identity for all $a \in \mathbb{R}$ to get it for all $A \in \mathcal{B}(\mathbb{R})$ (why?) and  then we can shortly write $\mathbb{S}(W \le a)$ for $\mathbb{S}(W \in A)$.
Additionally we have then $$f_W(h) = \frac{d}{da} \mathbb{S}(W \le a)$$
Now you consider a second random variable defined by $$X = W - c$$ and a related density $f_X$. So we get a new measure $$\mathbb{Q}(A) := \mathbb{S}(X \in A) = \int_A f_X(h) dh$$
Then again for all sets of the form $A = (-\infty,a)$ we have $$\mathbb{Q}(A) = \mathbb{S}(X \le a) = \int_{-\infty}^a f_X(h) dh$$
And is holds:
$$\begin{align} \int_{-\infty}^a f_X(h) dh &= \mathbb{Q}((-\infty,a))  &&= \mathbb{S}(X \le a)\\ &= \mathbb{S}(W-c \le a) &&= \mathbb{S}(W \le a+c) \\ &= \mathbb{P}((-\infty,a+c))  &&= \int_{-\infty}^{a+c} f_W(h) dh \end{align} $$
Additionally we know that the density functions above are nothing else then the R-N-density  related to the Lebesgue measure $\lambda$, so we get:
$$\frac{d\mathbb{Q}}{d\mathbb{P}} = \frac{\frac{d\mathbb{Q}}{d\lambda}}{\frac{d\mathbb{P}}{d\lambda}} = \frac{f_W}{f_X}$$
All the arguments above hold in general so if you deal with certain special densities just plug in and calculate…
