@Nate: I think your argument is fine. I didn't know the Ferinique's theorem. My argument on that part was this one.
Fix $T>0$ (as you did eventually $T+1$). Applying Jensen we get:
$$\exp\left(\frac{1}{2}\int_S^{S+\epsilon} X_t^2\,dt\right)=\exp\left(\frac{1}{\epsilon}\int_S^{S+\epsilon} \frac{\epsilon}{2}X_t^2\,dt\right)\leq \frac{1}{\epsilon}\int_S^{S+\epsilon} \exp\left(\frac{\epsilon}{2}X_t^2\right)\,dt \qquad a.s..$$
Now taking the expectation and using Tonelli's theorem we have to study:
$$\frac{1}{\epsilon}\int_S^{S+\epsilon} E\left[\exp\left(\frac{\epsilon}{2}X_t^2\right)\right]\,dt.$$
$X_t \sim N(\mu_t=X_0e^{-t}\,,\,\,\sigma_t^2=\frac{1-e^{-2t}}{2})$ so
$$E\left[\exp\left(\frac{\epsilon}{2}X_t^2\right)\right]=E\left[\exp\left(\frac{\epsilon}{2}(\mu_t+\sigma_tZ)^2\right)\right]=$$
$$=e^{\frac{\epsilon}{2}\mu_t^2}\int_{\mathbb{R}} \frac{1}{\sqrt{2 \pi}}\, \exp\left(-\frac{x^2}{2}(1-\sigma_t^2 \epsilon)+\epsilon \mu_t \sigma_t x\right)dx. $$
Setting $\lambda_t=1-\epsilon \sigma_t^2=\frac{1}{2}[2-\epsilon(1-e^{-2t})]$, if $\lambda_t>0$ (for example with $\epsilon < 1$) the last integral is convergent and its value is:$$\exp\left(\frac {\epsilon}{2}\mu_t^2\right)\,\exp\left(\frac{\epsilon^2}{2 \lambda_t}\mu_t^2 \sigma_t^2\right)\frac{1}{\sqrt{\lambda_t}}.$$
Finally all the functions involved are continuous and since $\epsilon < 1$, $\lambda_t$ is away from 0 and so the moment generating function is integrable.
My first idea was to use instead of $\epsilon$ $T$, but moment generating function of the chi squared is not define to away from 0 but only in a neighbourhood.
This is somehow lengthy, but I think you will better understand how Girsanov actually works. The theorem you stated are more an application of Girsanov. The motivation behind Girsanov is the following: You are interested in how semimartingales behave under a change of measure. Since the finite variation part do not change, the question reduces to how local martinagles behave under a change of measure. As I was taught, Girsanov answers this question:
Suppose you have $Q\approx P$ and assume for simplicity that the density process $Z$ is continuous. If you have a continuous local martingale $M$ null at zero (wrt $P$), i.e. $M\in \mathcal{M}_{0,loc}^c(P)$, then $$\bar{M}=M-\int\frac{1}{Z}d\langle Z, M\rangle = M-\langle L,M\rangle \in \mathcal{M}_{0,loc}^c(Q)$$
where we write $Z=Z_0\mathcal{E}(L)$.
This is what I would refer to Girsanov's theorem. Note that this implies that in particular $M$ is $Q$-Semimartingale. Of course there are generalizations of this theorem ($Z$ general etc.).
Both of your theorems are the same. It is a special case of Girsanov. Take $M=W$, where $W$ is $P$ Brownian Motion. As an application of Girsanov you get:
If $W$ is a $P$-Brownian Motion and $Q\approx P$ with density process of the form $Z=\mathcal{E}(\int \Theta_s dW_s)$, for a predictable process $\Theta$. Then $W$ is under $Q$ a Brownian Motion with drift, i.e.$$ W=\bar{W}+\int\Theta_s ds$$ for a $Q$-Brownian Motion $\bar{W}$.
This is an immediate consequence of Girsanov and the proof is straight forward using Lévy's characterization of Brownian Motion. However in most cases you have to go the other way around: Usually you do not have $Q\approx P$. This means you have a probability measure and want to construct an equivalent probability measure $Q$ such that the density process $Z$ is a stochastic exponential. Hence you start with $L\in\mathcal{M}_{0,loc}^c(P)$ and define $Z:=\mathcal{E}(L)$. You hope that $Z$ can be used to define an equivalent probability measure $Q$, as $\frac{dQ}{dP}=Z_\infty$. We have $\mathcal{E}(L)=Z$, hence $Z$ is a local martingale and strictly positive. Therefore it is a supermartingale on $[0,\infty)$ (use Fatou to prove that)! By the supermartingale convergence theorem $Z_t$ converges $P-a.s.$ to $Z_\infty$. The problem is, $Z_\infty$ can be $0$ or $E[Z_\infty]<1$ (or both together). As already mentioned you want do define $\frac{dQ}{dP}:=Z_\infty$. You want at least that this $Q$ is absolutely continuous w.r.t $P$, i.e. $Q\ll P$. Hence you need at least
- $Z_\infty >0$
- $E[Z_\infty]=1$.
A priori, as said before, $Z_\infty=0$ and/or $E[Z_\infty]<1$. Hence we must find some conditions, such that $1.$ and $2.$ are true. For $1.$ we must have $\langle L\rangle_\infty < \infty$ (by definition of $Z_\infty=e^{L_\infty -\frac{1}{2}\langle L \rangle_\infty}$). For $2.$ you can use: $E[Z_\infty]=1 $ if and only if $Z$ is a uniformly integrable $P$ martingale on $[0,\infty]$. Now there is a famous condtion, called Novikov's condition, which gives a sufficient condition of $Z=\mathcal{E}(L)$ to be a uniformly integrable martingale on $[0,\infty]$.
Looking at your question. The theorem from Wikipedia is exactly my second statement with $X:=\int\Theta_s dW$. Note that $[W,X]=\langle W,X\rangle = \langle W,\int \Theta_s dW \rangle = \int \Theta_s d\langle W,W\rangle = \int\Theta_s ds$. Furthermore $Z:=\mathcal{E}(X)$. The whole difference between the theorem from Wikipedia and Shreve is in specifying the process $X$ further. Shreve assumes that $X$ has a particular form, i.e. an integral w.r.t to a Brownian Motion. That is the only difference.
In finance you often work on finite time horizon, i.e. on $[0,T]$. You can easily extend everything to $[0,\infty)$ by setting everything equal zero outside $[0,T]$.
Best Answer
There is no difference with the case your drift is non-deterministic. Your calculation shows that $X$ is a $Q$-Brownian motion; since $Q$ and $P$ are equivalent and $Q(\{X_t > M\}) > 0$ it follows that $P(\{X_t > M\}) > 0$. Your argument for the second question is correct too, in that Girsanov's theorem only gives you a locally equivalent measure.