# Girsanov's theorem and absolutely continuous restrictions

Let $W$ be a Brownian motion on some probability space $(\Omega, \mathcal{F}, P)$. Let $\mathbb{F}^W$ be the filtration generated by $W$ and let $X$ be a process that is progressively measurable w.r.t. $\mathbb{F}^W$ such that$\int_0^T X^2_s ds < 1$ a.s. for all $T > 0$. Let $\mu = X \cdot W$, and assume that $Z = \mathcal{E}(\mu)$ (Doléans exponential) is a martingale. Then by a theorem I know that there exists a probability measure $Q$ on $\mathcal{F}_\infty^W$ such that $W^Q = W 􀀀- \int^\cdot_0 X_s ds$ is a Brownian Motion on $(\Omega, \mathcal{F}_\infty^W,Q)$. Similarly there exists a restriction $Q_T$ such that $W^Q$ is a Brownian Motion on $[0,T]$ under $Q_T$ on the filtration $\{\mathcal{F}_t \mid t\in [0,T]\}$.

I know it is tempting to think that the restriction of $Q$ to the $\sigma$-algebra $\mathcal{F}_T$ coincides with $Q_T$. But this is in general not true. I now want to prove this fact hence I guess what I have to prove it that we can find a set $F$ in $\mathcal{F}_T$ such that $P(F) = 0$ and hence $Q_T (F) = 0$ but $Q(F) > 0$(!).

I now considered the density process $Z=\mathcal{E}(W)$ and looked at the set $$F=\left\{ \lim_{t\to\infty} \frac{W_t}{t}=1\right\},$$ (is it $\mathcal{F}_T$-measurable?) but now how could I prove that $P(F)=0,Q_T(F)=0,Q(F)>0$ ? Some properties of a standard BM I know that $P(F)=0$ as under $P$ we have that $\lim_{t\to\infty} W_t/t = 0$ a.s.. But why are the probabilities under $Q_T$ and $Q$ of $F$ different? Thanks for any assistance!

As we consider $Z= \mathcal{E}(W)$ we have that $$Z_t = \exp \left( W_t-\frac12 \langle W \rangle_t\right) = \exp\left( W_t - \frac12 t \right).$$ $Z_t$ is the so called exponential Brownian martingale with $$\mathbb{E} Z_t = \mathbb{E} \left[ \exp \left( W_t - \frac12 t \right) \right] = \mathbb{E} \left[ \exp \left( W_0 - \frac12 \cdot 0 \right) \right] = 1,$$ by noting that $t$ is a deterministic stopping time and applying optional sampling. Hence $Z_t$ is a martingale under $P$.
Now applying the the statements above we see that $$W^Q = W - \int_0^\cdot X_s ds\quad \text{is a BM on }(\Omega,\mathcal{F}_\infty^W,Q),$$ where in our case $X=1$ constant. In other words we have that $$W^Q = W - t \quad \text{is a BM on }(\Omega,\mathcal{F}_\infty^W,Q).$$
By the law of large numbers for Brownian motion (I omit the proof as it is straightforward) we then have $$\lim_{t \to \infty} \frac{W_t}{t} = 0 \quad \text{P-a.s.,}$$ hence $P(F)=0$ and as $F$ is a null-event in $\mathcal{F}_T$ for all $0 \leq T < \infty$ we have that $Q_T(F)=0$. (is this true?)
But \begin{align*} Q(F) &= Q\left( \lim_{t \to \infty} \frac{W_t}{t} = 1 \right) = Q\left( \lim_{t \to \infty} W_t -t =0 \right) \\ &= Q\left( \lim_{t \to \infty} W^Q_t =0 \right) =1 (>0). \end{align*} By noting that $W^Q$ is a BM on $(\Omega,\mathcal{F}_\infty^W,Q)$ hence the LLN for Brownian motion applies here for $Q$-a.s.. We can now conclude that $Q$ and $P$ are mutually singular.