Let $B(t)$ be a BM on $[0,1]$ and let $X(t)=B(t)+t$. Let $P$ be the distribution of $B(t)$ and let $Q$ be that of $X(t)$. Is $Q<<P$?
Edit:
We have not discussed Girsanov's theorem yet... we have only just started with basic properties of Brownian Motion. My idea was the following: A generating set $G$ of the sigma-algebra on $C([0,1])$ is given by the projections $(\pi_{t_{i}})_{t_{i}\in[0,1]}$. So if we find that for all $\epsilon>0$ there exists a $\delta>0$ such that $\forall A \in G$ with $P(A)\le \delta$ we have $Q(A)\le \epsilon$, then $Q<<P$ (is that true?).
If $P(B(t_{i}\le a_{i})) \le \delta$, then also $P(B(t_{i})+t_{i}\le a_{i})) \le \epsilon$.
Does this make sense, and if so, how can I formalise it?
Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and $(B_t)_{0\leq t \leq 1}$ be a BM on $(\Omega,\mathcal{F},\mathbb{P})$. Set $L_t = -B_t$ and $D_t = \exp(L_t - \frac{1}{2}t) = \exp(-B_t - \frac{1}{2}t)$. Then $D$ is a strictly positive martingale. Define a new probability measure on $(\Omega,\mathcal{F})$ by $d\mathbb{Q} = D_1 d \mathbb{P}$. By Girsanov's theorem, the process $B_t - \langle B,L\rangle_t = B_t + t = X_t$ is a BM under $\mathbb{Q}$. In other words, the law of $X$ under $\mathbb{Q}$ is the same as the law of $B$ under $\mathbb{P}$. Therefore, if $\Phi \colon C([0,1],\mathbb{R}) \to [0,\infty)$ is measurable, we have $$\int D_1 \Phi(X) \, d \mathbb{P} = \int \Phi(X) \, d \mathbb{Q} = \int \Phi(B) \, d \mathbb{P}.$$ Since $D_1 = \exp(-B_1 - \frac{1}{2})=\exp(-X_1+\frac{1}{2})$, this can be rewritten as $$\int \exp\left(-x(1)+\frac{1}{2}\right) \Phi(x) \, Q(dx) = \int \Phi(x)P(dx),$$ where $x \in C([0,1],\mathbb{R})$. This proves that $Q \ll P$: more precisely, $Q(dx) = \exp(x(1)-\frac{1}{2})P(dx)$.