Cameron-Martin theorem for $C([0,\infty))$.

Question

Let B be a $1$-dimensional Brownian motion. I want to show that $B_t$ and $B_t + t$ are mutually singular on $C([0,\infty))$. I know that for $T > 0$ and \begin{equation} D[0,1] = \Bigl \lbrace F \in C[0,T] \hspace{1mm} \big| \hspace{1mm} \exists f \in L^2[0,T] : F(t) = \int^{t}_{0}{f(s) ds}, \forall t \in [0,T] \Bigr\rbrace \end{equation} the following statement holds:
Let $F \in C[0,T]$ satisfy $F(0) = 0$. If $F \not \in D[0,T]$ then $\mathbb{L}_F \perp \mathbb{L}_0$, where $\mathbb{L}_F$ or $\mathbb{L}_0$ are the law of $B + F$ or $B$ respectively.
I also know the proof of this statement. I tried to modify this proof for taking the limit $T \to \infty$ but I wasn't successful. I think the reason why $\mathbb{L}_{t} \perp \mathbb{L}_0$ is because the range of the domain isn't finite anymore, so the integral in our $D$ diverges if $F(t) = t$ and $f(t) = 1$, in other words $F(t) = t$ isn't absolutely continuous in $[0,\infty)$.

Answer 1

We prove

I want to show that $B_t$ and $B_t + t$ are mutually singular on $C([0,\infty))$.

We follow the Gubinelli notes, example 6. Using the process

$$Z_{t}\exp(B_{t}-\frac{1}{2}t),$$

we define the measure $\frac{dQ}{dP}|_{\mathcal{F}_{t}}=Z_{t}$. Under this $Q$-measure, we have that $B_{t}=W_{t}+t$ where $W_{t}$ is a standard Brownian motion. The measures $P$ and $Q$ are mutually singular because the event

$$A:=\left\lbrace \lim_{t\to +\infty}\frac{W_{t}+t}{t}=1\right\rbrace\in \mathcal{F}_{+\infty},$$

has full $Q$-measure $Q(A)=1$ (due to $W_{t}/t\to 0$ in $Q$) but has zero measure in $P$ because $B_{t}/t\to 0$ in $P$.