TITLE: Laws of $(B_t)_{t\in [0,T]}$ and $(2B_t)_{t\in [0,T]}$ : singular?
QUESTION [6 upvotes]: Are the laws of  $(B_t)_{t\in [0,T]}$ and $(2B_t)_{t\in [0,T]}$ mutually singular ?
More precisely, I know that the laws of two diffusion processes are mutually absolutely continuous if they share a common diffusion coefficient (application of Girsanov theorem). 
However, I cannot understand why it is no more the case when they have different diffusion coefficients. This is why I ask the question about $(B_t)_{t\in [0,T]}$ and $(2B_t)_{t\in [0,T]}$. If their laws are not mutually continuous, that would mean that one can find a measurable set of trajectories $\mathcal{A}$ for which $\mathbb{P}[(B_t)_{t\in [0,T]} \in \mathcal{A}]>0$ while $\mathbb{P}[(2B_t)_{t\in [0,T]} \in \mathcal{A}]=0$. The problem is : I don't know how to find such a set $\mathcal{A}$. Can anyone help me to understand this point ?
Thank you folks !

REPLY [3 votes]: You can also use the law of the iterated logarithm at $t=0$. For the set 
$$A:=\left\{\omega: \limsup_{t\to0} {|\omega_t|\over\sqrt{2t\log\log(1/t)}}=1\right\}$$
we have $\mu_B(A) = 1$, but $\mu_{2B}(A) = 0$. Here $\mu_B$ and $\mu_{2B}$ are the laws of Brownian motion and twice Brownian motion respectively.