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 !