How to show that the Wiener measure is singular with respect to a normal law

Question

We have a Gaussian process $X$, $X_t:=B_t - tB_1$, where $B$ is a $BM$, $t\in[0,1]$.
Let $\nu$ be the law of $X$ and $\mu$ the Wiener measure.

How can I show that $\mu$ is singular with respect to $\nu$?

I know that $X_t$ is distributed $N(0, t^2-t)$ but I don't know if this helps in any way. I presume I have to find a set so that $\nu$ on it is 0 and $\mu$ not, but I do not know how to continue.

Answer 1

$\nu(\{\omega:X_1(\omega)=0\})=1$ but $\mu(\{\omega:B_1(\omega)=0\})=0$.