Absolute continuity of distributions

Question

Let $B=(B_t)_{t}$ be a standard Brownian motion and let $X=(B_{t/2})_t$. I want to disprove that $\mu_X\ll\mu_B$ where $\mu_\cdot$. denotes the respective distribution. The standard way might be to note first that $\langle X\rangle_t=t/2$ whilst $\langle B\rangle_t=t$ and thus for $A=\{f\in C[0,1]\;|\;\langle f\rangle_1=1/2\}$ $$ \mu_X(A)=1,\qquad\mu_B(A)=0. $$ Is there another way of proving this?

Answer 1

Let $$D=\Bigl\{f\in C[0,1]: \lim_{n\to\infty}\sum_{k=1}^{2^n}[f(\tfrac{k}{2^n})-f(\tfrac{k-1}{2^n})]^2=1\Bigr\}.\qquad A=C[0,1]\setminus D. $$ Then \begin{gather} \mu_X(D)=0,\qquad \mu_B(D)=1.\\ \mu_X(A)=1, \qquad \mu_B(A)=0. \end{gather}