 Solution

Let $Z=\{t:B_t=0\}$. Since $A$ is the clock from part (b) and $\tau =A^{-1}$,For $\delta >1$, the absolutely continuous function $A$ has derivative $A^{\prime }(t)={\delta }^2|B_t{|}^{2\delta -2}=0$ on $Z$. The one-dimensional area bound for an absolutely continuous function therefore givesFor $\delta =1$, $A_t=t$ and the conclusion follows directly because the Brownian zero set has zero Lebesgue measure. Hence the zero set of $X$ has zero Lebesgue measure almost surely.