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The integrand $\mathrm{sign}({\beta }_s)$ is a bounded previsible process, so $B$ is a continuous local martingale. The quadratic variation of a stochastic integral isbecause the Brownian zero set has zero Lebesgue measure. Since $B_0=0$, the Lévy characterization of Brownian motion shows that $B$ is a standard Brownian motion.

LetThe clock $A$ is an absolutely continuous function and is strictly increasing: its derivative is positive away from the Brownian zero set, which has zero Lebesgue measure. It also tends to infinity. This is immediate for $\delta =1$; for $\delta >1$, recurrence and the Strong Markov property imply that the Brownian occupation time of, for example, $\{x:1<|x|<2\}$ is unbounded, while the integrand is bounded below there by ${\delta }^2$.

Thus $A$ is continuous, strictly increasing, and maps $[0,\infty )$ onto itself. Its inverse function ${\tau }_s=A^{-1}(s)$ is finite, continuous, and strictly increasing.

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.