= Solution
Let $Z=\{t:B_t=0\}$. Since $A$ is the clock from part (b) and $\tau=A^{-1}$,
$$
\{s:X_s=0\}=A(Z).
$$
For $\delta>1$, the <absolutely continuous function> $A$ has derivative $A'(t)=\delta^2|B_t|^{2\delta-2}=0$ on $Z$. The one-dimensional area bound for an absolutely continuous function therefore gives
$$
\lambda(A(Z))\leq\int_ZA'(t)dt=0.
$$
For $\delta=1$, $A_t=t$ 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.
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