Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 202 / 4 / b / Solution

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.