 Solution

Put $A_t=[M{]}_t$. Its continuous, strictly increasing paths start at zero and tend to infinity. Thus the inverse $T_s$ is finite and continuous, andso each $T_s$ is a stopping time. Set ${\mathcal{G}}_s={\mathcal{F}}_{T_s}$ and ${\beta }_s=M_{T_s}$. This is the inverse-clock construction in the Dambis-Dubins-Schwarz theorem.

To verify its martingale property, stop $M$ at times ${\tau }_n$ bounding its path, its bracket and time itself. The stopped $M$ and $M^2-A$ are bounded martingales. Optional sampling at $T_s\wedge {\tau }_n$ and $T_t\wedge {\tau }_n$ shows that the time-changed process, stopped at $A_{{\tau }_n}$, is a martingale and that its square minus the stopped clock $s\wedge A_{{\tau }_n}$ is a martingale. These clock stopping times tend to infinity because $A_{\infty }=\infty$. Therefore $\beta$ is a continuous local martingale with ${\beta }_0=0$ and quadratic variation $[\beta {]}_s=s$. The Lévy characterization of Brownian motion identifies it as a ${\mathcal{G}}_s$-Brownian motion.

Strict monotonicity and continuity give $A_{T_s}=s$ and $T_{A_t}=t$, soThe inverse-clock argument proves the asserted representation; the random clock need not be independent of the Brownian motion it produces.