 Solution

For $t>1$, $B_{t^2}$ is not measurable with respect to the original Brownian filtration ${\mathcal{F}}_t$: its conditional variance is $t^2-t>0$. Thus it is not even adapted on the full half-line. It also fails the martingale condition before time one. With $s=1/4$, $t=3/4$,The appropriate filtration isBrownian independent increments make $M_t=B_{t^2}$ a continuous ${\mathcal{G}}_t$-martingale, with quadratic variation $[M{]}_t=t^2$. Define the stochastic integralThe value of the integrand at zero is immaterial; it is square-integrable against the bracket because ${\int }_0^t(2s{)}^{-1}d(s^2)=t$. Consequently $[W{]}_t=t$, and the Lévy characterization of Brownian motion makes $W$ a ${\mathcal{G}}_t$-Brownian motion. Associativity of stochastic integrals gives the Brownian motion under a quadratic time change representationThe function $f$ is continuous at zero. One can first verify the inverse-integrand identity on $[\epsilon ,t]$ and then let $\epsilon \downarrow 0$ using the Itô isometry, since $\mathbb{E}{M}_{\epsilon }^2={\epsilon }^2$.