The assertion is false for a general continuous $f$. For example, take $f(t)=t^{1/3}$. The <Brownian Hölder regularity> gives $B_t/t^{1/3}\to0$ almost surely as $t\downarrow0$, while
$$
\frac{B_t+f(t)}{t^{1/3}}\longrightarrow1
$$
almost surely. The original and translated path laws therefore concentrate on disjoint measurable events and are <mutually singular measures>.

More generally, the <Cameron-Martin theorem> gives the exact criterion: translation by $f$ is absolutely continuous precisely when $f$ is an <absolutely continuous function>, $f(0)=0$, and $f'\in L^2[0,1]$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 202 / 5 / c / iii / Solution

{c}

On every compact time interval, a Brownian path is almost surely a <Hölder continuous function> of every exponent $\alpha<1/2$, and of no exponent $\alpha\geq1/2$. In particular, $B_t=o(t^\gamma)$ as $t\downarrow0$ for every $\gamma<1/2$ after choosing a Hölder exponent strictly between $\gamma$ and $1/2$.

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 Incoming links: Brownian Hölder regularity