 Solution

Monotonicity of $f$ gives, for each $n\ge 1$,Hence ${\sum }_{n\ge 1}f(2^{-n})<\infty$; the additional $n=0$ term is finite by continuity at one. Also $f(t)\to 0$ as $t\downarrow 0$, since a positive limiting lower bound would make the integral diverge.

By Brownian scaling, $S_{2^{-n-1}}$ has law $2^{-(n+1)/2}{S}_1$. Apply the small-value bound for the Brownian running maximum to obtainSumming yields the requested finite probability sum. The argument needs no independence of these nested-time events.