 Solution

Put $R_{\epsilon }={\int }_0^{\epsilon }(H_u-H_0)d{B}_u$ and $d_{\epsilon }={\epsilon }^{-1}{\int }_0^{\epsilon }\mathbb{E}|H_u-H_0{|}^{2}du$. Boundedness and path continuity of $H$ make $\mathbb{E}|H_u-H_0{|}^2\to 0$ by the dominated convergence theorem, so $d_{\epsilon }\to 0$. The Itô isometry gives $\mathbb{E}{R}_{\epsilon }^2=\epsilon d_{\epsilon }$.

We must not assume independence between the integral and its Brownian denominator. Instead, the Cauchy-Schwarz inequality givesConcavity, or Jensen inequality, bounds the first factor by $(\mathbb{E}{R}_{\epsilon }^2{)}^{1/8}$. Since $B_{\epsilon }\stackrel{d}{=}\sqrt{\epsilon }N$, the second factor is ${\epsilon }^{-1/8}(\mathbb{E}|N{|}^{-1/2}{)}^{1/2}$. This negative normal moment is finite: the density is bounded near zero, where ${\int }_0^1{u}^{-1/2}du<\infty$, and has Gaussian decay at infinity. In fact $\mathbb{E}|N{|}^{-1/2}=2^{-1/4}\Gamma (1/4)/\sqrt{\pi }$. The powers of $\epsilon$ cancel, yielding the fractional-moment control of Brownian increment ratiosThe ratio may be assigned any value on the null event $B_{\epsilon }=0$.