Let $B$ be a standard Brownian motion in one dimension and let $H$ be a continuous, adapted, bounded process. Prove that $$\frac{\int_t^{t+h}H_sdB_s}{B_{t+h}-B_t}\to H_t$$ in probability as $h\downarrow0$. The hints are to estimate $\mathbb{E}(|B_{t+h}-B_t|^{-1/2})$ and show that $$\mathbb{E}\left(\left|\int_t^{t+h}(H_s-H_t)dB_s\right|^{1/2}\right)\leq\mathbb{E}\left(\int_t^{t+h}(H_s-H_t)^2ds\right)^{1/4}.$$
My attempt so far: Firstly, in attempting to deduce how to use the hints, I am guessing that we want to use the Markov inequality (using the map $x\to x^{1/4}$), followed by the Cauchy-Schwarz inequality, leading to (for fixed $\epsilon>0$) $$\mathbb{P}\left(\left|\frac{\int_t^{t+h}H_sdB_s}{B_{t+h}-B_t}-H_t\right|\geq\epsilon\right)\leq\frac{1}{\epsilon^{1/4}}\left\{\mathbb{E}\left(\left|\int_t^{t+h}(H_s-H_t)dB_s\right|^{1/2}\right)\mathbb{E}(|B_{t+h}-B_t|^{-1/2})\right\}^{1/2},$$ but this appears very convoluted and I am worried that my approach is wrong. I have managed to show the property of the second hint using Jensen's inequality and Ito's isometry, but I do not know how to estimate $\mathbb{E}(|B_{t+h}-B_t|^{-1/2})$ with an upper bound: using the normal distribution of Brownian motion we have $$\mathbb{E}(|B_{t+h}-B_t|^{-1/2})=\frac{h^{-1/4}}{\sqrt{2\pi}}\int_{-\infty}^\infty\frac{1}{\sqrt{|z|}}e^{-z^2/2}dz,$$ but I'm not sure how to bound this as the integrand tends to infinity around $0$. Any advice regarding the problem would be greatly appreciated!