My goal is to prove that for all $t\geq 0$ \begin{equation} \frac{\int_{t}^{t+h} H_s dB_s} {B_{t+h}-B_{t}} \rightarrow H_t \end{equation} in probability, where $B$ is a Brownian motion and $H$ continuous and bounded.
Almost the same question was asked here: "Continuity" of stochastic integral wrt Brownian motion
When I follow the steps in the above thread, I get
\begin{equation} P \left( \left| \frac{\int_{t}^{t+h} H_s dB_s} {B_{t+h}-B_{t}} - H_t \right| > \epsilon \right) \leq P\left( \left| \frac{1}{\sqrt{h}} \int_{t}^{t+h} (H_s-H_t) dB_s \right| > \epsilon/K \right) + \frac{\epsilon}{4} \end{equation} Now in the above thread it is said that one should use the Markov-inequality and then the Ito-isometry. But I don't get how I can use the Ito isometry after I apply the Markov-inequality. Markov would give me that the probability on the right hand side is less or equal to: \begin{equation} E\left( \frac{1}{\sqrt{h}} \left| \int_{t}^{t+h} (H_s-H_t) dB_s \right| \right) \, \frac{K}{\epsilon} \ . \end{equation} Any ideas on how to show that the above converges to $0$ as $h\to 0$ (using the Ito isometry)?
Edit: The rest of the proof: Using Chebychev's inequality we have (ignoring constants) \begin{equation} E \left( \frac{1}{h} \left( \int_t^{t+h} (H_s -H_t) dB_s \right)^2 \right) = \frac{1}{h} \int_t^{t+h} E(H_s -H_t)^2 ds \to 0 \end{equation} as $h \to 0$ by the Ito isometry.