Question: Let $W$ be a Wiener process, and $H$, an adapted, right-continuous, bounded process. Prove that for a fixed $t$,
$$\lim_{h\rightarrow 0^+} \frac{1}{W_{t+h} - W_t}\int_t^{t+h} H_s dW_s = H_t$$
in probability.
My attempt: Since we need to prove convergence in the sense of probability, we need to consider
$$P(|\frac{1}{W_{t+h} - W_t}\int_t^{t+h} H_s dW_s - H_t| > \epsilon).$$
By Markov's inequality, we know that the above probability is less than or equal to
$$\frac{1}{\epsilon}E[|\frac{1}{W_{t+h} - W_t}\int_t^{t+h} H_s dW_s - H_t|].$$
Since $t$ is fixed, we can state the following
$$\frac{1}{\epsilon}E[|\frac{1}{W_{t+h} - W_t}\int_t^{t+h} H_s dW_s - H_t|] =\frac{1}{\epsilon}E[|\frac{1}{W_{t+h} - W_t}\int_t^{t+h} H_s - H_t dW_s|].$$
From here, I have considered using Cauchy-Schwarz inequality to state that
$$E[|\frac{1}{W_{t+h} - W_t}\int_t^{t+h} H_s - H_t dW_s|] \leq \bigg(E[\frac{1}{W_{t+h}-W_t}]^2\bigg)^{1/2}\bigg(E\big[\int_t^{t+h} H_s - H_t dW_s\big]^2\bigg)^{1/2}$$
$$\leq \bigg(E[\frac{1}{W_{t+h}-W_t}]^2\bigg)^{1/2}\bigg(E\big[\int_t^{t+h} (H_s - H_t)^2 ds\big]\bigg)^{1/2}.$$
At this point, I have come to problem of $\bigg(E[\frac{1}{W_{t+h}-W_t}]^2\bigg)^{1/2}$.
Any advice. Also, I have been given a hint to use Cauchy-Schwarz inequality on $E\big[|\frac{1}{W_\epsilon}\int_0^\epsilon(H_u-H_0)dW_u|^{1/4}\big]$.
I prefer to use a slightly different approach for this. First, it's enough to show $\lim_{t \rightarrow 0} \frac{1}{W_t}\int_0^t H_s dW_s = H_0$. For any $t > 0$ and $K > 0$ we have \begin{align*} P\left(\left|\frac{1}{W_t}\int_0^t H_s dW_s - H_0\right|>\varepsilon\right) &= P\left(\left|\frac{1}{W_t}\int_0^t (H_s - H_0) dW_s\right|>\varepsilon\right) \\ &= P\left(\left|\frac{1}{W_t}\int_0^t (H_s - H_0) dW_s\right|>\varepsilon, \frac{\sqrt{t}}{|W_t|} \le K \right) \\ &\qquad + P\left(\left|\frac{1}{W_t}\int_0^t (H_s - H_0) dW_s\right|>\varepsilon, \frac{\sqrt{t}}{|W_t|} > K \right) \\ &\le P\left(\left|\frac{K}{\sqrt t}\int_0^t (H_s - H_0) dW_s\right|>\varepsilon \right) + P\left( \frac{\sqrt{t}}{|W_t|} > K \right). \end{align*} The idea of doing this is that the $\frac 1{|W_t|}$ is the biggest problem because it can be very large, so we want to split this probability into where $\frac 1{|W_t|}$ is large and where $\frac 1{|W_t|}$ is small and address the two cases separately.
For the first term, by Markov's inequality we have \begin{align*} P\left(\left|\frac{K}{\sqrt t}\int_0^t (H_s - H_0) dW_s\right|>\varepsilon \right) &\le \frac{K^2}{\varepsilon t} \mathbb{E}\left[\left|\int_0^t (H_s - H_0) dW_s\right|^2\right] \\ &= \frac{K^2}{\varepsilon t} \mathbb{E}\left[\left|\int_0^t (H_s - H_0)^2 ds\right|\right] \\ &\le \frac{K^2}{\varepsilon t} \mathbb{E}\left[t \sup_{s \le t}|H_s - H_0|^2 \right] \\ &= \frac{K^2}{\varepsilon} \mathbb{E}\left[\sup_{s \le t}|H_s - H_0|^2 \right]. \end{align*} Since $H$ is right-continuous, $\lim_{t \rightarrow 0} \sup_{s \le t}|H_s(\omega) - H_0(\omega)|^2 = 0$ for all $\omega \in \Omega$, and since $H$ is bounded, the dominated convergence theorem gives $$\lim_{t \rightarrow 0} \frac{K^2}{\varepsilon} \mathbb{E}\left[\sup_{s \le t}|H_s - H_0|^2 \right] = 0.$$
For the second term, we have \begin{align*} P\left( \frac{\sqrt{t}}{|W_t|} > K \right) &= P\left(\frac{|W_t|}{\sqrt{t}} < \frac 1K \right) = P\left(|Z| < \frac 1K\right) \end{align*} where $Z \sim N(0,1)$. Thus, we have
\begin{align*} \lim_{t \rightarrow 0} P\left(\left|\frac{1}{W_t}\int_0^t H_s dW_s - H_0\right|>\varepsilon\right) &\le \lim_{t \rightarrow 0}\left(P\left(\left|\frac{K}{\sqrt t}\int_0^t (H_s - H_0) dW_s\right|>\varepsilon \right) + P\left( \frac{\sqrt{t}}{|W_t|} > K \right) \right) \\ &= 0 + P\left(|Z| < \frac 1K\right), \end{align*} so now sending $K \rightarrow \infty$ gives $\lim_{t \rightarrow 0} P\left(\left|\frac{1}{W_t}\int_0^t H_s dW_s - H_0\right|>\varepsilon\right) = 0$ as desired.