Lebesgue differentiation theorem at a stopping time Let $W$ be a standard Brownian motion, and $\mathcal F_t$ it’s natural filtration. Let $H$ be a continuous process, adapted to $\mathcal F_t$ and integrable with respect to $W$.
Question: Is it true that for all a.s. finite $\mathcal F_t$-stopping times $\sigma$ we have
$$\lim_{h \to 0+} \frac{1}{W_{\sigma + h} - W_{(\sigma - h) \vee 0}} \int_{(\sigma - h)\vee 0}^{\sigma + h} H_s \, dW_s = H_\sigma$$
in probability?
Note: By convention we set $\frac{1}{0} = \infty.$
 A: Yes. Let $h \in (0,1)$, $Q_h = \frac{1}{W_{\sigma + h} - W_{(\sigma - h) \vee 0}} \int_{(\sigma - h)\vee 0}^{\sigma + h} (H_s - H_{(\sigma - h) \vee 0}) \, dW_s$ and  set $M_h = \sqrt{-\log(h)}$.   Because $H$ is continuous, and hence, $\lim_{h \to 0} H_{(\sigma - h) \vee 0} = H_{\sigma} $ almost surely, the claim holds if we can prove that $Q_h$ converges to zero in probability.
For arbitrary $\epsilon>0$ \begin{align*}
& P( |Q_h| > \epsilon) = A_h + B_h \quad \text{where}  \\
&  A_h := P( |Q_h| > \epsilon , |W_{\sigma + h} -  W_{(\sigma - h) \vee 0} | > M_h \sqrt{\sigma + h - (\sigma - h) \vee 0 } )\\
& B_h := P( |Q_h| > \epsilon , |W_{\sigma + h} -  W_{(\sigma - h) \vee 0} | <  M_h \sqrt{\sigma + h - (\sigma - h) \vee 0} )
\end{align*}
By the strong Markov property of BM (given $\mathcal{F}_{(\sigma-h) \vee 0}$), $$
B_h \le P(|W_{\sigma + h} -  W_{(\sigma - h) \vee 0} | > M_h \sqrt{\sigma + h - (\sigma - h) \vee 0} ) = \mathcal{N}(0,1)[M_h, \infty] \le e^{-M_h^2 } \;.
$$
By Markov's inequality, the strong Markov property of BM (given $\mathcal{F}_{(\sigma-h) \vee 0}$), and Itô isometry, \begin{align*}
& A_h \le P\left(\left| \frac{1}{\sqrt{\sigma + h - (\sigma - h) \vee 0}} \int_{(\sigma - h)\vee 0}^{\sigma + h} (H_s - H_{(\sigma - h) \vee 0}) \, dW_s \right| > M_h \epsilon  \right) \\
& \le  E\left(\left| \frac{1}{\sqrt{\sigma + h - (\sigma - h) \vee 0}} \int_{(\sigma - h)\vee 0}^{\sigma + h} (H_s - H_{(\sigma - h) \vee 0}) \, dW_s \right|^2   \right) M_h^{-2}  \epsilon^{-2} \\
& \le E\left( \frac{1}{\sigma + h - (\sigma - h) \vee 0} \int_{(\sigma - h)\vee 0}^{\sigma + h} (H_s - H_{(\sigma - h) \vee 0})^2 ds    \right) M_h^{-2}  \epsilon^{-2}
\end{align*}
By continuity of $H$, we see that $A_h\searrow 0$ and $B_h \searrow 0$, and in turn, $P(|Q_h| > \epsilon) \searrow 0$ as $h \searrow 0$.
Thus, $Q_h$ converges to zero in probability, as required.
