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 $\frac{1}{0} = \infty.$*



Lebesgue differentiation theorem at a stopping time
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.


