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

• What does integrable with respect to $W$ mean here? For example, for all $T>0$, $E \int_{0 }^{T} H_s^2 ds < \infty$ or $P( \int_{0}^{T} H_s^2 ds < \infty) = 1$? Sep 17, 2022 at 13:07
• Ah, the latter, sorry. Sep 17, 2022 at 13:27
• This does not seem to be true even for deterministic $\sigma$: $W_{\sigma+h}-W_{\sigma-h} = 0$ along some sequence $h = h_n$ which converges to zero, and there is no reason to expect the integral to be zero for $h = h_n$. Or am I missing something? Sep 17, 2022 at 21:02
• Hmm I think it’s because the limit is taken in probability that it has a chance of existing. The sequence $h_n$ converges to $0$ but is itself random. @MateuszKwaśnicki Sep 18, 2022 at 0:33
• Ah, okay, sorry. Somehow I did not notice that the limit is in probability. Sep 18, 2022 at 8:49

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.
• Nice! One comment: I think one has to use the adapted process $(H_s - H_{(\sigma - h)\vee 0})$ rather than the non-adapted one $(H_s - H_\sigma)$ as the integrand. Sep 18, 2022 at 8:54