Convergence of stochastic integrals:$T_p \to t$ and $\int_0^{T_p}H_s^ndM_s \to \int_0^{T_p}H_sdM_s$ imply $\int_0^tH_s^n dM_s \to \int_0^tH_sdM_s$.

Question

I am struggling to understand a line from the proof below on convergence of stochastic integrals from Le Gall's Brownian Motion and Stochastic Calculus.

So the final displayed line shows that $\int_0^{T_p} H_s^n dM_s \to \int_0^{T_p} H_s dM_s$ in probability. Then, the author concludes the proof by stating that since $P(T_p=t) \to 1$ as $p \to \infty$, we get the desired result, i.e. $\int_0^t H_s^n dM_s \to \int_0^t H_s dM_s$.

How do we jump to this final result? I would greatly appreciate any help.

Answer 1

You want to show that $$\mathbb{P}\left(\left|\int_0^t (H_s^n - H_s) dM_s\right| \geq \delta\right) \to 0$$ as $n \to \infty$. Fix $\varepsilon > 0$ and pick $p$ large enough that $\mathbb{P}(T_p \neq t) \leq \varepsilon/2$. Then $$\mathbb{P}\left(\left|\int_0^t (H_s^n - H_s) dM_s\right| \geq \delta\right) \leq \mathbb{P}\left(\left|\int_0^{T_p} (H_s^n - H_s) dM_s\right| \geq \delta\right) + \mathbb{P}(T_p \neq t)$$

Since $\int_0^{T_p} H_s^n dM_s \to \int_0^{T_p} H_s dM_s$, there exists $N$ such that $n \geq N$ implies that $$\mathbb{P}\left(\left|\int_0^{T_p} (H_s^n - H_s) dM_s\right| \geq \delta\right) \leq \varepsilon/2$$ and hence for $n \geq N$ $$\mathbb{P}\left(\left|\int_0^t (H_s^n - H_s) dM_s\right| \geq \delta\right) \leq \varepsilon$$ which proves the desired result.