I am working through some notes on the construction of the Ito integral, and am stuck in the proof of a lemma. I have filled in some details of the proof, but am stuck in the step written in capital letters in the proof. I am working with the Ito integral constructed only for the Brownian motion (not a general semimartingale). $m_T$ denotes the Lebesgue measure restricted to the interval $[0,T]$, and a simple process $X$ is one such that for every $\omega$, $X_t(\omega)$ is piecewise constant with a finite number of jump points which do not depend on the choice of $\omega$. The notes I am reading from are by Prof. Ramon van Handel, and they are available on his website. The lemma is 4.2.9.
Lemma: Let $X_t$ be an $(F_t)$ adapted process in $\bigcap_{T < \infty} L^2(P \times m_T)$ and let $\tau$ be a stopping time. Then $I_{t \land \tau}(X) = I_t(X_{(\cdot)} I_{(\cdot) \leq \tau})$. Proof: Because $\tau$ is a stopping time on $(F_t)$ \begin{align} I_{t \leq \tau}(\{0\}) = \{\omega : \tau(\omega) < t \} = \bigcup_{n=1}^{\infty}\{\omega : \tau(\omega) \leq t - 1/n \} \in F_t \end{align} form which it follows that $\1_{t \leq \tau}$ is adapted so the argument in second Ito integral is admissible and is in $\bigcap_{T < \infty} L^2(P \times m_T)$. To prove the result, fix $T$ and let $(X^n)$ be a sequence of simple processes converging to $X$ in $L^2$ fast enough. Suppose that $\tau \leq T$ almost surely. Let $\tau^n$ be a random variable such that $\tau^n(\omega)$ is $\tau (\omega)$ rounded up to the next jump time of $X^n$. $\tau^n$ remains a stopping time. To see this, fix $t$ and let $t_k$ denote the largest jump time smaller than or equal to $t$. Then \begin{align} \{\tau^n \leq t \} = \{ \tau \leq t_k \} \in F_{t_k} \subseteq F_{t}. \end{align} WE ALSO HAVE THAT $(X^n I_{t \leq \tau^n})$ is an approximating sequence of simple processes converging to $X I_{t \leq \tau}$ in $L^2(P \times m_T)$. HERE IS MY ATTEMPT: \begin{align} \int(X^n I_{t \leq \tau^n} - X I_{t \leq \tau})^2d(P \times m_T) \\ = \int (X_t^n(\omega) - X_t(\omega))^2 I_{t \leq \tau(\omega)} dP(m_T \times P)(\omega,t) + \int (X^n_{t \leq \tau^n})^2 I_{\tau^n(\omega) \geq t > \tau(\omega)} d(m_T \times P) \\ \end{align} The first term goes to zero as $n \to \infty$ by definition. For the second term, I can't prove it. Does anyone have any suggestions? I can prove this if I assume that the simple processes have jump interval length converges to zero, and if $X$ has continuous sample paths, but I can't seem to do it in the general case.
Suggestion: Prove the assertion first in the special case that $\tau$ takes on only finitely many values. In this case $t\mapsto X^n_tI_{\{\tau\le t\}}$ is a simple process, and your approximation will succeed with no trouble, and there's no need to introduce $\tau^n$. In general, choose finitely-valued stopping times $(\tau_n)$ that decrease to $\tau$. Then $(t,\omega)\mapsto X_t(\omega)I_{\{t\le \tau_n(\omega)\}}$ converges in $L^2(m_T\times P)$ to $(t,\omega)\mapsto X_t(\omega)I_{\{t\le \tau(\omega)\}}$, and $I_{t\wedge\tau_n}(X)$ converges pointwise to $I_{t\wedge\tau}(X)$ by path-continuity of the stochastic integral.