I would like to show that if $X \in {\cal H}_2^{LOC}[0, T] $ (class of adapted processes such that $\int_0^T X_s\, ds < \infty $ with probablity one) is a continuous process, then for every sequence $\{\pi_n: n \in N\} $ of partitions $0=t_{n,0} < t_{n,1} < \cdots < t_{n, m_n} = T $ such that $||\pi_n|| = \max_{0 \le k \le m_n-1} |t_{n,k}-t_{n,k+1}|\, \to 0 $ as $n \to \infty $ then $$ \sum_{k=0}^{m_n-1} X_{t_{n,k}}(B_{t_{n,k+1}} - B_{t_{n,k}}) \to \int_0^T X_t\, dB_t \hskip 6pt \text {in probability as } \hskip 5pt n \to \infty. $$ I have seen some proofs of this fact that use some sort of localization.
Instead I wanted to avoid the localization to show that in some special, but useful cases, the Ito integral has a Riemann sum interpretatio, albeit as a convergence in probability, and tried a different approach. It seems correct to me, but I wonder if I am overlooking something and I would appreciate if someone could take a look at my proof below.
Let $X_n(u, \omega) = \sum_{k=0}^{m_n-1} X_{t_{n,k}}(\omega)\cdot 1_{(t_{n,k}, t_{n,k+1}]}(u) $ be a step process. Then, for a.a. $\omega, $ $X_n(u) \to X_u $ uniformly in $u \in [0, T] $ as $n \to \infty. $ Hence $\int_0^T |X_n(u) - X_u|^2\, du \to 0 $ a.s. In fact, if we let $\mu_h(X)(\omega):= \max\{|X_u(\omega)-X_v(\omega)|: |u-v|\, \le h\}, $ we have \begin{align*} \int_0^T &(X_n(u) - X_u)^2\, du = \int_0^T \sum_{k=0}^{m_n-1} (X_{n, t_{n,k}} - X_u)^2\cdot 1_{(t_{n,k}, t_{n,k+1}]}(u)\, du\\ &\le \int_0^T \sum_{k=0}^{m_n-1} (\mu_{||\pi_n||}(X)(\omega))^2\cdot 1_{(t_{n,k}, t_{n,k+1}]}(u)\, du = (\mu_{||\pi_n||}(X)(\omega))^2T \to 0 \hskip 4pt \text{a.s.} \end{align*} as $||\pi_n|| \to 0 $ while $n \to \infty $ using the fact that $X_u $ is uniformly continuous for each $\omega $ on the compact interval $[0, T]. $ At this point we use the result that states that if $$ \int_0^T (X_u - X_n(u))^2\, du \to 0 \hskip 5pt \text{ in probability as } \hskip 4pt n \to \infty, $$ then $$\int_0^T X_n(u)\, dB_u \to \int_0^T X_u\, dB_u \hskip 5pt \text{ in probability as } \hskip 4pt n \to \infty. $$ However, $$\int_0^T X_n(u)\, dB_u = \sum_{k=0}^{m_n-1} X_{t_{n,k}}(B_{t_{n,k+1}} - B_{t_{n,k}}) $$ which proves the statement.
Thank you
Maurice