Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. Let $(H_t, t \in [0, T])$ be an adapted process and $(B_t, t \ge 0)$ the standard Brownian motion w.r.t. the same filtration $(\mathcal F_t, t \ge 0)$. For $n \ge 1$, we define $(H^{(n)}_{t}, t \in [0, T])$ by $$ H^{(n)}_{t} := \sum_{i=1}^{2^n} H\left(\frac{(i-1) T}{2^n}\right) 1_{ \left (\frac{(i-1) T}{2^n}, \frac{i T}{2^n} \right]}(t). $$
So $H^{(n)}_{t}$ is a step function of time. We define $$ (H^{(n)} \cdot B)_T := \sum_{i=1}^{2^n} H\left(\frac{(i-1) T}{2^n}\right) \left ( B\left(\frac{i T}{2^n}\right) - B\left(\frac{(i-1) T}{2^n}\right) \right ) \quad \forall n \ge 1. $$
For ease of notation, let $t_i := iT/2^n$. Then $$ (H^{(n)} \cdot B)_T = \sum_{i=1}^{2^n} H_{t_{i-1}} (B_{t_i} - B_{t_{i-1}}). $$
It is mentioned (without proof) at page 37 of this note that
Theorem If $(H_t, t \in [0, T])$ has continuous trajectories and $\mathbb E[ \int_0^T H_t^2 \mathrm d t] < \infty$, then $$ (H^{(n)} \cdot B)_T \underset{n \rightarrow \infty}{\longrightarrow} (H \cdot B)_T \quad \text{in probability}, $$ where $(H \cdot B)_T \equiv \int_0^T H_t \mathrm d B_t$ is the Itô's stochastic integral.
This convergence result is then used in a version of Itô's lemma at page 39 of the same note.
My question Could you explain how to prove this theorem?
This is Proposition 5.9 in Le Gall's Brownian Motion, Martingales, and Stochastic Calculus.
Proposition 5.9 Let $X$ be a continuous semi-martingale, and let $H$ be an adapted process with continuous sample paths. Then, for every $t>0$, for every sequence $0=t_0^n<\cdots<t_{p_n}^n=t$ of subdivisions of $[0, t]$ whose mesh tends to $0$, we have $$ \lim _{n \rightarrow \infty} \sum_{i=0}^{p_n-1} H_{t_i^n}\left(X_{t_{i+1}^n}-X_{t_i^n}\right)=\int_0^t H_s \mathrm{~d} X_s, $$ in probability.