When studying Shreve's Volume 2, Chapter $4$ Stochastic Calculus, the author provided the following example $4.3.2$:
We compute $\int_0^T W(t)dW(t)$ where $W(t)$ is a Brownian motion. Choose a large integer $n$ and approximate the integrand $\Delta(t)= W(t)$ by the simple process $$ \Delta_n(t)= \begin{cases} W(0)=0 & \quad \text{if} \quad 0\leq t\leq \frac{T}{n}, \\ W\left(\frac{T}{n}\right)=0 & \quad \text{if} \quad \frac{T}{n}\leq t\leq \frac{2T}{n}, \\ \vdots \\ W\left( \frac{(n-1)T}{n} \right) & \quad \text{if}\quad \frac{(n-1)T}{n} \leq t\le T \end{cases}. $$ Then $\lim_{n\to\infty} \mathbb{E}\int_0^T|\Delta_n(t) -W(t)|^2dt = 0.$
I have trouble getting the last equality $\lim_{n\to\infty} \mathbb{E}\int_0^T|\Delta_n(t) -W(t)|^2dt = 0.$
Any hint is appreciated.
For almost all $\omega \in \Omega$ the Brownian motion $\{W_t; 0 \leq t \leq T\}$ is uniformly continuous (since B.M. is continuous and we are on a bounded interval [0,T]). Fix $\omega$ and let $\epsilon>0.$ Then, there exists $\delta>0$ such that $|t-s|< \delta$ implies $|W(\omega,t)-W(\omega,s)|<\epsilon.$ Take $n_0 > \frac{T}{\delta}.$ Then $\forall n \geq n_0,$ we have that for all $t \in [0,T],$ $|\Delta_n(\omega,t)-W(\omega,t)|< \epsilon.$ Therefore, $$\int^T_0 |\Delta_n(\omega,t)-W(\omega,t)|^2 dt \leq \epsilon^2 \int^T_0 dt=\epsilon^2 T.$$ This means that $$\int^T_0 |\Delta_n(\omega,t)-W(\omega,t)|^2 dt \rightarrow 0 \quad \text{a.s.}$$ And by the dominated convergence theorem $$E \int^T_0 |\Delta_n(\omega,t)-W(\omega,t)|^2 dt \rightarrow 0.$$