Quadratic variation of Ito integral : Relationship between Ito Integral and Brownian Motion This is theorem 4.2.3 from Shreve's Stochastic Calculus for finance II.
Statement of the theorem:
The quadratic variation accumulated up to time $t$ by the Ito integral is:
$$ [ I,I](t) = \int_{0}^{t} \Delta^{2}(u) du.$$
The proof revolves around computing quadratic variation accumulated by the Ito integral on one of the subintervals $[t_j, t_{j+1}]$ on which $\Delta(u)$ is constant. They choose the partition points:
$$t_{j} = s_{0} < s_{1} < \ldots < s_{m} = t_{j+1} $$
and compute
$$\sum_{i=0}^{m-1}[I(s_{i+1}) - I(s_{i})]^2 = \sum_{i=0}^{m-1}[\Delta(t_j)(W(s_{i+1}) - W(s_{i}))]^2 $$
I can't figure out how we got the aforementioned equality involving Ito integral and Brownian Motion. Any help appreciated.
Here we have the following definitions: if $t_{k} \leq t \leq t_{k+1}$ then
$$I(t) = \sum_{j=0}^{k-1}\Delta(t_{j})[W(t_{j+1} - W(t_j)] + \Delta(t_{k})[W(t) - W(t_{k})]$$
The process $I(t)$ is the Ito integral of the simple process $\Delta(t)$ that is written as
$I(t) = \int_{0}^{t}\Delta(u)dW(u).$
Suppose $\Delta(t)$ is an adapted stochastic process i.e. it is $\mathcal{F}(t)$ measurable for each $t \geq 0.$ Assume further that $\Delta(t)$ is constant in $t$ on each subinterval $[t_{j}, t_{j+1}).$ Then $\Delta(t)$ is a simple process.
Here u is just the dummy variable and W(t) is the Brownian Motion.
 A: Following the definitions, we have
$$I(s_{i+1})=I(t_j)+ \int_{t_j}^{s_{i+1}} \Delta(t_j)dW(u) = I(t_j)+\Delta(t_j)(W(s_{i+1})-W(t_j)).$$
Similarly
$$I(s_i)=I(t_j)+\Delta(t_j)(W(s_i)-W(t_j)).$$
Thus
$$I(s_{i+1})-I(s_i)=\Delta(t_j)(W(s_{i+1})-W(s_i))$$
and therefore
$$\sum_{i=0}^{m-1}[I(s_{i+1}) - I(s_{i})]^2 = \sum_{i=0}^{m-1}[\Delta(t_j)(W(s_{i+1}) - W(s_{i}))]^2. $$
EDIT: In the first expression I am using the linearity property of the Ito integral. Remember that
$$t_j = s_0 < s_1 < ... < s_m = t_{j+1},$$
so
\begin{align}
I(s_{i+1})=\int_0^{s_{i+1}} \Delta(u)dW(u)&=\int_0^{t_j} \Delta(u)dW(u)+\int_{t_j}^{s_{i+1}} \Delta(u)dW(u) \\
&=I(t_j)+\int_{t_j}^{s_{i+1}} \Delta(u)dW(u)
\end{align}
A: It's not clear if $\Delta (u)$ is deterministic or not. I'll suppose it is, and if it's not, I let you adapt the proof. I denote $\sum_{j}$ for $\sum_{j=0}^{m-1}$. Then,
\begin{align*}
\mathbb E\left[\left(\sum_{j}(I(t_{j+1})-I(t_j))^2-\int_{0}^t\Delta (u)^2\,\mathrm d u\right)^2\right]&=\mathbb E\left[\left(\sum_{j}\big(\Delta (t_j)(W_{t_{j+1}}-W_{t_j})\big)^2-\Delta (t_j)^2(t_{j+1}-t_j)\right)^2\right]\\
&=\mathbb E\left[\left(\sum_{j}\Delta (t_j)^2\big((W_{t_{j+1}}-W_{t_j})^2-(t_{j+1}-t_j)\big)\right)^2\right]\\
&=\sum_{i,j}\mathbb E[\Delta (t_i)^2\big((W_{t_{i+1}}-W_{t_i})^2-(t_{i+1}-t_i)\big)\Delta (t_j)^2\big((W_{t_{j+1}}-W_{t_j})^2-(t_{j+1}-t_j)\big)]\\
&\underset{(*)}{=}2\sum_{i}\Delta (t_i)^4(t_{i+1}-t_i)^2\\
&\leq 2\max_{i}|t_{i+1}-t_i|\int_0^t\Delta (u)^4\,\mathrm d u\underset{m\to \infty }{\longrightarrow }0,
\end{align*}
where in $(*)$ we used the fact that Brownian motion has independent increments and that if $X\sim \mathcal N(0,1)$, then $\mathbb E[X^4]=3$.
