An equality involving the Wiener process The equality below appears as a step in a proof in a chapter titled "Itô Stochastic Calculus" in Brzeźniak and Zatawniak's textbook "Basic Stochastic Processes", Springer 2005 (in a solution to exercise 7.10 on p. 216). I don't understand why the equality holds and would appreciate an explanation.
Let $W$ be a one dimensional Wiener process, let $T > 0$ be a strictly positive real number and for some natural number $n \in \{1, 2, \dots\}$ let $0 = t^n_0 < t^n_1 < \cdots < t^n_{n - 1} < t^n_n = T$ be a partition of the real interval $[0,T]$ into $n$ sub-intervals $[t^n_i, t^n_{i + 1}]$, each of length $T/n$. Then
$$
E\left(\left|\sum_{i=0}^{n-1}\int_{t^n_i}^{t^n_{i+1}}\left(W(t^n_i) - W(t)\right) dt\right|^2\right) = \sum_{i = 0}^{n - 1} \int_{t^n_i}^{t^n_{i+1}}E\left(\left|W(t^n_i)-W(t)\right|^2\right) dt
$$

EDIT
I'm not really interested in this equality per se. The ultimate purpose of the chain of equalities, of which this equality is one link, is to show that 
$$
E\left(\left|\sum_{i=0}^{n-1}\int_{t^n_i}^{t^n_{i+1}}\left(W(t^n_i) - W(t)\right) dt\right|^2\right) \underset{n \rightarrow \infty}{\rightarrow} 0
$$
 A: Edit: This answer shows the identity
$$\mathbb{E} \left( \left| \sum_{i=0}^{n-1} \int_{t_i^n}^{t_{i+1}^n} (W(t_i^n)-W(t)) \, dt \right|^2 \right) = \mathbb{E} \left( \sum_{i=0}^{n-1} \left[\int_{t_i^n}^{t_{i+1}^n} (W(t_i^n)-W(t)) \, dt\right]^2 \right). \tag{1}$$

Fix $i<j$. Then, by the tower property,
$$\begin{align*} &\mathbb{E}((W(t_i^n)-W(t))(W(t_j^n)-W(s))) \\ &= \mathbb{E} \bigg[ (W(t_i^n)-W(t)) \cdot \mathbb{E}(W(t_j^n)-W(s) \mid \mathcal{F}_{t_{i+1}^n}) \bigg] \tag{2}\end{align*}$$
for any $s \in [t_j^n,t_{j+1}^n]$ and $t \in [t_i^n,t_{i+1}^n]$. Since $(W_t)_{t \geq 0}$ is a martingale, we have
$$\mathbb{E}(W(t_j^n)-W(s) \mid \mathcal{F}_{t_{i+1}^n})=0.$$
Consequently, $(2)$ yields
$$ \mathbb{E}((W(t_i^n)-W(t))(W(t_j^n)-W(s)))=0. \tag{3}$$
Using the identity
$$\left( \sum_{i=1}^n a_i \right)^2 = 2 \sum_{j=1}^n \sum_{i<j} a_i a_j + \sum_{i=1}^n a_i^2$$
and Fubini's theorem, we conclude
$$\begin{align*} &\mathbb{E} \left( \left| \sum_{i=0}^{n-1} \int_{t_i^n}^{t_{i+1}^n} (W(t_i^n)-W(t)) \, dt \right|^2 \right) \\ &=2 \mathbb{E} \left( \sum_{j=0}^{n-1} \sum_{i=0}^{j-1} \int_{t_i^n}^{t_{i+1}^n} \int_{t_j^n}^{t_{j+1}^n} (W(t_i^n)-W(t))(W(t_j^n)-W(s)) \, ds \, dt \right) \\ &\quad + \mathbb{E} \left( \sum_{i=0}^{n-1} \left[ \int_{t_i^n}^{t_{i+1}^n} (W(t_i^n)-W(t)) \, dt \right]^2 \right) \\ &\stackrel{(3)}{=} \mathbb{E} \left( \sum_{i=0}^{n-1} \left[\int_{t_i^n}^{t_{i+1}^n} (W(t_i^n)-W(t)) \, dt\right]^2 \right). \end{align*}$$

Remark: Since
$$\left| \int_{t_i^n}^{t_{i+1}^n} (W(t_i^n)-W(t)) \, dt \right| \leq 2\sup_{t \in [0,T]} |W(t)| \cdot \frac{1}{n},$$
$(1)$ shows in particular
$$\begin{align*} \mathbb{E} \left( \left| \sum_{i=0}^{n-1} \int_{t_i^n}^{t_{i+1}^n} (W(t_i^n)-W(t)) \, dt \right|^2 \right) &\leq 4  \mathbb{E} \left( \sup_{t \in [0,T]} |W(t)|^2 \right) \frac{n}{n^2} \\ &\stackrel{n \to \infty}{\to} 0. \end{align*}$$
