I quote Øksendal (2003).
Itô integral. Let $\mathcal{V}=\mathcal{V}(S,T)$ be the class of functions $f(t,\omega):[0,\infty)\times\Omega\to\mathbb{R}$ such that $(t,\omega)\to f(t,\omega)$ is $\mathcal{B}\times\mathcal{F}$-measurable (where $\mathcal{B}$ denotes the Borel $\sigma$-algebra on $[0,\infty)$), $f(t,\omega)$ is $\mathcal{F}_t$-adapted and $\mathbb{E}\bigg[\int_{S}^T f(t,\omega)^2 dt\bigg]<\infty$.
[…] For functions $f\in\mathcal{V}$ we will now show how to define the Itô integral $$\mathcal{I}[f](\omega)=\int_{S}^{T}f(t,\omega)dB_t(\omega)$$
where $B_t$ is $1-$dimensional Brownian motion.
[…]
The idea is natual: First we define $\mathcal{I}[\phi]$ for a simple class of functions $\phi$. Then, we show that each $f\in\mathcal{V}$ can be approximated by such $\phi$'s and we use this to define $\int fdB$ as the limit of $\int\phi dB$ as $\phi\to f$.
Recall that a function $\phi\in\mathcal{V}$ is called elementary if it has the form $$\phi(t,\omega)=\sum_j e_j(\omega)\cdot\chi_{[t_j, t_{j+1}]}(t)\tag{1}$$
Lemma (Itô isometry). If $\phi(t,\omega)$ is bounded and elementary then $$\mathbb{E}\bigg[\bigg(\int_{S}^{T}\phi(t,\omega)dB_t(\omega)\bigg)^2\bigg]=\mathbb{E}\bigg[\int_{S}^T\phi(t,\omega)^2dt\bigg]\tag{2}$$Proof Set $\Delta B_j=B_{t_{j+1}}-B_{t_j}$. Then $$\mathbb{E}\left[e_ie_j\Delta B_i\Delta B_j\right]=\begin{cases}0\hspace{3.74cm}\text{if }i\ne j\\
\mathbb{E}\left[e_j^2\right]\cdot (t_{j+1}-t_j)\hspace{0.5cm}\text{if }i=j\end{cases}$$
using that $e_ie_j\Delta B_i$ and $\Delta B_j$ are independent if $i<j$. Thus:
$$\mathbb{E}\left[\left(\int_S^T \phi dB\right)^2\right]=\sum_{i,j}\mathbb{E}\left[e_ie_j\Delta B_i\Delta B_j\right]=\color{blue}{\sum_j\mathbb{E}\left[e_j^2\right]\cdot(t_{j+1}-t_j)}\\\color{red}{=}\mathbb{E}\left[\int_S^T\phi^2 dt\right]$$
My question refers to the $\color{red}{\text{red}}$ equality above:
starting from $(1)$, I would say that $$\mathbb{E}\left[\int_S^T \phi^2 dt\right]=\mathbb{E}\left[\int_S^T\left(\sum_j e_j(\omega)\cdot\chi_{[t_j,t_{j+1})}(t)\right)^2dt\right]=\color{orange}{\mathbb{E}\left[\left(\sum_j e_j(\omega)\right)^2(t_{j+1}-t_j)\right]}$$ So, why does it hold true that:
$$\color{blue}{\sum_j\mathbb{E}\left[e_j^2\right]\cdot(t_{j+1}-t_j)}=\color{orange}{\mathbb{E}\left[\left(\sum_j e_j(\omega)\right)^2(t_{j+1}-t_j)\right]}$$?
Best Answer
Note that with $I_j = [t_{j},t_{j+1})$,
$$\phi^2 =\left(\sum_{j}e_j \chi_{I_j} \right)^2 = \sum_{j}e_j \chi_{I_j}\sum_{k}e_k\chi_{I_k}= \sum_je_j^2 \chi^2_{I_j} + \underset{k\neq j}{\sum\sum}e_je_k \chi_{I_j} \chi_{I_k} \\= \sum_je_j^2 \chi_{I_j},$$
since $\chi^2_{I_j} = \chi_{I_j}$ and $\chi_{I_j} \chi_{I_k} = 0$ for disjoint intervals.