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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: $$\color{blue}{\sum_j\mathbb{E}\left[e_j^2\right]\cdot(t_{j+1}-t_j)}=\color{orange}{\mathbb{E}\left[\left(\sum_j e_j(\omega)\right)^2(t_{j+1}-t_j)\right]}$$?

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  • $\begingroup$ $\sum_j\mathbb{E}\left[e_j^2\right]\cdot(t_{j+1}-t_j)=\mathbb{E}\sum_j\left[e_j^2\right]\cdot(t_{j+1}-t_j)= \mathbb{E}\left[\int_S^T\phi^2 dt\right].$ The first equality is because of the linearity of the expectation and the second because of the definition of the integral. $\endgroup$ – UBM Oct 17 '20 at 15:00
  • $\begingroup$ As to your second equality: in my opinion, according to $(1)$, $\phi^2=\left[\sum_{j}e_j\right]^2$, and not $\phi^2=\sum_{j}\left[e_j^2\right]$. Isn't it? If so, your second equality would not hold true I think @UBM $\endgroup$ – Strictly_increasing Oct 17 '20 at 15:03
  • $\begingroup$ The intervals $[t_j,t_{j+1})$ are disjoint. $\endgroup$ – UBM Oct 17 '20 at 15:51
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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.

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    $\begingroup$ Easy to check that $\chi^2_{I_j}(\omega) = 1^2 = 1 = \chi_{I_j}(\omega)$ for $\omega \in I_j$, etc. $\endgroup$ – RRL Oct 17 '20 at 16:02

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