# Ito isometry proof

I have been reading Steven Shreve's Stochastic Calculus for Finance II. This question is from Chapter 4 Stochastic Calculus Page 129. This is theorem 4.2.2 (Proof of Ito isometry).

This theorem is about proving the following:

$$\mathbb E I^2(t)=\mathbb E\int_0^t \Delta^2(u)du$$

Here $$\Delta(t)$$ is an adapted stochastic process meaning that it is $$\mathcal{F}(t)$$- measurable for each $$t \geq 0.$$ I can't seem to understand how we got the following:

$$\Delta$$ is a simple process and is constant on $$[t_j,t_{j+1}),$$ i.e $$\Delta(t)=\Delta(t_j)$$ for all $$t \in [t_j,t_{j+1}),$$ Thus the integral $$\int_{t_j}^{t_{j+1}} \Delta^2(u)du$$ is just a Riemann integral and we have $$\int_{t_j}^{t_{j+1}} \Delta^2(u)du=\Delta^2(t_j) \int_{t_j}^{t_{j+1}} du =\Delta^2(t_j)(t_{j+1}-t_j).$$