I'm a stochastic calculus beginner. In every ito's integral presentation there is the following definition of a simple process:
$$H(t,ω)=∑_{i=1}^{n-1}h_{i}(ω)1_{(t_{i−1},t_{i}]}(t)$$
Then, the integral is defined as:
$$ \int_{0}^{T}H(t,ω)dB(t,ω) := ∑_{i=1}^{n-1}h_{i}(ω)[B(t_{j+1},ω)-B(t_{j},ω)] $$
But what I truly can't understand is if $H(t,w)$ is a sum itself why the ito's integral is not a integral of a sum? I mean, this:
$$ \int_{0}^{T}H(t,ω)dB(t,ω) := ∑_{t=0}^{T}∑_{i=1}^{n-1}h_{(i)}(ω)1_{(t_{i−1},t_{i}]}[B(t_{j+1},ω)-B(t_{j},ω)] $$
I think I'm just missing a very silly point but I wasn't able to figure it out by myself, so can someone enlighten me?
You could eventually define your integral as you propose but with slight modifications:
First, your summation cannot be from $t=0$ to $T$. The idea is to sum all over the partition. Second thing I've noticed is that there seems to be problem of indexing, your $h_i$ (in the expression for $H$ ) should be $h_{i-1}$ actually.
However you can write
$$\int_{0}^{T}H(t,ω)dB(t,ω) := ∑_{j=1}^{n}\left(∑_{i=1}^{n}h_{i-1}(ω)1_{(t_{i−1},t_{i}]}(s)\right)\bigg\rvert_{s=t_{j}}\cdot[B(t_{j},ω)-B(t_{j-1},ω)]$$
(Be aware that I am evaluating the simple process at a given point of the partition, otherwise the expression makes no sense!)
Now notice that $\left(∑_{i=1}^{n}h_{i-1}(ω)1_{(t_{i−1},t_{i}]}(s)\right)\bigg\rvert_{s=t_{j}}=h_{j-1}$ and hence the two expressions are actually equivalent. For the sake of simplicity we don't use the latter.