Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space, $(\mathcal F_t)$ a filtration and $B$ a Brownian motion adapted to the filtration. Let denote $\mathcal V(0,T)$ the space of function $f: [0,T]\times \Omega \to \mathbb R$ s.t.
$(i)$ $f(t,\cdot )$ is $\mathcal F_t$ for all $t\in [0,T]$,
$(ii)$ $f$ is progressively measurable, i.e. $f:[0,t]\times \Omega \to \mathbb R$ is $\mathcal B([0,t])\otimes \Omega $ measurable for all $0\leq t\leq T$
$(iii)$ $\mathbb E\left[\int_0^T f(t,\cdot )^2\,\mathrm d t\right]<\infty $
Then, stochastic integral is well defined for all $f\in \mathcal V(0,T)$.
Interpretation of these condition
As I understand (i), it comes from the fact that in the construction of the Itô integral, if $\Pi:0=t_0<t_1<...<t_n=T$ is a partition of $[0,T]$, in $$\sum_{k=0}^{n-1}f(t_i,\cdot )B_{t_{i+1}}-B_{t_i})\tag{*}$$ that $B_{t_{i+1}}-B_{t_i}$ and $f(t_i,\cdot )$ are independent.
Since the convergence of the sum $(*)$ is in $L^2(\Omega )$, so of course $\int_0^T f(t,\cdot )\,\mathrm d B_t\in L^2(\Omega )$. But in the other hand, $\int_0^T f(t,\cdot )\,\mathrm d B_t\in L^2(\Omega )$ if and only if $$\mathbb E\left[\left(\int_0^Tf(t,\infty )\,\mathrm d t\right)^2\right]=\mathbb E\left[\int_0^T f(t,\omega )^2\,\mathrm d t\right].$$ So, for $\int_0^T f(t,\cdot )\,\mathrm d B_t$ to be well define, we need $$\mathbb E\left[\int_0^T f(t,\cdot )^2\,\mathrm d t\right]<\infty,$$ that's why we need $(iii)$.
Question
1) Are my interpretations corrects ?
2) Now, why do we need $(ii)$ ? i.e. Why progressively measurable is important ?
P.S. I know that we can define Itô integral to a larger class of function that $\mathcal V(0,T)$, so no need to tell me that not only function of $\mathcal V(0,T)$ make the Itô integral well defined.