I am studying stochastic integrals w.r.t. brownian motion and for a generic integrand process $\{\Delta_t\}_{t\in[0,T]}$ the following condition is required: $E\big[\int_0^T\Delta_t^2(\omega)dt\big]<\infty$.
I don’t quite understand what kind of integral that is, and how would you assure the condition is satisfied for a specific process.
The same integral is also used to define the quadratic variation of the Itô integral: $[I,I]_t(\omega)=\int_0^t\Delta^2_u(\omega)du$, a.s.
We treat this integral as a path-by-path Riemann or Lebesgue integral. It is not an Itô integral. To get the quadratic variation, one fixes $\omega \in \Omega$, and obtains a single trajectory of the process: $$t \mapsto \Delta_t^2(\omega)$$ To this trajectory, we may assign an integral $$[I,I]_t(\omega) = \int_0^t \Delta_u^2(\omega )du$$ which is interpreted in the Riemann sense if possible, or Lebesgue integral (against the Lebesgue measure) if necessary. Observe now that $[I,I]$ is a stochastic process, so that its expectation is simply the integral against the background probability measure.