Definition of Ito integral: Progressively measurable vs. measurable/adapted

Question

Consider the probability space $(\Omega,\mathcal{F}\,P)$ and the filtration $\mathcal{F}_t$.

In Oksendal's book the Ito integral $I(f)(\omega)=\int_S^T f(t,\omega)dB_t(\omega)$ is defined on a space $\mathcal{V}(S,T)$. This is the space of functions such that

1. $(t,\omega)\to f(t,\omega)$ is $\mathcal{B}\times\mathcal{F}$ measurable where $\mathcal{B}$ is the borel sigma algebra on $[0,\infty)$.
2. $f(t,\cdot)$ is $\mathcal{F}_t$ adapted.
3. $E[\int_S^Tf^2(t,\omega)dt]<\infty$

Then the Ito integral of $f$ is defined by $$\int_S^T f(t,\omega)dB_t(\omega):=\lim_{n\to \infty}\int_S^T \phi_n (t,\omega) dB_t(\omega)\,\,in\,\, L^2(P)$$ where $\{\phi_n\}$ is sequence of elementary functions such that $$E\Big[\int_S^T (f(t,\omega)-\phi_n(t,\omega))^2 dt\Big]\to 0 \,\,as\,\,n\to\infty$$

In some other text books I found that instead of this space $\mathcal{V}$ the space (say $\mathcal{W}$) of progressively measurable processes such that $(3)$ holds. Now I am quite sure that $\mathcal{W}\subset \mathcal{V}$. But I am wondering why the difference in approaches? As far as I have noticed, they lead to the same assertions ( the integrals are martingales etc etc.). Or is there any striking difference between these two approaches?

Answer 1

In general, this problem has been addressed in this question on Math Overflow. In particular, I quote a comment:

If the process has continuous sample paths (actually just left continuous or right continuous is enough), then adapted and progressively measurable are the same thing. See Proposition 4.8 in Revuz & Yor.

The author of the comment is Martin Hairer, so I don't doubt it's correct.