TITLE: Why isn't the Ito integral just the Riemann-Stieltjes integral?
QUESTION [16 upvotes]: Why isn't the Ito integral just the Riemann-Stieltjes integral?
What I mean is, given a continuous function $f$, some path of standard brownian motion $B$, and the integral:
$$\int_0^Tf(t)\;dB(t).$$
So what if we can't apply the change of variables formula to make sense of
$$\int_0^Tf(t)B'(t)\;dt,$$
the Riemann-Stieltjes integral never required differentiability of the integrator anyways.  
Is there a reason to distinguish the Ito integral from the Riemann-Stieltjes integral above and beyond the need to develop a theory (Ito Calculus) to get around all the problems caused by the failure of change of variables?

REPLY [5 votes]: You can't simply integrate with regards to Brownian motion pathwise because it's not of bounded variation (because such functions are differentiable almost everywhere) and functions of bounded variation are precisely those with respect to which you can compute the Riemann–Stieltjes integrals.