I have an interesting question concerning those two integrals. Considering a Brownian motion $(B_t)_{t \geq 0}$ with start in $x$. We can choose an $\omega \in \Omega$ such that, $t \to B_t(\omega)$ is continuous. We consider now a continuous function $f$, over which we can define the Itô integral. We can then consider $\int_{0}^{T} f(t) dB(t,\omega)$ as an Riemann-Stieltjes integral. But if we take $f(t)=B(t,\omega)$ it holds that $\int_{0}^{T} B(t,\omega)dB(t,\omega)= \frac{1}{2} (B(T)^2 - T)$ so in this example it does not coincide with the Riemann-Stieltjes integral. I can't really find the error in my argumentation. It would be nice, if someone could help.
Yours,
Mark
Existence theorems for the Riemann-Stieltjes integral $\displaystyle\int f\,\mathrm dg$ on compact intervals usually assume that $f$ is continuous and $g$ of bounded variation. This excludes $g:t\mapsto B_t$ since the paths of Brownian motion have infinite variation, almost surely (but finite quadratic variation, almost surely, a fact which opens the way to Itô's integral).
To sum up, the integral $\displaystyle\int B_t\,\mathrm dB_t$ does not exist in the Riemann-Stieltjes sense.