Today my professor introduced the Ito integral as a way to make sense of
$$\int \sigma(u) \cdot "noise"du$$
where noise is modeled as Brownian motion. He then said:
With Riemann integrals you can approximate with rectangles to get the area. But take a look at the plot of a brownian motion; you can't do it. It's a fractal (it does not get smoothed if you "zoom in"), so good luck drawing rectangles with that
Which struck me weird as the Brownian path is continuous and hence Riemann integrable.
I guess it was just a simplification, but I wonder now what exactly goes horribly wrong if we define $$\int \sigma(u) B(\omega, u) du$$ for a fixed $\omega$ (so that the integral depends on $\omega$) as the usual Riemann integral
Thank you!
There is no problem integrating a brownian motion pathwise with respect to a continuous (say) integrand. This is because a brownian motion has continuous almost everywhere sample paths.
In the Ito integral the brownian motion is used as an integrand, i.e. it has the form $\int_{[0,T]} f(s,\omega) dW(s,\omega)$. One can show that if every continuous function is Riemann integrable with respect to some integrand $g$, then $g$ has to be of bounded variation. But it is a well known fact that the sample paths of brownian motion do not have bounded variation.