Pathwise measurability of Ito integral under supremum norm

191 Views Asked by At

I'm doing my first research project on Stochastic Analysis and in order to prove something which is crucial, I need to prove the following claim:

LEMMA:

Denote by $(C_{0}[0,\,T],\,||\centerdot||_{\infty},\,\mathcal{B})$ the classical Wiener space of continuous paths defined on $[0,\,T]$ and started at zero, equipped the supremum norm $||\centerdot||_{\infty}$ and the corresponding Borel sigma algebra $\mathcal{B}$.

For an $m-$dimensional Brownian motion, consider the n-dimensional SDE: $ X_{t}=X_{0}+\int_{0}^{t}a(s,\,X_{s})ds+\int_{0}^{t}b(s,\,X_{s})dB_{s},\,0\leq t\leq T$

where the random variable $X_{0}$ and the coefficients $a,\,b$ have a very good regularity (as good as we want).

For each Brownian path $B_{.}:=\{B_{t}\}_{0\leq t\leq T}$ and realization $x$ of the random variable $X_{0}$, denote by $f(B_{.},\,x)$ the corresponding path of the pathwise unique strong solution to the above SDE, which is given by Ito's theorem.

Then the map $f$ which maps the product space $(C_{0}[0,\,T])^{m}\times\mathbb{R}^{n}$ to the product space $(C_{0}[0,\,T])^{n}$ is measurable with the product topologies.

. . . . .

I have searched a lot, but I have been unable to find something relevant. Working with Rough path theory, we find that this result is true if we replace the supremum norm with the a-Holer norm. However this does not help, because I want the preimages to be measurable with the Wiener probability measure, and again this is something I'm unable to prove or disprove. Can anyone help?

Thank you in advance.