Set a probability space $(\Omega, \mathcal F,P)$, and a one dimension wiener process $W$ on it, with filtration ${\mathcal F}_t=\sigma({W(s)|0 \le s \le t})$ which is generated by the wiener process.
When we consider such equation \begin{equation} (1)\left\{\begin{array}{ll} du= u(1+W)dt \\ u(0)=1 \end{array}\right. \end{equation}
Obviously, we can solve this question on some ODE environment, for fixed $\omega \in \Omega$, we can solve \begin{equation} (2)\left\{\begin{array}{ll} \displaystyle\frac{du}{dt}= u(1+W(\cdot,\omega)) \\ u(0)=1 \end{array}\right. \end{equation}
on $[0,T^*]$, here $T^*$ is the constant depend on $\omega$ because the right side of the equation, \begin{equation} f:R^1 \times R^+ \times \Omega \rightarrow R^1 \\ f(x,t,w)=x(1+W(t,w)) \end{equation}
$f$ satisfies some locally lipschitz condition $$ |f(x_1,t,w)-f(x_2,t,w)| \le (1+\sup_{0\le s \le t}W(s,w))|x_1-x_2| $$ and we can extend this solution from $[0,T^*]$ to $[0,+\infty)$, by a priori estimate $$ \displaystyle\frac{d|u|^2}{dt}= 2|u|^2(1+W(\cdot,\omega)), $$ and the gronwall inequality.
So by the above steps, we can obtain a binary mapping \begin{equation} u:[0,+\infty) \times \Omega \rightarrow R^1 \end{equation} where $u(\cdot,\omega)$ is the solution of $(2)$, after adjust a zero measure set in $\Omega$.
${\bf What \, I \, want \, to \, ask \, is}$: Whether the mapping $ u(t,\cdot): \Omega \rightarrow R^1 $ is ${\mathcal F}_t$ measurable?
The problem arise when we want to approach it by iteration method, even in a inteval suffcient close to $0$. For instance, if we set, for $t \in [0,T']$ \begin{equation} \begin{array}{ll} u_0(t,w) = 1 \\ u_{k+1}(t,w) = 1 + \int_0^t f(u_k(s,w),s,w) ds \end{array} \end{equation} Then, for arbitrary small T' we choose, the convergence property of $u_k$ can't be ensure for each $\omega$, because the lipschitz constant of $f$ vary with $\omega$, and it could be arbitrary large.
@ zhoraster point out that we can solve this question explicitly, because u satisfies \begin{equation} \left\{\begin{array}{ll} \displaystyle \frac{du}{u}= (1+W)dt \\ u(0)=1 \end{array}\right. \end{equation} Integral both sides, we can get that u is ${\mathcal F}_t$ measurable.
But I wonder on some more complex condition, for example \begin{equation} \left\{\begin{array}{ll} du= u(1+W)+Wdt \\ u(0)=1 \end{array}\right. \end{equation} or other condition that we can't get an explicit expression of u.
That is:
- for a.e $\omega$, we can solve the ODE like $(2)$, and extend it to $\infty$
- the locally lipschitz constant vary with $\omega$, and could be arbitrary large
- term of $W$ on right side of the equation is ${\mathcal F}_t$ adapted
Can the ${\mathcal F}_t$ measurable property of u be hold?
