Maybe the question should be "Understanding the measurability: Hilbert space valued process V.S. random field"
Consider the SPDE $${\rm d}u+\cdots{\rm d}t=\sigma(t,u){\rm d}W.$$
Consider the Cauchy problem with $H^s$ initial data $u(0)=u_0(\omega,x)\in L^2(\Omega;H^s)$ ($H^s$ is Sobolev space), then for some $\tau>0$, there is an $\mathcal{F}_t$ predictable $H^s$-valued process $u$ such that $u\in C([0,\tau];H^s)$, ${\mathbb P}-a.s.$ and $u$ solves the equation almost surely. We consider the path wise solution so a probability space $(\Omega, \mathcal{F},{\mathbb P},\{\mathcal{F}_t\}_{t\geq0}, W)$ is fixed in advance.
Then we move to the global existence issue and then the question comes.
For the deterministic PDE, some conditions on $u_0$ can give a global solution. And that condition is not on the size of $\|u_0\|_{H^s}$, but on the point behavior of $u_0(x)$ (for example, $-3<\partial_xu_0<3$ for all $x$). One can say that the condition is given when $u_0$ is viewed as a random field.
To extend the deterministic results to SPDE, how can we extent the probability space such that the solution $u$ can be also measured as a random field??? Or we need to solve the equation again when $u$ is viewed as a random field?
For example, ${\mathbb P}\{\|u\|_{H^s}<100\}$ makes sense because $u$ is a $H^s$-valued process. But ${\mathbb P}\{\partial_xu<100\}$ is not clear because $\{\partial_xu<100\}$ is not a Borel set in $H^s$ (am I right???) and can we extent the probability space to overcome this??
This question also comes when we consider the non-negativity of the solution. For example, many papers considered stochastic reaction-diffusion equations. The solution is considered as an Hilbert valued process. But the authors just suddenly assumed that ${\mathbb P}\{u_0\ge0\}=1$ and then it can be proved that ${\mathbb P}\{u\ge0\}=1$. But $\{ u_0\ge0\}$ does not belong to $\mathcal{F}_0$.
Why can we do this???
Many thanks!!!!
