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Oct 31, 2014 at 13:18 comment added yada Thank you! I'm not an expert in this area, but just to recapitulate: For the domain for $g$ you use that the path $X^\omega\in M([0, \infty), E)$ the set of measurable maps. By fixing the Lebesgue measure $\lambda$ on $[0, \infty)$ one can define the topology of convergence in measure on $M$ which results in $L^0$. This one is Polish, by going from $\lambda$ to an equivalent finite measure. $g$ is then defined as $g(t,f) := g_t(f|_{[0,t]})$ for $f \in L^0$. Why is $g$ jointly measurable on its whole domain? Is there something special in the choice of $\lambda$ for generating a topology on $M$?
Oct 31, 2014 at 12:39 vote accept yada
Oct 30, 2014 at 13:22 history answered Dan CC BY-SA 3.0