Let
- $(\Omega,\mathcal A)$ be a measurable space;
- $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A,\operatorname P)$;
- $(E,\mathcal E)$ be a measurable space;
- $(X_t)_{t\ge0}$ be an $(E,\mathcal E)$-valued process on $(\Omega,\mathcal A,\operatorname P)$;
- $c:E\to[0,\infty)$ be $\mathcal E$-measurable;
- $\xi$ be a real-valued random variable on $(\Omega,\mathcal A,\operatorname P)$.
How can we shot that $$\tau:=\inf\left\{t\ge0:\int_0^tc(X_s)\:{\rm d}s\ge\xi\right\}$$ is an $(\mathcal F_t)_{t\ge0}$-stopping time (imposing suitable additional assumptions, if necessary)?
I guess we need to assume that $F:=c\circ X$ is (pathwisely) locally Lebesgue integrable so that we can infer that $$G_t:=\int_0^tF_s\:{\rm d}s\;\;\;\text{for }t\ge0$$ is continuous. From this continuity, we can infer that $$\tau\le t\Leftrightarrow G_t\ge\xi\tag1$$ for all $t\ge0$.
Next, I guess we need to assume that $(X_t)_{t\ge0}$ is $(\mathcal F_t)_{t\ge0}$-progressive, since this ensures that $(G_t)_{t\ge0}$ is $(\mathcal F_t)_{t\ge0}$-adapted.
Now it seems like we need to assume that $\xi$ is $\mathcal F_0$-measurable in order to conclude from $(1)$ (using the fact that if $Y,Z$ are real-valued random variables on $(\Omega,\mathcal A)$, then $\{Y\le Z\}\in\sigma(Y,Z)$).
EDIT: In practice, I would like to take $$F_t:=\sigma(X_s,s\le t)\;\;\;\text{for }t\ge0$$ and $\xi$ being independent of $(X_t)_{t\ge0}$ and exponentially distributed. If we really need to assume that $\xi$ is measurable with respect to $\mathcal F_0=\sigma(X_0)$, how do we need to argue that the $((X_t)_{t\ge0},\xi)$ exist on a common probability space $(\Omega,\mathcal A,\operatorname P)$?