let $f : [0, +\infty[\to\mathbb R$ be bounded and continuous, and $(X_t)_{t\geq 0}$ be an adapted process such that X$_0 = 0$ and $X_t = \int_0^t f(s, X_s){\rm d}\!s$ for all $t\geq0$, also define $\pmb t = \inf\{t\ge 0 \;;\; X_t\gt1\}$.
I have to prove that $(X_t)_{t\geq0}$ is almost surely continuous and that $\pmb t$ is a stopping time.
The first thing pops up in my brain is that $f$ is bounded meaning ${\rm d}\!X_t$ is bounded, so even $\frac{{\rm d}\!X}{{\rm d}\!t}$ might be discontinuous, we can still see that integrating a small amount of time will have an upper and lower bound, this implies $t\mapsto X_t$ is continuous as its change for a given change in t can be bounded.
But I am not sure if this technic is applicable for stochastic process or adapted process or not. also I am very confused by the term "almost surely continuous", I know its a property in Brownian motion, but have no idea to prove it in this question. I searched a bit, got the idea that almost sure continuity could be understood as sample continuity.. https://en.wikipedia.org/wiki/Sample-continuous_process
I also tried to take advantages of an theorem about almost sure continuity: https://en.wikipedia.org/wiki/Kolmogorov_continuity_theorem But still didn't figure this out.
Sincerely hope any expert could give me some hints! Many thanks in advance!
The process $X$ is differentiable, thus continuous.
Now, $]-\infty,1[$ is an open set. But the first instant of leaving an open set for a continuous process is a stopping time.
Therefore, $\mathbf t$ is a stopping time.
(If the set is closed, it suffices that $X$ is right-continuous.)