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!

• Something is missing in $X_t = \int_0^t f(s,X_s)\,d s$. – Sayantan Sep 15 at 6:02
• @Sayantan Hi Sayantan! Thanks for your comment! I am sure I didn't miss any information here, the context is $X_0$=$0$ and $X_t$ is defined as $\int_0^t$f(s, X$_s$)ds for all t$\geq$$0. – Wendy Zhang Sep 15 at 6:10 • In that case X solves the ODE$$ \frac{\, d X_t }{\,d t} = f(t,X_t), X_0 =0.$\$ Where does the randomness come from? – Sayantan Sep 15 at 7:12
• @Sayantan Thank you so much for your explanation! I presumably think that would be the solution, but I am still got stuck on how to use bounded derivative to prove the almost sure continuity. – Wendy Zhang Sep 15 at 7:18

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.)