Let $X_0 = \frac12$. Let $a : \mathbb R_+ \to \{-1,0,1\}$ be a measurable function.
A particle moves in the interval $[0,1]$ as follows: $$ dX_t = a_t dt \implies X_t = X_0 + \int_0^t a_s ds$$ where the integral is the Lebesgue integral. Both $0$ and $1$ are absorbing, i.e. once $X_t$ hits $0$ or $1$ it remains there.
Suppose $a_t$ is Markovian. That is, $\exists$ a function $g: [0,1] \to \{-1,0,1\}$ such that, $a_s = g(X_s)$ for all $s$. Therefore,
$$X_t = X_0 + \int_0^t g(X_s) ds$$
I would like to claim the following:
Claim: $X_t$ moves only in one direction. That is, if $X_t = x >X_0$ for some $t$. Then, $\nexists s > t$ such that $X_s \in (X_0,x)$.
The reasoning is the following. Since $X_t$ reached $x$ and $a_t$ is Markovian, for any $y \in (X_0,x)$, $a(y) = 1$ (with some abuse of notation). So, it should be impossible for the particle to return back to $y$ once it moves forward.
How do I argue this formally?
Can we write the claim for when $a_t$ is everywhere Markovian?
Thanks.