Conditions for SDE to become constant?

Question

Suppose that $(X_t)$ is some process that satisfies the SDE $$dX_t=f(X_t)d t+g(X_t)d B_t,\qquad X_0=0,$$ where $f,g$ are bounded Lipschitz maps.

Suppose then that $X_t$ reaches some point $x$ where $f(x)=g(x)=0$. Then, it seems intuitive that $X_t$ should become constant and equal to $x$ for all subsequent times, as the two functions that drive the dynamics of the process vanish. However, I'm having trouble understanding this in mathematical terms.

Suppose that we set $\tau$ as the hitting time of $x$, and we assume for simplicity that $\tau<\infty$ almost surely. Then we would like to say that the process $(X_{\tau+t})_t$ is constant and equal to $x$ (i suspect the strong Markov property might be of use, but all my attempts have failed).

Answer 1

Hints:

1. Show that $(X_{t+\tau})_{t \geq 0}$ is a solution to the SDE $$dY_t = f(Y_t) \, dt + g(Y_t) \, dW_t, \qquad Y_0 = x \tag{1}$$ for a Brownian motion $(W_t)_{t \geq 0}$.
2. Show that $Y_t := x$ is a solution to the SDE $(1)$.
3. Since there exists a (pathwise) unique solution to (1), it follows from step 1 and 2 that $X_{t+\tau}=x$ for all $t \geq 0$ almost surely.