When $t\to X_t$ is an absolutely continuous process ($X_t= X_0+ \int_0^t Y_s dt$ for some measurable process $Y_t$) we have $$\sigma(Y_t) \subset \cap_{\epsilon >0}\sigma(X_{s}, s\in [t,t+\epsilon))$$ Is the converse inclusion true ? If yes can you give me a proof or a reference ? I was not able to find a counterexemple. Thank you.

When $t\to X_t$ is an absolutely continuous process ($X_t= X_0+ \int_0^t Y_s dt$ for some measurable process $Y_t$) we have for all $t$ $$\sigma(Y_t) \subset \cap_{\epsilon >0}\sigma(X_{s}, s\in [t,t+\epsilon))$$ Is the converse inclusion true ? If yes can you give me a proof or a reference ? I was not able to find a counterexemple. Thank you.