Let $A$ be a strictly increasing continuous process which is adapted to a right-continuous filtration $(\mathcal F_t)_{t\ge0}$. We also suppose that $A_\infty=\infty$. For $t\ge0$, let $C_t=\inf\{s\ge0:A_s>t\}$. Under the preceding conditions, it is known that $\forall t\ge0$, $C_t$ is an $(\mathcal F_s)_{s\ge0}$ stopping time, $C$ is a strictly increasing continuous process with $C_\infty=\infty$, and that $A_{C_t}=C_{A_t}=t$ for $t\ge0$. We can therefore define $\hat{\mathcal F}_t=\mathcal F_{C_t}$ for $t\ge0$. Again, it is known that for $t\ge0$, $A_t$ is an $(\hat{\mathcal F}_s)_{s\ge0}$ stopping time, and we define $\tilde{\mathcal F}_t=\hat{\mathcal F}_{A_t}$. Is it true that $\forall t\ge0$, $\mathcal F_t=\tilde{\mathcal F}_t$?\
Firstly, we can start off by noticing that the fact that $\mathcal F$ is right-continuous implies that $\hat{\mathcal F}$ and $\tilde{\mathcal F}$ are also right-continuous. Furthermore, we have that $C_\infty=C_{A_\infty}=\infty$ hence $\mathcal F_\infty=\hat{\mathcal F}_\infty=\tilde{\mathcal F}_\infty$. It's now tempting to use the formal definition of $\tilde{\mathcal F}_t=\{B\in\hat{\mathcal F}_\infty:B\cap\{A_t\le s\}\in\hat{\mathcal F}_s\}$ and the fact that we have $\{A_t\le s\}=\{t\le C_s\}$, but this doesn't seem to yield anything interesting. Any help would be appreciated.
How about $\tilde{\mathcal F}_t=\hat{\mathcal F}_{A_t}=\mathcal F_{C_{A_t}}=\mathcal F_t$, since $C_{A_t}=t$ almost surely ?