In optimal control theory, we often need a filtration do be right continuous. Consider a filtered probability space $(\Omega, \mathcal F, \mathbb P)$ equipped with a right continuous filtration $\mathcal F\_t$. Let $y\_t$ be an $\mathcal F\_t$ measurable process. Define, $a\_t := \mathbb{1}\_{y\_t = 0}$. 


Let $\zeta\_t := \int\_0^t (1-a\_s) ds$ and $\lambda(s) := \sup\{t \ge 0: \zeta\_t \le s\}$. (Assume that $y\_t = 1$ for all $t \ge T$ for some finite $T$ so that this sup is well-defined and finite for all $s$.) 


Consider a filtration $\mathcal G\_s := \mathcal F\_{\lambda(s)}$. Is $\mathcal G\_s$ a right continuous filtration in $s$?


I want to say it is because $\lambda(s)$ is right continuous. So, $\cap\_{\epsilon >0} \mathcal G\_{s+\epsilon} = \mathcal G\_s$. But, continuity of filtrations seems sufficiently abstract for me to not have any confidence in this reasoning. 



Right continuous filtration
Just use the definitions.


Consider first the case of an arbitrary filtration $\mathcal F\_t$ (not necessarily right-continuous). For every $s$, the random variable $\lambda(s)$ is a Markov time with respect to $\mathcal F\_{t+}$, in the sense that $$\{\lambda(s) < t\} \in \mathcal F\_t \quad \text{for every $t$.}$$ By definition, $E \in \mathcal F\_{\lambda(s)+}$ if and only if $$E \in \mathcal F\_\infty \quad \text{and} \quad E \cap \{\lambda(s) < t\} \in \mathcal F\_t \quad \text{for every $t$.}$$ Similarly, $E \in \mathcal F\_{\lambda(s+\varepsilon)+}$ if and only if $$E \in \mathcal F\_\infty \quad \text{and} \quad E \cap \{\lambda(s + \varepsilon) < t\} \in \mathcal F\_t \quad \text{for every $t$.}$$ It follows that if $E \in \bigcap\_{\varepsilon > 0} \mathcal F\_{\lambda(s+\varepsilon)+}$, then $E \in \mathcal F\_\infty$ and $$E \cap \{\lambda(s) < t\} = \bigcup\_n E \cap \{\lambda(s + \tfrac{1}{n}) < t\} \in \mathcal F\_t$$ for every $t$, and thus $E \in \mathcal F\_{\lambda(s)+}$ (here indeed we use right-continuity of $\lambda$). Of course, $\bigcap\_{\varepsilon > 0} \mathcal F\_{\lambda(s+\varepsilon)+}$ contains $\mathcal F\_{\lambda(s)+}$, and consequently the two $\sigma$-algebras are equal.


If $\mathcal F\_t$ is right-continuous, then $\mathcal F\_{t+} = \mathcal{F}\_t$, and consequently $\lambda(s)$ is a Markov time with respect to $\mathcal F\_t$ (in the sense that $\{\lambda(s) \leqslant t\} \in \mathcal F\_t$), and $\mathcal F\_{\lambda(s)+} = \mathcal F\_{\lambda(s)}$. Here $E \in \mathcal F\_{\lambda(s)}$ if and only if $$E \in \mathcal F\_\infty \quad \text{and} \quad E \cap \{\lambda(s) \leqslant t\} \in \mathcal F\_t \quad \text{for every $t$.}$$ It follows that indeed $\mathcal G\_s = \mathcal F\_{\lambda(s)}$ is right-continuous.


