Countable intersection consequence of right-continuity of filtrations

Question

Statement: $(F_t)_{t\in[0,\infty)}$ is a right-continuous filtration, i.e. $\cap_{u>t}F_u=F_t$ for all $t$. Then $\cap_{k=1}^\infty F_{t+1/k}=F_t$.

It suffices to show that $\cap_{u>t}F_u=\cap_{k=1}^\infty F_{t+1/k}$. I've shown $\subset$, but am having trouble showing $\supset$ due to cardinality issues.

Pick $A\in \cap_{u>t} F_u$, and $k\in\mathbb{N}$. Then $tt} F_u\subset \cap_{k=1}^\infty F_{u_k}\subset \cap_{k=1}^\infty F_{t+1/k}$.

I'd like to argue analogously for the other direction, but that would require picking uncountably many integers $k_u$ so that $t+1/k_u\in(t,u)$.

Any help would be greatly appreciated! Maybe this isn't the best way to go about proving the original statement after all.

Answer 1

Using the monotonous of filtration, $F_{t+u}\supset F_{t+1/n}, \forall u\ge 1/n$, hence $$\bigcap_{u\ge 1/n}F_{t+u}\supset F_{t+1/n}\qquad \text{and}\qquad \bigcap_{u>t}F_u\supset \bigcap_{n\ge1} F_{t+1/n}$$.