Let $\{W_t, t\geq 0\}$ be a Brownian motion, and has a.s. continuous sample paths.
Let $\{\mathcal{F}^W_t, t\geq 0\}$ be the canonical filtration, i.e. $\mathcal{F}^W_t=\sigma(W_s, 0\leq s\leq t)$.
So why is $\{\mathcal{F}^W_t, t\geq 0\}$ left-continuous? i.e. $\displaystyle{\bigcup_{s<t}}\mathcal{F}^W_s= \mathcal{F}^W_t$.
Thank you so much!
Best Answer
If you want to keep using a.s. continuous sample paths, you need to augment the filtration.
If you don't want to augment the filtration, suppose $W_t$ is always continuous.
Then apply $\lim_{s\uparrow t}W_s = W_t$, almost surely(if you complete the filtration) or always(if you suppose $W_t$ is always continuous)