# Why is the canonical filtration of a Brownian motion left-continuous?

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!

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)
• @Tony.Liu We only need to confirm that the information contained in $\cup_{s<t}\mathcal{F}_s^W$ can make the value of $W_t$ available, which is true because the above limit. – Petite Etincelle Oct 14 '14 at 22:44