# Filtration from a Brownian Motion

The textbook I am reading defines the filtration induced from a Brownian Motion as follows. Let $\{B(t): t \geq 0\}$ be a Brownian Motion defined on some probability space, then we can define a filtration $(\mathcal F^0(t): t\geq0)$ by letting

$$\mathcal F^0(t) := \sigma(B(s): 0\leq s\leq t)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$$

be the $\sigma$-algebra generated by the random variables $B(s)$ for $0 \leq s\leq t$.

My question is whether I can understand the above definition $(*)$ as follows.

$$\mathcal F^0(t) := \sigma(B(s): 0\leq s\leq t) = \sigma\left(\cup_{0\leq s\leq t} \sigma(B(s))\right).$$

Thank you!

Yes, and this is quite general since, for any collection $(X_s)_{0\leqslant s\leqslant t}$ of random variables, $$\sigma(X_s;0\leqslant s\leqslant t)=\sigma\left(\bigcup_{0\leqslant s\leqslant t}\sigma(X_s)\right).$$ Proof: Double inclusion.